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Theorem 2.2.1. Let \( L = K\left( \theta \right) \) be an extension of number fields with \( T\left( \theta \right) = \) 0 as above, and let \( n = \deg \left( T\right) = \left\lbrack {L : K}\right\rbrack \) . Let \( \sigma \) be an embedding (an injective field homomorphism) of \( K \) into an arbitrary field \( \Omega \) (not necessarily a number field). Assume that the polynomial \( {T}^{\sigma }\left( X\right) \) has \( n \) roots in \( \Omega \), where \( {T}^{\sigma } \) denotes the polynomial obtained from \( T \) by applying \( \sigma \) to all the coefficients. Then \( \sigma \) can be extended to exactly \( n \) embeddings of \( L \) into \( \Omega \) .
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Proof. Indeed, let \( \alpha = A\left( \theta \right) \in L = K\left( \theta \right) \) . If \( \phi \) is an extension of \( \sigma \) to \( L \) , we must have \( \phi \left( \alpha \right) = \phi \left( {A\left( \theta \right) }\right) = {A}^{\sigma }\left( {\phi \left( \theta \right) }\right) \) . Since \( T\left( \theta \right) = 0 \), we must have \( \begin{matrix} {T}^{\sigma }\left( {\phi \left( \theta \right) }\right) = 0,\text{hence}\;\phi \left( \theta \right) \;\text{must be one of the}\;n\;\text{roots}\;{\beta }_{i}\;\text{of}\;{T}^{\sigma }\;\text{in}\;\mathit{Ω},\text{so there are} \end{matrix} \) at most \( n \) embeddings. Conversely, if we set \( \phi \left( \alpha \right) = {A}^{\sigma }\left( {\beta }_{i}\right) \) and if \( \alpha = {A}_{1}\left( \theta \right) \) for some other polynomial \( {A}_{1} \), we have \( {A}_{1}\left( X\right) - A\left( X\right) = T\left( X\right) U\left( X\right) \), hence\n\n\[ \n{A}_{1}^{\sigma }\left( {\beta }_{i}\right) = {A}^{\sigma }\left( {\beta }_{i}\right) + {T}^{\sigma }\left( {\beta }_{i}\right) {U}^{\sigma }\left( {\beta }_{i}\right) = {A}^{\sigma }\left( {\beta }_{i}\right) ,\n\]\n\nso \( \phi \) is a well-defined embedding of \( L \) extending \( \sigma \) .
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Yes
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Proposition 2.2.3. Let \( L = K\left( \theta \right) \) be a relative extension of number fields, and let \( T \) be the minimal monic polynomial of \( \theta \), as above. Let \( m = \left\lbrack {K : \mathbb{Q}}\right\rbrack \) and \( n = \left\lbrack {L : K}\right\rbrack \), so that \( \left\lbrack {L : \mathbb{Q}}\right\rbrack = {nm} \) . For each of the \( m \) embeddings \( {\tau }_{i} \) of \( K \) into \( \mathbb{C} \), denote by \( {T}^{{\tau }_{i}} \) the polynomial obtained from \( T \) by applying \( {\tau }_{i} \) on the coefficients. Then we have the following.\n\n(1) Each of the \( m \) embeddings \( {\tau }_{i} \) of \( K \) into \( \mathbb{C} \) extends to exactly \( n \) embeddings of \( L \) into \( \mathbb{C} \), given by \( \theta \mapsto {\theta }_{i, j} \), where the \( {\theta }_{i, j} \) are the roots of \( {T}^{{\tau }_{i}} \) for \( 1 \leq j \leq n \) .\n\n(2) There exist exactly \( n \) embeddings \( {\sigma }_{j, K} \) of \( L \) into \( \mathbb{C} \) which are \( K \) -linear, given by \( \theta \mapsto {\theta }_{j} \), where the \( {\theta }_{j} \) are the roots of the polynomial \( T \) in \( \mathbb{C} \) .
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Proof. If \( \sigma \) is an embedding of \( L \) into \( \mathbb{C} \), then \( {\left. \sigma \right| }_{K} \) is an embedding of \( K \) into \( \mathbb{C} \), so \( {\left. \sigma \right| }_{K} = {\tau }_{i} \) for a certain \( i \) . If \( \sigma \left( \theta \right) = {\theta }^{\prime } \), applying \( \sigma \) to the equality \( T\left( \theta \right) = 0 \) we obtain \( {T}^{{\tau }_{\imath }}\left( {\theta }^{\prime }\right) = 0 \), and hence \( {\theta }^{\prime } = {\theta }_{i, j} \) for a certain index \( j \) . Conversely, it is clear that\n\n\[ \sigma \left( {\mathop{\sum }\limits_{k}{a}_{k}{\theta }^{k}}\right) = \mathop{\sum }\limits_{k}{\tau }_{i}\left( {a}_{k}\right) {\theta }_{i, j}^{k} \]\n\ndefines an embedding of \( L \) into \( \mathbb{C} \), which extends \( {\tau }_{i} \) .\n\nThis embedding will be \( K \) -linear if and only if \( {\tau }_{i}\left( a\right) = a \) for all \( a \in K \)\n\n- in other words, if \( {\tau }_{i} \) is the identity map (recall that we have explicitly embedded \( K \) into \( \overline{\mathbb{Q}} \subset \mathbb{C} \) ) - hence there are exactly \( {nK} \) -linear embeddings, namely those that extend the identity of \( K \) .
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Yes
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Proposition 2.2.5. Let \( L/K \) be a relative extension of relative degree \( n \) . Denote by \( \left( {{r}_{1},{r}_{2}}\right) \) (resp., \( \left( {{R}_{1},{R}_{2}}\right) \) ) the signature of the number field \( K \) (resp., \( L) \) . If all the embeddings \( \tau \) of \( K \) are unramified in \( L \), we have \( \left( {{R}_{1},{R}_{2}}\right) = \) \( \left( {n{r}_{1}, n{r}_{2}}\right) \) .
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Proof. If \( \tau \) is a nonreal embedding of \( K \), any extension of \( \tau \) to \( L \) must also be nonreal since \( L \) is an extension of \( K \) . On the other hand, if \( \tau = {\tau }_{i} \) is a real embedding, the polynomial \( {T}^{{\tau }_{\imath }} \) has \( {R}_{1, i} \) real and \( 2{R}_{2, i} \) nonreal roots for some nonnegative integers \( {R}_{1, i} \) and \( {R}_{2, i} \) such that \( {R}_{1, i} + 2{R}_{2, i} = n \) . Hence the signature of \( L \) is equal to \( \left( {{R}_{1},{R}_{2}}\right) \) with\n\n\[{R}_{1} = \mathop{\sum }\limits_{{1 \leq i \leq {r}_{1}}}{R}_{1, i}\;\text{ and }\;{R}_{2} = \mathop{\sum }\limits_{{1 \leq i \leq {r}_{1}}}{R}_{2, i} + n{r}_{2},\]\n\nas claimed. In the special case where all the \( {\tau }_{i} \) are unramified, we have \( {R}_{1, i} = \) \( n \) and \( {R}_{2, i} = 0 \), proving the formulas of the proposition.
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Yes
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Corollary 2.2.6. Keep the notation of Proposition 2.2.5, and assume in addition that \( L/K \) is a Galois extension.\n\n(1) If \( k \) is the number of ramified real places of \( K \) in \( L/K \), we have \( \left( {{R}_{1},{R}_{2}}\right) = \) \( \left( {n\left( {{r}_{1} - k}\right), n\left( {{r}_{2} + k/2}\right) }\right) \) .\n\n(2) If \( n \) is odd, we have \( k = 0 \), so \( \left( {{R}_{1},{R}_{2}}\right) = \left( {n{r}_{1}, n{r}_{2}}\right) \) .
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Proof. Let \( \tau \) be a real embedding of \( K \) . If \( \tau \) has a real extension to \( L \), then since \( L/K \) is Galois, the roots of the defining polynomial \( {T}^{\tau } \) can be expressed as polynomials with coefficients in \( \tau \left( K\right) \) of any one of them. Hence if one root is real, all of them are, and if one is nonreal, all of them are. Thus, either \( \tau \) is unramified, or all the extensions of \( \tau \) to \( L \) are nonreal. In the case where \( n \) is odd, \( {T}^{\tau } \) is an odd-degree polynomial with real coefficients — hence has at least one real root - so all real places \( \tau \) are unramified, thus proving the corollary.
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Yes
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Proposition 2.2.8. Let \( M \) be a \( {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \) -module that is projective of rank \( n \) as a \( {\mathbb{Z}}_{K} \) -module, and let \( {\left( {\omega }_{i},{\mathfrak{a}}_{i}\right) }_{1 \leq i \leq n} \) be a pseudo-basis of \( M \) in relative HNF on the basis \( \left( {1,\theta ,\ldots ,{\theta }^{n - 1}}\right) \), where \( \theta \) is assumed to be an algebraic integer.\n\n(1) The ideals \( {\mathfrak{q}}_{i} = {\mathfrak{a}}_{i}^{-1} \) are divisible by \( {\mathfrak{q}}_{1} \), and we have\n\n\[{\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \subset \cdots \subset {\mathfrak{a}}_{n}\]\n\nin other words,\n\n\[{q}_{1}\left| {q}_{2}\right| \cdots \mid {q}_{n}\]\n\n(2) For all \( i \leq n \) we have \( {\omega }_{i} \in {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \) ; in other words, if \( \left( {H,\left( {\mathfrak{a}}_{i}\right) }\right) \) is the pseudo-matrix representing the pseudo-basis \( \left( {{\omega }_{i},{\alpha }_{i}}\right) \), then the entries of \( H \) are in \( {\mathbb{Z}}_{K} \) .
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Proof. We will prove (1) and (2) simultaneously by showing by induction on \( j \) that \( {\omega }_{j} \in {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \) and \( {\mathfrak{a}}_{j - 1} \subset {\mathfrak{a}}_{j} \) for \( j > 1 \) . Since \( {\omega }_{1} = 1 \), this is trivially true for \( j = 1 \) . Assume that it is true up to \( j - 1 \), and let \( a \) be any element of \( {\mathfrak{a}}_{j - 1} \) . Since \( M \) is a \( {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \) -module, we have \( {\mathfrak{a}}_{j - 1}\theta {\omega }_{j - 1} \subset M \) ; hence in particular,\n\n\[{a\theta }{\omega }_{j - 1} = \mathop{\sum }\limits_{{1 \leq i \leq n}}{x}_{i}{\omega }_{i}\;\text{ with }\;{x}_{i} \in {\mathfrak{a}}_{i}.\]\n\nSince the matrix of the \( {\omega }_{i} \) is upper-triangular with 1 on the diagonal, we obtain \( {x}_{i} = 0 \) for \( i > j \) and \( {x}_{j} = a \) . Since this is true for any \( a \in {\mathfrak{a}}_{j - 1} \), we therefore have \( {\mathfrak{a}}_{j - 1} \subset {\mathfrak{a}}_{j} \) . In addition,\n\n\[a{\omega }_{j} = {a\theta }{\omega }_{j - 1} - \mathop{\sum }\limits_{{1 \leq i < j}}{x}_{i}{\omega }_{i}\]\n\nSince \( {x}_{i} \in {\mathfrak{a}}_{i} \), and by induction we have \( {\mathfrak{a}}_{1} \subset \cdots \subset {\mathfrak{a}}_{j - 1} \) and \( {\omega }_{i} \in {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \) for \( i < j \), we have for all \( a \in {\mathfrak{a}}_{j - 1} \)\n\n\[a\left( {{\omega }_{j} - \theta {\omega }_{j - 1}}\right) \in {\mathfrak{a}}_{j - 1}{\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack\]\n\nHence\n\n\[{\mathfrak{a}}_{j - 1}\left( {{\omega }_{j} - \theta {\omega }_{j - 1}}\right) \subset {\mathfrak{a}}_{j - 1}{\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack ,\]\nfrom which we deduce that \( {\omega }_{j} - \theta {\omega }_{j - 1} \in {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \), hence that \( {\omega }_{j} \in {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \) , proving our induction hypothesis. Thus \( {\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \subset \cdots \subset {\mathfrak{a}}_{n} \) ; hence by taking inverses, \( {\mathfrak{q}}_{1} \supset {\mathfrak{q}}_{2} \supset \cdots \supset {\mathfrak{q}}_{n} \), showing that all the \( {\mathfrak{q}}_{i} \) are divisible by \( {\mathfrak{q}}_{1} \) and proving the proposition.
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Yes
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Corollary 2.2.9. Let \( \left( {{\omega }_{i},{\mathfrak{a}}_{i}}\right) \) be an integral pseudo-basis of \( {\mathbb{Z}}_{L} \) in HNF on \( \left( {1,\theta ,\ldots ,{\theta }^{n - 1}}\right) \), where \( \theta \) is assumed to be an algebraic integer.\n\n(1) The ideals \( {\mathfrak{q}}_{i} = {\mathfrak{a}}_{i}^{-1} \) are integral ideals, \( {\mathfrak{a}}_{1} = {\mathfrak{q}}_{1} = {\mathbb{Z}}_{K} \), and\n\n\[{\mathbb{Z}}_{K} = {\mathfrak{q}}_{1}\left| {\mathfrak{q}}_{2}\right| \cdots \mid {\mathfrak{q}}_{n}.\n\]\n\n(2) For all \( i \leq n \) we have \( {\omega }_{i} \in {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \) .\n\n(3) For every \( i \leq j \), we have\n\n\[{\mathfrak{q}}_{i}{\mathfrak{q}}_{j + 1 - i} \mid {\mathfrak{q}}_{j}\n\]
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Proof. Since \( {\mathfrak{a}}_{1} = {\mathbb{Z}}_{L} \cap K = {\mathbb{Z}}_{K} \), we have \( {\mathfrak{a}}_{1} = {\mathfrak{q}}_{1} = {\mathbb{Z}}_{K} \) and (1) and (2) are restatements of Proposition 2.2.8. The proof of (3) is very similar to the proof of the proposition: since the leading term of \( {\omega }_{i}{\omega }_{j + 1 - i} \) is \( {\theta }^{j - 1} \), we must have \( {\mathfrak{a}}_{i}{\mathfrak{a}}_{j + 1 - i} \subset {\mathfrak{a}}_{j} \), so \( {\mathfrak{q}}_{i}{\mathfrak{q}}_{j + 1 - i} \mid {\mathfrak{q}}_{j} \) . Note that (3) combined with the fact that the \( {q}_{i} \) are integral implies (1).
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Yes
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Proposition 2.2.10. Let \( \mathcal{B} = \left( {{\omega }_{j},{\mathfrak{a}}_{j}}\right) \) be a relative pseudo-basis. Let \( T = \) \( \left( {{\operatorname{Tr}}_{L/K}\left( {{\omega }_{i}{\omega }_{j}}\right) }\right), I = \left( {{\mathfrak{a}}_{1}^{-1},\ldots ,{\mathfrak{a}}_{n}^{-1}}\right) \), and \( J = \left( {{\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{n}}\right) \) . Then \( \left( {T, I, J}\right) \) is an integral pseudo-matrix. For \( 1 \leq i \leq n \), let \( {\mathfrak{d}}_{i} \) be the elementary divisors of this pseudo-matrix in the sense of Theorem 1.7.2. The \( {\mathfrak{d}}_{i} \) are independent of the chosen pseudo-basis \( \mathcal{B} \) (hence are invariants of the field extension), and their product is equal to the relative discriminant ideal.
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Proof. The proof is straightforward and left to the reader (see Exercises 24 and 25).
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No
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Proposition 2.2.12. (1) If \( I \) and \( J \) are two integral ideals of \( {\mathbb{Z}}_{L} \), we have\n\n\[ \n{\mathcal{N}}_{L/K}\left( {IJ}\right) = {\mathcal{N}}_{L/K}\left( I\right) {\mathcal{N}}_{L/K}\left( J\right) .\n\]
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Proof. The proof of (1) is exactly as in the absolute case (see, for example, [Coh0, Proposition 4.6.8]), replacing the index \( \left\lbrack {M : N}\right\rbrack \) by the index-ideal \( \left\lbrack {M : N}\right\rbrack \) (that is, by the order-ideal of \( M/N \) ; see Definition 1.2.33).
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No
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Proposition 2.2.13. Let \( I \) be a fractional ideal of \( L \) . Then \( {\mathcal{N}}_{L/K}\left( I\right) \) is the ideal of \( K \) generated by all the \( {\mathcal{N}}_{L/K}\left( \alpha \right) \) for \( \alpha \in I \) . More precisely, there exist \( \alpha \) and \( \beta \) in \( I \) such that \( {\mathcal{N}}_{L/K}\left( I\right) = {\mathcal{N}}_{L/K}\left( \alpha \right) {\mathbb{Z}}_{K} + {\mathcal{N}}_{L/K}\left( \beta \right) {\mathbb{Z}}_{K} \) .
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Proof. Clearly, if \( \alpha \in I \) then \( {\mathcal{N}}_{L/K}\left( \alpha \right) \in {\mathcal{N}}_{L/K}\left( I\right) \) . To prove the converse, we proceed in two steps. First, by the approximation theorem in Dedekind domains, we can find \( \alpha \in L \) such that \( {v}_{\mathfrak{P}}\left( \alpha \right) = {v}_{\mathfrak{P}}\left( I\right) \) for all \( \mathfrak{P} \) above the prime ideals \( \mathfrak{p} \) of \( K \) such that \( {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( I\right) }\right) \neq 0 \), and \( {v}_{\mathfrak{P}}\left( \alpha \right) \geq 0 \) for all other \( \mathfrak{P} \) . With this choice of \( \alpha \), it is clear that \( \alpha \in I \) and that \( {\mathcal{N}}_{L/K}\left( \alpha \right) = {\mathcal{N}}_{L/K}\left( I\right) \mathfrak{a} \) with \( \mathfrak{a} \) an integral ideal of \( {\mathbb{Z}}_{K} \) coprime to \( {\mathcal{N}}_{L/K}\left( I\right) \) . Applying once again the approximation theorem, we can find \( \beta \in {\mathbb{Z}}_{L} \) such that \( {v}_{\mathfrak{P}}\left( \beta \right) = {v}_{\mathfrak{P}}\left( I\right) \) for all \( \mathfrak{P} \) above the prime ideals \( \mathfrak{p} \) of \( K \) such that \( {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( \alpha \right) }\right) \neq 0 \), and \( {v}_{\mathfrak{P}}\left( \beta \right) \geq 0 \) for all other \( \mathfrak{P} \) . It is clear that \( \beta \in I \) and that \( {\mathcal{N}}_{L/K}\left( \beta \right) = {\mathcal{N}}_{L/K}\left( I\right) \mathfrak{b} \), where \( \mathfrak{b} \) is an integral ideal coprime to \( {\mathcal{N}}_{L/K}\left( \alpha \right) \), hence in particular to \( \mathfrak{a} \) . Thus, the ideal generated by \( {\mathcal{N}}_{L/K}\left( \alpha \right) \) and \( {\mathcal{N}}_{L/K}\left( \beta \right) \) is equal to \( {\mathcal{N}}_{L/K}\left( I\right) \), proving the proposition.
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Yes
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Lemma 2.2.14. Let \( \\mathfrak{P} \) be a prime ideal of \( L \) above \( \\mathfrak{p} \), and let \( f = f\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) \) be its residual degree. Then\n\n\\[ \n{\\mathcal{N}}_{L/K}\\left( \\mathfrak{P}\\right) = {\\mathfrak{p}}^{f} \n\\]
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Proof. We have \( {\\mathbb{Z}}_{L}/\\mathfrak{P} \\simeq {\\left( {\\mathbb{Z}}_{K}/\\mathfrak{p}\\right) }^{f} \) as \( {\\mathbb{Z}}_{K}/\\mathfrak{p} \) -modules, hence also as \( {\\mathbb{Z}}_{K} \) - modules, thus we conclude by Proposition 1.2.34. Note that the elementary divisors \( {\\mathfrak{d}}_{i} \) of \( {\\mathbb{Z}}_{L}/\\mathfrak{P} \) are given by \( {\\mathfrak{d}}_{i} = \\mathfrak{p} \) for \( 1 \\leq i \\leq f \) and \( {\\mathfrak{d}}_{i} = {\\mathbb{Z}}_{K} \) for \( f < i \\leq n \) .
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Yes
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Proposition 2.3.1. Keep the above notation, let \( H = {H}_{Z}^{-1}{H}_{I} \) (which is the matrix giving the \( {\gamma }_{j} \) on the basis of the \( \left. {\omega }_{i}\right) \), and write \( H = \left( {h}_{i, j}\right) \). (1) Let \( {\mathfrak{q}}_{i} = {\mathfrak{a}}_{i}^{-1} \) (which are integral ideals by Corollary 2.2.9). Then for all \( i \leq j \) we have \[ {\mathfrak{c}}_{j}\left| {{\mathfrak{c}}_{j}{\mathfrak{q}}_{j + 1 - i}}\right| {\mathfrak{c}}_{i} \] and, in particular, \[ {c}_{n}\left| {c}_{n - 1}\right| \cdots \mid {c}_{1} \]
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The proof of (1) is essentially identical to the proof of Corollary 2.2.9 (3): since the leading term of \( {\gamma }_{i}{\omega }_{j + 1 - i} \) is \( {\theta }^{j - 1} \) and \( I \) is an ideal, we must have \( {\mathfrak{c}}_{i}{\mathfrak{a}}_{j + 1 - i} \subset {\mathfrak{c}}_{j} \), in other words \( {\mathfrak{c}}_{j}{\mathfrak{q}}_{j + 1 - i} \mid {\mathfrak{c}}_{i} \), and (1) follows since \( {\mathfrak{q}}_{j + 1 - i} \) is an integral ideal.
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Yes
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Proposition 2.3.2. Keep the above notation. Assume that \( I \) is an integral ideal, so that in particular by Proposition 2.3.1, we have for all \( i,{\mathfrak{c}}_{i} \subset {\mathfrak{a}}_{i} \) (or, equivalently, \( {\mathfrak{a}}_{i} \mid {\mathfrak{c}}_{i} \) ). Let \( {S}_{i} \) be a system of representatives of \( {\mathbb{Z}}_{K}/\left( {{\mathfrak{c}}_{i}{\mathfrak{a}}_{i}^{-1}}\right) \) . For all \( i \) and \( j \) such that \( i < j \), choose \( {c}_{i, j} \in {\mathfrak{a}}_{i}{\mathfrak{c}}_{j}^{-1} \) such that \( {v}_{\mathfrak{p}}\left( {c}_{i, j}\right) = \) \( {v}_{\mathfrak{p}}\left( {{\mathfrak{a}}_{i}{\mathfrak{c}}_{j}^{-1}}\right) \) for all prime ideals \( \mathfrak{p} \) of \( {\mathbb{Z}}_{K} \) such that \( {v}_{\mathfrak{p}}\left( {\mathfrak{a}}_{i}\right) < {v}_{\mathfrak{p}}\left( {\mathfrak{c}}_{i}\right) \) . Then for \( i < j \) we may choose \( {h}_{i, j} \in {c}_{i, j}{S}_{i} \), and the pseudo-matrix \( \left( {H,{\mathfrak{c}}_{j}}\right) \) is then uniquely determined by the ideal \( I \) .
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Proof. According to Corollary 1.4.11, to obtain a unique pseudo-matrix \( \left( {H,{\mathfrak{c}}_{j}}\right) \) we must choose \( {h}_{i, j} \in {S}_{i, j} \), where \( {S}_{i, j} \) is a system of representatives of \( K/{\mathfrak{c}}_{i}{\mathfrak{c}}_{j}^{-1} \) . Since \( I \) is an integral ideal, \( {h}_{i, j} \in {\mathfrak{a}}_{i}{\mathfrak{c}}_{j}^{-1} \), so it is enough to define a system of representatives of \( {\mathfrak{a}}_{i}{\mathfrak{c}}_{j}^{-1}/{\mathfrak{c}}_{i}{\mathfrak{c}}_{j}^{-1} \) . If \( {c}_{i, j} \) satisfies the conditions of the proposition, it is easy to check that the map \( x \mapsto {c}_{i, j}x \) induces an isomorphism from \( {\mathbb{Z}}_{K}/\left( {{c}_{i}{\mathfrak{a}}_{i}^{-1}}\right) \) to \( {\mathfrak{a}}_{i}{c}_{j}^{-1}/{c}_{i}{c}_{j}^{-1} \) (see Exercise 27), proving the proposition. Note that by Proposition 2.3.1, we have for \( i \leq j,{\mathfrak{a}}_{i}{\mathrm{c}}_{j}^{-1} \subset \) \( {\mathfrak{a}}_{i}{\mathfrak{c}}_{i}^{-1} \subset {\mathbb{Z}}_{K} \) ; hence \( {\mathfrak{a}}_{i}{\mathfrak{c}}_{j}^{-1} \) is an integral ideal.
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Yes
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Proposition 2.3.5. Keep the notation of Proposition 2.3.1, in particular that \( I \) is an ideal of \( L,\left( {{\gamma }_{j},{\mathfrak{c}}_{j}}\right) \) is a pseudo-basis of \( I \), and the matrix \( \left( {h}_{i, j}\right) \) of the \( {\gamma }_{j} \) on the \( {\omega }_{i} \) is in HNF.\n\n(1) The content \( c\left( I\right) \) of \( I \) is given by\n\n\[ c\left( I\right) = \mathop{\sum }\limits_{{1 \leq i \leq j \leq n}}{h}_{i, j}{\mathfrak{c}}_{j}{\mathfrak{q}}_{i} = \mathop{\sum }\limits_{{1 \leq i \leq j \leq n}}{h}_{i, j}{\mathfrak{c}}_{j}{\mathfrak{a}}_{i}^{-1}. \]\n\n(2) The ideal \( I \) is an integral ideal of \( {\mathbb{Z}}_{L} \) if and only if \( c\left( I\right) \) is an integral ideal of \( {\mathbb{Z}}_{K} \) .\n\n(3) The ideal \( I \) is primitive if and only if \( c\left( I\right) = {\mathbb{Z}}_{K} \) .
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Proof. The proof follows immediately from Proposition 2.3.1 and is left to the reader (Exercise 28).
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No
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Lemma 2.3.7. Let \( I \) be an integral ideal of \( {\mathbb{Z}}_{L} \). Let \( \alpha \in K \), let \( \mathfrak{a} \) be a fractional ideal of \( {\mathbb{Z}}_{K} \) such that \( \alpha \mathfrak{a} \subset I \), and assume that \[ {\mathcal{N}}_{L/K}\left( I\right) + {\mathcal{N}}_{L/K}\left( {\alpha \mathfrak{a}}\right) {\left( {\mathcal{N}}_{L/K}\left( I\right) \right) }^{-1} = {\mathbb{Z}}_{K}. \] Then \( I = {\mathcal{N}}_{L/K}\left( I\right) {\mathbb{Z}}_{L} + \alpha \mathfrak{a}{\mathbb{Z}}_{L} \); in other words, \[ \left( {\left( {1,{\mathcal{N}}_{L/K}\left( I\right) }\right) ,\left( {\alpha ,\mathfrak{a}}\right) }\right) \] is a pseudo-two-element representation of \( I \).
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Proof. We have a \( {\mathbb{Z}}_{K} \) -module isomorphism \[ {\mathbb{Z}}_{L}/I \simeq {\bigoplus }_{i}{\mathbb{Z}}_{K}/{\mathfrak{d}}_{i} \] with \( {\mathcal{N}}_{L/K}\left( I\right) = \mathop{\prod }\limits_{i}{\mathfrak{d}}_{i} \). Since the \( {\mathfrak{d}}_{i} \) are integral ideals, we have \[ {\mathcal{N}}_{L/K}\left( I\right) \cdot \left( {{\mathbb{Z}}_{K}/{\mathfrak{d}}_{i}}\right) = \{ \overline{0}\} \] for all \( i \), hence \( {\mathcal{N}}_{L/K}\left( I\right) \cdot \left( {{\mathbb{Z}}_{L}/I}\right) = \{ \overline{0}\} \), in other words \( {\mathcal{N}}_{L/K}\left( I\right) \in I \). Since \( \alpha \mathfrak{a} \subset I \) and \( I \) is an ideal, it follows that \( {\mathcal{N}}_{L/K}\left( I\right) {\mathbb{Z}}_{L} + \alpha \mathfrak{a}{\mathbb{Z}}_{L} \subset I \). Conversely, let \( J = {\mathcal{N}}_{L/K}\left( I\right) {\mathbb{Z}}_{L} + \alpha \mathfrak{a}{\mathbb{Z}}_{L} \), let \( \mathfrak{P} \) be a prime ideal of \( {\mathbb{Z}}_{L} \), and let \( \mathfrak{p} = \mathfrak{P} \cap {\mathbb{Z}}_{K} \) be the prime ideal of \( {\mathbb{Z}}_{K} \) below \( \mathfrak{P} \). We will show that \( {v}_{\mathfrak{P}}\left( J\right) \leq {v}_{\mathfrak{P}}\left( I\right) \) for all \( \mathfrak{P} \), which will show that \( I \subset J \), and hence the equality \( I = J \), as claimed in the lemma. If \( {v}_{\mathfrak{P}}\left( J\right) = 0 \), there is nothing to prove since \( I \) is an integral ideal, so assume that \( {v}_{\mathfrak{P}}\left( J\right) > 0 \). Since \[ {v}_{\mathfrak{P}}\left( J\right) = \min \left( {{v}_{\mathfrak{P}}\left( {{\mathcal{N}}_{L/K}\left( I\right) }\right) ,{v}_{\mathfrak{P}}\left( {\alpha \mathfrak{a}}\right) }\right) , \] we have \( \mathfrak{P} \mid {\mathcal{N}}_{L/K}\left( I\right) \) and hence \( \mathfrak{p} \mid {\mathcal{N}}_{L/K}\left( I\right) \). By assumption, this implies that \( \mathfrak{p} \nmid {\mathcal{N}}_{L/K}\left( {\alpha \mathfrak{a}}\right) {\left( {\mathcal{N}}_{L/K}\left( I\right) \right) }^{-1} \), or in other words that \( {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( {\alpha \mathfrak{a}{I}^{-1}}\right) }\right) = \) 0. This means that \( {v}_{\mathfrak{Q}}\left( {\alpha \mathfrak{a}}\right) = {v}_{\mathfrak{Q}}\left( I\right) \) for all prime ideals \( \mathfrak{Q} \) above \( \mathfrak{p} \), and in particular for \( \mathfrak{Q} = \mathfrak{P} \). Thus, when \( \mathfrak{P} \mid J \), \[ {v}_{\mathfrak{P}}\left( J\right) = \min \left( {{v}_{\mathfrak{P}}\left( {{\mathcal{N}}_{L/K}\left( I\right) }\right) ,{v}_{\mathfrak{P}}\left( I\right) }\right) \leq {v}_{\mathfrak{P}}\left( I\right) , \] proving the lemma.
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Yes
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Lemma 2.3.10. Let \( \mathfrak{P} \) be a prime ideal of \( L \) above a prime ideal \( \mathfrak{p} \) of \( K \) , let \( f = f\left( {\mathfrak{P}/\mathfrak{p}}\right) = {\dim }_{{\mathbb{Z}}_{K}/\mathfrak{p}}\left( {{\mathbb{Z}}_{L}/\mathfrak{P}}\right) \) be the residual degree of \( \mathfrak{P} \), and finally let \( \alpha \in L \) and a be a fractional ideal of \( K \) such that \( \alpha \mathfrak{a} \subset \mathfrak{P} \) . Let \( \pi \) be any element of \( {\mathfrak{{pa}}}^{-1} \) such that \( {v}_{\mathfrak{p}}\left( {\pi \mathfrak{a}}\right) = 1 \) . Then
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\[ \mathfrak{P} = \left( {\left( {1,\mathfrak{p}}\right) ,\left( {\alpha ,\mathfrak{a}}\right) }\right) = \mathfrak{p}{\mathbb{Z}}_{L} + \alpha \mathfrak{a}{\mathbb{Z}}_{L} \] if and only if \( {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( {\alpha \mathfrak{a}}\right) }\right) = f \) or \( {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( {\left( {\alpha + \pi }\right) \mathfrak{a}}\right) = f}\right. \) . Proof. Assume first that \( {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( {\alpha \mathfrak{a}}\right) }\right) = f \) . Since \( \mathfrak{P} \mid \alpha \mathfrak{a} \), we can write \( \alpha \mathfrak{a} = \mathfrak{P}I \) for some integral ideal \( I \) of \( {\mathbb{Z}}_{L} \) . Since \( {\mathcal{N}}_{L/K}\left( \mathfrak{P}\right) = {\mathfrak{p}}^{f} \) by Lemma 2.2.14, we have \( {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( I\right) }\right) = 0 \) . This means that \( {v}_{\mathfrak{Q}}\left( I\right) = 0 \) for all prime ideals \( \mathfrak{Q} \) above \( \mathfrak{p} \), including \( \mathfrak{P} \) . Thus \( {v}_{\mathfrak{P}}\left( {\alpha \mathfrak{a}}\right) = 1 \) and \( {v}_{\mathfrak{Q}}\left( {\alpha \mathfrak{a}}\right) = 0 \) for the other \( \mathfrak{Q} \) above \( \mathfrak{p} \) . Since \( {v}_{\mathfrak{P}}\left( \mathfrak{p}\right) \geq 1 \), this implies that \( \min \left( {{v}_{\mathfrak{Q}}\left( \mathfrak{p}\right) ,{v}_{\mathfrak{Q}}\left( {\alpha \mathfrak{a}}\right) }\right) = 0 \) for all prime ideals \( \mathfrak{Q} \) different from \( \mathfrak{P} \) (and not only for those above \( \mathfrak{p} \) ) and is equal to 1 for \( \mathfrak{Q} = \mathfrak{P} \), thus showing the equality \( \mathfrak{P} = {\mathfrak{{pZ}}}_{L} + \alpha \mathfrak{a}{\mathbb{Z}}_{L} \) . Assume now that \( {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( {\left( {\alpha + \pi }\right) \mathfrak{a}}\right) = f}\right. \) . Since \( \left( {\alpha + \pi }\right) \mathfrak{a} \subset \alpha \mathfrak{a} + \pi \mathfrak{a} \subset \) \( \mathfrak{P} + \mathfrak{p} \subset \mathfrak{P} \), we conclude as above that \( \mathfrak{P} = \mathfrak{p}{\mathbb{Z}}_{L} + \left( {\alpha + \pi }\right) \mathfrak{a}{\mathbb{Z}}_{L} \), and this is equal to \( \mathfrak{p}{\mathbb{Z}}_{L} + \alpha \mathfrak{a}{\mathbb{Z}}_{L} \) since \( \pi \mathfrak{a} \subset \mathfrak{p} \) .
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Yes
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Lemma 2.3.12. We have \( \mathfrak{p}{\mathfrak{P}}^{-1} = \mathfrak{p}{\mathbb{Z}}_{L} + \beta \mathfrak{b}{\mathbb{Z}}_{L} \) if and only if \( \beta \mathfrak{b}\mathfrak{P} \subset \mathfrak{p}{\mathbb{Z}}_{L} \) and \( \beta \mathfrak{b} ⊄ \mathfrak{p}{\mathbb{Z}}_{L} \) .
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Proof. Assume first that \( \mathfrak{p}{\mathfrak{P}}^{-1} = \mathfrak{p}{\mathbb{Z}}_{L} + \beta \mathfrak{b}{\mathbb{Z}}_{L} \) . Thus \( \beta \mathfrak{b} \subset \mathfrak{p}{\mathfrak{P}}^{-1} \) ; hence \( \beta \mathfrak{{bP}} \subset \mathfrak{p}{\mathbb{Z}}_{L} \) . Furthermore, if we had \( \beta \mathfrak{b} \subset \mathfrak{p}{\mathbb{Z}}_{L} \), we would have \( \mathfrak{p}{\mathfrak{P}}^{-1} = \mathfrak{p}{\mathbb{Z}}_{L} \) , hence \( {\mathfrak{P}}^{-1} = {\mathbb{Z}}_{L} \), which is impossible since \( \mathfrak{P} \neq {\mathbb{Z}}_{L} \) . Thus, \( \beta \mathfrak{b} ⊄ \mathfrak{p}{\mathbb{Z}}_{L} \) .\n\nConversely, assume that \( \beta \mathfrak{{bP}} \subset \mathfrak{p}{\mathbb{Z}}_{L} \) and \( \beta \mathfrak{b} ⊄ \mathfrak{p}{\mathbb{Z}}_{L} \) . Then \( \mathfrak{P} \subset \mathfrak{P} + \) \( \beta \mathfrak{{bP}}{\mathfrak{p}}^{-1} \subset {\mathbb{Z}}_{L} \), and since \( \mathfrak{P} \) is a maximal ideal, it follows that \( \mathfrak{P} + \beta \mathfrak{{bP}}{\mathfrak{p}}^{-1} \) is equal either to \( \mathfrak{P} \) or to \( {\mathbb{Z}}_{L} \) . But \( \mathfrak{P} + \beta \mathfrak{b}\mathfrak{P}{\mathfrak{p}}^{-1} = \mathfrak{P} \) implies \( \beta \mathfrak{b}\mathfrak{P}{\mathfrak{p}}^{-1} \subset \mathfrak{P} \) ; hence \( \beta \mathfrak{b} \subset \mathfrak{p}{\mathbb{Z}}_{L} \) since \( \mathfrak{P} \) is invertible, contrary to our assumption. Thus, \( \mathfrak{P} + \beta \mathfrak{{bP}}{\mathfrak{p}}^{-1} = {\mathbb{Z}}_{L} \), and hence \( \mathfrak{p}{\mathbb{Z}}_{L} + \beta \mathfrak{b}{\mathbb{Z}}_{L} = \mathfrak{p}{\mathfrak{P}}^{-1} \), as claimed.
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Yes
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Proposition 2.3.15. Let \( I = \alpha \mathfrak{a}{\mathbb{Z}}_{L} + \beta \mathfrak{b}{\mathbb{Z}}_{L} \) be a pseudo-two-element representation of an ideal \( I \), and let \( k \) be a nonnegative integer. Then \[ {I}^{k} = {\alpha }^{k}{\mathfrak{a}}^{k}{\mathbb{Z}}_{L} + {\beta }^{k}{\mathfrak{b}}^{k}{\mathbb{Z}}_{L} \] In the special case where \( \mathfrak{P} = \mathfrak{p}{\mathbb{Z}}_{L} + \alpha \mathfrak{a}{\mathbb{Z}}_{L} \) is a prime ideal, we even have \[ {\mathfrak{P}}^{k} = {\mathfrak{p}}^{s}{\mathbb{Z}}_{L} + {\alpha }^{k}{\mathfrak{a}}^{k}{\mathbb{Z}}_{L}\;\text{ with }\;s = \left\lceil \frac{k}{e\left( {\mathfrak{P}/\mathfrak{p}}\right) }\right\rceil . \]
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Proof. The equality \( I = \alpha {\mathbb{Z}}_{L} + \beta {\mathbb{{bZ}}}_{L} \) is equivalent to \[ {v}_{\mathfrak{P}}\left( I\right) = \min \left( {{v}_{\mathfrak{P}}\left( {\alpha \mathfrak{a}}\right) ,{v}_{\mathfrak{P}}\left( {\beta \mathfrak{b}}\right) }\right) \] for all prime ideals \( \mathfrak{P} \) of \( L \), and hence \[ {v}_{\mathfrak{P}}\left( {I}^{k}\right) = k{v}_{\mathfrak{P}}\left( I\right) = \min \left( {{v}_{\mathfrak{P}}\left( {{\alpha }^{k}{\mathfrak{a}}^{k}}\right) ,{v}_{\mathfrak{P}}\left( {{\beta }^{k}{\mathfrak{b}}^{k}}\right) }\right) , \] proving our first claim. In the case of a prime ideal \( \mathfrak{P} = \mathfrak{p}{\mathbb{Z}}_{L} + \alpha \mathfrak{a}{\mathbb{Z}}_{L} \), we have \( \min \left( {{v}_{\mathfrak{Q}}\left( \mathfrak{p}\right) ,{v}_{\mathfrak{Q}}\left( {\alpha \mathfrak{a}}\right) }\right) = 0 \) for any prime ideal \( \mathfrak{Q} \) different from \( \mathfrak{P} \), while \( \min \left( {{v}_{\mathfrak{P}}\left( \mathfrak{p}\right) ,{v}_{\mathfrak{P}}\left( {\alpha \mathfrak{a}}\right) }\right) = \min \left( {e\left( {\mathfrak{P}/\mathfrak{p}}\right) ,{v}_{\mathfrak{P}}\left( {\alpha \mathfrak{a}}\right) }\right) = 1. \) It follows that \( \min \left( {{v}_{\mathfrak{Q}}\left( {\mathfrak{p}}^{s}\right) ,{v}_{\mathfrak{Q}}\left( {{\alpha }^{k}{\mathfrak{a}}^{k}}\right) }\right) = 0 \) for any such prime ideal \( \mathfrak{Q} \) and any strictly positive \( s \) . For the prime ideal \( \mathfrak{P} \), we may assume that \( \mathfrak{P} \) is ramified; otherwise the result is not any stronger than the general claim. If \( e\left( {\mathfrak{P}/p}\right) > 1 \), we necessarily have \( {v}_{\mathfrak{P}}\left( {\alpha \mathfrak{a}}\right) = 1 \), and so \[ \min \left( {{v}_{\mathfrak{P}}\left( {\mathfrak{p}}^{s}\right) ,{v}_{\mathfrak{P}}\left( {{\alpha }^{k}{\mathfrak{a}}^{k}}\right) }\right) = \min \left( {s \cdot e\left( {\mathfrak{P}/p}\right), k}\right) = k \] if and only if \( s \geq k/e\left( {\mathfrak{P}/\mathfrak{p}}\right) \), proving the proposition.
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Yes
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Proposition 2.3.17 (Transitivity of the Different). Let \( L/K \) be a relative extension of number fields, and let \( k \) be a subfield of \( K \) (for example, \( k = \mathbb{Q}) \) . Then \( \mathfrak{D}\left( {L/k}\right) = \mathfrak{D}\left( {L/K}\right) \mathfrak{D}\left( {K/k}\right) \) .
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Proof. Let \( \mathfrak{a} \) be an ideal of \( L \) . Using the transitivity of the trace, by definition of the codifferent, we have\n\n\[ \mathfrak{a} \subset \mathfrak{D}{\left( L/K\right) }^{-1} \Leftrightarrow {\operatorname{Tr}}_{L/K}\left( \mathfrak{a}\right) \subset {\mathbb{Z}}_{K} \]\n\n\[ \Leftrightarrow \mathfrak{D}{\left( K/k\right) }^{-1}{\operatorname{Tr}}_{L/K}\left( \mathfrak{a}\right) \subset \mathfrak{D}{\left( K/k\right) }^{-1} \]\n\n\[ \Leftrightarrow {\operatorname{Tr}}_{K/k}\left( {\mathfrak{D}{\left( K/k\right) }^{-1}{\operatorname{Tr}}_{L/K}\left( \mathfrak{a}\right) }\right) \subset {\mathbb{Z}}_{k} \]\n\n\[ \Leftrightarrow {\mathrm{{Tr}}}_{K/k}\left( {{\mathrm{{Tr}}}_{L/K}\left( {\mathfrak{D}{\left( K/k\right) }^{-1}\mathfrak{a}}\right) }\right) \subset {\mathbb{Z}}_{k} \]\n\n\[ \Leftrightarrow {\operatorname{Tr}}_{L/k}\left( {\mathfrak{D}{\left( K/k\right) }^{-1}\mathfrak{a}}\right) \subset {\mathbb{Z}}_{k} \]\n\n\[ \Leftrightarrow \mathfrak{D}{\left( K/k\right) }^{-1}\mathfrak{a} \subset \mathfrak{D}{\left( L/k\right) }^{-1} \]\n\n\[ \Leftrightarrow \mathfrak{a} \subset \mathfrak{D}\left( {K/k}\right) \mathfrak{D}{\left( L/k\right) }^{-1}. \]\n\nIt follows that \( \mathfrak{D}{\left( L/K\right) }^{-1} = \mathfrak{D}\left( {K/k}\right) \mathfrak{D}{\left( L/k\right) }^{-1} \), proving the proposition.
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Yes
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Proposition 2.3.18. Let \( \left( {{\omega }_{i},{\mathfrak{a}}_{i}}\right) \) be an integral pseudo-basis of \( {\mathbb{Z}}_{L} \), and let \( I \) be an ideal of \( {\mathbb{Z}}_{L} \) given by a pseudo-matrix \( \left( {M,{\mathfrak{c}}_{i}}\right) \), where the columns of \( \left( {M,{\mathfrak{c}}_{i}}\right) \) give the coordinates of a pseudo-basis \( \left( {{\gamma }_{i},{\mathfrak{c}}_{i}}\right) \) on the \( {\omega }_{i} \). If \( T = \left( {{\operatorname{Tr}}_{L/K}\left( {{\omega }_{i}{\omega }_{j}}\right) }\right) \), the pseudo-matrix \( \left( {{\left( {M}^{t}T\right) }^{-1},{\mathfrak{c}}_{i}^{-1}}\right) \) represents a pseudo-basis of the ideal \( {I}^{-1}\mathfrak{D}{\left( L/K\right) }^{-1} \) on the \( {\omega }_{i} \).
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Proof. The proof is almost identical to the absolute case. By definition of \( M \), the entry of row \( i \) and column \( j \) in \( {M}^{t}T \) is equal to \( {\operatorname{Tr}}_{L/K}\left( {{\gamma }_{i}{\omega }_{j}}\right) \). If \( V = \left( {v}_{i}\right) \) is a column vector with \( {v}_{i} \in K \), then \( V \) belongs to the image of the pseudo-matrix \( \left( {{\left( {M}^{t}T\right) }^{-1},{\mathfrak{c}}_{i}^{-1}}\right) \) if and only if \( {M}^{t}{TV} \) is a vector \( \left( {x}_{i}\right) \) with \( {x}_{i} \in {\mathfrak{c}}_{i}^{-1} \). This implies that for all \( i \), \[ {\operatorname{Tr}}_{L/K}\left( {{\gamma }_{i}{\mathfrak{c}}_{i}\left( {\mathop{\sum }\limits_{j}{v}_{j}{\omega }_{j}}\right) }\right) \subset {\mathbb{Z}}_{K} \] hence that \( {\operatorname{Tr}}_{L/K}\left( {xI}\right) \subset {\mathbb{Z}}_{K} \) with \( x = \mathop{\sum }\limits_{j}{v}_{j}{\omega }_{j} \). Since \( {xI} = {xI}{\mathbb{Z}}_{L} \), the proposition follows. Note that, in the same way that \( \mathfrak{D}{\left( L/K\right) }^{-1} \) is the dual of \( {\mathbb{Z}}_{L} \) for the trace form, the ideal \( {I}^{-1}\mathfrak{D}{\left( L/K\right) }^{-1} \) is the dual of the ideal \( I \) for the trace form.
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Yes
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Lemma 2.3.20. Let \( I = \alpha \mathfrak{a}{\mathbb{Z}}_{L} + \beta \mathfrak{b}{\mathbb{Z}}_{L} \) be a pseudo-two-element representation of an ideal of \( L \) . Then \[ {I}^{-1} = \left( {{\alpha }^{-1}{\mathfrak{a}}^{-1}{\mathbb{Z}}_{L}}\right) \cap \left( {{\beta }^{-1}{\mathfrak{b}}^{-1}{\mathbb{Z}}_{L}}\right) . \]
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Proof. Indeed, by looking at valuations, it is clear that for any two ideals \( I \) and \( J \) in a Dedekind domain we have \( \left( {I + J}\right) \left( {I \cap J}\right) = {IJ} \) (which is the generalization to ideals of the formula \( \gcd \left( {a, b}\right) \operatorname{lcm}\left( {a, b}\right) = {ab} \) ). Applying this to the two ideals \( {\alpha }^{-1}{\mathfrak{a}}^{-1}{\mathbb{Z}}_{L} \) and \( {\beta }^{-1}{\mathfrak{b}}^{-1}{\mathbb{Z}}_{L} \) of \( {\mathbb{Z}}_{L} \) and multiplying by \( {\alpha \beta }\mathfrak{{ab}} \) immediately gives the desired result.
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Yes
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Proposition 2.4.3. Let \( p \) be the prime number below \( \mathfrak{p} \), assume that \( p > \) \( n = \left\lbrack {L : K}\right\rbrack \), and let \( \alpha \in \mathcal{O} \) . The following three properties are equivalent.\n\n(1) \( \alpha \in {I}_{\mathfrak{p}} \) .\n\n(2) The characteristic polynomial \( {C}_{\alpha }\left( X\right) \) of \( \alpha \) over \( K \) satisfies \( {C}_{\alpha }\left( X\right) \equiv {X}^{n} \) \( \left( {\;\operatorname{mod}\;\mathfrak{p}}\right) \) .\n\n(3) For all \( \beta \in \mathcal{O} \) we have \( {\operatorname{Tr}}_{L/K}\left( {\alpha \beta }\right) \in \mathfrak{p} \) .
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Proof. The proof that (1) implies (2) is the same as the proof of Proposition 2.4.2 (see [Coh0], Lemma 6.1.6): if \( \alpha \in {I}_{\mathfrak{p}} \), multiplication by \( \alpha \) induces a nilpotent map from the \( {\mathbb{Z}}_{K}/\mathfrak{p} \) -vector space \( \mathcal{O}/\mathfrak{p}\mathcal{O} \) to itself, hence its eigenvalues are all equal to 0, so its characteristic polynomial is equal to \( {X}^{n} \) modulo p. Conversely, (2) implies (1) by the Cayley-Hamilton theorem (note that the equivalence of (1) and (2) does not use the condition \( p > n \) ).\n\nAssume now that \( \alpha \in {I}_{\mathfrak{p}} \) . Then for all \( \beta \in \mathcal{O} \) we have \( {\alpha \beta } \in {I}_{\mathfrak{p}} \) ; hence by what we have just proved, \( {C}_{\alpha \beta }\left( X\right) \equiv {X}^{n}\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) \), and in particular \( {\operatorname{Tr}}_{L/K}\left( {\alpha \beta }\right) \in \mathfrak{p} \) .\n\nConversely, assume (3). If we apply (3) to \( \beta = {\alpha }^{k - 1} \) for \( k \geq 1 \), we deduce that \( {\operatorname{Tr}}_{L/K}\left( {\alpha }^{k}\right) \in \mathfrak{p} \) for all \( k \geq 1 \) . Let \( {C}_{\alpha }\left( X\right) = {X}^{n} + \mathop{\sum }\limits_{{j = 1}}^{n}{\left( -1\right) }^{j}{a}_{j}{X}^{n - j} \) be the characteristic polynomial of \( \alpha \), where the \( {a}_{j} \in {\mathbb{Z}}_{K} \) are the elementary symmetric functions of \( \alpha \) . Newton’s relations between elementary symmetric functions and sums of powers give the recursion\n\n\[ k{a}_{k} = \mathop{\sum }\limits_{{j = 1}}^{k}{\left( -1\right) }^{j - 1}{a}_{k - j}{\operatorname{Tr}}_{L/K}\left( {\alpha }^{j}\right) . \]\n\nSince \( {\operatorname{Tr}}_{L/K}\left( {\alpha }^{j}\right) \in \mathfrak{p} \) for \( j \geq 1 \), it follows by induction that \( {a}_{k} \in \mathfrak{p} \) for \( k < p \) , since \( k < p \) implies \( k \notin \mathfrak{p} \) . Since \( p \) has been assumed to be larger than \( n \), it follows that \( {a}_{k} \in \mathfrak{p} \) for \( 1 \leq k \leq n \), proving (2).
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Yes
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Proposition 2.4.6. Let \( \left( {{\omega }_{1},\ldots ,{\omega }_{m}}\right) \) be a \( \mathbb{Z} \) -integral basis of a number field \( K \), let \( \mathfrak{p} \) be a prime ideal of degree \( f \) of \( {\mathbb{Z}}_{K} \), and let \( A = {\left( {a}_{i, j}\right) }_{1 \leq i, j \leq m} \) be its Hermite normal form on the integral basis. Let \( {D}_{1} \) (resp., \( {D}_{p} \) ) be the set of indices \( i \in \left\lbrack {1, m}\right\rbrack \) such that \( {a}_{i, i} = 1 \) (resp., \( {a}_{i, i} = p \) ). Then\n\n(1) \( \left| {D}_{p}\right| = f \) and \( \left| {D}_{1}\right| = m - f \) ;\n\n(2) if \( i \in {D}_{1} \), then \( {a}_{i, j} = 0 \) for \( j > i \) (each off-diagonal entry of row \( i \) is equal to zero);\n\n(3) if \( j \in {D}_{p} \), then \( {a}_{i, j} = 0 \) for \( i < j \) (each off-diagonal entry of column \( j \) is equal to zero).
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Proof. Call \( {\alpha }_{j} \) the HNF basis elements of \( \mathfrak{p} \) given by the matrix \( A \) . The determinant of \( A \) is equal to the index of \( \mathfrak{p} \) in \( {\mathbb{Z}}_{K} \), hence is equal to \( \mathcal{N}\left( \mathfrak{p}\right) = {p}^{f} \) , so the diagonal entries of the HNF matrix must be powers of \( p \) . Assume that \( {a}_{j, j} = {p}^{k} \) . Since \( p{\mathbb{Z}}_{K} \subset \mathfrak{p} \), we have \( p{\omega }_{j} = \mathop{\sum }\limits_{{1 < i < m}}{x}_{i}{\alpha }_{i} \) for some integers \( {x}_{i} \) . Since the matrix \( A \) is triangular, we deduce that \( {x}_{i} = 0 \) for \( i > j \), and in addition \( p = {x}_{j}{a}_{j, j} = {x}_{j}{p}^{k} \), and it follows that \( k = 0 \) or \( k = 1 \), so the diagonal entries are equal to 1 or \( p \) . Since the determinant is equal to \( {p}^{f} \), we have \( f \) diagonal entries equal to \( p \), proving (1). (2) is a trivial consequence of the definition of the HNF.\n\nFor (3), let \( j \) be such that \( {a}_{j, j} = p \) . Then\n\n\[ \mathop{\sum }\limits_{{i < j}}{a}_{i, j}{\omega }_{i} = {\alpha }_{j} - p{\omega }_{j} \in \mathfrak{p} \]\n\nLet \( {i}_{0} \) be the largest index \( i \) in the sum (if it exists) such that \( {a}_{i, j} \neq 0 \) . Thus,\n\n\[ \beta = {a}_{{i}_{0}, j}{\omega }_{{i}_{0}} + \mathop{\sum }\limits_{{i < {i}_{0}}}{a}_{i, j}{\omega }_{i} \in \mathfrak{p}. \]\n\nSince \( {a}_{{i}_{0}, j} \neq 0 \), by definition of the HNF we have \( {a}_{{i}_{0},{i}_{0}} > 1 \), so \( {a}_{{i}_{0},{i}_{0}} = p \) . But then, once again writing \( \beta = \sum {x}_{i}{\alpha }_{i} \), we obtain \( {x}_{i} = 0 \) for \( i > {i}_{0} \) and \( {x}_{{i}_{0}}{a}_{{i}_{0},{i}_{0}} = {x}_{{i}_{0}}p = {a}_{{i}_{0}, j} \) . Since \( 0 \leq {a}_{{i}_{0}, j} < p \), this implies that \( {a}_{{i}_{0}, j} = 0 \), a contradiction. It follows that \( {\alpha }_{j} = p{\omega }_{j} \), which is (3).
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Yes
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Corollary 2.4.7. Keep the notation of the preceding proposition. The classes modulo \( \mathfrak{p} \) of the \( {\omega }_{i} \) for \( i \in {D}_{p} \) form an \( {\mathbb{F}}_{p} \) -basis of \( {\mathbb{Z}}_{K}/\mathfrak{p} \) .
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Proof. Since \( \left| {D}_{p}\right| = f = {\dim }_{{\mathbb{F}}_{p}}\left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) \), we must simply show that the classes \( \overline{{\omega }_{i}} \) for \( i \in {D}_{p} \) are \( {\mathbb{F}}_{p} \) -linearly independent. Assume that \( \mathop{\sum }\limits_{{i \in {D}_{p}}}\overline{{x}_{i}{\omega }_{i}} = \) \( \overline{0} \), in other words that \( \mathop{\sum }\limits_{{i \in {D}_{p}}}{x}_{i}{\omega }_{i} \in \mathfrak{p} \), where we can assume that \( 0 \leq {x}_{i} < p. \) Using the same method as in the proof of the proposition, letting \( {i}_{0} \) be the largest index \( i \in {D}_{p} \) (if it exists) such that \( {x}_{i} \neq 0 \), the triangular form of the HNF implies that \( {a}_{{i}_{0},{i}_{0}} \mid {x}_{{i}_{0}} \) . Since \( 0 \leq {x}_{{i}_{0}} < p \) and \( {a}_{{i}_{0},{i}_{0}} = p \), we have \( {x}_{{i}_{0}} = 0 \), which is absurd, proving the corollary.
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Yes
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Theorem 2.4.8. Let \( L/K \) be a relative extension, with \( L = K\left( \theta \right) \) and \( \theta \) an algebraic integer whose minimal monic polynomial in \( K\left\lbrack X\right\rbrack \) is denoted \( T\left( X\right) \) ; let \( \mathfrak{p} \) be a prime ideal of \( {\mathbb{Z}}_{K} \), and let \( \beta \) be a uniformizer of \( {\mathfrak{p}}^{-1} \), so that \( \beta \in {\mathbb{Z}}_{K} \smallsetminus {\mathfrak{p}}^{-1} \) .\n\nLet \( \;\overline{T\left( X\right) } = \mathop{\prod }\limits_{{1 \leq i \leq k}}{\overline{{T}_{i}\left( X\right) }}^{{e}_{i}}\; \) be the factorization of \( \;\overline{T\left( X\right) }\; \) in \( \;\left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) \left\lbrack X\right\rbrack \) with the \( {T}_{i} \) monic. Set\n\n\[ g\left( X\right) = \mathop{\prod }\limits_{{1 \leq i \leq k}}{T}_{i}\left( X\right) ,\;h\left( X\right) = \mathop{\prod }\limits_{{1 \leq i \leq k}}{T}_{i}{\left( X\right) }^{{e}_{i} - 1}, \]\n\nso that \( g\left( X\right) h\left( X\right) - T\left( X\right) \in \mathfrak{p}\left\lbrack X\right\rbrack \) . Set\n\n\[ f\left( X\right) = \beta \cdot \left( {g\left( X\right) h\left( X\right) - T\left( X\right) }\right) \in {\mathbb{Z}}_{K}\left\lbrack X\right\rbrack , \]\n\nand let \( U \) be a monic lift of \( \bar{T}/\left( {\bar{f},\bar{g},\bar{h}}\right) \) to \( {\mathbb{Z}}_{K}\left\lbrack X\right\rbrack \) . The order given by Zassen-haus’s theorem starting with \( \mathcal{O} = {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \) is equal to\n\n\[ {\mathcal{O}}^{\prime } = {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack + {\mathfrak{p}}^{-1}U\left( \theta \right) {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \]\n\nIn particular, \( \mathcal{O} \) is \( \mathfrak{p} \) -maximal if and only if \( \left( {\bar{f},\bar{g},\bar{h}}\right) = 1 \) in \( \left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) \left\lbrack X\right\rbrack \) .
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## Remarks\n\n(1) The proof of this theorem is essentially identical to the one in the absolute case and is left to the reader (Exercise 33).\n\n(2) The result does not depend on the uniformizer \( \beta \) that we choose.\n\n(3) A more direct construction of \( {\mathcal{O}}^{\prime } \) can be obtained by generalizing [Coh0 (third printing), Exercise 3 of Chapter 6]; see Exercise 34.\n\n(4) Because of the presence of the ideal \( {\mathfrak{p}}^{-1},{\mathcal{O}}^{\prime } \) is not free in general. Using an HNF algorithm in Dedekind domains, we can obtain a pseudo-basis for \( {\mathcal{O}}^{\prime } \) if desired.
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No
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Theorem 2.5.1. Let \( L/K \) be an extension, and as usual let \( \mathfrak{d}\left( {L/K}\right) \) be the discriminant ideal of \( L/K \) . Denote by \( \left( {{r}_{1},{r}_{2}}\right) \) (resp., \( \left. \left( {{R}_{1},{R}_{2}}\right) \right) \) the signature of \( K \) (resp., \( L \) ). The absolute discriminant \( d\left( L\right) \) of \( L \) is given by the following formula:\n\n\[ d\left( L\right) = {\left( -1\right) }^{{R}_{2} - \left\lbrack {L : K}\right\rbrack {r}_{2}}d{\left( K\right) }^{\left\lbrack L : K\right\rbrack }{\mathcal{N}}_{K/\mathbb{Q}}\left( {\mathfrak{d}\left( {L/K}\right) }\right) ,\]\n\nwhere \( {R}_{2} - \left\lbrack {L : K}\right\rbrack {r}_{2} \) is given by Proposition 2.2.5.
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Proof. This theorem immediately follows from the transitivity of the different. Indeed, by Proposition 2.3.17, we have \( \mathfrak{D}\left( {L/\mathbb{Q}}\right) = \mathfrak{D}\left( {L/K}\right) \mathfrak{D}\left( {K/\mathbb{Q}}\right) \) . Taking norms, and using \( \mathfrak{d}\left( {L/K}\right) = {\mathcal{N}}_{L/K}\left( {\mathfrak{D}\left( {L/K}\right) }\right) \) for any extension \( L/K \) , we obtain\n\n\[ d\left( L\right) \mathbb{Z} = \mathfrak{d}\left( {L/\mathbb{Q}}\right) = {\mathcal{N}}_{L/\mathbb{Q}}\left( {\mathfrak{D}\left( {L/\mathbb{Q}}\right) }\right) = {\mathcal{N}}_{K/\mathbb{Q}}\left( {{\mathcal{N}}_{L/K}\left( {\mathfrak{D}\left( {L/K}\right) \mathfrak{D}\left( {K/\mathbb{Q}}\right) }\right) }\right) \]\n\n\[ = {\mathcal{N}}_{K/\mathbb{Q}}\left( {\mathfrak{d}\left( {L/K}\right) \mathfrak{D}{\left( K/\mathbb{Q}\right) }^{\left\lbrack L : K\right\rbrack }}\right) = {\mathcal{N}}_{K/\mathbb{Q}}\left( {\mathfrak{d}\left( {L/K}\right) }\right) d{\left( K\right) }^{\left\lbrack L : K\right\rbrack }.\]\n\nIt follows that \( d\left( L\right) = \pm d{\left( K\right) }^{\left\lbrack L : K\right\rbrack }{\mathcal{N}}_{K/\mathbb{Q}}\left( {\mathfrak{d}\left( {L/K}\right) }\right) .\; \) By [Coh 0, Proposition 4.8.11], we know that the sign of \( d\left( K\right) \) is \( {\left( -1\right) }^{{r}_{2}} \) and that of \( d\left( L\right) \) is \( {\left( -1\right) }^{{R}_{2}} \) , from which the theorem follows.
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Yes
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Proposition 2.6.3. Let \( I = \mathfrak{n}\left( {\mathfrak{a} \oplus {\mathfrak{q}}^{-1}\left( {\sqrt{D} - b}\right) }\right) \) as in Proposition 2.6.2.\n\n(1) The content of \( I \) in the sense of Definition 2.3.4 is the ideal \( \mathfrak{n} \).\n\n(2) The ideal \( I \) is an integral ideal of \( {\mathbb{Z}}_{L} \) if and only if \( \mathfrak{n} \) is an integral ideal of \( {\mathbb{Z}}_{K} \).\n\n(3) The ideal \( I \) is primitive in \( L/K \) if and only if \( \mathfrak{n} = {\mathbb{Z}}_{K} \).\n\n(4) \( {\mathcal{N}}_{L/K}\left( I\right) = {\mathfrak{{an}}}^{2} \).
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Proof. Note that with the notation of Proposition 2.3.5 we have \( {h}_{1,2} = \) \( \delta - b \), hence by that proposition \( c\left( I\right) = \left( {\mathfrak{n},\mathfrak{n}\mathfrak{a},\left( {\delta - b}\right) \mathfrak{n}{\mathfrak{q}}^{-1}}\right) = \mathfrak{n} \) since by Proposition 2.6.2, \( \mathfrak{a} \) is an integral ideal and \( \delta - b \in \mathfrak{q} \), proving (1), and (2) and (3) are immediate consequences. Statement (4) is a restatement of Proposition 2.3.1 (4).
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Yes
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Proposition 2.6.7. For \( i = 1,2 \), and 3, let \( {I}_{i} = {\mathfrak{n}}_{i}\left( {{\mathfrak{a}}_{i} \oplus {\mathfrak{q}}^{-1}\left( {\sqrt{D} - {b}_{i}}\right) }\right) \) be three ideals of \( L \), and assume that \( {I}_{3} = {I}_{1}{I}_{2} \) . Then \( {\mathfrak{n}}_{3},{\mathfrak{a}}_{3},{b}_{3} \) are given by the following formulas. Set \( \mathfrak{d} = {\mathfrak{a}}_{1} + {\mathfrak{a}}_{2} + {\mathfrak{q}}^{-1}\left( {{b}_{1} + {b}_{2}}\right) \) and let \( {a}_{1} \in {\mathfrak{a}}_{1}{\mathfrak{d}}^{-1} \) , \( {a}_{2} \in {\mathfrak{a}}_{2}{\mathfrak{d}}^{-1} \), and \( q \in {\mathfrak{q}}^{-1}{\mathfrak{d}}^{-1} \) such that \( {a}_{1} + {a}_{2} + q\left( {{b}_{1} + {b}_{2}}\right) = 1 \) . Then
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Proof. The proof is identical to that of the absolute case. We have \( {I}_{1}{I}_{2} = \) \( {\mathrm{n}}_{1}{\mathrm{n}}_{2}J \) with\n\n\[ J = {\mathfrak{a}}_{1}{\mathfrak{a}}_{2} + {\mathfrak{a}}_{1}{\mathfrak{q}}^{-1}\left( {\sqrt{D} - {b}_{2}}\right) + {\mathfrak{a}}_{2}{\mathfrak{q}}^{-1}\left( {\sqrt{D} - {b}_{1}}\right) + {\mathfrak{q}}^{-2}\left( {D + {b}_{1}{b}_{2} - \left( {{b}_{1} + {b}_{2}}\right) \sqrt{D}}\right) . \]\n\nThe ideal of coefficients of \( \sqrt{D} \) is\n\n\[ {q}^{-1}\left( {{\mathfrak{a}}_{1} + {\mathfrak{a}}_{2} + {q}^{-1}\left( {{b}_{1} + {b}_{2}}\right) }\right) = {q}^{-1}\mathfrak{d} = \left( {{\mathfrak{n}}_{3}/\left( {{\mathfrak{n}}_{1}{\mathfrak{n}}_{2}}\right) }\right) {q}^{-1}, \]\n\nhence \( {n}_{3} = {n}_{1}{n}_{2}\delta \), and by multiplicativity of the norm we know that \( {\mathcal{N}}_{L/K}\left( J\right) = {\mathfrak{a}}_{1}{\mathfrak{a}}_{2} = {\mathfrak{d}}^{2}{\mathfrak{a}}_{3} \), so \( {\mathfrak{a}}_{3} = {\mathfrak{a}}_{1}{\mathfrak{a}}_{2}{\mathfrak{d}}^{-2} \), as claimed.\n\nFinally, if \( {a}_{1} \in {\mathfrak{a}}_{1}{\mathfrak{d}}^{-1},{a}_{2} \in {\mathfrak{a}}_{2}{\mathfrak{d}}^{-1} \) and \( q \in {\mathfrak{q}}^{-1}{\mathfrak{d}}^{-1} \) are such that \( {a}_{1} + {a}_{2} + \) \( q\left( {{b}_{1} + {b}_{2}}\right) = 1 \), then\n\n\[ {a}_{1}\left( {\sqrt{D} - {b}_{2}}\right) + {a}_{2}\left( {\sqrt{D} - {b}_{1}}\right) - q\left( {D + {b}_{1}{b}_{2} - \left( {{b}_{1} + {b}_{2}}\right) \sqrt{D}}\right) \in J\mathfrak{q}, \]\n\nwhich is equal to \( \sqrt{D} - {b}_{3} \) with\n\n\[ {b}_{3} = {a}_{1}{b}_{2} + {a}_{2}{b}_{1} + q\left( {D + {b}_{1}{b}_{2}}\right) \]\n\n\[ = {b}_{2}\left( {1 - {a}_{2} - q\left( {{b}_{1} + {b}_{2}}\right) }\right) + {a}_{2}{b}_{1} + q\left( {D + {b}_{1}{b}_{2}}\right) \]\n\n\[ = {b}_{2} + {a}_{2}\left( {{b}_{1} - {b}_{2}}\right) + q\left( {D - {b}_{2}^{2}}\right) ,\]\n\nproving the proposition.
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Yes
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Corollary 2.6.9. If \( I = \mathfrak{n}\left( {\mathfrak{a} \oplus {\mathfrak{q}}^{-1}\left( {\sqrt{D} - b}\right) }\right) \), then \( {I}^{-1} = {\mathfrak{n}}^{-1}{\mathfrak{a}}^{-1}(\mathfrak{a} \oplus \) \( {\mathfrak{q}}^{-1}\left( {\sqrt{D} + b}\right) ) \) . In other words, in terms of pseudo-quadratic forms we have\n\n\[{\left( \mathfrak{a}, b,\mathfrak{c};\mathfrak{n}\right) }^{-1} = \left( {\mathfrak{a}, - b,\mathfrak{c};{\mathfrak{n}}^{-1}{\mathfrak{a}}^{-1}}\right) .\]
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Proof. By the above proposition, we have\n\n\[ \left( {\mathfrak{a} \oplus {\mathfrak{q}}^{-1}\left( {\sqrt{D} - b}\right) }\right) \left( {\mathfrak{a} \oplus {\mathfrak{q}}^{-1}\left( {\sqrt{D} + b}\right) }\right) = \mathfrak{d}\left( {{\mathfrak{a}}_{3} \oplus {\mathfrak{q}}^{-1}\left( {\sqrt{D} - {b}_{3}}\right) }\right) \]\n\nwith \( \mathfrak{d} = \mathfrak{a} + \mathfrak{a} + \left( {b - b}\right) {\mathfrak{q}}^{-1} = \mathfrak{a},{\mathfrak{a}}_{3} = {\mathfrak{a}}^{2}{\mathfrak{d}}^{-2} = {\mathbb{Z}}_{K} \) . In addition, we may choose \( {a}_{1} = 1 \in \mathfrak{a}{\mathfrak{d}}^{-1},{a}_{2} = 0, q = 0 \), so that \( {b}_{3} = {b}_{2} \equiv \delta \) (mod \( \mathfrak{q} \) ) by Proposition 2.6.2; hence\n\n\[ \left( {\mathfrak{a} \oplus {\mathfrak{q}}^{-1}\left( {\sqrt{D} - b}\right) }\right) \left( {\mathfrak{a} \oplus {\mathfrak{q}}^{-1}\left( {\sqrt{D} + b}\right) }\right) = \mathfrak{a}\left( {{\mathbb{Z}}_{K} \oplus {\mathfrak{q}}^{-1}\left( {\sqrt{D} - \delta }\right) }\right) = \mathfrak{a}{\mathbb{Z}}_{L}, \]\n\nand the first formula of the corollary follows. The second follows from the trivial observation that \( {\left( -b\right) }^{2} = {b}^{2} \) .
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Yes
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Proposition 2.6.10. Let \( \\left( {\\mathfrak{a}, b,\\mathfrak{c};\\mathfrak{n}}\\right) \) be a pseudo-quadratic form and \( I = \) \( \\mathfrak{n}\\left( {\\mathfrak{a} \\oplus {\\mathfrak{q}}^{-1}\\left( {\\sqrt{D} - b}\\right) }\\right) \) the corresponding ideal. We have the equality\n\n\\[ \n\\operatorname{naq}\\left( {\\mathfrak{c} \\oplus {\\mathfrak{q}}^{-1}\\left( {\\sqrt{D} + b}\\right) }\\right) = \\left( {\\sqrt{D} + b}\\right) I.\n\\]\n\nIn particular, the ideal class in \( {Cl}\\left( L\\right) \) of the ideal corresponding to \( \\left( {\\mathfrak{a}, b,\\mathfrak{c};\\mathfrak{n}}\\right) \) is equal to ideal class of the ideal corresponding to \( \\left( {\\mathfrak{c}, - b,\\mathfrak{a};\\mathfrak{{naq}}}\\right) \) .
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Proof. This follows trivially from the equality\n\n\\[ \n\\mathfrak{c} \\oplus {\\mathfrak{q}}^{-1}\\left( {\\sqrt{D} + b}\\right) = \\left( {\\sqrt{D} + b}\\right) {\\left( \\mathfrak{a}\\mathfrak{q}\\right) }^{-1}\\left( {\\mathfrak{a} \\oplus {\\mathfrak{q}}^{-1}\\left( {\\sqrt{D} - b}\\right) }\\right) .\n\\]
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Yes
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Proposition 3.2.3. We have the following five-term exact sequence\n\n\[ 1 \rightarrow {U}_{\mathfrak{m}}\left( K\right) \rightarrow U\left( K\right) \overset{\rho }{ \rightarrow }{\left( {\mathbb{Z}}_{K}/\mathfrak{m}\right) }^{ * }\overset{\psi }{ \rightarrow }C{l}_{\mathfrak{m}}\left( K\right) \overset{\phi }{ \rightarrow }{Cl}\left( K\right) \rightarrow 1 \]
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Proof. The kernel of \( \rho \) is by definition the set of units congruent to 1 (mod *m) and so is equal to \( {U}_{\mathrm{m}}\left( K\right) \) . Furthermore, the map \( \psi \) that we have just described is well-defined (for the other maps this is clear). Indeed, if \( \rho \left( \alpha \right) = \rho \left( \beta \right) \), this means that \( \alpha \equiv \beta \left( {\;\operatorname{mod}\;{\mathfrak{m}}_{0}}\right) \), that \( \alpha \) and \( \beta \) are coprime to \( {\mathfrak{m}}_{0} \), and that \( \operatorname{sign}\left( {{\sigma }_{i}\left( \alpha \right) }\right) = \operatorname{sign}\left( {{\sigma }_{i}\left( \beta \right) }\right) \) for \( {\sigma }_{i} \in {\mathfrak{m}}_{\infty } \) . These conditions mean precisely that \( \alpha /\beta \equiv 1\left( {{\;\operatorname{mod}\;{}^{ * }}\mathfrak{m}}\right) \), and so the principal ideals \( \alpha {\mathbb{Z}}_{K} \) and \( \beta {\mathbb{Z}}_{K} \) are in the same ideal class modulo \( {P}_{\mathrm{m}}\left( K\right) \) or, equivalently, have the same image in \( C{l}_{\mathfrak{m}}\left( K\right) \) .\n\nAssume now that \( \rho \left( \alpha \right) \in {\left( {\mathbb{Z}}_{K}/\mathfrak{m}\right) }^{ * } \) is sent to the unit element of \( C{l}_{\mathfrak{m}}\left( K\right) \) . This means that \( \alpha {\mathbb{Z}}_{K} \in {P}_{\mathfrak{m}}\left( K\right) \), so there exists \( \beta \equiv 1\left( {{\;\operatorname{mod}\; * }\mathfrak{\;m}}\right) \) such that \( \alpha {\mathbb{Z}}_{K} = \beta {\mathbb{Z}}_{K} \), hence \( u = \alpha /\beta \) is a unit; in other words, it belongs to \( U\left( K\right) \) . Since \( \beta \equiv 1\left( {{\;\operatorname{mod}\;{}^{ * }}\mathfrak{m}}\right) \), we have \( \bar{u} = \bar{\alpha } \) in \( {\left( {\mathbb{Z}}_{K}/{\mathfrak{m}}_{0}\right) }^{ * } \), but also sign \( \left( {{\sigma }_{i}\left( u\right) }\right) = \) \( \operatorname{sign}\left( {{\sigma }_{i}\left( \alpha \right) }\right) \) . Thus, \( \rho \left( u\right) = \rho \left( \alpha \right) \), and so the kernel of \( \psi \) is indeed equal to the image of \( \rho \) .\n\nNow let \( \overline{\mathfrak{a}} \) be an ideal class in \( C{l}_{\mathrm{m}}\left( K\right) \) which is sent to the trivial class in \( {Cl}\left( K\right) \) . This simply means that \( \mathfrak{a} = \alpha {\mathbb{Z}}_{K} \) is a principal ideal coprime to \( \mathfrak{m} \), hence \( \alpha \) is also coprime to \( \mathfrak{m} \), and this shows that the kernel of \( \phi \) is the image of \( \psi \) .\n\nFinally, the surjectivity of \( \phi \) follows from the approximation theorem in Dedekind domains, more precisely from Corollary 1.2.11, since one can choose as representative of an ideal class an ideal coprime to \( m \) .
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Yes
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Corollary 3.2.4. The ray class group is finite. Its cardinality, which we will denote by \( {h}_{\mathrm{m}}\left( K\right) \) (or simply by \( {h}_{\mathrm{m}} \) when the field is understood), is given by \[ {h}_{\mathfrak{m}}\left( K\right) = h\left( K\right) \frac{\phi \left( \mathfrak{m}\right) }{\left\lbrack U\left( K\right) : {U}_{\mathfrak{m}}\left( K\right) \right\rbrack }.\]
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Proof. The proof is clear from the proposition.
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No
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Lemma 3.3.1. Let \( {\mathfrak{m}}_{1} \) and \( {\mathfrak{m}}_{2} \) be two arbitrary moduli, and let \( \mathfrak{a} \in {I}_{{\mathfrak{m}}_{1}} \) . There exists \( \alpha \equiv 1\left( {{\;\operatorname{mod}\;{}^{ * }}{\mathfrak{m}}_{1}}\right) \) such that \( \alpha \mathfrak{a} \) is an integral ideal coprime to \( {m}_{1}{m}_{2} \) .
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Proof. For the infinite places, we of course simply ask that \( \sigma \left( \alpha \right) > 0 \) for all \( \sigma \mid {\mathfrak{m}}_{1} \) . For the finite places, let \( \mathfrak{p} \) be a prime ideal. If \( \mathfrak{p} \mid {\mathfrak{m}}_{1} \), we ask that \( {v}_{\mathfrak{p}}\left( {\alpha - 1}\right) \geq {v}_{\mathfrak{p}}\left( {\mathfrak{m}}_{1}\right) \) . Then, since \( \mathfrak{a} \in {I}_{{\mathfrak{m}}_{1}} \) we necessarily have \( {v}_{\mathfrak{p}}\left( \alpha \right) = 0 = \) \( - {v}_{\mathfrak{p}}\left( \mathfrak{a}\right) \) . If \( \mathfrak{p} \nmid {\mathfrak{m}}_{1} \) and \( \mathfrak{p} \mid {\mathfrak{m}}_{2} \), we ask that \( {v}_{\mathfrak{p}}\left( \alpha \right) = - {v}_{\mathfrak{p}}\left( \mathfrak{a}\right) \) . Finally, if \( \mathfrak{p} \nmid {\mathfrak{m}}_{1}{\mathfrak{m}}_{2} \) , we ask that \( {v}_{\mathfrak{p}}\left( \alpha \right) \geq - {v}_{\mathfrak{p}}\left( \mathfrak{a}\right) \) . Thus, the conditions are compatible. The strong approximation theorem (more precisely, Corollary 1.2.9) shows the existence of an \( \alpha \) with the desired properties, and it is clear that such an \( \alpha \) satisfies the conditions of the lemma.
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Yes
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Corollary 3.3.2. Let \( {\mathfrak{m}}_{1} \) and \( {\mathfrak{m}}_{2} \) be two arbitrary moduli.\n\n(1) We have \( {I}_{{\mathfrak{m}}_{2}} \subset {I}_{{\mathfrak{m}}_{1}}{P}_{{\mathfrak{m}}_{2}} \) (and, of course, also \( {I}_{{\mathfrak{m}}_{1}} \subset {I}_{{\mathfrak{m}}_{2}}{P}_{{\mathfrak{m}}_{1}} \) ).\n\n(2) If \( {\mathfrak{m}}_{2} \mid {\mathfrak{m}}_{1} \) and \( {C}_{2} \) is a congruence subgroup modulo \( {\mathfrak{m}}_{2} \) (for example, \( \left. {{C}_{2} = {P}_{{\mathfrak{m}}_{2}}}\right) \), then we have the equality \( {I}_{{\mathfrak{m}}_{2}} = {I}_{{\mathfrak{m}}_{1}}{C}_{2} \) .
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Proof. If \( \mathfrak{a} \in {I}_{{\mathfrak{m}}_{2}} \), by the above lemma, we can find \( \alpha \equiv 1\left( {{\;\operatorname{mod}\;{}^{ * }}{\mathfrak{m}}_{2}}\right) \) such that \( \alpha \mathfrak{a} \in {I}_{{\mathfrak{m}}_{1}} \) . Since \( \alpha {\mathbb{Z}}_{K} \in {P}_{{\mathfrak{m}}_{2}} \), we thus have \( \mathfrak{a} \in {I}_{{\mathfrak{m}}_{1}}{P}_{{\mathfrak{m}}_{2}} \), so\n\n\[ {I}_{{\mathfrak{m}}_{2}} \subset {I}_{{\mathfrak{m}}_{1}}{P}_{{\mathfrak{m}}_{2}} \subset {I}_{{\mathfrak{m}}_{1}}{C}_{2} \]\n\nfor any congruence subgroup \( {C}_{2} \) modulo \( {\mathfrak{m}}_{2} \) . If \( {\mathfrak{m}}_{2} \mid {\mathfrak{m}}_{1} \), then \( {I}_{{\mathfrak{m}}_{1}} \subset {I}_{{\mathfrak{m}}_{2}} \) and \( {C}_{2} \subset {I}_{{\mathfrak{m}}_{2}} \), so the reverse inclusion is also valid, thus proving the corollary.
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Yes
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Proposition 3.3.4. (1) The relation \( \sim \) defined above between congruence subgroups is an equivalence relation.
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Proof. (1). The reflexivity and symmetry are trivial, so the only thing to prove is the transitivity. Assume that \( \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \sim \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \) and \( \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \sim \) \( \left( {{\mathfrak{m}}_{3},{C}_{3}}\right) \) ; in other words, that \( {I}_{{\mathfrak{m}}_{2}} \cap {C}_{1} = {I}_{{\mathfrak{m}}_{1}} \cap {C}_{2} \) and \( {I}_{{\mathfrak{m}}_{3}} \cap {C}_{2} = {I}_{{\mathfrak{m}}_{2}} \cap {C}_{3} \) . We must prove that \( \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \sim \left( {{\mathfrak{m}}_{3},{C}_{3}}\right) \) or, equivalently, that \( {I}_{{\mathfrak{m}}_{3}} \cap {C}_{1} = {I}_{{\mathfrak{m}}_{1}} \cap {C}_{3} \) .\n\nLet \( \mathfrak{a} \in {I}_{{\mathfrak{m}}_{3}} \cap {C}_{1} \) . Since \( \mathfrak{a} \in {C}_{1} \subset {I}_{{\mathfrak{m}}_{1}} \) we must only show that \( \mathfrak{a} \in {C}_{3} \) . By Lemma 3.3.1, since \( \mathfrak{a} \in {I}_{{\mathfrak{m}}_{1}{\mathfrak{m}}_{3}} \), we can find \( \alpha \equiv 1\left( {{\;\operatorname{mod}\;{}^{ * }}{\mathfrak{m}}_{1}{\mathfrak{m}}_{3}}\right) \) such that \( \alpha \mathfrak{a} \) is an integral ideal coprime to \( {\mathfrak{m}}_{1}{\mathfrak{m}}_{2}{\mathfrak{m}}_{3} \) . Since \( \alpha {\mathbb{Z}}_{K} \in {P}_{{\mathfrak{m}}_{1}} \subset {C}_{1} \) and \( {C}_{1} \) is a group, it follows that \( \alpha \mathfrak{a} \in {C}_{1} \), and since \( \alpha \mathfrak{a} \) is coprime to \( {\mathfrak{m}}_{2} \) we have \( \alpha \mathfrak{a} \in {I}_{{\mathfrak{m}}_{2}} \cap {C}_{1} = {I}_{{\mathfrak{m}}_{1}} \cap {C}_{2} \), so \( \alpha \mathfrak{a} \in {C}_{2} \) . But since \( \alpha \mathfrak{a} \) is also coprime to \( {\mathfrak{m}}_{3} \) , we have \( \alpha \mathfrak{a} \in {I}_{{\mathfrak{m}}_{3}} \cap {C}_{2} = {I}_{{\mathfrak{m}}_{2}} \cap {C}_{3} \), so \( \alpha \mathfrak{a} \in {C}_{3} \) . Finally, since \( \alpha {\mathbb{Z}}_{K} \in {P}_{{\mathfrak{m}}_{3}} \) and \( {C}_{3} \) is a group containing \( {P}_{{\mathfrak{m}}_{3}} \), we deduce that \( \mathfrak{a} = \alpha \mathfrak{a}{\alpha }^{-1} \in {C}_{3} \), as was to be proved. We have proved the inclusion \( {I}_{{\mathfrak{m}}_{3}} \cap {C}_{1} \subset {I}_{{\mathfrak{m}}_{1}} \cap {C}_{3} \), and the reverse inclusion follows by symmetry.
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Yes
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Proposition 3.3.5. (1) Let \( \\left( {{\\mathfrak{m}}_{1},{C}_{1}}\\right) \) be a congruence subgroup, and let \( {\\mathfrak{m}}_{2} \) be a divisor of \( {\\mathfrak{m}}_{1} \) (see Definition 3.2.1). There exists a congruence subgroup \( {C}_{2} \) modulo \( {\\mathfrak{m}}_{2} \) such that \( \\left( {{\\mathfrak{m}}_{1},{C}_{1}}\\right) \\sim \\left( {{\\mathfrak{m}}_{2},{C}_{2}}\\right) \) if and only if
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Proof. (1). If \( {\\mathfrak{m}}_{2} \\mid {\\mathfrak{m}}_{1} \), we have \( \\left( {{\\mathfrak{m}}_{1},{C}_{1}}\\right) \\sim \\left( {{\\mathfrak{m}}_{2},{C}_{2}}\\right) \) if and only if \( {I}_{{\\mathfrak{m}}_{1}} \\cap {C}_{2} = \) \( {C}_{1} \) . Thus, since \( {P}_{{\\mathfrak{m}}_{2}} \\subset {C}_{2} \), \[ {I}_{{\\mathfrak{m}}_{1}} \\cap {P}_{{\\mathfrak{m}}_{2}} \\subset {I}_{{\\mathfrak{m}}_{1}} \\cap {C}_{2} = {C}_{1} \] Furthermore, \[ {C}_{1}{P}_{{\\mathfrak{m}}_{2}} = \\left( {{I}_{{\\mathfrak{m}}_{1}} \\cap {C}_{2}}\\right) {P}_{{\\mathfrak{m}}_{2}} \\subset {C}_{2}{P}_{{\\mathfrak{m}}_{2}} = {C}_{2}. \] Set \( {C}_{2}^{\\prime } = {C}_{1}{P}_{{\\mathfrak{m}}_{2}} \) . Then I claim that \( \\left( {{\\mathfrak{m}}_{1},{C}_{1}}\\right) \\sim \\left( {{\\mathfrak{m}}_{2},{C}_{2}^{\\prime }}\\right) \) . Indeed, this means that \( {C}_{1} = {I}_{{\\mathfrak{m}}_{2}} \\cap {C}_{1} = {I}_{{\\mathfrak{m}}_{1}} \\cap {C}_{1}{P}_{{\\mathfrak{m}}_{2}} \) or, equivalently (since the other inclusion is obvious), that \( {I}_{{\\mathfrak{m}}_{1}} \\cap {C}_{1}{P}_{{\\mathfrak{m}}_{2}} \\subset {C}_{1} \) . But this follows from the inclusion \( {I}_{{\\mathfrak{m}}_{1}} \\cap {P}_{{\\mathfrak{m}}_{2}} \\subset {C}_{1} \) by multiplying both sides by the group \( {C}_{1} \) . Thus, since our equivalence relation is transitive, we have \( \\left( {{\\mathfrak{m}}_{2},{C}_{2}}\\right) \\sim \) \( \\left( {{\\mathfrak{m}}_{2},{C}_{2}^{\\prime }}\\right) \), which of course means that \( {C}_{2} = {C}_{2}^{\\prime } \), and so that \( {C}_{2} = {C}_{1}{P}_{{\\mathfrak{m}}_{2}} \), as claimed. Conversely, if we assume \( {I}_{{\\mathfrak{m}}_{1}} \\cap {P}_{{\\mathfrak{m}}_{2}} \\subset {C}_{1} \) and \( {C}_{2} = {C}_{1}{P}_{{\\mathfrak{m}}_{2}} \), then by multiplication by \( {C}_{1} \) we get as above \( {I}_{{\\mathfrak{m}}_{1}} \\cap {C}_{2} \\subset {C}_{1} \) ; since the reverse inclusion is trivial, we have equality, proving (1). Statement (2) is a trivial consequence of the definition.
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Yes
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Proposition 3.3.6. Let \( {\mathfrak{m}}_{1} \) and \( {\mathfrak{m}}_{2} \) be two moduli such that \( {\mathfrak{m}}_{2} \mid {\mathfrak{m}}_{1} \), and let \( {C}_{1} \) and \( {C}_{2} \) be two congruence subgroups modulo \( {\mathfrak{m}}_{1} \) and \( {\mathfrak{m}}_{2} \), respectively, such that \( {C}_{1} \subset {C}_{2} \). (1) We have a canonical isomorphism \[ {I}_{{\mathfrak{m}}_{2}}/{C}_{2} \simeq \frac{{I}_{{\mathfrak{m}}_{1}}/{C}_{1}}{\left( {{I}_{{\mathfrak{m}}_{1}} \cap {C}_{2}}\right) /{C}_{1}}. \] In particular, we have \[ \frac{{h}_{{\mathfrak{m}}_{1},{C}_{1}}}{{h}_{{\mathfrak{m}}_{2},{C}_{2}}} = \left| {\left( {{I}_{{\mathfrak{m}}_{1}} \cap {C}_{2}}\right) /{C}_{1}}\right| . \] (2) We have \( {h}_{{\mathfrak{m}}_{1},{C}_{1}} = {h}_{{\mathfrak{m}}_{2},{C}_{2}} \) if and only if \( \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \sim \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \).
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Proof. Applying Corollary 3.3.2, we have \[ \frac{{I}_{{\mathrm{m}}_{2}}}{{C}_{2}} \simeq \frac{{I}_{{\mathrm{m}}_{1}}{C}_{2}}{{C}_{2}} \simeq \frac{{I}_{{\mathrm{m}}_{1}}}{{I}_{{\mathrm{m}}_{1}} \cap {C}_{2}} \simeq \frac{{I}_{{\mathrm{m}}_{1}}/{C}_{1}}{\left( {{I}_{{\mathrm{m}}_{1}} \cap {C}_{2}}\right) /{C}_{1}}, \] proving (1). (2). We have already seen in Proposition 3.3.4 that if \( \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \sim \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \) , then \( {h}_{{\mathfrak{m}}_{1},{C}_{1}} = {h}_{{\mathfrak{m}}_{2},{C}_{2}} \) even when \( {\mathfrak{m}}_{2} \) does not divide \( {\mathfrak{m}}_{1} \) . Conversely, assume that we have this equality. By (1) we have \( {I}_{{\mathfrak{m}}_{1}} \cap {C}_{2} = {C}_{1} \), and in particular \( {I}_{{\mathfrak{m}}_{1}} \cap {P}_{{\mathfrak{m}}_{2}} \subset {C}_{1} \), so by Proposition 3.3.5 we deduce that \( \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \sim \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \).
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Yes
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Corollary 3.3.7. Let \( {\mathfrak{m}}_{1} \) and \( {\mathfrak{m}}_{2} \) be two moduli such that \( {\mathfrak{m}}_{2} \mid {\mathfrak{m}}_{1} \), let \( {C}_{1} \) be a congruence subgroup modulo \( {\mathfrak{m}}_{1} \), and let \( {C}_{2} = {C}_{1}{P}_{{\mathfrak{m}}_{2}} \) . Then \( \left| {\left( {{I}_{{\mathfrak{m}}_{1}} \cap {C}_{2}}\right) /{C}_{1}}\right| = {h}_{{\mathfrak{m}}_{1},{C}_{1}}/{h}_{{\mathfrak{m}}_{2},{C}_{2}} \) divides \( \phi \left( {\mathfrak{m}}_{1}\right) /\phi \left( {\mathfrak{m}}_{2}\right) \) .
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Proof. By the above proposition and Corollary 3.2.4, we have \[ \left| \frac{{I}_{{\mathfrak{m}}_{1}} \cap {C}_{2}}{{C}_{1}}\right| = \frac{{h}_{{\mathfrak{m}}_{1},{C}_{1}}}{{h}_{{\mathfrak{m}}_{2},{C}_{2}}} = \frac{\phi \left( {\mathfrak{m}}_{1}\right) }{\phi \left( {\mathfrak{m}}_{2}\right) }\frac{1}{\left\lbrack {{U}_{{\mathfrak{m}}_{2}}\left( K\right) : {U}_{{\mathfrak{m}}_{1}}\left( K\right) }\right\rbrack \left( {\left| \overline{{C}_{1}}\right| /\left| \overline{{C}_{2}}\right| }\right) }, \] where \( \overline{{C}_{i}} = {C}_{i}/{P}_{{\mathfrak{m}}_{i}} \) for \( i = 1 \) and \( i = 2 \) . Using the same proof as in (1) of the above proposition and the hypothesis \( {C}_{2} = {C}_{1}{P}_{{\mathrm{m}}_{2}} \) instead of Corollary 3.3.2, we find a canonical isomorphism \[ \overline{{C}_{2}} = {C}_{2}/{P}_{{\mathfrak{m}}_{2}} \simeq \frac{{C}_{1}/{P}_{{\mathfrak{m}}_{1}}}{\left( {{C}_{1} \cap {P}_{{\mathfrak{m}}_{2}}}\right) /{P}_{{\mathfrak{m}}_{1}}} = \frac{\overline{{C}_{1}}}{\left( {{C}_{1} \cap {P}_{{\mathfrak{m}}_{2}}}\right) /{P}_{{\mathfrak{m}}_{1}}}, \] showing in particular that \( \left| \overline{{C}_{2}}\right| \) divides \( \left| \overline{{C}_{1}}\right| \), and the corollary follows.
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Yes
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Proposition 3.3.9. Let \( \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \) and \( \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \) be two congruence subgroups such that \( \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \sim \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \), and let \( \mathfrak{n} = \gcd \left( {{\mathfrak{m}}_{1},{\mathfrak{m}}_{2}}\right) \) . There exists a unique \( \textit{congruence subgroup C}\textit{modulo}\;\mathfrak{n}\textit{ such that}\;\left( {\mathfrak{n}, C}\right) \sim \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \sim \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) ,\textit{ and } \) \( C \) is given by \( C = {C}_{1}{P}_{\mathrm{n}} = {C}_{2}{P}_{\mathrm{n}} \) . The congruence subgroup \( \left( {\mathrm{n}, C}\right) \) will be called the GCD of the congruence subgroups \( \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \) and \( \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \) (note that the GCD is defined only when the congruence subgroups are equivalent).
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Proof. Set \( \mathfrak{m} = {\mathfrak{m}}_{1}{\mathfrak{m}}_{2} \) . By Proposition 3.3.5 (2), if we set \( D = {I}_{\mathfrak{m}} \cap {C}_{1} = \) \( {I}_{\mathfrak{m}} \cap {C}_{2} \), we have \( \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \sim \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \sim \left( {\mathfrak{m}, D}\right) \) . Applying part (1) of the same proposition, we deduce that\n\n\[ \n{P}_{{\mathfrak{m}}_{1}} \cap {I}_{\mathfrak{m}} \subset D\;\text{ and }\;{P}_{{\mathfrak{m}}_{2}} \cap {I}_{\mathfrak{m}} \subset D.\n\]\n\nBy the same proposition, to show the existence of \( C \), we must show that \( {P}_{\mathfrak{n}} \cap {I}_{\mathfrak{m}} \subset D \) . Thus, let \( \mathfrak{a} \in {P}_{\mathfrak{n}} \cap {I}_{\mathfrak{m}} \) . Since \( \mathfrak{a} \in {P}_{\mathfrak{n}} \), there exists \( \alpha \equiv 1 \) (mod * \( n \) ) such that \( \mathfrak{a} = \alpha {\mathbb{Z}}_{K} \) . By Lemma 3.3.8, this implies the existence of \( \beta \in {K}^{ * } \) such that \( \beta \equiv \alpha {\;(\operatorname{mod}\;{}^{ * }{\mathfrak{m}}_{1})} \) and \( \beta \equiv 1{\;(\operatorname{mod}\;{}^{ * }{\mathfrak{m}}_{2})} \) . Since \( \mathfrak{a} \in {I}_{\mathfrak{m}},\;\alpha \) is coprime to \( \mathfrak{m} \) ; hence \( \beta \) is coprime both to \( {\mathfrak{m}}_{1} \) and to \( {\mathfrak{m}}_{2} \) and hence to \( \mathfrak{m} \) . It follows that \( \left( {\beta /\alpha }\right) {\mathbb{Z}}_{K} \in {P}_{{\mathfrak{m}}_{1}} \cap {I}_{\mathfrak{m}} \subset D \) . Since \( \beta {\mathbb{Z}}_{K} \in {P}_{{\mathfrak{m}}_{2}} \cap {I}_{\mathfrak{m}} \subset D \) and \( D \) is a group, we obtain \( \alpha {\mathbb{Z}}_{K} = \left( {\beta {\mathbb{Z}}_{K}}\right) {\left( \left( \beta /\alpha \right) {\mathbb{Z}}_{K}\right) }^{-1} \in D \), as was to be proved. Proposition 3.3.5 thus shows the existence of a unique congruence subgroup \( C \) modulo \( \mathfrak{n} \) such that \( \left( {\mathfrak{n}, C}\right) \sim \left( {\mathfrak{m}, D}\right) \), hence by transitivity \( \left( {\mathfrak{n}, C}\right) \sim \left( {{\mathfrak{m}}_{1},{C}_{1}}\right) \sim \) \( \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \), and the uniqueness statement of the same proposition implies that \( C = {C}_{1}{P}_{\mathrm{n}} = {C}_{2}{P}_{\mathrm{n}} \)
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Yes
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Corollary 3.3.10. Let \( \mathcal{C} \) be an equivalence class of congruence subgroups. There exists a congruence subgroup \( \left( {\mathfrak{f},{C}_{\mathfrak{f}}}\right) \in \mathcal{C} \) (called the conductor of the class) such that \( \mathcal{C} \) consists exactly of all congruence subgroups of the form \( \left( {\mathfrak{m},{C}_{\mathfrak{f}} \cap {I}_{\mathfrak{m}}}\right) \) for all multiples \( \mathfrak{m} \) of \( \mathfrak{f} \) .
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Proof. This immediately follows from the proposition by taking for \( \mathfrak{f} \) the GCD of all moduli in the class \( \mathcal{C} \) (which will, in fact, be the GCD of only a finite number of moduli) and applying the proposition inductively.
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Yes
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Proposition 3.3.12. (1) If a modulus \( \mathfrak{f} \) is equal to the conductor of \( \left( {\mathfrak{f}, C}\right) \) , then for all congruence subgroups \( D \subset C \) modulo \( \mathfrak{f} \), the conductor of \( \left( {\mathfrak{f}, D}\right) \) is also equal to \( \mathfrak{f} \) .
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Proof. (1). Assume that \( \mathfrak{f} \) is equal to the conductor of \( \left( {\mathfrak{f}, C}\right) \), let \( D \subset C \) , and let \( \mathfrak{n} \) be the conductor of \( \left( {\mathfrak{f}, D}\right) \), so that \( \mathfrak{n} \mid \mathfrak{f} \) . By Proposition 3.3.5, we have \( {I}_{\mathfrak{f}} \cap {P}_{\mathfrak{n}} \subset D \subset C \) . Thus, \( \left( {\mathfrak{n}, C{P}_{\mathfrak{n}}}\right) \sim \left( {\mathfrak{f}, C}\right) \), and since \( \mathfrak{f} \) is the conductor of \( \left( {\mathfrak{f}, C}\right) \) and \( \mathfrak{n} \mid \mathfrak{f} \), we must have \( \mathfrak{n} = \mathfrak{f} \), proving (1).
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Yes
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Corollary 3.3.13. A modulus \( \mathfrak{f} \) is the conductor of the equivalence class of \( \left( {\mathfrak{f}, C}\right) \; \) if and only if for any \( \;\mathfrak{n} \mid \mathfrak{f},\;\mathfrak{n} \neq \mathfrak{f},\; \) we have \( \;{h}_{\mathfrak{n}, C{P}_{\mathfrak{n}}} < {h}_{\mathfrak{f}, C}.\; \) In particular, \( \mathfrak{f} \) is a conductor if and only if for all \( \mathfrak{n} \mid \mathfrak{f},\mathfrak{n} \neq \mathfrak{f} \), we have \( {h}_{\mathfrak{n}} < {h}_{\mathfrak{f}} \) .
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Proof. This is an immediate consequence of Proposition 3.3.6 and the above proposition.
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Yes
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Proposition 3.3.16. Let \( \chi \) be a character modulo \( {\mathfrak{m}}_{1} \). (1) If \( {\mathfrak{m}}_{2} \mid {\mathfrak{m}}_{1} \), then \( \chi \) can be defined modulo \( {\mathfrak{m}}_{2} \) if and only if \( {I}_{{\mathfrak{m}}_{1}} \cap {P}_{{\mathfrak{m}}_{2}} \subset \operatorname{Ker}\left( \chi \right) \), if and only if there exists a congruence subgroup \( {C}_{2} \) modulo \( {\mathfrak{m}}_{2} \) such that \( \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \sim \left( {{\mathfrak{m}}_{1},\operatorname{Ker}\left( \chi \right) }\right) \) .
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Proof. By Proposition 3.3.6, we have the following exact sequence: \[ 1 \rightarrow \left( {{I}_{{\mathfrak{m}}_{1}} \cap {P}_{{\mathfrak{m}}_{2}}}\right) /{P}_{{\mathfrak{m}}_{1}} \rightarrow C{l}_{{\mathfrak{m}}_{1}} \rightarrow C{l}_{{\mathfrak{m}}_{2}} \rightarrow 1 \] Thus, if \( \chi \) can be defined modulo \( {\mathfrak{m}}_{2} \), then \( \bar{\chi } \) factors through \( C{l}_{{\mathfrak{m}}_{2}} \) ; hence it is trivial on the kernel of the map from \( C{l}_{{\mathfrak{m}}_{1}} \) to \( C{l}_{{\mathfrak{m}}_{2}} \), that is, on \( \left( {{I}_{{\mathfrak{m}}_{1}} \cap }\right. \left. {P}_{{\mathrm{m}}_{2}}\right) /{P}_{{\mathrm{m}}_{1}} \) . Conversely, if \( \bar{\chi } \) is trivial on this kernel, then clearly \( \bar{\chi } \) can be lifted to a map from \( C{l}_{{\mathfrak{m}}_{2}} \) to \( {\mathbb{C}}^{ * } \) ; hence \( \chi \) can be defined modulo \( {\mathfrak{m}}_{2} \) . Since \( {P}_{{\mathfrak{m}}_{1}} \subset \operatorname{Ker}\left( \chi \right) \), we see that \( \chi \) can be defined modulo \( {\mathfrak{m}}_{2} \) if and only if \( {I}_{{\mathfrak{m}}_{1}} \cap {P}_{{\mathfrak{m}}_{2}} \subset \operatorname{Ker}\left( \chi \right) \) . By Proposition 3.3.5, this is equivalent to the existence of a congruence subgroup \( {C}_{2} \) modulo \( {\mathfrak{m}}_{2} \) such that \( \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \sim \left( {{\mathfrak{m}}_{1},\operatorname{Ker}\left( \chi \right) }\right) \) . Statements (2) and (3) are trivial consequences of (1) and of the definitions.
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Yes
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Proposition 3.3.17. Let \( \\left( {\\mathfrak{m}, C}\\right) \) be a congruence subgroup, and let \( \\mathfrak{f} \) be the conductor of \( \\left( {\\mathfrak{m}, C}\\right) \). Then\n\n(1) we have \( C = \\mathop{\\bigcap }\\limits_{\\chi }\\operatorname{Ker}\\left( \\chi \\right) \), where the intersection is taken over the characters of the congruence subgroup \( \\left( {\\mathfrak{m}, C}\\right) \); (2) we have\n\n\[ \n\\mathfrak{f} = \\operatorname{lcm}\\{ \\mathfrak{f}\\left( \\chi \\right) /C \\subset \\operatorname{Ker}\\left( \\chi \\right) \\}\n\]\n\nIn other words, the conductor of \( \\left( {\\mathfrak{m}, C}\\right) \) is the \( {LCM} \) (or the intersection) of the conductors of the characters of the congruence subgroup \( \\left( {\\mathfrak{m}, C}\\right) \).
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Proof. (1). By assumption, \( C \) is included in the intersection. Conversely, if \( C \) was not equal to the intersection, we could find an \( \\mathfrak{a} \\in {I}_{\\mathfrak{m}},\\mathfrak{a} \\notin C \), such that \( \\chi \\left( \\mathfrak{a}\\right) = 1 \) for all characters \( \\chi \) of the congruence subgroup \( \\left( {\\mathfrak{m}, C}\\right) \). But in the finite quotient group \( {I}_{\\mathrm{m}}/C \), this means that \( \\bar{\\chi }\\left( \\overline{\\mathfrak{a}}\\right) = 1 \) for all characters of the group, hence that \( \\overline{\\mathfrak{a}} = \\overline{1} \), so that \( \\mathfrak{a} \\in C \), a contradiction.\n\n(2). If \( \\chi \) is a character of the congruence subgroup \( \\left( {\\mathfrak{m}, C}\\right) \), then \( C \\subset \) \( \\operatorname{Ker}\\left( \\chi \\right) \). By definition of the conductor of a congruence subgroup, we have \( {I}_{\\mathfrak{m}} \\cap {P}_{\\mathfrak{f}} \\subset C \\subset \\operatorname{Ker}\\left( \\chi \\right) \). Hence by Proposition 3.3.16, \( \\chi \) can be defined modulo \( f \), and so \( f\\left( \\chi \\right) \\mid f \).\n\nConversely, let \( \\mathfrak{n} \) be a multiple of all the \( \\mathfrak{f}\\left( \\chi \\right) \) for \( \\chi \) a character of the congruence subgroup \( \\left( {\\mathfrak{m}, C}\\right) \). Let \( \\chi \) be such a character. Then by Proposition 3.3.16, since \( \\chi \) can be defined modulo \( \\mathfrak{f}\\left( \\chi \\right) \), we have \( {I}_{\\mathfrak{m}} \\cap {P}_{\\mathfrak{f}\\left( \\chi \\right) } \\subset \\operatorname{Ker}\\left( \\chi \\right) \). Therefore, since \( \\mathfrak{f}\\left( \\chi \\right) \\mid \\mathfrak{n} \), we have in particular \( {I}_{\\mathfrak{m}} \\cap {P}_{\\mathfrak{n}} \\subset \\operatorname{Ker}\\left( \\chi \\right) \), so\n\n\[ \n{I}_{\\mathfrak{m}} \\cap {P}_{\\mathfrak{n}} \\subset \\mathop{\\bigcap }\\limits_{\\chi }\\operatorname{Ker}\\left( \\chi \\right)\n\]\n\nThus, by (1) we have \( {I}_{\\mathfrak{m}} \\cap {P}_{\\mathfrak{n}} \\subset C \), so by Proposition 3.3.5, there exists a congruence subgroup \( {C}^{\\prime } \) modulo \( \\mathfrak{n} \) such that \( \\left( {\\mathfrak{n},{C}^{\\prime }}\\right) \\sim \\left( {\\mathfrak{m}, C}\\right) \), and hence \( \\mathfrak{f} \\mid \\mathfrak{n} \), proving the proposition.
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Yes
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Proposition 3.3.18. Let \( \mathfrak{f} \) be a conductor (in other words, the conductor of some equivalence class of congruence subgroups). Then \( \mathfrak{f} \) satisfies the following properties.\n\n(1) If \( \mathfrak{p} \mid \mathfrak{f} \) and \( \mathcal{N}\left( \mathfrak{p}\right) = 2 \), then \( {\mathfrak{p}}^{2} \mid \mathfrak{f} \) .
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Proof. (1). Assume \( \mathfrak{p} \mid \mathfrak{f},\mathcal{N}\left( \mathfrak{p}\right) = 2 \), and \( {\mathfrak{p}}^{2} \nmid \mathfrak{f} \) . Then \( \mathfrak{f}/\mathfrak{p} \) and \( \mathfrak{p} \) are coprime ideals, so\n\n\[ \phi \left( \mathfrak{f}\right) = \phi \left( {\mathfrak{f}/\mathfrak{p}}\right) \phi \left( \mathfrak{p}\right) = \phi \left( {\mathfrak{f}/\mathfrak{p}}\right) \left( {\mathcal{N}\left( \mathfrak{p}\right) - 1}\right) = \phi \left( {\mathfrak{f}/\mathfrak{p}}\right) . \]\n\nHowever, \( {U}_{\mathfrak{f}}\left( K\right) \subset {U}_{\mathfrak{f}/\mathfrak{p}}\left( K\right) \), so \( \left\lbrack {U\left( K\right) : {U}_{\mathfrak{f}}\left( K\right) }\right\rbrack \geq \left\lbrack {U\left( K\right) : {U}_{\mathfrak{f}/\mathfrak{p}}\left( K\right) }\right\rbrack \) . Thus, Corollary 3.2.4 implies that \( {h}_{\mathfrak{f}/\mathfrak{p}} \geq {h}_{\mathfrak{f}} \) (and, since \( {h}_{\mathfrak{f}/\mathfrak{p}} \mid {h}_{\mathfrak{f}} \), that \( {h}_{\mathfrak{f}/\mathfrak{p}} = {h}_{\mathfrak{f}} \) ), so by Corollary 3.3.13 we deduce that \( f \) is not a conductor.
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Yes
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Proposition 3.3.19. The moduli \( \infty ,3\mathbb{Z},4\mathbb{Z} \), and \( m\mathbb{Z} \) and \( \left( {m\mathbb{Z}}\right) \infty \) for \( m \equiv \) \( 2\left( {\;\operatorname{mod}\;4}\right) \) are not conductors. All other moduli are conductors.
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Proof. This is an easy consequence of Proposition 3.3.18 and the properties of the \( \phi \) -function, and the details are left to the reader (Exercise 8).
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No
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Proposition 3.3.20. Denote by \( {\mathfrak{p}}_{\ell } \) (resp., \( {\mathfrak{p}}_{\ell } \) and \( {\mathfrak{p}}_{\ell }^{\prime } \) ) the prime ideal(s) above \( \ell \) when \( \ell \) is ramified (resp., split) in a quadratic field \( K \) . If \( K \) is an imaginary quadratic field, all moduli are conductors with the following exceptions, given in completely factored form:\n\n(1) If \( K = \mathbb{Q}\left( \sqrt{-3}\right) \), the moduli \( {\mathfrak{p}}_{3},2{\mathbb{Z}}_{K},{\mathfrak{p}}_{7},{\mathfrak{p}}_{7}^{\prime },{\mathfrak{p}}_{3}^{2},2{\mathfrak{p}}_{3} \) ;\n\n(2) if \( K = \mathbb{Q}\left( \sqrt{-1}\right) \), the moduli \( {\mathfrak{p}}_{2}^{2},{\mathfrak{p}}_{2}^{3},{\mathfrak{p}}_{5},{\mathfrak{p}}_{5}^{\prime } \), and \( {\mathfrak{p}}_{2}\mathfrak{n} \), where \( \mathfrak{n} \) is not divisible by \( {\mathfrak{p}}_{2} \) ;\n\n(3) in all other cases, the excluded moduli are exactly those given by Proposition 3.3.18: in other words, \( {\mathfrak{p}}_{2}^{2} \) and \( {\mathfrak{p}}_{2}^{\prime 2} \) if \( {\mathfrak{p}}_{2} \) and \( {\mathfrak{p}}_{2}^{\prime } \) are the unramified ideals of degree 1 above 2 when \( d\left( K\right) \equiv 1\left( {\;\operatorname{mod}\;8}\right) ,{\mathfrak{p}}_{2}^{3} \) when \( d\left( K\right) \equiv 0 \) \( \left( {\;\operatorname{mod}\;4}\right) ,{\mathfrak{p}}_{3} \) or \( {\mathfrak{p}}_{3} \) and \( {\mathfrak{p}}_{3}^{\prime } \) if \( {\mathfrak{p}}_{3} \) and \( {\mathfrak{p}}_{3}^{\prime } \) are ideals of degree 1 above 3 when \( d\left( K\right) ≢ 2\left( {\;\operatorname{mod}\;3}\right) \), and \( {\mathfrak{p}}_{2}\mathfrak{n} \), where \( \mathfrak{n} \) is not divisible by \( {\mathfrak{p}}_{2} \), where \( {\mathfrak{p}}_{2} \) is an ideal of degree 1 above 2, when \( d\left( K\right) ≢ 5\left( {\;\operatorname{mod}\;8}\right) \) .
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Proof. Once again the proof is left to the reader (Exercise 9).
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No
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Proposition 3.3.21. Let \( \left( {\mathfrak{m}, C}\right) \) be a congruence subgroup, let \( \mathfrak{f} \) be its conductor, let \( n = {h}_{\mathfrak{m}, C} \), let \( \mathfrak{p} \) be a prime ideal dividing \( \mathfrak{f} \), and finally let \( \ell \) be the prime number below \( \mathfrak{p} \) . (1) If \( {v}_{\mathfrak{p}}\left( \mathfrak{f}\right) \geq 2 \), we necessarily have \( \ell \mid n \) . In other words, if \( \ell \nmid n \), then \( {v}_{\mathfrak{p}}\left( \mathfrak{f}\right) = 1 \) .
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Proof. Since \( {h}_{\mathfrak{f}, C{P}_{\mathfrak{f}}} = {h}_{\mathfrak{m}, C} \), replacing \( \left( {\mathfrak{m}, C}\right) \) by the equivalent congruence subgroup \( \left( {\mathfrak{f}, C{P}_{\mathfrak{f}}}\right) \), we may assume that \( \mathfrak{f} = \mathfrak{m} \). For (1), let \( \mathfrak{p} \) be such that \( {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) \geq 2 \), and assume that \( \ell \nmid n \). If we set \( \mathfrak{n} = \mathfrak{m}/\mathfrak{p} \), it follows in particular that \( {I}_{\mathfrak{n}} = {I}_{\mathfrak{m}} \). Set \( G = C{P}_{\mathfrak{n}}/C \). We have \[ G \subset C{I}_{\mathrm{n}}/C = {I}_{\mathrm{m}}/C \simeq C{l}_{\mathrm{m}}/\bar{C}, \] so \( \left| G\right| \mid {h}_{\mathfrak{m}, C} = n \). On the other hand, \( {P}_{\mathrm{n}} \subset {I}_{\mathrm{n}} = {I}_{\mathrm{m}} \), so \( {I}_{\mathrm{m}} \cap C{P}_{\mathrm{n}} = C{P}_{\mathrm{n}} \). Hence Corollary 3.3.7 tells us that \( \left| G\right| = {h}_{\mathfrak{m}, C}/{h}_{\mathfrak{n}, C{P}_{\mathfrak{n}}} \) divides \( \phi \left( \mathfrak{m}\right) /\phi \left( \mathfrak{n}\right) \). Since \( {\mathfrak{p}}^{2} \mid \mathfrak{m} \), we have \( \phi \left( \mathfrak{m}\right) /\phi \left( \mathfrak{n}\right) = \mathcal{N}\left( \mathfrak{p}\right) \), and since we have assumed that \( \ell \nmid n \), it follows that \( \left| G\right| \) divides \( \gcd \left( {n,\mathcal{N}\left( \mathfrak{p}\right) }\right) = 1 \). It follows that \( C{P}_{\mathrm{n}} = C \), so \( {P}_{\mathrm{n}} \subset C \). Thus \( {I}_{\mathrm{m}} \cap {P}_{\mathrm{n}} = {P}_{\mathrm{n}} \subset C \), so Proposition 3.3.5 shows that the conductor divides \( \mathfrak{n} = \mathfrak{m}/\mathfrak{p} \), which is absurd since we have assumed that \( \mathfrak{m} \) is the conductor.
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Yes
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Theorem 3.5.1. (1) The map that sends an equivalence class of Abelian extensions \( L/K \) to the equivalence class of the congruence subgroup \( \left( {\mathfrak{m},{A}_{\mathfrak{m}}\left( {L/K}\right) }\right) \) for any suitable \( \mathfrak{m} \) for the extension \( L/K \) is a bijection (by Theorem 3.4.6, this equivalence class is independent of \( \mathfrak{m} \) ).
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The proof that the map is injective is not very difficult. However, the proof of the surjectivity is an existence proof and is very hard, like almost all such existence proofs in mathematics. In fact, we will see that this phenomenon is also reflected in algorithmic practice. The difficulty with the proof lies mainly in the very few tools that we have available to construct Abelian extensions. The known proofs all rely on the method of Kummer extensions (see Chapter 10), which is elementary but heavy to use, and we will do the same in algorithmic practice in Chapter 5.
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No
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Theorem 3.5.3. Let \( L/K \) be an Abelian extension of degree \( n \) corresponding to a congruence subgroup \( \left( {\mathfrak{m}, C}\right) \) under the Takagi map (with \( \mathfrak{m} \) a multiple of the conductor of \( \left( {\mathfrak{m}, C}\right) \) but not necessarily equal to it), and let \( \mathfrak{p} \) be a prime ideal of \( K \) . Let \( \mathfrak{n} = {\mathfrak{{mp}}}^{-{v}_{\mathfrak{p}}\left( \mathfrak{m}\right) } \) be the prime to \( \mathfrak{p} \) part of the modulus \( \mathfrak{m} \) . If we let \( \mathfrak{p}{\mathbb{Z}}_{L} = \mathop{\prod }\limits_{{1 < i \leq g}}{\mathfrak{P}}_{i}^{e} \) be the prime ideal decomposition of \( \mathfrak{p} \) in the extension \( L/K \) , we have\n\n\[ e = e\left( {{\mathfrak{P}}_{i}/\mathfrak{p}}\right) = \frac{n}{\left| {I}_{\mathfrak{n}}/C{P}_{\mathfrak{n}}\right| } = \left| \frac{{I}_{\mathfrak{m}} \cap C{P}_{\mathfrak{n}}}{C}\right| = \left| \frac{{I}_{\mathfrak{m}} \cap {P}_{\mathfrak{n}}}{C \cap {P}_{\mathfrak{n}}}\right| ,\]\n\n\( f = f\left( {{\mathfrak{P}}_{i}/\mathfrak{p}}\right) \) is the order of the class of \( \mathfrak{p} \) in \( {I}_{\mathfrak{n}}/C{P}_{\mathfrak{n}} \) (equivalently, it is the least positive integer \( f \) such that \( \left. {{\mathfrak{p}}^{f} \in C{P}_{\mathfrak{n}}}\right) \), hence \( g = n/{ef} \) is equal to the index of the cyclic subgroup generated by the class of \( \mathfrak{p} \) in the group \( {I}_{\mathfrak{n}}/C{P}_{\mathfrak{n}} \) .
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In particular, if \( \mathfrak{p} \) is unramified in \( L/K \), then the common residual degree \( f \) is the smallest positive integer such that \( {\mathfrak{p}}^{f} \in C \), and \( g = n/f \) .
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Yes
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Proposition 3.5.7. A modulus \( \mathrm{m} \) is the conductor of \( L/K \) if and only if for all places \( \mathfrak{p} \mid \mathfrak{m} \) (including the places at infinity) we have \( {h}_{\mathfrak{m}/\mathfrak{p}, C} < {h}_{\mathfrak{m}, C} \).
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Proof. Indeed, by Corollary 3.3.13 the condition is necessary; but conversely, if this condition is satisfied and if \( n \mid m, n \neq m \), then if \( p \mid m/n \), we have \( {h}_{\mathfrak{n}, C} \leq {h}_{\mathfrak{m}/\mathfrak{p}, C} < {h}_{\mathfrak{m}, C} \), so we conclude again by Corollary 3.3.13.
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No
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Proposition 3.5.8. Let \( \left( {{R}_{1},{R}_{2}}\right) \) be the signature of \( L \), so that \( {R}_{1} + 2{R}_{2} = \) \( \left\lbrack {L : \mathbb{Q}}\right\rbrack = \left\lbrack {K : \mathbb{Q}}\right\rbrack \cdot {h}_{\mathfrak{m}, C} \) . Write \( {m}_{\infty } \) for \( \left| {\mathfrak{m}}_{\infty }\right| \) . We have
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Proof. Since \( L/K \) is normal, \( {R}_{1} \) is equal to \( \left\lbrack {L : K}\right\rbrack = {h}_{\mathfrak{m}, C} \) times the number of real places of \( K \) unramified in \( L \) . By definition of the ray class group, the \( {r}_{1} - {m}_{\infty } \) real places not in the modulus \( \mathfrak{m} \) are unramified. On the other hand, let \( v \in {\mathfrak{m}}_{\infty } \) . If \( {h}_{\mathfrak{m}/v, C} = {h}_{\mathfrak{m}, C} \), then \( v \) does not divide the conductor of \( L \), hence \( v \) is unramified in \( L \) . On the contrary, if \( {h}_{\mathfrak{m}/v, C} < {h}_{\mathfrak{m}, C} \) , then \( v \) divides the conductor of \( L \), so \( v \) is ramified in \( L \) . This gives the first formula of the proposition. The second follows immediately.
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Yes
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Theorem 3.5.10. Let \( L/K \) be an Abelian extension, and denote by \( \widehat{G} \) the group of characters of \( L/K \) in the sense of Definition 3.5.9.\n\n(1) The conductor of \( L/K \) is given by \( \mathfrak{f}\left( {L/K}\right) = {\operatorname{lcm}}_{\chi \in \widehat{G}}\left( {\mathfrak{f}\left( \chi \right) }\right) \), where \( f\left( \chi \right) \) is the conductor of the character \( \chi \) (see Definition 3.3.15).\n\n(2) The discriminant ideal is given by \( \mathfrak{d}\left( {L/K}\right) = \mathop{\prod }\limits_{{\chi \in \widehat{G}}}\mathfrak{f}{\left( \chi \right) }_{0} \), where as usual \( \mathfrak{f}{\left( \chi \right) }_{0} \) denotes the finite part of the modulus \( \mathfrak{f}\left( \chi \right) \).\n\n(3) We have \( \mathfrak{f}\left( {L/K}\right) \mid \mathfrak{d}\left( {L/K}\right) \), and both ideals are divisible by exactly the same prime ideals: the prime ideals of \( K \) ramified in \( L/K \) .
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Proof. (1). Theorem 3.5.10 tells us that \( \mathfrak{d}\left( {L/K}\right) = \mathop{\prod }\limits_{{\chi \in \widehat{G}}}\mathfrak{f}{\left( \chi \right) }_{0} \) . Set \( D\left( {L/K}\right) = \mathop{\prod }\limits_{{\chi \in \widehat{G}}}\mathfrak{f}\left( \chi \right) \), so that \( \mathfrak{d}\left( {L/K}\right) \) is the finite part of \( \widehat{D}\left( {L/K}\right) \) . Note that this can also be taken as the definition of an extended discriminant ideal if desired. Since it is just as simple, we will in fact compute a formula for \( D\left( {L/K}\right) \) .\n\nFor each \( \mathfrak{n} \mid \mathfrak{m} \), denote by \( a\left( \mathfrak{n}\right) \) the number of characters of the congruence subgroup \( \left( {\mathfrak{m}, C}\right) \) of conductor exactly equal to \( \mathfrak{n} \) . Since the total number of characters is equal to the order of the group, we have the equation\n\n\[ \mathop{\sum }\limits_{{\mathfrak{n} \mid \mathfrak{m}}}a\left( \mathfrak{n}\right) = \left| {C{l}_{\mathfrak{m}}/\bar{C}}\right| = {h}_{\mathfrak{m}, C}. \]\n\nBy Möbius inversion, it follows that\n\n\[ a\left( \mathfrak{n}\right) = \mathop{\sum }\limits_{{\mathfrak{q} \mid \mathfrak{n}}}\mu \left( {\mathfrak{n}/\mathfrak{q}}\right) {h}_{\mathfrak{q}, C} \]\n\nwhere \( \mu \left( \mathfrak{n}\right) \) is defined as in the case of ordinary integers (this is valid since a modulus can be written as a product of finite or infinite primes in essentially onl
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Yes
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Theorem 3.5.11. Let \( \left( {\mathfrak{m}, C}\right) \) be a congruence subgroup, and let \( L/K \) be the Abelian extension associated to \( \left( {\mathfrak{m}, C}\right) \) by class field theory (defined up to \( K \) -isomorphism). Set \( n = \left\lbrack {L : K}\right\rbrack = {h}_{\mathfrak{m}, C} \). (1) The relative discriminant ideal \( \mathfrak{d}\left( {L/K}\right) \) is given by \( \mathfrak{d}\left( {L/K}\right) = \mathop{\prod }\limits_{{\mathfrak{p} \mid \mathfrak{m}}}{\mathfrak{p}}^{{a}_{\mathfrak{p}}} \) with \[ {a}_{\mathfrak{p}} = {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) {h}_{\mathfrak{m}, C} - \mathop{\sum }\limits_{{1 \leq k \leq {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) }}{h}_{\mathfrak{m}/{\mathfrak{p}}^{k}, C}. \]
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Proof. (1). Theorem 3.5.10 tells us that \( \mathfrak{d}\left( {L/K}\right) = \mathop{\prod }\limits_{{\chi \in \widehat{G}}}\mathfrak{f}{\left( \chi \right) }_{0} \) . Set \( D\left( {L/K}\right) = \mathop{\prod }\limits_{{\chi \in \widehat{G}}}\mathfrak{f}\left( \chi \right) \), so that \( \mathfrak{d}\left( {L/K}\right) \) is the finite part of \( \widehat{D}\left( {L/K}\right) \) . Note that this can also be taken as the definition of an extended discriminant ideal if desired. Since it is just as simple, we will in fact compute a formula for \( D\left( {L/K}\right) \) . For each \( \mathfrak{n} \mid \mathfrak{m} \), denote by \( a\left( \mathfrak{n}\right) \) the number of characters of the congruence subgroup \( \left( {\mathfrak{m}, C}\right) \) of conductor exactly equal to \( \mathfrak{n} \) . Since the total number of characters is equal to the order of the group, we have the equation \[ \mathop{\sum }\limits_{{\mathfrak{n} \mid \mathfrak{m}}}a\left( \mathfrak{n}\right) = \left| {C{l}_{\mathfrak{m}}/\bar{C}}\right| = {h}_{\mathfrak{m}, C}. \] By Möbius inversion, it follows that \[ a\left( \mathfrak{n}\right) = \mathop{\sum }\limits_{{\mathfrak{q} \mid \mathfrak{n}}}\mu \left( {\mathfrak{n}/\mathfrak{q}}\right) {h}_{\mathfrak{q}, C} \] where \( \mu \left( \mathfrak{n}\right) \) is defined as in the case of ordinary integers (this is valid since a modulus can be written as a product of finite or infinite primes in essentially only one way). Thus, we have \[ D\left( {L/K}\right) = \mathop{\prod }\limits_{{\mathfrak{n} \mid \mathfrak{m}}}\mathop{\prod }\limits_{{\mathfrak{f}\left( \chi \right) = \mathfrak{n}}}\mathfrak{f}\left( \chi \right) = \mathop{\prod }\limits_{{\mathfrak{n} \mid \mathfrak{m}}}{\mathfrak{n}}^{a\left( \mathfrak{n}\right) } = \mathop{\prod }\limits_{{\mathfrak{n} \mid \mathfrak{m}}}{\mathfrak{n}}^{\mathop{\sum }\limits_{{\mathfrak{q} \mid \mathfrak{n}}}\mu \left( {\mathfrak{n}/\mathfrak{q}}\right) {h}_{\mathfrak{q}, C}} \] \[ = \mathop{\prod }\limits_{{\mathfrak{q} \mid \mathfrak{m}}}{\left( \mathop{\prod }\limits_{{\mathfrak{c} \mid \left( {\mathfrak{m}/\mathfrak{q}}\right) }}{\left( \mathfrak{{cq}}\right) }^{\mu \left( \mathfrak{c}\right) }\right) }^{{h}_{\mathfrak{q}, C}} = \mathop{\prod }\limits_{{\mathfrak{q} \mid \mathfrak{m}}}{\left( {p}_{1}\left( \mathfrak{q}\right) {p}_{2}\left( \mathfrak{q}\right) \right) }^{{h}_{\mathfrak{q}, C}}, \] where \[ {p}_{1}\left( \mathfrak{q}\right) = \mathop{\prod }\limits_{{\mathfrak{c} \mid \left( {\mathfrak{m}/\mathfrak{q}}\right) }}{\mathfrak{c}}^{\mu \left( \mathfrak{c}\right) }\;\text{ and }\;{p}_{2}\left( \mathfrak{q}\right) = \mathop{\prod }\limits_{{\mathfrak{c} \mid \left( {\mathfrak{m}/\mathfrak{q}}\right) }}{\mathfrak{q}}^{\mu \left( \mathfrak{c}\right) } \]
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Yes
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Corollary 3.5.12. Assume that \( \\left( {\\mathfrak{m}, C}\\right) \) is the conductor of the Abelian extension \( L/K \) and that \( \\ell = \\left\\lbrack {L : K}\\right\\rbrack \) is prime.
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Proof. (1). We always have \( {h}_{\\mathfrak{n}, C} \\mid {h}_{\\mathfrak{m}, C} \) for all \( \\mathfrak{n} \\mid \\mathfrak{m} \), and when \( \\mathfrak{m} \) is the conductor, we also have \( {h}_{\\mathfrak{n}, C} < {h}_{\\mathfrak{m}, C} \) for all \( \\mathfrak{n} \\mid \\mathfrak{m} \) different from \( \\mathfrak{m} \). Thus, when \( \\ell = {h}_{\\mathfrak{m}, C} \) is prime, we must have \( {h}_{\\mathfrak{n}, C} = 1 \) for all \( \\mathfrak{n} \\mid \\mathfrak{m} \) other than \( \\mathfrak{m} \), and (1) easily follows from the theorem. Note, however, that it can also easily be proved directly (see Exercise 14).\n\nStatement (2) is simply a reformulation of Proposition 3.3.21.
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No
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Proposition 4.1.6. Let \( \mathcal{B} = \left( {B,{D}_{B}}\right) \) be a finite Abelian group given in SNF, where \( B = {\left( {\beta }_{i}\right) }_{1 \leq i \leq n} \) . There is a natural one-to-one correspondence between subgroups \( \mathcal{A} \) of \( \mathcal{B} \) and integral matrices \( H \) in Hermite normal form satisfying \( {H}^{-1}{D}_{B} \in {\mathcal{M}}_{n}\left( \mathbb{Z}\right) \) . The correspondence is as follows.\n\n(1) The subgroup \( \mathcal{A} \) associated to such a matrix \( H \) is the group given by generators and relations (not necessarily in SNF), as \( \mathcal{A} = \left( {{BH},{H}^{-1}{D}_{B}}\right) \).\n\n(2) Conversely, if \( \mathcal{A} \) is a subgroup of \( \mathcal{B} \) and \( {B}^{\prime } \) is a row vector of generators of \( \mathcal{A} \), we can write \( {B}^{\prime } = {BP} \) for some integer matrix \( P \) . The corresponding matrix \( H \) is the Hermite normal form of the matrix \( \left( {P \mid {D}_{B}}\right) \).\n\n(3) Let \( H \) be a matrix in \( {HNF} \), and let \( \mathcal{A} \) be the corresponding subgroup. Then \( \left| \mathcal{A}\right| = \left| \mathcal{B}\right| /\det \left( H\right) \) or, equivalently, \( \left| {\mathcal{B}/\mathcal{A}}\right| = \left\lbrack {\mathcal{B} : \mathcal{A}}\right\rbrack = \det \left( H\right) \).
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Proof. Let \( B = {\left( {\beta }_{i}\right) }_{1 \leq i \leq n} \) and let \( {D}_{B} = \operatorname{diag}\left( {\left( {b}_{i}\right) }_{1 \leq i \leq n}\right) \), where \( \operatorname{diag}\left( {\left( {b}_{i}\right) }_{i}\right) \) denotes the diagonal matrix whose diagonal entries are the \( {b}_{i} \) . By definition, the following sequence is exact:\n\n\[ 1 \rightarrow {\bigoplus }_{i = 1}^{n}{b}_{i}\mathbb{Z} \rightarrow {\mathbb{Z}}^{n}\overset{\phi }{ \rightarrow }\mathcal{B} \rightarrow 1 \]\n\nwhere\n\n\[ \phi \left( {{x}_{1},\ldots ,{x}_{n}}\right) = \mathop{\prod }\limits_{{1 \leq i \leq n}}{\beta }_{i}^{{x}_{i}}. \]\n\nLet \( {\left( {\varepsilon }_{i}\right) }_{1 \leq i \leq n} \) be the canonical basis elements of \( {\mathbb{Z}}^{n} \), and let \( \Lambda \) be the lattice defined by \( \Lambda = {\bigoplus }_{i}{b}_{i}{\varepsilon }_{i} \) . We thus have a canonical isomorphism \( \mathcal{B} \simeq {\mathbb{Z}}^{n}/\Lambda \) , obtained by sending the \( i \) th generator \( {\beta }_{i} \) of \( \mathcal{B} \) to the class of \( {\varepsilon }_{i} \) .\n\nSubgroups of \( {\mathbb{Z}}^{n}/\Lambda \) are of the form \( {\Lambda }^{\prime }/\Lambda \), where \( {\Lambda }^{\prime } \) is a lattice such that \( \Lambda \subset {\Lambda }^{\prime } \subset {\mathbb{Z}}^{n} \) . Such a lattice \( {\Lambda }^{\prime } \) can be uniquely defined by a matrix \( H \) in Hermite normal form so that the columns of this matrix express a \( \mathbb{Z} \) -basis of \( {\Lambda }^{\prime } \) on the \( {\varepsilon }_{i} \) . The condition \( {\Lambda }^{\prime } \subset {\mathbb{Z}}^{n} \) means that \( H \) has integer entries, and the condition \( \Lambda \subset {\Lambda }^{\prime } \) means that \( {H}^{-1}{D}_{B} \) also has integer entries, since it is the matrix that expresses the given basis of \( \Lambda \) in terms of that of \( {\Lambda }^{\prime } \) . In terms of generators, this correspondence translates into the equality \( {B}^{\prime } = {BH} \) . Furthermore, \( {B}^{\prime }X = {1}_{\mathcal{B}} \) if and only if \( {BHX} = {1}_{\mathcal{B}} \), hence \( {HX} = {D}_{B}Y \), or \( X = {H}^{-1}{D}_{B}Y \), and so if \( \mathcal{A} \) is the subgroup of \( \mathcal{B} \) corresponding to \( {\Lambda }^{\prime }/\Lambda \), it is given in terms of generators and relations by \( \left( {{BH},{H}^{-1}{D}_{B}}\right) \), proving (1).\n\nFor (2), we note that \( B{D}_{B} = {\mathbf{1}}_{\mathcal{B}} \), hence if \( {B}^{\prime \prime } = B\left( {P \mid {D}_{B}}\right) \), we have simply added some \( {1}_{\mathcal{A}} \) ’s to the generators of \( \mathcal{A} \) . Thus, the group can be defined by the generators
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Yes
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Lemma 4.1.12. Assume that we have a matrix equality of the form\n\n\\[ \n\\left( {P \\mid {H}_{1}}\\right) U = \\left( {0 \\mid H}\\right) \n\\]\n\nwhere \\( U \\) is invertible and \\( {H}_{1} \\) is a square matrix with nonzero determinant. Let \\( r \\) be the number of columns of \\( P \\) or, equivalently, the number of 0 columns on the right-hand side. Then the upper-left \\( r \\times r \\) submatrix \\( {U}_{1} \\) of \\( U \\) has nonzero determinant equal to \\( \\pm \\det \\left( {H}_{1}\\right) /\\det \\left( H\\right) \\), where the sign is equal to the determinant of \\( U \\) .
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Proof. Write \\( U = \\left( \\begin{array}{ll} {U}_{1} & {U}_{2} \\\\ {U}_{3} & {U}_{4} \\end{array}\\right) \\) . One easily checks the block matrix identity\n\n\\[ \n\\left( \\begin{matrix} {H}_{1} & 0 \\\\ - P & H \\end{matrix}\\right) \\left( \\begin{matrix} {U}_{1} & {U}_{2} \\\\ 0 & I \\end{matrix}\\right) = \\left( \\begin{matrix} {H}_{1} & 0 \\\\ 0 & {H}_{1} \\end{matrix}\\right) \\left( \\begin{matrix} {U}_{1} & {U}_{2} \\\\ {U}_{3} & {U}_{4} \\end{matrix}\\right) . \n\\]\n\nSince \\( \\det \\left( {H}_{1}\\right) \\neq 0 \\), it follows that \\( \\det \\left( {U}_{1}\\right) \\det \\left( H\\right) = \\pm \\det \\left( {H}_{1}\\right) \\), where the sign is equal to \\( \\det \\left( U\\right) \\) . This proves the lemma.
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Yes
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Proposition 4.1.16. Let \( \mathcal{C} = \left( {C,{D}_{C}}\right) \) be a group given in \( {SNF} \), with \( C = \) \( {\left( {\gamma }_{i}\right) }_{1 \leq i \leq n} \) and \( {D}_{C} = \operatorname{diag}\left( {\left( {c}_{i}\right) }_{1 \leq i \leq n}\right) \), and let \( p \) be a prime number. Let \( {r}_{c} \) be the largest index \( i \leq n \) such that \( p \mid {c}_{i}\left( {{r}_{c} = 0}\right. \) if none exist). Then \( {\mathcal{C}}_{p} \) is given in SNF by \( {\mathcal{C}}_{p} = \left( {{C}_{p},{D}_{C, p}}\right) \), where \[ {C}_{p} = {\left( {\gamma }_{i}^{{c}_{i}/\left( {{p}^{\infty },{c}_{i}}\right) }\right) }_{1 \leq i \leq {r}_{c}}\;\text{ and }\;{D}_{C, p} = \operatorname{diag}\left( {\left( {p}^{\infty },{c}_{i}}\right) }_{1 \leq i \leq {r}_{c}}\right) .
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Proof. Let \( g \in {\mathcal{C}}_{p} \) . There exists \( a \geq 0 \) such that \( {g}^{{p}^{a}} = 1 \) . Let \( g = \) \( \mathop{\prod }\limits_{{1 \leq i \leq n}}{\gamma }_{i}^{{x}_{i}} \) . Thus, \( {c}_{i}\left| {{p}^{a}{x}_{i}\text{, hence}\left( {{c}_{i}/\left( {{p}^{a},{c}_{i}}\right) }\right) }\right| {x}_{i} \), which implies that \( \left( {{c}_{i}/\left( {{p}^{\infty },{c}_{i}}\right) }\right) \mid {x}_{i} \) . Hence, if we set \( {\gamma }_{i, p} = {\gamma }_{i}^{{c}_{i}/\left( {{p}^{\infty },{c}_{i}}\right) } \), the \( {\gamma }_{i, p} \) are generators of \( {\mathcal{C}}_{p} \), and we can restrict to \( i \leq {r}_{c} \), since otherwise the \( {\gamma }_{i, p} \) are equal to 1 . It is clear that the matrix of relations between the \( {\gamma }_{i, p} \) is given by \( {D}_{C, p} = \operatorname{diag}\left( \left( {{p}^{\infty },{c}_{i}}\right) \right) \), and since this is already in SNF, this proves the proposition.
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Yes
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Proposition 4.1.17. (1) Let \( 1 \rightarrow \mathcal{A}\overset{\psi }{ \rightarrow }\mathcal{B}\overset{\phi }{ \rightarrow }\mathcal{C} \) be an exact sequence of Abelian groups, which are exceptionally not assumed to be finite. Then \( 1 \rightarrow {\mathcal{A}}_{p}\overset{{\psi }_{p}}{ \rightarrow }{\mathcal{B}}_{p}\overset{{\phi }_{p}}{ \rightarrow }{\mathcal{C}}_{p} \) is also an exact sequence, where the maps are simply the restrictions of the corresponding maps.
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Proof. For (1), we first note that if \( \psi \) is a group homomorphism from \( \mathcal{A} \) to \( \mathcal{B} \), then clearly \( \psi \left( {\mathcal{A}}_{p}\right) \subset {\mathcal{B}}_{p} \), so the restricted maps are well-defined. Exactness at \( \mathcal{A} \) is also clear since the restriction of an injective map is injective. In addition, the identity \( \phi \circ \psi = 0 \) is preserved by restriction. Thus, we must simply show that \( \operatorname{Ker}\left( {\phi }_{p}\right) \subset \operatorname{Im}\left( {\psi }_{p}\right) \) . Let \( x \in \operatorname{Ker}\left( {\phi }_{p}\right) \) . This means first that \( \phi \left( x\right) = 1 \) in \( \mathcal{C} \), hence by the exactness of the initial sequence, that \( x = \psi \left( y\right) \) for some \( y \in \mathcal{A} \) . It also means that \( {x}^{{p}^{a}} = 1 \) for some \( a \geq 0 \) . But then \( \psi \left( {y}^{{p}^{a}}\right) = 1 \) , hence \( {y}^{{p}^{a}} = 1 \) since \( \psi \) is injective, and so \( y \in {\mathcal{A}}_{p} \), and \( x \in \operatorname{Im}\left( {\psi }_{p}\right) \) as desired.
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Yes
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Proposition 4.1.19. Let \( \mathcal{C} = \left( {C,{D}_{C}}\right) \) be an Abelian group given in SNF, with \( {D}_{C} = \operatorname{diag}\left( {\left( {c}_{i}\right) }_{i}\right) \), and let \( \ell \) be a prime number. Let \( {r}_{c} \) be the largest index \( i \) such that \( \ell \mid {c}_{i}\left( {{r}_{c} = 0}\right. \) if none exist), so that \( {r}_{c} \) is the \( \ell \) -rank of \( \mathcal{C} \) .\n\nThe subgroups of \( \mathcal{C} \) of index \( \ell \) correspond under Proposition 4.1.6 to matrices \( H = \left( {h}_{i, j}\right) \) such that there exists a row index \( k \) (necessarily unique) satisfying the following properties.\n\n(1) We have \( k \leq {r}_{c} \) .\n\n(2) For \( i \neq k \), then \( {h}_{i, j} = 0 \) for \( j \neq i \) and \( {h}_{i, i} = 1 \) .\n\n(3) We have \( {h}_{k, k} = \ell ,{h}_{k, j} = 0 \) if \( j < k \) or \( j > {r}_{c} \), and \( 0 \leq {h}_{k, j} < \ell \) if \( k < j \leq {r}_{c}. \)\n\nIn particular, there are \( \left( {{\ell }^{{r}_{c}} - 1}\right) /\left( {\ell - 1}\right) \) subgroups of index \( \ell \) (this is, of course, a well-known and easy result).
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Proof. The proof of this proposition is easy and is left to the reader (Exercise 11).
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No
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Lemma 4.2.1. Let \( \mathfrak{a} \) and \( \mathfrak{c} \) be two coprime integral ideals of \( K \), and set \( b = {ac} \). (1) We can find in polynomial time elements \( a \) and \( c \) such that \( a \in \mathfrak{a}, c \in \mathfrak{c} \), and \( a + c = 1 \). (2) We have a split exact sequence \[ 1 \rightarrow {\left( {\mathbb{Z}}_{K}/\mathfrak{a}\right) }^{ * }\overset{\psi }{ \rightarrow }{\left( {\mathbb{Z}}_{K}/\mathfrak{b}\right) }^{ * }\overset{\phi }{ \rightarrow }{\left( {\mathbb{Z}}_{K}/\mathfrak{c}\right) }^{ * } \rightarrow 1 \] where \( \psi \left( \bar{\alpha }\right) = \overline{{c\alpha } + a},\phi \left( \bar{\beta }\right) = \bar{\beta } \), and a section \( \sigma \) of \( \phi \) is given by \( \sigma \left( \bar{\gamma }\right) = \overline{{a\gamma } + c} \) . (Here - denotes the classes in the respective groups, but using the same notation for each will not lead to any confusion as long as we know in which group we work.)
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Proof. The proof is a little tedious but straightforward. (1). This is a restatement of Proposition 1.3.1. (2) a). The map \( \psi \) is well-defined: if \( \bar{\alpha } = \overline{{\alpha }^{\prime }} \), then \( {\alpha }^{\prime } - \alpha \in \mathfrak{a} \) ; hence, \( \left( {c{\alpha }^{\prime } + a}\right) - \left( {{c\alpha } + a}\right) = c\left( {{\alpha }^{\prime } - \alpha }\right) \in \mathfrak{{ac}} = \mathfrak{b} \) since \( c \in \mathfrak{c} \). b). The map \( \psi \) is a group homomorphism. Indeed, this follows from the fact that \( a \) and \( c \) are orthogonal idempotents modulo \( \mathfrak{b} \) ; in other words, that \( {ac} \in \mathfrak{b} \) and \[ {a}^{2} - a = - a\left( {1 - a}\right) = - {ac} = - c\left( {1 - c}\right) = {c}^{2} - c \in \mathfrak{b}. \] Hence, \[ \psi \left( \bar{\alpha }\right) \psi \left( \overline{{\alpha }^{\prime }}\right) = \overline{\left( {{c\alpha } + a}\right) \left( {c{\alpha }^{\prime } + a}\right) } = \overline{{c\alpha }{\alpha }^{\prime } + a} = \psi \left( \overline{\alpha {\alpha }^{\prime }}\right) . \] c). The map \( \psi \) is injective. Indeed, \[ \psi \left( \bar{\alpha }\right) = \overline{1} \Leftrightarrow \overline{{c\alpha } + a} \equiv 1\left( {\;\operatorname{mod}\;\mathfrak{{ac}}}\right) \Rightarrow {c\alpha } \equiv 1\left( {\;\operatorname{mod}\;\mathfrak{a}}\right) \] \[ \Rightarrow \alpha \equiv 1\left( {\;\operatorname{mod}\;\mathfrak{a}}\right) \; \Leftrightarrow \bar{\alpha } = 1 \] since \( c \equiv 1\left( {\;\operatorname{mod}\;\mathfrak{a}}\right) \). d). The map \( \phi \) is clearly well-defined and is a group homomorphism. e). By symmetry with a), b), and c), the map \( \sigma \) is well-defined and is an injective group homomorphism. Furthermore, since \( a \equiv 1\left( {\;\operatorname{mod}\;\mathfrak{c}}\right) ,\phi \circ \sigma \) is the identity map, which implies that \( \sigma \) is a section of \( \phi \) and in particular that \( \phi \) is surjective. Statement (3) is an immediate consequence of (2).
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Yes
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Proposition 4.2.4. Let \( \mathfrak{p} \) be a prime ideal of degree \( f \), and let \( q = {p}^{f} = \) \( \left| {{\mathbb{Z}}_{K}/\mathfrak{p}}\right| \) . Set \( G = {\left( {\mathbb{Z}}_{K}/{\mathfrak{p}}^{k}\right) }^{ * } \) . Let\n\n\[ \nW = \left\{ {x \in G/{x}^{q - 1} = 1}\right\} \;\text{ and }\;{G}_{\mathfrak{p}} = \left( {1 + \mathfrak{p}}\right) /\left( {1 + {\mathfrak{p}}^{k}}\right) .\n\]\n\nThen\n\n(1) \( W \simeq {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \), and in particular \( W \) is a cyclic subgroup of order \( q - 1 \) of \( G \) . More precisely, if \( \overline{{g}_{0}} \) is a generator of \( {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \), then \( \left\lceil {{\log }_{2}\left( k\right) }\right\rceil \) iterations of \( g \leftarrow g - \left( {{g}^{q - 1} - 1}\right) /\left( {\left( {q - 1}\right) {g}^{q - 2}}\right) {\;\operatorname{mod}\;{\mathfrak{p}}^{k}} \) applied to \( {g}_{0} \) gives a generator of \( W \) .\n\n(2) \( {G}_{\mathfrak{p}} \) is a p-subgroup of \( G \) of order \( {q}^{k - 1} \) .\n\n(3) \( G = W \times {G}_{\mathfrak{p}} \) .
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Proof. (1). All nonzero elements of \( {\mathbb{Z}}_{K}/\mathfrak{p} \) are roots of the polynomial equation \( {X}^{q - 1} - 1 = 0 \) ; hence this equation has exactly \( q - 1 \) distinct solutions in the field \( {\mathbb{Z}}_{K}/\mathfrak{p} \) . Thus\n\n\[ \n{X}^{q - 1} - 1 \equiv \mathop{\prod }\limits_{{a \in {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * }}}\left( {X - a}\right) \left( {\;\operatorname{mod}\;\mathfrak{p}}\right) .\n\]\n\nIt follows from Hensel's lemma that this factorization can be lifted to a factorization modulo any power of \( \mathfrak{p} \) . Thus there exists a group isomorphism between \( {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \) and solutions to \( {X}^{q - 1} - 1 \equiv 0\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k}}\right) \) ; in other words, between \( {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \) and \( W \) .\n\nIt follows that \( W \) is a cyclic group of order \( q - 1 = {p}^{f} - 1 \) and that a generator of \( W \) can be obtained by Hensel lifting a generator of \( {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \) . This is done using the Newton-Hensel iteration given in the proposition.\n\n(2). If we send the class of \( 1 + x \) to the class of \( x \), it is clear that, as \( a \) set, \( {G}_{\mathfrak{p}} \) is isomorphic to \( \mathfrak{p}/{\mathfrak{p}}^{k} \) . In fact, as we will see in more detail below, the whole difficulty of the structure problem for \( {\left( {\mathbb{Z}}_{K}/{\mathfrak{p}}^{k}\right) }^{ * } \) comes from the fact that this is only a set isomorphism, and not always a group isomorphism.\n\nIn any case, it follows that\n\n\[ \n\left| {G}_{\mathfrak{p}}\right| = \left| {\mathfrak{p}/{\mathfrak{p}}^{k}}\right| = \mathcal{N}\left( {\mathfrak{p}}^{k}\right) /\mathcal{N}\left( \mathfrak{p}\right) = {q}^{k - 1},\n\]\n\nso \( {G}_{\mathfrak{p}} \) is a \( p \) -subgroup of \( G \) of order \( {q}^{k - 1} = {p}^{f\left( {k - 1}\right) } \) .\n\n(3). Consider the map \( \phi \) from \( W \times {G}_{\mathfrak{p}} \) to \( G \) defined by \( \phi \left( \left( {x, y}\right) \right) = x \cdot y \) . It is clearly a group homomorphism, and it is an isomorphism since an element of \( W \) is characterized by its residue modulo \( \mathfrak{p} \), and each nonzero residue is attained.\n\nIt follows from this proposition that to compute the structure of \( {\left( {\mathbb{Z}}_{K}/\mathfrak{m}\right) }^{ * } \) it is sufficient to compute the structure of \( {G}_{\mathfrak{p}} \), which is of course the \( p \) -Sylow subgroup of \( G \) .
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Yes
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Proposition 4.2.7. Let \( \mathfrak{p} \) be a prime ideal above a prime number \( p \), and let \( e = e\left( {\mathfrak{p}/p}\right) = {v}_{\mathfrak{p}}\left( p\right) \) be its ramification index.\n\n(1) The expansion for \( {\log }_{\mathfrak{p}}\left( {1 + x}\right) \) converges \( \mathfrak{p} \) -adically if and only if \( {v}_{\mathfrak{p}}\left( x\right) \geq 1 \) .
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Proof. (1). It is easily shown that a series \( \mathop{\sum }\limits_{i}{u}_{i} \) converges \( \mathfrak{p} \) -adically if and only if \( {u}_{i} \) tends to zero \( \mathfrak{p} \) -adically; in other words, if and only if the \( \mathfrak{p} \) -adic valuation \( {v}_{\mathfrak{p}}\left( {u}_{i}\right) \) tends to infinity as \( i \rightarrow \infty \) .\n\nWe have\n\n\[ {v}_{\mathfrak{p}}\left( \frac{{x}^{i}}{i}\right) = i{v}_{\mathfrak{p}}\left( x\right) - {v}_{\mathfrak{p}}\left( i\right) = i{v}_{\mathfrak{p}}\left( x\right) - e{v}_{p}\left( i\right) .\n\]\n\nThus, if \( {v}_{\mathfrak{p}}\left( x\right) \geq 1 \), we have \( {v}_{\mathfrak{p}}\left( {{x}^{i}/i}\right) \geq i \) - \( e{v}_{p}\left( i\right) \geq i \) - \( e\log \left( i\right) /\log \left( p\right) \rightarrow \infty \) as \( i \rightarrow \infty \) ; hence the series converges \( \mathfrak{p} \) -adically. On the other hand, if \( {v}_{\mathfrak{p}}\left( x\right) \leq 0 \) , then \( {v}_{\mathfrak{p}}\left( {{x}^{i}/i}\right) \leq - e{v}_{p}\left( i\right) \), which does not tend to \( + \infty \) as \( i \rightarrow \infty \) .
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Yes
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Corollary 4.2.8. Let \( \mathfrak{p} \) be a prime ideal above a prime number \( p \), let \( e = \) \( e\left( {\mathfrak{p}/p}\right) = {v}_{\mathfrak{p}}\left( p\right) \) be its ramification index, and set \( {k}_{0} = 1 + \lfloor e/\left( {p - 1}\right) \rfloor \) . For any integers \( a \) and \( b \) such that \( b > a \geq {k}_{0} \), the functions \( {\log }_{\mathfrak{p}} \) and \( {\exp }_{\mathfrak{p}} \) induce inverse isomorphisms between the multiplicative group \( \left( {1 + {\mathfrak{p}}^{a}}\right) /\left( {1 + {\mathfrak{p}}^{b}}\right) \) and the additive group \( {\mathfrak{p}}^{a}/{\mathfrak{p}}^{b} \) . In particular, if \( e < p - 1 \) and \( k \geq 2 \), they induce inverse isomorphisms between \( {G}_{\mathfrak{p}} = \left( {1 + \mathfrak{p}}\right) /\left( {1 + {\mathfrak{p}}^{k}}\right) \) and \( \mathfrak{p}/{\mathfrak{p}}^{k} \) .
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Proof. Set \( v = {v}_{\mathfrak{p}}\left( x\right) \) . We have seen above that if \( v > e/\left( {p - 1}\right) \), then \( {v}_{\mathfrak{p}}\left( {{\log }_{\mathfrak{p}}\left( {1 + x}\right) }\right) = {v}_{\mathfrak{p}}\left( x\right) = v \) . Hence \( {\log }_{\mathfrak{p}} \) sends \( \left( {1 + {\mathfrak{p}}^{a}}\right) /\left( {1 + {\mathfrak{p}}^{b}}\right) \) to \( {\mathfrak{p}}^{a}/{\mathfrak{p}}^{b} \) , and it is a group homomorphism because of the additive property of the logarithm.\n\nOn the other hand, since \( v > e/\left( {p - 1}\right) \), the function \( {\exp }_{\mathfrak{p}}\left( x\right) \) converges for \( x \in {\mathfrak{p}}^{a} \) . Furthermore, since \( {v}_{p}\left( {i!}\right) \leq \left( {i - 1}\right) /\left( {p - 1}\right) \), when \( v = {v}_{\mathfrak{p}}\left( x\right) \geq {k}_{0} \) we have\n\n\[ \n{v}_{\mathfrak{p}}\left( \frac{{x}^{i}}{i!}\right) = {iv} - e{v}_{p}\left( {i!}\right) \geq v + \left( {i - 1}\right) \left( {v - \frac{e}{p - 1}}\right) .\n\]\n\nTherefore, if \( i > 1 \), we have \( {v}_{\mathfrak{p}}\left( {{x}^{i}/i!}\right) > v \) ; hence \( {v}_{\mathfrak{p}}\left( {{\exp }_{\mathfrak{p}}\left( x\right) - 1}\right) = v \) . It follows that \( {\exp }_{\mathfrak{p}} \) sends \( {\mathfrak{p}}^{a}/{\mathfrak{p}}^{b} \) to \( \left( {1 + {\mathfrak{p}}^{a}}\right) /\left( {1 + {\mathfrak{p}}^{b}}\right) \), and it is the inverse map of \( {\log }_{\mathfrak{p}} \) by the proposition, proving the corollary.
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Yes
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Lemma 4.2.9. Let \( \mathfrak{a} \) and \( \mathfrak{b} \) be (nonzero) integral ideals of \( {\mathbb{Z}}_{K} \) . The additive group \( \mathfrak{b}/\mathfrak{a}\mathfrak{b} \) is isomorphic to the additive group \( {\mathbb{Z}}_{K}/\mathfrak{a} \) .
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Proof. By the approximation theorem for Dedekind domains, there exists \( \alpha \in {\mathbb{Z}}_{K} \) such that \( {v}_{\mathfrak{p}}\left( \alpha \right) = {v}_{\mathfrak{p}}\left( \mathfrak{b}\right) \) for all \( \mathfrak{p} \) dividing \( \mathfrak{a} \) and \( {v}_{\mathfrak{p}}\left( \alpha \right) \geq {v}_{\mathfrak{p}}\left( \mathfrak{b}\right) \) for all \( \mathfrak{p} \) dividing \( \mathfrak{b} \) . In particular, \( \alpha \in \mathfrak{b} \) . Thus the map \( x \mapsto {\alpha x} \) induces a well-defined additive group homomorphism from \( {\mathbb{Z}}_{K}/\mathfrak{a} \) to \( \mathfrak{b}/\mathfrak{a}\mathfrak{b} \) . Since\n\n\[ \overline{\alpha x} = \overline{0} \Leftrightarrow {\alpha x} \in \mathfrak{a}\mathfrak{b} \Leftrightarrow \forall \mathfrak{p}\;{v}_{\mathfrak{p}}\left( \alpha \right) + {v}_{\mathfrak{p}}\left( x\right) \geq {v}_{\mathfrak{p}}\left( \mathfrak{a}\right) + {v}_{\mathfrak{p}}\left( \mathfrak{b}\right) ,\]\n\nit follows from our choice of \( \alpha \) that, for all \( \mathfrak{p} \) dividing \( \mathfrak{a} \), we have \( {v}_{\mathfrak{p}}\left( x\right) \geq {v}_{\mathfrak{p}}\left( \mathfrak{a}\right) \) , and hence \( x \in \mathfrak{a} \) so \( \bar{x} = \overline{0} \) . Thus our map is an injective group homomorphism. Since the norm is multiplicative in \( {\mathbb{Z}}_{K} \), we have\n\n\[ \left| {\mathfrak{b}/\mathfrak{a}\mathfrak{b}}\right| = \mathcal{N}\left( {\mathfrak{a}\mathfrak{b}}\right) /\mathcal{N}\left( \mathfrak{b}\right) = \mathcal{N}\left( \mathfrak{a}\right) = \left| {{\mathbb{Z}}_{K}/\mathfrak{a}}\right| . \]\n\nand hence our map is also surjective, proving the lemma.
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Yes
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Corollary 4.2.11. Let \( \mathfrak{p} \) be a prime ideal above \( p \), with ramification index \( e = e\left( {\mathfrak{p}/p}\right) \) and residual degree \( f = f\left( {\mathfrak{p}/p}\right) \), and let \( k \geq 2 \) be an integer. Write\n\n\[ k + e - 2 = {eq} + r\;\text{ with }\;0 \leq r < e. \]\n\nAssume that \( p \geq \min \left( {e + 2, k}\right) \) . Then\n\n\[ {\left( {\mathbb{Z}}_{K}/{\mathfrak{p}}^{k}\right) }^{ * } \simeq \left( {\mathbb{Z}/\left( {{p}^{f} - 1}\right) \mathbb{Z}}\right) \times {\left( \mathbb{Z}/{p}^{q}\mathbb{Z}\right) }^{\left( {r + 1}\right) f} \times {\left( \mathbb{Z}/{p}^{q - 1}\mathbb{Z}\right) }^{\left( {e - r - 1}\right) f}\;. \]
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Proof. Assume first that \( k \geq e + 2 \) . Then \( p \geq e + 2 \) or, in other words, \( e < \left( {p - 1}\right) \) . We can thus apply Corollary 4.2.8, and Lemma 4.2.9, Theorem 4.2.10, together with Proposition 4.2.4, imply the result in this case.\n\nAssume now that \( k \leq e + 1 \) . Then \( e \leq k + e - 2 \leq {2e} - 1 \) ; hence \( q = 1 \) and \( r = k - 2 \) . Thus, we must prove that\n\n\[ \left( {1 + \mathfrak{p}}\right) /\left( {1 + {\mathfrak{p}}^{k}}\right) \simeq {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{\left( {k - 1}\right) f}, \]\n\nand since these groups have the same cardinality, we must simply show that \( \left( {1 + \mathfrak{p}}\right) /\left( {1 + {\mathfrak{p}}^{k}}\right) \) is killed by \( p \) . Since\n\n\[ {\left( 1 + x\right) }^{p} = 1 + {x}^{p} + \mathop{\sum }\limits_{{1 \leq i \leq p - 1}}\left( \begin{matrix} p \\ i \end{matrix}\right) {x}^{i} \]\n\nwhen \( x \in \mathfrak{p} \) and \( 1 \leq i \leq p - 1 \), we have\n\n\[ {v}_{\mathfrak{p}}\left( {\left( \begin{matrix} p \\ i \end{matrix}\right) {x}^{i}}\right) = e + i{v}_{\mathfrak{p}}\left( x\right) \geq e + 1 \geq k \]\n\nand \( {v}_{\mathfrak{p}}\left( {x}^{p}\right) = p{v}_{\mathfrak{p}}\left( x\right) \geq p \geq k \) by assumption. Hence, if \( x \in \mathfrak{p} \), we have \( {\left( 1 + x\right) }^{p} \equiv 1\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k}}\right) \), and so \( \left( {1 + \mathfrak{p}}\right) /\left( {1 + {\mathfrak{p}}^{k}}\right) \) is killed by \( p \), as claimed, proving the corollary.
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Yes
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Proposition 4.2.12. Let \( \mathfrak{p} \) be a prime ideal above \( p \), of ramification index \( e = e\left( {\mathfrak{p}/p}\right) \) and degree \( f \), and let \( k \geq 1 \) be an integer. We have\n\n\[ \n{\left( {\mathbb{Z}}_{K}/{\mathfrak{p}}^{k}\right) }^{ * } \simeq \left( {\mathbb{Z}/\left( {{p}^{f} - 1}\right) \mathbb{Z}}\right) \times {G}_{\mathfrak{p}},\text{ where }\;{G}_{\mathfrak{p}} = \mathop{\prod }\limits_{{i = 1}}^{{k - 1}}{\left( \mathbb{Z}/{p}^{i}\mathbb{Z}\right) }^{{a}_{i}} \n\] \n\nfor certain nonnegative integers \( {a}_{i} \) . For \( 2 \leq k \leq 4 \), they are given by the following table.\n\n(1) If \( k = 2 \), then \( \left( {a}_{1}\right) = \left( f\right) \) .\n\n(2) If \( k = 3 \), then \( \left( {{a}_{1},{a}_{2}}\right) \) is given by\n\n\[ \n\left( {0, f}\right) \;\text{ if }p \geq 3\text{ and }e = 1 \n\] \n\n\[ \n\left( {{2f},0}\right) \;\text{if}\;p \geq 3\;\text{and}\;e \geq 2; \n\] \n\n\[ \n\left( {2, f - 1}\right) \text{if}p = 2\text{and}e = 1\text{;} \n\] \n\n\[ \n\left( {0, f}\right) \;\text{ if }p = 2\text{ and }e \geq 2. \n\] \n\n(3) If \( k = 4 \), then \( \left( {{a}_{1},{a}_{2},{a}_{3}}\right) \) is given by\n\n\( \left( {0,0, f}\right) \)\n\n\[ \n\text{if}p \geq 5\text{and}e = 1\text{;} \n\] \n\n\( \left( {f, f,0}\right) \)\n\n\[ \n\text{if}p \geq 5\text{and}e = 2\text{;} \n\] \n\n\[ \n\left( {{3f},0,0}\right) \;\text{if}\;p \geq 5\;\text{and}\;e \geq 3; \n\] \n\n\[ \n\left( {0,0, f}\right) \;\text{if}\;p = 3\;\text{and}\;e = 1; \n\] \n\n\( \left( {f + {2a}, f - a,0}\right) \) if \( p = 3 \) and \( e = 2 \) ;\n\n\[ \n\left( {f, f,0}\right) \;\text{if}\;p = 3\;\text{and}\;e \geq 3; \n\] \n\n\( \left( {1,1, f - 1}\right) \; \) if \( \;p = 2\; \) and \( \;e = 1; \)\n\n\( \left( {f, f,0}\right) \)\n\nIn the above, \( a = 1 \) if there exists \( x \in {\mathbb{Z}}_{K} \) such that \( {x}^{2} \equiv - 3\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{3}}\right) \) , and \( a = 0 \) otherwise.\n\n(Note that in this proposition, we have, as usual, mixed multiplicative and additive notation.)
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Proof. We must first prove that \( {G}_{\mathfrak{p}} \) is killed by \( {p}^{k - 1} \) ; in other words, that \( {\left( 1 + x\right) }^{{p}^{k - 1}} \equiv 1\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k}}\right) \) for all \( x \in \mathfrak{p} \) . We prove this by induction on \( k \) . The statement is trivially true for \( k = 1 \), so assume that it is true for \( k \) . Thus \( {\left( 1 + x\right) }^{{p}^{k - 1}} = 1 + y \) with \( y \in {\mathfrak{p}}^{k} \) . Hence\n\n\[ \n{\left( 1 + x\right) }^{{p}^{k}} = {\left( 1 + y\right) }^{p} = 1 + \mathop{\sum }\limits_{{1 \leq j \leq p - 1}}\left( \begin{array}{l} p \\ j \end{array}\right) {y}^{j} + {y}^{p}.\n\]\n\nSince \( p \mid \left( \begin{array}{l} p \\ j \end{array}\right) \) for \( 1 \leq j \leq p - 1 \), we have\n\n\[ \n{v}_{\mathfrak{p}}\left( {\left( \begin{array}{l} p \\ j \end{array}\right) {y}^{j}}\right) = e + j{v}_{\mathfrak{p}}\left( y\right) \geq 1 + k\n\]\n\nOn the other hand,\n\n\[ \n{v}_{\mathfrak{p}}\left( {y}^{p}\right) = p{v}_{\mathfrak{p}}\left( y\right) \geq {pk} \geq k + 1\n\]\n\nsince \( p \geq 2 \), from which our assertion follows by induction. Note that one can prove a much more precise statement than this (see Exercise 20).\n\nCorollary 4.2.11 gives us directly a number of special cases. Specifically, it gives the cases \( k = 2, k = 3 \) and \( p \geq 3;k = 4 \) and \( p \geq 5 \) ; and \( k = 4, p = 3 \) , and \( e = 1 \) .\n\nLet us look at the remaining cases. The easiest way is probably to use the following lemma, similar to the proof of Theorem 4.2.10.\n\nLemma 4.2.13. With the notation of the above proposition, let \( {p}^{k} \), be the cardinality of the kernel of the map \( x \mapsto {x}^{{p}^{j}} \) from \( {G}_{\mathfrak{p}} \) into itself. The exponents \( {a}_{i} \) are given by \( {a}_{i} = 0 \) for \( i \geq k \) and the following backwards recursion:\n\n\[ \n{a}_{j} = \left( {k - 1}\right) f - \mathop{\sum }\limits_{{i = j + 1}}^{{k - 1}}\left( {i - j + 1}\right) {a}_{i} - {k}_{j - 1}.\n\]\n\nThe reader is invited t
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Yes
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Proposition 4.2.14. (1) Let \( a \leq b \leq c \) be integers. We have the exact sequence \[ 1 \rightarrow \left( {1 + {\mathfrak{p}}^{b}}\right) /\left( {1 + {\mathfrak{p}}^{c}}\right) \rightarrow \left( {1 + {\mathfrak{p}}^{a}}\right) /\left( {1 + {\mathfrak{p}}^{c}}\right) \rightarrow \left( {1 + {\mathfrak{p}}^{a}}\right) /\left( {1 + {\mathfrak{p}}^{b}}\right) \rightarrow 1. \] (2) Assume that \( b \leq {2a} \) . Then the map from the multiplicative group \( (1 + \) \( \left. {\mathfrak{p}}^{a}\right) /\left( {1 + {\mathfrak{p}}^{b}}\right) \) to the additive group \( {\mathfrak{p}}^{a}/{\mathfrak{p}}^{b} \), which sends the class of \( 1 + x \) modulo \( 1 + {\mathfrak{p}}^{b} \) to the class of \( x \) modulo \( {\mathfrak{p}}^{b} \), is well-defined and is a group isomorphism.
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Proof. The existence of the exact sequence is trivial. For (2), the definition of \( \left( {1 + {\mathfrak{p}}^{a}}\right) /\left( {1 + {\mathfrak{p}}^{b}}\right) \) (Definition 4.2.3) shows that the map \( \overline{1 + x} \mapsto \bar{x} \) is a bijection. However, it is not a group homomorphism in general (otherwise, \( {G}_{\mathfrak{p}} \) would always be isomorphic to \( \mathfrak{p}/{\mathfrak{p}}^{k} \), and we have seen in Proposition 4.2.12 that this is not always the case). If, however, \( b \leq {2a} \) and \( x \) and \( y \) belong to \( {\mathfrak{p}}^{a} \), we have \( {\mathfrak{p}}^{b}\left| {\mathfrak{p}}^{2a}\right| {xy} \), and hence \( \left( {1 + x}\right) \left( {1 + y}\right) = 1 + x + y + {xy} \equiv 1 + x + y \) \( \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{b}}\right) \), and so the map is a group homomorphism.
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No
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Proposition 4.2.19. Assume \( \\mathfrak{p} \) is a prime ideal above \( p \) of ramification index \( e \) and residual degree \( f \), and assume that \( e = p - 1 \) . Let \( {\\omega }_{i} \) be such that \( \\left( {{\\mathbb{Z}}_{K}/\\mathfrak{p}}\\right) = {\\bigoplus }_{i \\in {D}_{p}}\\left( {\\mathbb{Z}/p\\mathbb{Z}}\\right) \\overline{{\\omega }_{i}}\\; \) with the notation of Proposition 2.4.6 and Corollary 2.4.7, and let \( \\pi \) be a uniformizer of \( \\mathfrak{p} \) (in other words, \( \\pi \\in \\mathfrak{p} \\smallsetminus {\\mathfrak{p}}^{2} \) ). Finally, let \( \\overline{{\\gamma }_{j}} \) as output by Algorithm 4.2.15 be such that \( {\\mathfrak{p}}^{2}/{\\mathfrak{p}}^{p} = \\bigoplus \\left( {\\mathbb{Z}/{p}^{{c}_{j}}\\mathbb{Z}}\\right) \\overline{{\\gamma }_{j}} \) with \( {c}_{j} \\geq 1 \) (after removing the trivial components).
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Proof. Since \( \\pi \) is a uniformizer of \( \\mathfrak{p} \), we have \( \\mathfrak{p}/{\\mathfrak{p}}^{2} = {\\bigoplus }_{i \\in {D}_{p}}\\left( {\\mathbb{Z}/p\\mathbb{Z}}\\right) \\overline{\\pi {\\omega }_{i}} \) ; hence\n\n\[\\left( {1 + \\mathfrak{p}}\\right) /\\left( {1 + {\\mathfrak{p}}^{2}}\\right) = {\\bigoplus }_{i \\in {D}_{p}}\\left( {\\mathbb{Z}/p\\mathbb{Z}}\\right) \\overline{\\left( 1 + \\pi {\\omega }_{i}\\right) }.\]\n\nNote that these are equalities, and not only isomorphisms.\n\nOn the other hand, \( {\\mathfrak{p}}^{2}/{\\mathfrak{p}}^{p} \) is clearly killed by \( p \), so \( {c}_{j} = 1 \) for all \( j \) such that \( {c}_{j} > 0 \) . Since \( e = p - 1 \), by Corollary 4.2.8, we also have\n\n\[\\left( {1 + {\\mathfrak{p}}^{2}}\\right) /\\left( {1 + {\\mathfrak{p}}^{p}}\\right) = {\\bigoplus }_{1 \\leq j \\leq \\left( {p - 2}\\right) f}\\left( {\\mathbb{Z}/p\\mathbb{Z}}\\right) \\overline{{\\exp }_{\\mathfrak{p}}\\left( {\\gamma }_{j}\\right) }.\]\n\nFinally, I claim that the exact sequence\n\n\[1 \\rightarrow \\left( {1 + {\\mathfrak{p}}^{2}}\\right) /\\left( {1 + {\\mathfrak{p}}^{p}}\\right) \\rightarrow \\left( {1 + \\mathfrak{p}}\\right) /\\left( {1 + {\\mathfrak{p}}^{p}}\\right) \\rightarrow \\left( {1 + \\mathfrak{p}}\\right) /\\left( {1 + {\\mathfrak{p}}^{2}}\\right) \\rightarrow 1\]\n\nis split, which will prove the proposition. To prove this, instead of giving a direct proof (which is easy; see Exercise 23), we will use Algorithm 4.1.8.\n\nUsing the same symbol to mean the class modulo different subgroups, we have \( A = {\\left( \\overline{{\\exp }_{\\mathfrak{p}}\\left( {\\gamma }_{j}\\right) }\\right) }_{j},{D}_{A} = p{I}_{f\\left( {e - 1}\\right) }, C = {\\left( \\overline{1 + \\pi {\\omega }_{i}}\\right) }_{i \\in {D}_{p}} \), and \( {D}_{C} = p{I}_{f} \) (where \( {I}_{n} \) always denotes the \( n \\times n \) identity matrix). Thus, following the algorithm, we take \( {B}^{\\prime } = {\\left( \\overline{1 + \\pi {\\omega }_{i}}\\right) }_{i \\in {D}_{p}} \) and \( \\psi \\left( A\\right) = \\left( \\overline{{\\exp }_{\\mathfrak{p}}\\left( {\\gamma }_{j}\\right) }\\right) \).\n\nSet \( \\alpha = 1 + \\pi {\\omega }_{i} \) . By the binomial theorem, we have\n\n\[{\\alpha }^{p} = 1 + \\mathop{\\sum }\\limits_{{1 \\leq k \\leq p - 1}}\\left( \\begin{array}{l} p \\\\ k \\end{array}\\right) {\\pi }^{k}{\\omega }_{i}^{k} + {\\pi }^{p}{\\omega }_{i}^{p}.\]\n\nNow \( {v}_{\\mathfrak{p}}\\left( {{\\pi }^{p}{\\omega }_{i}^{p}}\\right)
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Yes
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Proposition 4.2.23. Let \( \beta \in {K}^{ * } \) be such that \( {v}_{\mathfrak{p}}\left( \beta \right) \geq 0 \) for all \( \mathfrak{p} \mid m{\mathbb{Z}}_{K} \) (this is the case, in particular, if \( \beta \) is coprime to \( m{\mathbb{Z}}_{K} \) ). Then the least common multiple of the denominators occurring in the representation of \( \beta \) on an integral basis is coprime to \( m \) . In other words, there exist \( d \in \mathbb{Z} \) and \( \alpha \in {\mathbb{Z}}_{K} \) such that \( \beta = \alpha /d \) and \( \left( {d, m}\right) = 1 \) .
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Proof. Write \( \beta = {\alpha }_{0}/{d}_{0} \) for \( {\alpha }_{0} \in {\mathbb{Z}}_{K} \) and \( {d}_{0} \in \mathbb{Z} \), for the moment arbitrary. Let \[ g = \left( {{d}_{0},{m}^{\infty }}\right) = \mathop{\prod }\limits_{{p\left| {{d}_{0}, p}\right| m}}{p}^{{v}_{p}\left( {d}_{0}\right) }. \] By definition, we have \( \left( {{d}_{0}/g, m}\right) = 1 \) . On the other hand, let \( \mathfrak{p} \) be a prime ideal and \( p \) be the prime number below \( \mathfrak{p} \) . Then either \( p \nmid g \), in which case \( {v}_{\mathfrak{p}}\left( {{\alpha }_{0}/g}\right) = {v}_{\mathfrak{p}}\left( {\alpha }_{0}\right) \geq 0 \), or \( p \mid g \), in which case we have \( p\left| {{d}_{0}, p}\right| m \), and \( {v}_{p}\left( g\right) = {v}_{p}\left( {d}_{0}\right) \) ; hence \( {v}_{\mathfrak{p}}\left( g\right) = {v}_{\mathfrak{p}}\left( {d}_{0}\right) = {v}_{p}\left( g\right) e\left( {\mathfrak{p}/p}\right) \), so \[ {v}_{\mathfrak{p}}\left( \frac{{\alpha }_{0}}{g}\right) = {v}_{\mathfrak{p}}\left( {\alpha }_{0}\right) - {v}_{\mathfrak{p}}\left( {d}_{0}\right) = {v}_{\mathfrak{p}}\left( \frac{{\alpha }_{0}}{{d}_{0}}\right) = {v}_{\mathfrak{p}}\left( \beta \right) \geq 0. \] It follows that \( {\alpha }_{0}/g \in {\mathbb{Z}}_{K} \), so \( \beta = \left( {{\alpha }_{0}/g}\right) /\left( {{d}_{0}/g}\right) \) is a suitable representation, proving the proposition.
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Yes
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Proposition 5.1.1. Keep all the above notation. Let \( {C}_{j} \) be the congruence subgroup modulo \( \mathfrak{m} \) generated by \( C \) and by the \( {\mathfrak{b}}_{i} \) for \( i \neq j \), and let \( {L}_{j} \) be the subfield of \( K\left( \mathfrak{m}\right) \) corresponding to the congruence subgroup \( \left( {\mathfrak{m},{C}_{j}}\right) \) under the Takagi correspondence.\n\n(1) The group \( \operatorname{Gal}\left( {{L}_{j}/K}\right) \) is isomorphic via the Artin map to \( C{l}_{\mathfrak{m}}\left( K\right) /\overline{{C}_{j}} = \) \( \left( {\mathbb{Z}/{b}_{j}\mathbb{Z}}\right) \overline{{\mathfrak{b}}_{j}} \), and in particular \( {L}_{j}/K \) is a cyclic extension of prime power degree \( {b}_{j} \) .\n\n(2) The compositum in \( K\left( \mathfrak{m}\right) \) of the \( {L}_{j} \) is equal to the class field \( L \) corresponding to the congruence subgroup \( \left( {\mathfrak{m}, C}\right) \) .
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Proof. By Galois theory we have \( {L}_{j} = K{\left( \mathfrak{m}\right) }^{\operatorname{Art}\left( {C}_{j}\right) } \), hence via the Artin map, \( \operatorname{Gal}\left( {{L}_{j}/K}\right) \) is isomorphic to \( C{l}_{\mathrm{m}}/\overline{{C}_{j}} = \left( {\mathbb{Z}/{b}_{j}\mathbb{Z}}\right) \overline{{\mathfrak{b}}_{j}} \), so \( {L}_{j}/K \) is a cyclic extension of prime power degree \( {b}_{j} \), proving the first statement. Furthermore, by Galois theory the compositum of the \( {L}_{j} \) in \( K\left( \mathfrak{m}\right) \) corresponds to the congruence subgroup \( \left( {\mathfrak{m},\mathop{\bigcap }\limits_{j}{C}_{j}}\right) \) . But since the only relations satisfied by the \( \overline{{\mathfrak{b}}_{i}} \) in \( C{l}_{\mathrm{m}}\left( K\right) /\bar{C} \) are \( {\overline{\mathfrak{b}}}_{i}^{{b}_{i}} = \overline{1} \), it follows that \( {\mathfrak{b}}_{j} \notin {C}_{j} \), hence that \( \mathop{\bigcap }\limits_{j}{C}_{j} = C \) , proving the proposition.
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Yes
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Theorem 5.2.2. With the above notation, the field \( L = K\left( \sqrt[\ell ]{\alpha }\right) \) with \( \alpha \in \) \( {K}^{ * } \smallsetminus {K}^{*\ell } \) is a cyclic extension of \( K \) of conductor equal to \( \mathfrak{m} \) and degree \( \ell \) if and only if the following ten conditions hold.\n\n(1) \( {S}_{\mathrm{m},\ell ,3} = \varnothing \) .\n\n(2) If \( \mathfrak{p} \in {S}_{\mathfrak{m},\ell ,2} \), then \( {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) ≢ 1\left( {\;\operatorname{mod}\;\ell }\right) \) and, in particular, \( {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) \geq 2 \) .\n\n(3) If \( \mathfrak{p} \in {S}_{\mathfrak{m}} \), then \( {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) = 1 \) .\n\n(4) If \( \mathfrak{p} \in {S}_{\mathfrak{m},\ell ,1} \), then \( \ell \nmid {v}_{\mathfrak{p}}\left( \alpha \right) \) .\n\n(5) If \( \mathfrak{p} \in {S}_{\mathfrak{m},\ell ,2} \), then \( \ell \mid {v}_{\mathfrak{p}}\left( \alpha \right) \) and the largest \( k \) such that the congruence\n\n\[ \alpha \equiv {x}^{\ell }\;\left( {\text{mod}\;{\mathfrak{p}}^{{v}_{\mathfrak{p}}\left( \alpha \right) + k}}\right) \]\n\nhas a solution must be equal to \( z\left( {\mathfrak{p},\ell }\right) - {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) \) .\n\n(6) If \( \mathfrak{p} \in {S}_{\ell } \), then \( \ell \mid {v}_{\mathfrak{p}}\left( \alpha \right) \) and the congruence\n\n\[ \alpha \equiv {x}^{\ell }\left( {{\;\operatorname{mod}\;{\mathfrak{p}}^{{v}_{\mathfrak{p}}\left( \alpha \right) + z\left( {\mathfrak{p},\ell }\right) - 1}}}\right) \]\n\nhas a solution.\n\n(7) If \( \mathfrak{p} \in {S}_{\mathfrak{m}} \), then \( \ell \nmid {v}_{\mathfrak{p}}\left( \alpha \right) \) .\n\n(8) If \( \mathfrak{p} \in {S}_{\varnothing } \), then \( \ell \mid {v}_{\mathfrak{p}}\left( \alpha \right) \) .\n\n(9) If \( \sigma \in {\mathfrak{m}}_{\infty } \), then \( \sigma \left( \alpha \right) < 0 \) .\n\n(10) If \( \sigma \) is a real embedding that is not in \( {\mathfrak{m}}_{\infty } \), then \( \sigma \left( \alpha \right) > 0 \) .
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Proof. Assume first that \( L/K \) is of conductor equal to \( \mathfrak{m} \) . Then by Corollary 3.5.12 (1), we know that \( \mathfrak{d}\left( {L/K}\right) = {\mathfrak{m}}_{0}^{\ell - 1} \), where as usual \( {\mathfrak{m}}_{0} \) is the finite part of \( m \) . By Theorem 10.2.9, we thus have the following.\n\n(1) If \( \ell \nmid {v}_{\mathfrak{p}}\left( \alpha \right) \), then \( {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) = z\left( {\mathfrak{p},\ell }\right) \).\n\n(2) If \( \mathfrak{p} \nmid \ell \) and \( \ell \mid {v}_{\mathfrak{p}}\left( \alpha \right) \), then \( {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) = 0 \).\n\n(3) If \( \mathfrak{p}\left| {\ell ,\mathfrak{p}}\right| \mathfrak{m} \), and \( \ell \mid {v}_{\mathfrak{p}}\left( \alpha \right) \), then \( {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) = 0 \) if \( a \geq z\left( {\mathfrak{p},\ell }\right) - 1,{v}_{\mathfrak{p}}\left( \mathfrak{m}\right) = \) \( z\left( {\mathfrak{p},\ell }\right) - a \) if \( a < z\left( {\mathfrak{p},\ell }\right) \), where \( a \) is the largest value of \( k \) for which the congruence\n\n\[ {x}^{\ell } \equiv \alpha \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k + {v}_{\mathfrak{p}}\left( \alpha \right) }}\right) \]\n\nhas a solution.\n\nBy Theorem 10.2.9, we also know that \( a \geq 1 \) and \( \ell \nmid a \) . Since we want all the places of \( K \) dividing \( \mathfrak{m} \), and only those, to ramify, this implies immediately all the necessary conditions on \( \alpha \) . It also implies that \( {S}_{\mathfrak{m},\ell ,3} = \varnothing \) . Finally, the two other conditions (2) and (3) on the modulus \( m \) are immediate consequences of Corollary 3.
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Yes
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Proposition 5.2.3. Let \( \gamma \in {K}^{ * } \) . The following two properties are equivalent.\n\n(1) There exists an ideal \( \mathfrak{q} \) such that \( \gamma {\mathbb{Z}}_{K} = {\mathfrak{q}}^{\ell } \).\n\n(2) The element \( \gamma \) belongs to the group generated by the units, the \( {\alpha }_{i} \) defined above for \( 1 \leq i \leq {r}_{c} \), and the \( \ell \) th powers of elements of \( {K}^{ * } \) .
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Proof. Since for \( i \leq h \), we have \( {\alpha }_{i}{\mathbb{Z}}_{K} = {\left( {\mathfrak{a}}_{i}^{{d}_{ * }/\ell }\right) }^{\ell } \), it is clear that if \( \gamma \) belongs to the group mentioned in the proposition, then \( \gamma {\mathbb{Z}}_{K} \) is the \( \ell \) th power of an ideal. Conversely, assume that \( \gamma {\mathbb{Z}}_{K} = {\mathfrak{q}}^{\ell } \) . Then, if \( \mathfrak{q} = \beta \mathop{\prod }\limits_{{1 \leq i \leq {g}_{c}}}{\mathfrak{a}}_{i}^{{x}_{i}} \), we have \( \gamma {\mathbb{Z}}_{K} = {\beta }^{\ell }\mathop{\prod }\limits_{{1 \leq i \leq {g}_{c}}}{\mathfrak{a}}_{i}^{\ell {x}_{i}} \), hence \( {d}_{i} \mid \ell {x}_{i} \) for all \( i \), so \( {d}_{i} \mid {x}_{i} \) for \( i > {r}_{c} \) , while \( \left( {{d}_{i}/\ell }\right) \mid {x}_{i} \) for \( \bar{i} \leq {r}_{c} \) . It follows that\n\n\[ \gamma {\mathbb{Z}}_{K} = {\beta }^{\ell }\mathop{\prod }\limits_{{1 \leq i \leq {r}_{c}}}{\alpha }_{i}^{{n}_{i}}\mathop{\prod }\limits_{{{r}_{c} < i \leq {g}_{c}}}{\alpha }_{i}^{{n}_{i}\ell } \]\n\nwith \( {n}_{i} = {x}_{i}/{d}_{i} \) for \( i > {r}_{c} \) and \( {n}_{i} = {x}_{i}/\left( {{d}_{i}/\ell }\right) \) for \( i \leq {r}_{c} \), thus proving the proposition.
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Yes
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(1) The quotient group \( U\left( K\right) /U{\left( K\right) }^{\ell } \) is a \( \mathbb{Z}/\ell \mathbb{Z} \) -vector space of dimension \( {r}_{u} + 1 \), a basis consisting of the classes of the \( {\varepsilon }_{j} \) for \( 0 \leq j \leq {r}_{u} \) .
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Proof. Since \( {V}_{\ell }\left( K\right) \) is generated by the \( {\alpha }_{i} \), the \( {\varepsilon }_{j} \), and \( {K}^{*\ell } \), we must simply find the dependencies between the \( {\alpha }_{i} \) and \( {\varepsilon }_{j} \) in \( {V}_{\ell }\left( K\right) /{K}^{*\ell } \) as a \( \mathbb{Z}/\ell \mathbb{Z} \) -vector space. Hence assume that\n\n\[ \mathop{\prod }\limits_{{0 \leq j \leq {r}_{u}}}{\varepsilon }_{j}^{{x}_{j}}\mathop{\prod }\limits_{{1 \leq i \leq {r}_{c}}}{\alpha }_{i}^{{n}_{i}} = {\gamma }^{\ell } \]\n\nfor some \( \gamma \in {K}^{ * } \) . By definition of \( {\alpha }_{i} \), this implies\n\n\[ \mathop{\prod }\limits_{{1 \leq i \leq {r}_{c}}}{\mathfrak{a}}_{i}^{{d}_{i}{n}_{i}} = {\gamma }^{\ell }{\mathbb{Z}}_{K} \]\n\nhence \( {\mathfrak{b}}^{\ell } = {\gamma }^{\ell }{\mathbb{Z}}_{K} = {\left( \gamma {\mathbb{Z}}_{K}\right) }^{\ell } \) with\n\n\[ \mathfrak{b} = \mathop{\prod }\limits_{{1 \leq i \leq {r}_{c}}}{\mathfrak{a}}_{i}^{\left( {{d}_{i}/\ell }\right) {n}_{i}} \]\n\nThus, \( \mathfrak{b} = \gamma {\mathbb{Z}}_{K} \) is a principal ideal; hence \( {d}_{i} \mid \left( {{d}_{i}/\ell }\right) {n}_{i} \) for \( 1 \leq i \leq {r}_{c} \) or, equivalently, \( {n}_{i} \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) \), so the virtual units \( {\alpha }_{i} \) do not enter into our dependency. Thus, it is enough to prove (1). So assume that\n\n\[ \mathop{\prod }\limits_{{0 \leq j \leq {r}_{u}}}{\varepsilon }_{j}^{{x}_{j}} = {\varepsilon }^{\ell } \]\n\nfor some \( \varepsilon \in {K}^{ * } \) . Since \( \varepsilon \) is a root of the monic polynomial \( {X}^{\ell } - {\varepsilon }^{\ell } = 0 \) with coefficients in \( {\mathbb{Z}}_{K} \), it follows that it is an algebraic integer (see, for example, [Coh0, Corollary 4.1.5]), and since \( \varepsilon \in {K}^{ * } \), we have \( \varepsilon \in {\mathbb{Z}}_{K} \) . Furthermore, the absolute norm of \( \varepsilon \) is in \( \mathbb{Z} \), and its \( \ell \) th power is equal to \( \pm 1 \), from which it follows that \( \mathcal{N}\left( \varepsilon \right) = \pm 1 \), hence \( \varepsilon \) is a unit of \( K \) . For future reference, we isolate this as a lemma.\n\nLemma 5.2.6. We have \( U\left( K\right) \cap {K}^{*\ell } = U{\left( K\right) }^{\ell } \) .\n\nThus, if \( \varepsilon = \mathop{\prod }\limits_{{0 < j < {r}_{u}}}{\varepsilon }_{j}^{{y}_{j}} \), we have \( {x}_{j} = \ell {y}_{j} \) for \( j \geq 1 \), while \( {x}_{0} \equiv \ell {y}_{0} \) \( \left( {{\;\operatorname{mod}\;w}\left( K\right) }\right) \) . Thus, for \( j \geq 1,{x}_{j} \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) \) so the fundamental units \( {\varepsilon }_{j} \) do not enter into our dependency. Furthermore, since \( {\zeta }_{\ell } \in K \), we have \( \ell \mid w\left( K\right) \) , so we also have \( {x}_{0} \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) \), proving the proposition.
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Yes
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Lemma 5.2.6. We have \( U\left( K\right) \cap {K}^{*\ell } = U{\left( K\right) }^{\ell } \) .
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Thus, if \( \varepsilon = \mathop{\prod }\limits_{{0 < j < {r}_{u}}}{\varepsilon }_{j}^{{y}_{j}} \), we have \( {x}_{j} = \ell {y}_{j} \) for \( j \geq 1 \), while \( {x}_{0} \equiv \ell {y}_{0} \) \( \left( {{\;\operatorname{mod}\;w}\left( K\right) }\right) \) . Thus, for \( j \geq 1,{x}_{j} \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) \) so the fundamental units \( {\varepsilon }_{j} \) do not enter into our dependency. Furthermore, since \( {\zeta }_{\ell } \in K \), we have \( \ell \mid w\left( K\right) \) , so we also have \( {x}_{0} \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) \), proving the proposition.
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No
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Proposition 5.2.8. We have the following exact sequence:\n\n\[ 1 \rightarrow {\mathbf{\mu }}_{\ell }\left( K\right) \rightarrow U\left( K\right) \overset{\left\lbrack \ell \right\rbrack }{ \rightarrow }U\left( K\right) \rightarrow \frac{{V}_{\ell }\left( K\right) }{{K}^{*\ell }} \]\n\n\[ \overset{\phi }{ \rightarrow }{Cl}\left( K\right) \overset{\left\lbrack \ell \right\rbrack }{ \rightarrow }{Cl}\left( K\right) \rightarrow \frac{{Cl}\left( K\right) }{{Cl}{\left( K\right) }^{\ell }} \rightarrow 1. \]
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Proof. The proof is straightforward and is left to the reader (Exercise 3).
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No
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Theorem 5.2.9. Keep the above notation, and in particular recall that we write \( z\left( {\mathfrak{p},\ell }\right) = \ell e\left( {\mathfrak{p}/\ell }\right) /\left( {\ell - 1}\right) + 1 \) and \( S = {S}_{\mathfrak{m}} \cup {S}_{\mathfrak{m},\ell ,1} \). Let \( L/K \) be a cyclic extension of \( K \) of degree \( \ell \) and of conductor equal to \( \mathfrak{m} \). Then \( \mathfrak{m} \) satisfies conditions (1),(2), and (3) of Theorem 5.2.2, and up to Kummer-equivalence, we can choose \( L = K\left( \sqrt[\ell ]{\alpha }\right) \) with \( \alpha \) of the following form: \[ \alpha = \mathop{\prod }\limits_{{\mathfrak{p} \in S}}{\beta }_{\mathfrak{p}}^{{x}_{\mathfrak{p}}}\mathop{\prod }\limits_{{i = 1}}^{{{r}_{c} + {r}_{u} + 1}}{v}_{i}^{{n}_{i}} \] with the following additional conditions. (1) For all \( \mathfrak{p} \in S \), we have \( 1 \leq {x}_{\mathfrak{p}} \leq \ell - 1 \) ; for all \( i \), we have \( 0 \leq {n}_{i} \leq \ell - 1 \). (2) For \( \mathfrak{p} \in {S}_{\mathfrak{m},\ell ,2} \), the largest \( k \) such that the congruence \[ {x}^{\ell } \equiv \alpha \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k + {v}_{\mathfrak{p}}\left( \alpha \right) }}\right) \] has a solution must be equal to \( z\left( {\mathfrak{p},\ell }\right) - {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) \). (3) If \( S \) is not empty, we may fix any one (but only one) of the \( {x}_{\mathfrak{p}} \) equal to 1. (4) For each \( \mathfrak{p} \in {S}_{\ell } \), the congruence \[ {x}^{\ell } \equiv \alpha \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{z\left( {\mathfrak{p},\ell }\right) - 1 + {v}_{\mathfrak{p}}\left( \alpha \right) }}\right) \] has a solution. (5) For each \( i \leq {r}_{c} \), we must have \[ \mathop{\sum }\limits_{{\mathfrak{p} \in S}}{x}_{\mathfrak{p}}{p}_{i,\mathfrak{p}} \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) \] (6) For each \( \sigma \in {\mathfrak{m}}_{\infty },\sigma \left( \alpha \right) < 0 \), while for each real embedding \( \sigma \notin {\mathfrak{m}}_{\infty } \), we have \( \sigma \left( \alpha \right) > 0 \). Conversely, if \( \mathfrak{m} \) satisfies conditions (1),(2), and (3) of Theorem 5.2.2, if the above conditions are satisfied, and if \( \alpha \neq 1 \), then \( L = K\left( \sqrt[\ell ]{\alpha }\right) \) is a cyclic extension of degree \( \ell \) and of conductor \( \mathfrak{m} \).
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Proof. Since by Theorem 5.2.2, we have \( {S}_{\mathfrak{m},\ell ,3} = \varnothing \), we can write \[ \alpha {\mathbb{Z}}_{K} = \mathop{\prod }\limits_{{\mathfrak{p} \in S}}{\mathfrak{p}}^{{x}_{\mathfrak{p}}}\mathop{\prod }\limits_{{\mathfrak{p} \in {S}_{\mathfrak{m},\ell ,2}}}{\mathfrak{p}}^{{x}_{\mathfrak{p}}}\mathop{\prod }\limits_{{\mathfrak{p} \in {S}_{\ell }}}{\mathfrak{p}}^{{x}_{\mathfrak{p}}}\mathop{\prod }\limits_{{\mathfrak{p} \in {S}_{\varnothing }}}{\mathfrak{p}}^{{x}_{\mathfrak{p}}}. \] By Theorem 5.2.2, when \( \mathfrak{p} \in {S}_{\mathfrak{m},\ell ,2},\mathfrak{p} \in {S}_{\ell } \), or \( \mathfrak{p} \in {S}_{\varnothing } \), we must have \( \ell \mid {v}_{\mathfrak{p}}\left( \alpha \right) = {x}_{\mathfrak{p}} \). By the approximation theorem, we can find \( \gamma \in {K}^{ * } \) such that \( {v}_{\mathfrak{p}}\left( \gamma \right) = - {x}_{\mathfrak{p}}/\ell \) for \( \mathfrak{p} \in {S}_{\mathfrak{m},\ell ,2} \) and \( \mathfrak{p} \in {S}_{\ell },{v}_{\mathfrak{p}}\left( \gamma \right) = - \left\lfloor {{x}_{\mathfrak{p}}/\ell }\right\rfloor \) for \( \mathfrak{p} \in \) \( S = {S}_{\mathfrak{m}} \cup {S}_{\mathfrak{m},\ell ,1} \), and no special conditions for \( \mathfrak{p} \in {S}_{\varnothing } \). Since \( \alpha \) is Kummer-equivalent to \( \alpha {\gamma }^{\ell } \), we may thus replace \( \alpha \) by \( \alpha {\gamma }^{\ell } \), hence for this new \( \alpha \) we will have \( {x}_{\mathfrak{p}} = 0 \) for \( \mathfrak{p} \in {S}_{\mathfrak{m},\ell ,2} \) and \( \mathfrak{p} \in {S}_{\ell } \), and also \( 1 \leq {x}_{\mathfrak{p}} \le
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No
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Let \( \mathfrak{p} \) be a prime ideal, let \( k \geq 1 \) be an integer, and let \( {\left( {\mathbb{Z}}_{K}/{\mathfrak{p}}^{k}\right) }^{ * } = {\bigoplus }_{1 < i < s}\left( {\mathbb{Z}/{c}_{i}\mathbb{Z}}\right) \overline{{g}_{i}} \) as above. Let \( t \) be the largest index \( i \) such that \( \ell \mid {c}_{i} \) . If \( \alpha \in {K}^{ * } \), the congruence\n\n\[ \n{x}^{\ell } \equiv \alpha \left( {{\;\operatorname{mod}\;{}^{ * }}{\mathfrak{p}}^{k + {v}_{\mathfrak{p}}\left( \alpha \right) }}\right)\n\]\n\nhas a solution if and only if\n\n\[ \n\left( {{v}_{\mathfrak{p}},\left( {{a}_{1},\ldots ,{a}_{t}}\right) }\right) \equiv 0\left( {\;\operatorname{mod}\;\ell }\right)\n\]\n\nwhere \( \left( {{v}_{\mathfrak{p}},\left( {{a}_{1},\ldots ,{a}_{s}}\right) }\right) \) is the discrete logarithm of \( \alpha \) as defined above.
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Write \( \alpha = \beta {\pi }_{\mathfrak{p}}^{{v}_{\mathfrak{p}}}\mathop{\prod }\limits_{{1 \leq i \leq s}}{g}_{i}^{{a}_{i}} \) and \( x = \gamma {\pi }_{\mathfrak{p}}^{w}\mathop{\prod }\limits_{{1 \leq i \leq s}}{g}_{i}^{{x}_{\imath }} \) as above. Since \( k \geq 1,{x}^{\ell } \equiv \alpha \left( \right. \) mod \( \left. {\mathfrak{p}}^{k + {\bar{v}}_{\mathfrak{p}}}\right) \) implies that \( {v}_{\mathfrak{p}} = w\ell \), hence that \( {v}_{\mathfrak{p}} \equiv 0 \) \( \left( {\;\operatorname{mod}\;\ell }\right) \) . In addition, since \( \beta \equiv \gamma \equiv 1\left( {{\;\operatorname{mod}\;{}^{ * }}{\mathfrak{p}}^{k}}\right) \), we must have \( {a}_{i} \equiv \ell {x}_{i} \) (mod \( {c}_{i} \) ) for all \( i \) . For \( i \leq t \), the existence of \( {x}_{i} \) is equivalent to \( \ell \mid {a}_{i} \), while for \( i > t,{x}_{i} \) always exists since \( {c}_{i} \) is coprime to \( \ell \), proving the proposition.
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Yes
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Proposition 5.3.1. Let \( L \) be a number field, let \( {L}_{1} \) and \( {L}_{2} \) be two extensions of \( L \) included in a fixed algebraic closure \( \bar{L} \) of \( L \), and let \( {L}_{1}{L}_{2} \) be the compositum of \( {L}_{1} \) and \( {L}_{2} \) in \( \bar{L} \). (1) If \( {L}_{1}/L \) and \( {L}_{2}/L \) are normal extensions, then \( {L}_{1}{L}_{2}/L \) is also a normal extension.
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Proof. Let \( N \) be the normal closure of \( {L}_{1}{L}_{2} \) (or any field containing \( N \) and normal over \( L \) ), and let \( \mathcal{G} = \operatorname{Gal}\left( {N/L}\right) \) be the Galois group of \( N/L \) . For \( i = 1,2 \), let \( {\mathcal{G}}_{i} = \operatorname{Gal}\left( {N/{L}_{i}}\right) \) so that \( {\mathcal{G}}_{i} \) is a normal subgroup of \( \mathcal{G} \) with \( \operatorname{Gal}\left( {{L}_{i}/L}\right) \simeq \mathcal{G}/{\mathcal{G}}_{i} \) . By Galois theory, subfields of \( N \) containing both \( {L}_{1} \) and \( {L}_{2} \) are in one-to-one correspondence with subgroups of \( \mathcal{G} \) contained in \( {\mathcal{G}}_{1} \cap {\mathcal{G}}_{2} \) , hence \( \operatorname{Gal}\left( {N/{L}_{1}{L}_{2}}\right) = {\mathcal{G}}_{1} \cap {\mathcal{G}}_{2} \) . Since \( {\mathcal{G}}_{1} \cap {\mathcal{G}}_{2} \) is the intersection of two normal subgroups, it is also a normal subgroup, hence \( {L}_{1}{L}_{2}/L \) is normal with Galois group isomorphic to \( \mathcal{G}/{\mathcal{G}}_{1} \cap {\mathcal{G}}_{2} \), proving (1).
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Yes
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Proposition 5.3.2. The extension \( {K}_{z}/K = K\left( {\zeta }_{\ell }\right) /K \) is a cyclic extension of degree \( d = \left( {\ell - 1}\right) /m \) for some divisor \( m \) of \( \ell - 1 \) such that \( m < \ell - 1 \) . The Galois group \( \operatorname{Gal}\left( {{K}_{z}/K}\right) \) is generated by the automorphism \( \tau \) of order \( d \) defined by \( \tau \left( {\zeta }_{\ell }\right) = {\zeta }_{\ell }^{g} \) and \( \tau \left( x\right) = x \) for \( x \in K \), where \( g = {g}_{0}^{m} \) .
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Proof. We apply Proposition 5.3.1 (3) to the case \( L = \mathbb{Q},{L}_{1} = \mathbb{Q}\left( {\zeta }_{\ell }\right) ,{L}_{2} = \) \( K \), hence \( {L}_{1}{L}_{2} = K\left( {\zeta }_{\ell }\right) = {K}_{z} \) . Thus \( {K}_{z}/K \) is normal, and its Galois group can be identified with a subgroup of \( \mathrm{{Gal}}\left( {\mathbb{Q}\left( {\zeta }_{\ell }\right) /\mathbb{Q}}\right) \simeq {\left( \mathbb{Z}/\ell \mathbb{Z}\right) }^{ * } \) . Since this is a cyclic group of order \( \ell - 1,\operatorname{Gal}\left( {{K}_{z}/K}\right) \) is a cyclic group of order dividing \( \ell - 1 \) , hence of order \( \left( {\ell - 1}\right) /m \) for some \( m < \ell - 1 \), since we have assumed \( {\zeta }_{\ell } \notin K \) . Since \( {\left( \mathbb{Z}/\ell \mathbb{Z}\right) }^{ * } \) has a unique subgroup of given order \( \left( {\ell - 1}\right) /m \), generated by \( g = {g}_{0}^{m} \), the proposition follows.
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Yes
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Corollary 5.3.4. With the above notation, the eigenspace \( {W}_{k} \) is equal to \( {e}_{k}W = \left\{ {{x}^{{e}_{k}}/x \in W}\right\} . \)
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Proof. We have\n\n\[ \tau \left( {x}^{{e}_{k}}\right) = {x}^{\tau {e}_{k}} = {x}^{{g}^{k}{e}_{k}} = {x}^{{e}_{k}{g}^{k}}. \]\n\nThus, \( {x}^{{e}_{k}} \in {W}_{k} \) ; hence \( {e}_{k}W \subset {W}_{k} \) . It follows that\n\n\[ {\bigoplus }_{0 \leq k < d}{e}_{k}W \subset {\bigoplus }_{0 \leq k < d}{W}_{k} = W \]\n\nSince the \( {e}_{k} \) form a complete set of orthogonal idempotents, we have \( W = \) \( {\bigoplus }_{0 < k < d}{e}_{k}W \) (since \( x = \mathop{\prod }\limits_{{0 \leq k < d}}{x}^{{e}_{k}} \) ), and so we must have the equality \( {e}_{k}\bar{W} = {W}_{k} \) for all \( k \), proving the corollary.
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Yes
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Theorem 5.3.5. Let \( K \) be a number field, and let \( L \) be a cyclic extension of \( K \) of degree \( \ell \) . Assume that \( K \) does not contain \( {\zeta }_{\ell } \), and let \( {K}_{z} = K\left( {\zeta }_{\ell }\right) \) and \( {L}_{z} = L\left( {\zeta }_{\ell }\right) \) . Let \( {g}_{0} \) be a primitive root modulo \( \ell \), let \( d = \left\lbrack {{K}_{z} : K}\right\rbrack = \left( {\ell - 1}\right) /m \) , and \( g = {g}_{0}^{m} \) as above. Finally, let \( W = {K}_{z}^{ * }/{K}_{z}^{*\ell } \).\n\n(1) Any \( \alpha \) such that \( {L}_{z} = {K}_{z}\left( \sqrt[\ell ]{\alpha }\right) \) belongs to the eigenspace \( {e}_{1}W = {W}_{1} \) of \( W \) (and such \( \alpha \) exist by Kummer theory).
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Proof. The extensions \( {K}_{z}/K \) and \( L/K \) are cyclic and have coprime degrees, hence by Proposition 5.3.1 \( {L}_{z}/K \) is an Abelian extension and \( \operatorname{Gal}\left( {{L}_{z}/L}\right) \simeq \operatorname{Gal}\left( {{K}_{z}/K}\right) = \langle \tau \rangle \), and \( \operatorname{Gal}\left( {{L}_{z}/{K}_{z}}\right) \simeq \operatorname{Gal}\left( {L/K}\right) = \langle \sigma \rangle \) for some \( \sigma \) of order \( \ell \) . For any \( K \) -automorphism \( s \) of \( L \) and \( K \) -automorphism \( t \) of \( {K}_{z} \), there exists a unique \( K \) -automorphism of \( {L}_{z} \) that extends both \( s \) and \( t \), and it is defined in a natural way.\n\nBy a natural abuse of notation, we will denote by \( \tau \) the unique \( K \) - automorphism of \( {L}_{z} \) that extends the \( K \) -automorphism \( \tau \) of \( {K}_{z} \) and is the identity on \( L \), and similarly we will denote by \( \sigma \) the unique \( K \) -automorphism of \( {L}_{z} \) that extends the \( K \) -automorphism \( \sigma \) of \( L \) and is the identity on \( {K}_{z} \) .\n\nThis being noted, let us prove the theorem. Since \( {L}_{z} \) is a cyclic extension of \( {K}_{z} \) of degree \( \ell \), by Kummer theory (Corollary 10.2.7 in this case) we know that there exists \( \alpha \in {K}_{z}^{ * } \) not in \( {K}_{z}^{*\ell } \) such that \( {L}_{z} = {K}_{z}\left( \theta \right) \) with \( \theta = \sqrt[\ell ]{\alpha } \) . Since \( \tau \) is an automorphism of \( {L}_{z} \), we have \( \tau \left( \theta \right) \in {L}_{z} \) and \( {\tau }^{-1}\left( \theta \right) \in {L}_{z} \), from which it follows that \( {L}_{z} = {K}_{z}\left( {\tau \left( \theta \right) }\right) \) . Since \( \tau {\left( \theta \right) }^{\ell } = \tau \left( \alpha \right) \), Corollary 10.2.7 (2) tells us that there exists \( j \) coprime to \( \ell \) (hence of the form \( {g}_{0}^{a} \) ) and \( \gamma \in {K}_{z} \) such that\n\n\[ \tau \left( \alpha \right) = {\alpha }^{j}{\gamma }^{\ell } = {\alpha }^{{g}_{0}^{a}}{\gamma }^{\ell } \]
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Yes
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Proposition 5.3.6. Let \( \alpha \in {K}_{z}^{ * } \) be such that \( \tau \left( \alpha \right) = {\alpha }^{g}{\gamma }^{\ell } \) for some \( \gamma \in {K}_{z}^{ * } \) . Then \( \alpha = {\beta }^{\lambda }{\delta }^{\ell } \), where\n\n\[ \beta = {\gamma }^{-a}{\zeta }_{\ell }^{k}\;\text{ with }\;a = {\left( \frac{{g}^{d} - 1}{\ell }\right) }^{-1}\left( {\;\operatorname{mod}\;\ell }\right) \]\n\nfor some integer \( k \) and some \( \delta \in {K}_{z}^{ * } \) .\n\nConversely, let \( \beta \) be given such that \( \alpha = {\beta }^{\lambda } \), and let \( \lambda = {\lambda }_{0} + \nu \ell \) for \( \nu \in \mathbb{Z}\left\lbrack G\right\rbrack \) with \( {\lambda }_{0} \) as above. Then \( \tau \left( \alpha \right) = {\alpha }^{g}{\gamma }^{\ell } \) with\n\n\[ \gamma = {\beta }^{-\left( {{g}^{d} - 1}\right) /\ell + \left( {\tau - g}\right) \nu }.\n\]
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Proof. It is easy to prove by induction on \( a \) that for any \( a \geq 0 \) ,\n\n\[ {\tau }^{a}\left( \alpha \right) = {\alpha }^{{g}^{a}}{\gamma }^{\ell \left( {{g}^{a} - {\tau }^{a}}\right) /\left( {g - \tau }\right) }.\]\n\nApplying this formula to \( a = d \), we obtain\n\n\[ \alpha = {\alpha }^{{g}^{d}}{\gamma }^{{\lambda }_{0}\ell } \]\n\nwith \( {\lambda }_{0} \) as above. Thus,\n\n\[ {\alpha }^{\left( {{g}^{d} - 1}\right) /\ell } = {\gamma }^{-{\lambda }_{0}}{\zeta }_{\ell }^{u} = {\gamma }^{-\lambda }{\varepsilon }^{\ell }{\zeta }_{\ell }^{u} \]\n\nfor some integer \( u \) and some \( \varepsilon \in {K}_{z}^{ * } \) .\n\nWe now use for the first and only time the technical condition \( {g}^{d} ≢ 1 \) \( \left( {\;\operatorname{mod}\;{\ell }^{2}}\right) \) imposed above on the generator \( g \) . This implies that \( \left( {{g}^{d} - 1}\right) /\ell ≢ 0 \) \( \left( {\;\operatorname{mod}\;\ell }\right) \), so we can find integers \( a \) and \( b \) such that \( a\left( {{g}^{d} - 1}\right) /\ell + b\ell = 1 \) . It follows that\n\n\[ \alpha = {\alpha }^{a\left( {{g}^{d} - 1}\right) /\ell }{\alpha }^{b\ell } = {\gamma }^{-{a\lambda }}{\delta }^{\ell }{\zeta }_{\ell }^{au} \]\n\nfor some \( \delta \in {K}_{z}^{ * } \) .\n\nFinally, note that \( {\zeta }_{\ell }^{\lambda } = {\zeta }_{\ell }^{d{g}^{d - 1}} \), so that \( {\zeta }_{\ell }^{-{gm\lambda }} = {\zeta }_{\ell } \) . We thus obtain\n\n\[ \alpha = {\left( {\gamma }^{-a}{\zeta }_{\ell }^{-{gmau}}\right) }^{\lambda }{\delta }^{\ell } \]\n\nproving the first part of the proposition.\n\nFor the converse, we have\n\n\[ \frac{\tau \left( \alpha \right) }{{\alpha }^{g}} = {\alpha }^{\tau - g} = {\beta }^{\left( {\tau - g}\right) \lambda } = {\beta }^{\left( {\tau - g}\right) \left( {{\lambda }_{0} + \ell \nu }\right) }.\]\n\nSince \( \left( {\tau - g}\right) {\lambda }_{0} = 1 - {g}^{d} \), we have \( \tau \left( \alpha \right) /{\alpha }^{g} = {\gamma }^{\ell } \) with \( \gamma = {\beta }^{-\left( {{g}^{d} - 1}\right) /\ell + \left( {\tau - g}\right) \nu } \) , as claimed.
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Yes
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Corollary 5.3.7. Keep the notation of Theorem 5.3.5 and Proposition 5.3.6. Set \( \mu = - \left( {{g}^{d} - 1}\right) /\ell + \left( {\tau - g}\right) \nu \) . Then if \( \alpha = {\beta }^{\lambda } \), we can take \( \tau \left( \theta \right) = {\theta }^{g}{\beta }^{\mu } \) .
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Proof. We have \( {\theta }^{\ell } = \alpha \) . With any initial choice of \( \tau \), we have \( \tau {\left( \theta \right) }^{\ell } = \) \( \tau \left( \alpha \right) = {\alpha }^{g}{\gamma }^{\ell } \) with \( \gamma = {\beta }^{\mu } \) . Thus, \( \tau \left( \theta \right) = {\theta }^{g}{\beta }^{\mu }{\zeta }_{\ell }^{k} \) for some integer \( k \) . If we set \( {\tau }^{\prime } = {\sigma }^{-{g}^{-1}k}\tau \), then \( {\tau }^{\prime } \) also extends \( \tau \), and we have\n\n\[ \n{\tau }^{\prime }\left( \theta \right) = {\theta }^{g}{\zeta }_{\ell }^{-g{g}^{-1}k}{\zeta }_{\ell }^{k}{\beta }^{\mu } = {\theta }^{g}{\beta }^{\mu },\n\]\n\nso the result follows. Of course, the fixed field of \( {L}_{z} \) by \( {\tau }^{\prime } \) is still equal to \( L \) since \( L \) is normal over \( K \) .
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Yes
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Lemma 5.3.8. Keep all the above notation, and choose\n\n\[ \lambda = \mathop{\sum }\limits_{{0 \leq a < d}}{r}_{d - 1 - a}{\tau }^{a} \]\n\nThen \( {\tau }^{b}\left( \theta \right) = {\theta }^{{r}_{b}}{\beta }^{{\mu }_{b}} \) with\n\n\[ {\mu }_{b} = - \mathop{\sum }\limits_{{0 \leq a < d}}\left\lfloor \frac{{r}_{b}{r}_{d - 1 - a}}{\ell }\right\rfloor {\tau }^{a}. \]
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Proof. This follows from a direct computation. Note that \( {r}_{a} \) is periodic of period \( d \) for \( a \geq 0 \), hence it is reasonable to extend it by periodicity to all integral \( a \) . By the above corollary, we have \( \tau \left( \theta \right) = {\theta }^{g}{\beta }^{\mu } \), with\n\n\[ \mu = - \frac{{g}^{d} - 1}{\ell } + \left( {\tau - g}\right) \frac{\lambda - {\lambda }_{0}}{\ell } = - \mathop{\sum }\limits_{{0 \leq a < d}}\frac{g{r}_{d - 1 - a} - {r}_{d - a}}{\ell }{\tau }^{a}. \]\n\nBy induction, we immediately obtain\n\n\[ {\tau }^{b}\left( \theta \right) = {\theta }^{{g}^{b}}{\beta }^{\left( {{g}^{b} - {\tau }^{b}}\right) /\left( {g - \tau }\right) \mu } = {\theta }^{{r}_{b}}{\beta }^{{\mu }_{b}} \]\n\nwith\n\n\[ {\mu }_{b} = \frac{{g}^{b} - {r}_{b}}{\ell }\lambda + \frac{{g}^{b} - {\tau }^{b}}{g - \tau }\mu . \]\n\nSince\n\n\[ {\tau }^{k}\mu = - \mathop{\sum }\limits_{a}\frac{g{r}_{d - 1 - a + k} - {r}_{d - a + k}}{\ell }{\tau }^{a}, \]\n\nthe series \( \mathop{\sum }\limits_{{0 \leq k \leq b - 1}}{g}^{b - 1 - k}{\tau }^{k}\mu \) telescopes and we obtain\n\n\[ \frac{{g}^{b} - {\tau }^{b}}{g - \tau }\mu = \mathop{\sum }\limits_{{0 \leq k \leq b - 1}}{g}^{b - 1 - k}{\tau }^{k}\mu = - \mathop{\sum }\limits_{a}\frac{{g}^{b}{r}_{d - 1 - a} - {r}_{d - 1 - a + b}}{\ell }{\tau }^{a}. \]\n\nIt follows that\n\n\[ {\mu }_{b} = - \mathop{\sum }\limits_{a}\frac{{r}_{b}{r}_{d - 1 - a} - {r}_{d - 1 - a + b}}{\ell }{\tau }^{a} = - \mathop{\sum }\limits_{a}\left\lfloor \frac{{r}_{b}{r}_{d - 1 - a}}{\ell }\right\rfloor {\tau }^{a}, \]\n\nas claimed.
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Yes
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Proposition 5.3.9. Keep the notation of Theorem 5.3.5. For \( 2 \leq k \leq \ell \), set\n\n\[ t\left( {{b}_{1},\ldots ,{b}_{k - 1}}\right) = \frac{1}{\ell }\mathop{\sum }\limits_{{0 \leq a < d}}{\tau }^{a}\left( {{r}_{d - 1 - a} - \ell + \mathop{\sum }\limits_{{1 \leq i \leq k - 1}}{r}_{d - 1 - a + {b}_{\imath }}}\right) \]\n\nand define \( {\gamma }_{k} \) by the formula\n\n\[ {\gamma }_{k} = \mathop{\sum }\limits_{\substack{{{b}_{1} \leq \cdots \leq {b}_{k - 1}} \\ {{r}_{{b}_{1}} + \cdots + {r}_{{b}_{k - 1}} + 1 \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) } }}\frac{\left( {k - 1}\right) !}{\prod {m}_{j}!}{\beta }^{t\left( {{b}_{1},\ldots ,{b}_{k - 1}}\right) }, \]\n\nwhere the \( {m}_{j} \) denote the multiplicities of the \( {b}_{i} \). Set \( e = {\mathcal{N}}_{{K}_{z}/K}\left( \beta \right) \) and \( {S}_{k} = e\ell {\operatorname{Tr}}_{{K}_{z}/K}\left( {\gamma }_{k}\right) \). If we define \( {\mathfrak{S}}_{k} \) by the usual Newton recursion\n\n\[ k{\mathfrak{S}}_{k} = \mathop{\sum }\limits_{{1 \leq i \leq k}}{\left( -1\right) }^{i + 1}{S}_{i}{\mathfrak{S}}_{k - i} \]\n\na defining polynomial \( P\left( X\right) \) for the \( L/K \) is given by\n\n\[ P\left( X\right) = \mathop{\sum }\limits_{{0 \leq k \leq \ell }}{\left( -1\right) }^{k}{\mathfrak{S}}_{k}{X}^{\ell - k}. \]
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Proof. By Theorem 5.3.5 and Lemma 5.3.8, the kth power sum of the roots of the polynomial \( P\left( X\right) \) is given by\n\n\[ {S}_{k} = \mathop{\sum }\limits_{{0 \leq j < \ell }}{\left( \mathop{\sum }\limits_{{0 \leq b < d}}{\zeta }_{\ell }^{j{g}^{b}}{\theta }^{{r}_{b}}{\beta }^{{\mu }_{b}}\right) }^{k} \]\n\n\[ = \mathop{\sum }\limits_{{0 \leq j < \ell }}\left( {\mathop{\sum }\limits_{{{b}_{1},\ldots ,{b}_{k}}}{\zeta }_{\ell }^{j\left( {{g}_{1}^{b} + \cdots {g}^{{b}_{k}}}\right) }{\theta }^{{r}_{{b}_{1}} + \cdots + {r}_{{b}_{k}}}{\beta }^{{\mu }_{{b}_{1}} + \cdots + {\mu }_{{b}_{k}}}}\right) \]\n\n\[ = \ell \mathop{\sum }\limits_{{{r}_{{b}_{1}} + \cdots + {r}_{{b}_{k}} \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) }}{\beta }^{e\left( {{b}_{1},\ldots ,{b}_{k}}\right) }, \]\n\nwhere\n\n\[ e\left( {{b}_{1},\ldots ,{b}_{k}}\right) = \mathop{\sum }\limits_{{0 \leq a < d}}{\tau }^{a}\left( {\mathop{\sum }\limits_{{1 \leq i \leq k}}\frac{{r}_{{b}_{i}}{r}_{d - 1 - a}}{\ell } - \mathop{\sum }\limits_{{1 \leq i \leq k}}\left\lfloor \frac{{r}_{{b}_{\imath }}{r}_{d - 1 - a}}{\ell }\right\rfloor }\right) \]\n\n\[ = \frac{1}{\ell }\mathop{\sum }\limits_{{0 \leq a < d}}{\tau }^{a}\left( {\mathop{\sum }\limits_{{1 \leq i \leq k}}{r}_{d - 1 - a + {b}_{i}}}\right) . \]\n\nIt follows in particular from this formula that\n\n\[ {\tau e}\left( {{b}_{1},\ldots ,{b}_{k}}\right) = e\left( {{b}_{1} + 1,\ldots ,{b}_{k} + 1}\right) , \]\n\nhence that\n\n\[ {S}_{k} = \ell {\operatorname{Tr}}_{{K}_{z}/K}\left( {\mathop{\sum }\limits_{{{r}_{{b}_{1}} + \cdots + {r}_{{b}_{k - 1}} + {r}_{0} \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) }}{\beta }^{e\left( {{b}_{1},\ldots ,{b}_{k - 1},0}\right) }}\right) , \]\n\nso that\n\n\[ {S}_{k} = \ell {\operatorname{Tr}}_{{K}_{z}/K}\left( {\mathop{\sum }\limits_{{{r}_{{b}_{1}} + \cdots + {r}_{{b}_{k - 1}} + 1 \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) }}{\beta }^{{t}^{\prime }\left( {{b}_{1},\ldots ,{b}_{k - 1}}\right) }}\right) \]\n\nwith\n\n\[ {t}^{\prime }\left( {{b}_{1},\ldots ,{b}_{k - 1}}\right) = \frac{1}{\ell }\mathop{\sum }\limits_{{0 \leq a < d}}{\tau }^{a}\left( {{r}_{d - 1 - a} + \mathop{\sum }\limits_{{1 \leq i \leq k - 1}}{r}_{d - 1 - a + {b}_{i}}}\right) . \]\n\nTo finish the proof, we simply notice that each coefficient of \( \tau \) is strictly positive, hence we can factor out the norm \( {N}_{{K}_{z}/K}\left( \beta \right) = {\beta }^{\mathop{\sum }\limits_{{0 \leq a < d}}{\tau }^{a}} \), and furthermore the summands are symmetrical in the \( {b}_{i} \), so it is enough to sum for \( {b}_{1} \leq \cdots \leq {b}_{k - 1} \), except that we must compensate by the multinomial coefficient that counts the number of \( \left( {k - 1}\right) \)
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Yes
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Corollary 5.3.14. Let\n\n\[ \mathfrak{p} = \beta {\mathfrak{q}}^{\ell }\mathop{\prod }\limits_{{1 \leq i \leq {r}_{c}}}{\mathfrak{b}}_{i}^{{x}_{i}} \]\n\nbe the decomposition of some ideal \( \mathfrak{p} \) on the \( {\mathfrak{b}}_{i} \), and let \( X \) be the column vector of the \( {x}_{i} \). Then\n\n\[ {\mathfrak{p}}^{\lambda } = {\beta }^{\lambda }{\mathfrak{q}}_{1}^{\ell }\mathop{\prod }\limits_{{1 \leq i \leq \delta }}{\mathfrak{c}}_{i}^{{y}_{i}} \]\n\nwhere, if \( Y \) is the column vector of the \( {y}_{i} \) we have \( Y = d{g}^{-1}R{Q}^{t}X \) .
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Proof. In matrix terms, we have \( \mathfrak{p} \equiv \beta + \mathfrak{B}X\left( {\;\operatorname{mod}\;\ell }\right) \). Thus, by Proposition 5.3.13 we have\n\n\[ {\mathfrak{p}}^{\lambda } \equiv {\beta }^{\lambda } + {\mathfrak{B}}^{\lambda }X \equiv {\beta }^{\lambda } + d{g}^{-1}\mathfrak{C}\left( {R{Q}^{t}X}\right) \left( {\;\operatorname{mod}\;\ell }\right) ,\]\n\nproving the corollary.
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Yes
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Theorem 5.3.15. Keep the notation of the preceding section. In particular, \( K \) is a number field and \( \ell \) is a prime number such that \( {\zeta }_{\ell } \notin K \). Let \( L/K \) be a cyclic extension of degree \( \ell \) corresponding to the congruence subgroup \( \left( {\mathfrak{m}, C}\right) \), let \( {K}_{z} = K\left( {\zeta }_{\ell }\right) \) and \( {L}_{z} = L\left( {\zeta }_{\ell }\right) \), and let \( \mathfrak{f} = \mathfrak{f}\left( {{L}_{z}/{K}_{z}}\right) \) be the conductor of the extension \( {L}_{z}/{K}_{z} \) (or, equivalently, of the congruence subgroup \( \left( {\mathfrak{m}{\mathbb{Z}}_{{K}_{z}},{\mathcal{N}}_{{K}_{z}/K}^{-1}\left( C\right) }\right) \) of \( \left. {K}_{z}\right) \). Let \( {W}_{1} = {e}_{1}{V}_{\ell }\left( {K}_{z}\right) /{K}_{z}^{*\ell } \) be the \( {g}^{1} \) -eigenspace of the \( \ell \) -Selmer group of \( {K}_{z} \) for the action of \( \tau \), let \( {d}_{v} \) be its dimension, and let \( {\left( \overline{{w}_{i}}\right) }_{1 \leq i \leq {d}_{v}} \) be an \( {\mathbb{F}}_{\ell } \) -basis of \( {W}_{1} \) computed as explained in Section 5.3.4. For each prime ideal \( \mathfrak{p} \in S/\langle \tau \rangle \), write \[ \mathfrak{p} = {\beta }_{\mathfrak{p}}{\mathfrak{q}}_{\mathfrak{p}}^{\ell }\mathop{\prod }\limits_{{1 \leq i \leq {r}_{c}}}{\mathfrak{b}}_{i}^{{p}_{i,\mathfrak{p}}}. \] Then \( \mathfrak{f} \) satisfies the following conditions: (1) \( {S}_{\mathrm{f},\ell ,3} = \varnothing \) ; (2) if \( \mathfrak{p} \in {S}_{\mathfrak{f},\ell ,2}/\langle \tau \rangle \), then \( {v}_{\mathfrak{p}}\left( \mathfrak{f}\right) ≢ 1\left( {\;\operatorname{mod}\;\ell }\right) \) ; (3) if \( \mathfrak{p} \in {S}_{\mathfrak{f}}/\langle \tau \rangle \), then \( {v}_{\mathfrak{p}}\left( \mathfrak{f}\right) = 1 \). In addition, up to Kummer-equivalence, we can take \( \alpha = {\beta }^{\lambda } \) with \( \beta \) of the following form: \[ \beta = \mathop{\prod }\limits_{{1 \leq i \leq {d}_{v}}}{w}_{i}^{{n}_{i}}\mathop{\prod }\limits_{{\mathfrak{p} \in S/\langle \tau \rangle }}{\beta }_{\mathfrak{p}}^{{x}_{\mathfrak{p}}}, \] with the following additional conditions. (1) For all \( \mathfrak{p} \in S/\langle \tau \rangle \), we have \( 1 \leq {x}_{\mathfrak{p}} \leq \ell - 1 \), and for all \( i \), we have \( 0 \leq {n}_{i} \leq \ell - 1 \) (2) For \( \mathfrak{p} \in {S}_{\mathfrak{f},\ell ,2}/\langle \tau \rangle \), the largest \( k \) such that the congruence \[ {x}^{\ell } \equiv \alpha \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k + {v}_{\mathfrak{p}}\left( \alpha \right) }}\right) \] has a solution must be equal to \( z\left( {\mathfrak{p},\ell }\right) - {v}_{\mathfrak{p}}\left( \mathfrak{f}\right) \) . (3) If \( S \) is not empty, we may fix any one (but only one) of the \( {x}_{\mathfrak{p}} \) equal to 1. (4) For all \( \mathfrak{p} \in {S}_{\ell }/\langle \tau \rangle \), the congruence \[ {x}^{\ell } \equiv \alpha \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{z\left( {\mathfrak{p},\ell }\right) - 1 + {v}_{\mathfrak{p}}\left( \alpha \right) }}\right) \] has a solution. (5) For all \( \mathfrak{p} \in S/\langle \tau \rangle \), let \( {X}_{\mathfrak{p}} \) be the column vector of the \( {p}_{i,\mathfrak{p}} \) for \( 1 \leq i \leq {r}_{c} \) . We have \[ {Q}^{t}\mathop{\sum }\limits_{{\mathfrak{p} \in S/\langle \tau \rangle }}{x}_{\mathfrak{p}}{X}_{\mathfrak{p}} \equiv 0\left( {\;\operatorname{mod}\;\ell }\right) \] where \( Q \) is as in Proposition 5.3.13. Conversely, if all the above conditions are satisfied, if \( \alpha \neq 1 \), and if the norm group \( {T}_{\mathrm{m}}\left( {L/K}\right) \) is equal to \( C \), then \( L = K\left( {{\operatorname{Tr}}_{{L}_{z}/L}\left( \sqrt[L]{\alpha }\right) }\right) \) is a cyclic extension of degree \( \ell \) corresponding to the congruence subgroup \( \left( {\mathfrak{m}, C}\right) \) .
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Proof. It follows from Theorem 5.3.5 and Proposition 5.3.6 that, up to Kummer-equivalence, we ca
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No
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Lemma 5.3.16. Let \( \mathfrak{p} \in S = {S}_{\mathfrak{f}} \cup {S}_{\mathfrak{f},\ell ,1} \), and assume that there exists a cyclic extension \( L/K \) such that the corresponding extension \( {L}_{z}/{K}_{z} \) is of degree \( \ell \) and conductor \( \mathfrak{f} \) . Then the ideals \( {\tau }^{j}\left( \mathfrak{p}\right) \) for \( 0 \leq j < d \) are distinct; in other words, the prime ideal of \( K \) below \( \mathfrak{p} \) is totally split in \( {K}_{z} \) .
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Proof. Let \( j \) be some index such that \( {\tau }^{j}\left( \mathfrak{p}\right) = \mathfrak{p} \) . Applying the recursion for \( {x}_{\mathfrak{p}} \), we obtain\n\n\[ \n{x}_{\mathfrak{p}} \equiv {g}^{j}{x}_{{\tau }^{j}\left( \mathfrak{p}\right) } = {g}^{j}{x}_{\mathfrak{p}}\left( {\;\operatorname{mod}\;\ell }\right) \n\]\n\nand since \( \mathfrak{p} \in S \), we have \( {x}_{\mathfrak{p}} ≢ 0\left( {\;\operatorname{mod}\;\ell }\right) \), so \( {g}^{j} \equiv 1\left( {\;\operatorname{mod}\;\ell }\right) \) and \( j \) is a multiple of \( d \), as claimed.
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Yes
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Proposition 5.4.1. Let \( n > 1 \) be an arbitrary integer (not necessarily a prime power), let \( K \) be a number field such that \( {\zeta }_{n} \in K \), let \( N = K\left( \theta \right) \) with \( {\theta }^{n} = \alpha \in {\mathbb{Z}}_{K} \) of degree \( n \), and let \( \mathfrak{p} \) be a prime ideal of \( K \) not dividing \( \alpha \) and \( n \) (in particular, unramified in \( N/K \) ). Finally, let \( g \) be a generator of the multiplicative group \( {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \). (1) There exists an integer \( q \geq 1 \) such that \( \mathcal{N}\left( \mathfrak{p}\right) = {qn} + 1 \) . (2) In the group \( {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \), there exist integers \( y \) and \( z \) with \( \left( {z, n}\right) = 1 \) such that \( \overline{{\zeta }_{n}} = {g}^{qz} \) and \( \bar{\alpha } = {g}^{y} \) . (3) We have \[ {\sigma }_{\mathfrak{p}}\left( \theta \right) = {\zeta }_{n}^{y{z}^{-1}}\theta \] where \( {z}^{-1} \) denotes an inverse of \( z \) modulo \( n \) .
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Proof. Since \( {\sigma }_{\mathfrak{p}}\left( \theta \right) \) is a conjugate of \( \theta \), we must have \( {\sigma }_{\mathfrak{p}}\left( \theta \right) = {\zeta }_{n}^{{s}_{\mathfrak{p}}}\theta \) for some integer \( {s}_{\mathfrak{p}} \) . Let \( \mathcal{N}\left( \mathfrak{p}\right) = {qn} + t \) with \( 0 \leq t < n \) be the Euclidean division of \( \mathcal{N}\left( \mathfrak{p}\right) \) by \( n \) . We obtain \[ {\sigma }_{\mathfrak{p}}\left( \theta \right) \equiv {\theta }^{\mathcal{N}\left( \mathfrak{p}\right) } = {\left( {\theta }^{n}\right) }^{q}{\theta }^{t} = {\alpha }^{q}{\theta }^{t}\left( {{\;\operatorname{mod}\;\mathfrak{p}}{\mathbb{Z}}_{N}}\right) , \] hence \[ {\zeta }_{n}^{{s}_{\mathfrak{p}}}\theta \equiv {\alpha }^{q}{\theta }^{t}\left( {{\;\operatorname{mod}\;\mathfrak{p}}{\mathbb{Z}}_{N}}\right) \] Now \[ \operatorname{disc}\left( {1,\theta ,\ldots ,{\theta }^{n - 1}}\right) = \operatorname{disc}\left( {{X}^{n} - \alpha }\right) = {\left( -1\right) }^{\left( {n - 1}\right) \left( {n - 2}\right) /2}{n}^{n}{\alpha }^{n - 1}, \] and since we have assumed that \( \mathfrak{p} \nmid \alpha \) and \( \mathfrak{p} \nmid n \), we have disc \( \left( {1,\theta ,\ldots ,{\theta }^{n - 1}}\right) ≢ \) \( 0\left( {{\;\operatorname{mod}\;\mathfrak{p}}{\mathbb{Z}}_{N}}\right) \), so the classes of \( 1,\theta ,\ldots ,{\theta }^{n - 1} \) are \( {\mathbb{Z}}_{K}/\mathfrak{p} \) -linearly independent in \( {\mathbb{Z}}_{N}/\mathfrak{p}{\mathbb{Z}}_{N} \) . It follows that \( t = 1 \) and that \( {\zeta }_{n}^{{s}_{\mathfrak{p}}} \equiv {\alpha }^{q}\left( {{\;\operatorname{mod}\;\mathfrak{p}}{\mathbb{Z}}_{N}}\right) \) . Since \( {\zeta }_{n} \) and \( \alpha \) belong to \( {\mathbb{Z}}_{K} \) and \( \mathfrak{p}{\mathbb{Z}}_{N} \cap {\mathbb{Z}}_{K} = \mathfrak{p} \), we therefore have the congruence \[ {\zeta }_{n}^{{s}_{\mathfrak{p}}} \equiv {\alpha }^{q}\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) \] in \( {\mathbb{Z}}_{K} \) . Since \( t = 1 \), the group \( {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \) is of order \( {qn} \) . If \( g \) is a generator, we have \( \bar{\alpha } = {g}^{y} \) for some \( y \) since \( \mathfrak{p} \nmid \alpha \) . On the other hand, \( \overline{{\zeta }_{n}} \) is of order dividing \( n \) , hence we have \( \overline{{\zeta }_{n}} = {g}^{qz} \) for some integer \( z \) . I claim that \( \overline{{\zeta }_{n}} \) is of order exactly \( n \) . Indeed, the discriminant of the set of elements \( {\zeta }_{n}^{a} \) for \( \left( {a, n}\right) = 1 \) is equal to the discriminant of the cyclotomic polynomial \( {\Phi }_{n}\left( X\right) \), which is only divisible by primes dividing \( n \) (see Exercise 9) hence not divisible by \( \mathfrak{p} \) by assumption, proving my claim. Thus we have \( \left( {z, n}\right) = 1 \), and the congruence for \( {\zeta }_{n}^{{s}_{p}} \) that we have obtained gives \[ {g}^{{qz}
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Yes
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Lemma 5.4.3. Let \( L = K\left( \theta \right) \) with \( {\theta }^{n} = \alpha \in K \) and \( n = {\ell }^{r} \) be a Kummer extension as above. If \( \mathfrak{p} \) is a prime ideal of \( K \) that satisfies \( {\ell }^{r} \nmid {v}_{\mathfrak{p}}\left( \alpha \right) \), then \( \mathfrak{p} \) is ramified in \( L/K \) . In other words, if \( \mathfrak{p} \) is unramified, then \( {\ell }^{r} \mid {v}_{\mathfrak{p}}\left( \alpha \right) \) .
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Proof. Multiplying \( \alpha \) by a suitable \( {\ell }^{r} \) th power, we may assume that \( \alpha \in \) \( {\mathbb{Z}}_{K} \) . Indeed, this does not change the field \( L \) and does not change the condition \( {\ell }^{r} \nmid {v}_{\mathfrak{p}}\left( \alpha \right) . \n\nLet \( {v}_{\mathfrak{p}}\left( \alpha \right) = {\ell }^{a}w \) with \( \ell \nmid w \) . By assumption, we have \( a < r \), and we can find nonnegative integers \( x \) and \( y \) such that \( - x{\ell }^{r - a} + {yw} = 1 \) . If \( \pi \in {\mathfrak{p}}^{-1} \smallsetminus {\mathfrak{p}}^{-2} \) , then \( \beta = {\pi }^{x{\ell }^{r}}{\alpha }^{y} \) satisfies \( {v}_{\mathfrak{p}}\left( \beta \right) = - x{\ell }^{r} + {yw}{\ell }^{a} = {\ell }^{a} \), it is such that \( \beta \in {\mathbb{Z}}_{K} \) , and by Corollary 10.2.7, its \( n \) th root defines the same field as \( \theta \) since \( \ell \nmid y \) . Thus, we may assume that \( \alpha \in {\mathbb{Z}}_{K} \) and that \( w = 1 \), so that \( {v}_{\mathfrak{p}}\left( \alpha \right) = {\ell }^{a} \) for \( a < r \) .\n\nThe ideal \( \mathfrak{P} = \mathfrak{p}{\mathbb{Z}}_{L} + \theta {\mathbb{Z}}_{L} \) is not necessarily a prime ideal, but since \( \alpha \in {\mathbb{Z}}_{K} \) it is an integral ideal of \( {\mathbb{Z}}_{L} \) . Using Proposition 2.3.15 we know that\n\n\[ \n{\mathfrak{P}}^{{\ell }^{r}} = {\mathfrak{p}}^{{\ell }^{r}}{\mathbb{Z}}_{L} + \alpha {\mathbb{Z}}_{L} = \left( {{\mathfrak{p}}^{{\ell }^{r}} + \alpha {\mathbb{Z}}_{K}}\right) {\mathbb{Z}}_{L}.\n\]\n\nSince \( \mathfrak{p} \) is a prime ideal of \( K,{v}_{\mathfrak{q}}\left( {{\mathfrak{p}}^{{\ell }^{r}} + \alpha {\mathbb{Z}}_{K}}\right) = 0 \) if \( \mathfrak{q} \neq \mathfrak{p} \) while\n\n\[ \n{v}_{\mathfrak{p}}\left( {{\mathfrak{p}}^{{\ell }^{r}} + \alpha {\mathbb{Z}}_{K}}\right) = \min \left( {{\ell }^{r},{\ell }^{a}}\right) = {\ell }^{a},\n\]\n\nhence \( {\mathfrak{p}}^{{\ell }^{r}} + \alpha {\mathbb{Z}}_{K} = {\mathfrak{p}}^{{\ell }^{a}} \) so that \( {\mathfrak{P}}^{{\ell }^{r}} = {\left( \mathfrak{p}{\mathbb{Z}}_{L}\right) }^{{\ell }^{a}} \) . It follows that \( {\mathfrak{P}}^{{\ell }^{r - a}} = \mathfrak{p}{\mathbb{Z}}_{L} \) , and since \( a < r \) this shows that \( \mathfrak{p} \) is ramified in \( L/K \) with ramification exponent equal to \( {\ell }^{b} \) for some \( b \) such that \( r - a \leq b \leq r \), proving the lemma.
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Yes
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Proposition 5.4.4. Let \( {Cl}\left( K\right) = {\bigoplus }_{i}\left( {\mathbb{Z}/{d}_{i}\mathbb{Z}}\right) \overline{{\mathfrak{a}}_{i}} \) be the SNF of the ordinary class group of \( K \) . Let \( S \) be the set of prime ideals of \( K \) dividing \( \mathfrak{m} \) and the \( {\mathfrak{a}}_{i} \) . We may choose \( \alpha \in {\mathbb{Z}}_{K} \) such that \( L = K\left( \theta \right) \) with \( {\theta }^{n} = \alpha \) with \( \alpha \) an \( S \) -unit; in other words, \( \alpha \) divisible only by prime ideals of \( S \) (see Definition 7.4.1).
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Proof. Since \( L/K \) is a cyclic Kummer extension of degree \( n \), we know that there exists \( \alpha \in K \) such that \( L = K\left( \theta \right) \) with \( {\theta }^{n} = \alpha \) . We do not assume for the moment that \( \alpha \in {\mathbb{Z}}_{K} \) . We are going to modify \( \alpha \) so that it satisfies the required properties. We prove the result by induction on the number \( k \) of prime ideals occurring in the prime decomposition of \( \alpha \) and not belonging to \( S \) . If \( k = 0 \), we are done. Otherwise let \( k \geq 1 \), assume the proposition proved up to \( k - 1 \), let \( \alpha \) have exactly \( k \) prime ideals not in \( S \) in its prime ideal decomposition, and let \( \mathfrak{p} \) be such an ideal. Then \( \mathfrak{p} \nmid \mathfrak{m} \) so \( \mathfrak{p} \) is unramified, hence by the above lemma we know that \( n \mid {v}_{\mathfrak{p}}\left( \alpha \right) \) . Let \( \mathfrak{a} = {\mathfrak{p}}^{{v}_{\mathfrak{p}}\left( \alpha \right) /n} \) . By definition of the \( {\mathfrak{a}}_{i} \) we can write \( \mathfrak{a} = \beta \mathop{\prod }\limits_{i}{\mathfrak{a}}_{i}^{{x}_{i}} \) for some \( \beta \in K \) . Thus,\n\n\[ \alpha {\beta }^{-n}{\mathbb{Z}}_{K} = \alpha {\mathfrak{p}}^{-{v}_{\mathfrak{p}}\left( \alpha \right) }\mathop{\prod }\limits_{i}{\mathfrak{a}}_{i}^{n{x}_{i}} \]\n\nso \( \gamma = \alpha {\beta }^{-n} \) also defines the extension \( L/K \) and has exactly \( k - 1 \) prime ideals not belonging to \( S \) in its prime ideal decomposition, proving our induction hypothesis.\n\nWe have thus shown that \( \alpha \in K \) exists with the desired properties. Finally, let us prove that we can choose such an \( \alpha \) belonging to \( {\mathbb{Z}}_{K} \) . Indeed, we can write in a unique way \( \alpha {\mathbb{Z}}_{K} = \mathfrak{a}/\mathfrak{b} \) with \( \mathfrak{a} \) and \( \mathfrak{b} \) coprime integral ideals. Since \( \alpha \) is an \( S \) -unit, \( \mathfrak{a} \) and \( \mathfrak{b} \) are divisible only by prime ideals of \( S \) . If \( h\left( K\right) = \left| {{Cl}\left( K\right) }\right| \) is the class number of \( K \), then \( {\mathfrak{b}}^{h\left( K\right) } = \gamma {\mathbb{Z}}_{K} \) is a principal ideal. It is clear that \( \alpha {\gamma }^{n} \) also defines \( L/K \), is still an \( S \) -unit, and is in \( {\mathbb{Z}}_{K} \) , proving the proposition.
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Yes
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Proposition 5.4.5. Let \( \mathfrak{a} \) be an integral ideal of \( K \) coprime to the prime ideals of \( S \), and let \( {\sigma }_{\mathfrak{a}} = {\operatorname{Art}}_{N/K}\left( \mathfrak{a}\right) \in \operatorname{Gal}\left( {N/K}\right) \) be the automorphism corresponding to a by the Artin map in \( \operatorname{Gal}\left( {N/K}\right) \). Then \( {\sigma }_{\mathfrak{a}} \) is determined by \( {\sigma }_{\mathfrak{a}}\left( {\theta }_{j}\right) = {\zeta }_{n}^{{s}_{\mathfrak{a}, j}}{\theta }_{j} \) with\n\n\[ \n{s}_{\mathfrak{a}, j} = \mathop{\sum }\limits_{\mathfrak{p}}{y}_{\mathfrak{p}, j}{v}_{\mathfrak{p}}\left( \mathfrak{a}\right) {z}_{\mathfrak{p}}^{-1}\n\]\n\nwhere if for each prime ideal \( \mathfrak{p} \) we denote by \( {g}_{\mathfrak{p}} \) a generator of \( {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \), we set in this last group\n\n\[ \n\overline{{\zeta }_{n}} = {g}_{\mathfrak{p}}^{{z}_{\mathfrak{p}}\left( {\mathcal{N}\left( \mathfrak{p}\right) - 1}\right) /n}\n\]\n\nand\n\n\[ \n\overline{{\theta }_{j}} = {g}_{\mathfrak{p}}^{{y}_{\mathfrak{p}, j}}\n\]
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Proof. By Proposition 3.5.6, the computation of \( {\sigma }_{\mathfrak{a}}\left( {\theta }_{j}\right) \) can be done in the subextension \( {N}_{j}/K \). Proposition 5.4.1 then tells us that \( {\sigma }_{a}\left( {\theta }_{j}\right) = {\zeta }_{n}^{{s}_{a, j}}{\theta }_{j} \) with\n\n\[ \n{s}_{\mathfrak{a}, j} = \mathop{\sum }\limits_{\mathfrak{p}}{y}_{\mathfrak{p}, j}{v}_{\mathfrak{p}}\left( \mathfrak{a}\right) {z}_{\mathfrak{p}}^{-1},\n\]\n\nwhere \( {y}_{\mathfrak{p}, j} \) and \( {z}_{\mathfrak{p}} \) are as in the proposition.
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Yes
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Lemma 5.4.7. Let \( \bar{D} \) be the kernel of the canonical surjection \( s \) from \( {\mathrm{{Cl}}}_{{\mathrm{m}}_{N}}\left( K\right) \) to \( {\mathrm{{Cl}}}_{\mathrm{m}}\left( K\right) /\bar{C} \) . Then\n\n\[ C{l}_{{\mathfrak{m}}_{N}}{\left( K\right) }^{n} \subset \operatorname{Ker}\left( {\operatorname{Art}}_{N/K}\right) \subset \bar{D}, \]\n\nand if we set\n\n\[ \mathcal{H} = {\operatorname{Art}}_{N/K}\left( \bar{D}\right) = {\operatorname{Art}}_{N/K}\left( {\bar{D}/C{l}_{{\mathfrak{m}}_{N}}{\left( K\right) }^{n}}\right) \]\n\nthen \( L = {N}^{\mathcal{H}} \) .
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Proof. Note that by abuse of notation, we denote also by \( {\operatorname{Art}}_{N/K} \) the Artin map at the level of ideals or of ideal classes.\n\nThe Artin map \( {\operatorname{Art}}_{N/K} \) is a surjective homomorphism from the ideals of \( K \) coprime to \( {\mathfrak{m}}_{N} \) onto \( \operatorname{Gal}\left( {N/K}\right) \) with kernel containing \( {P}_{{\mathfrak{m}}_{N}} \), hence also from \( C{l}_{{\mathfrak{m}}_{N}}\left( K\right) \) onto \( \operatorname{Gal}\left( {N/K}\right) \), which is isomorphic to \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{u} \) . Let us first show the two inclusions. By class field theory, we have\n\n\[ C{l}_{{\mathfrak{m}}_{N}}\left( K\right) /\operatorname{Ker}\left( {\operatorname{Art}}_{N/K}\right) \simeq \operatorname{Gal}\left( {N/K}\right) \simeq {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{u}. \]\n\nIt follows that \( C{l}_{{\mathfrak{m}}_{N}}\left( K\right) /\operatorname{Ker}\left( {\operatorname{Art}}_{N/K}\right) \) is of exponent \( n \), hence \( C{l}_{{\mathfrak{m}}_{N}}{\left( K\right) }^{n} \subset \) \( \operatorname{Ker}\left( {\operatorname{Art}}_{N/K}\right) \) .\n\nFor the second inclusion, we note that since \( \mathfrak{m} \) is the conductor of \( L/K \) we have \( \mathfrak{m} \mid {\mathfrak{m}}_{N} \) . Thus \( \bar{D} \) is equal to the kernel of \( {\operatorname{Art}}_{L/K} \) viewed as a map from \( C{l}_{{\mathfrak{m}}_{N}}\left( K\right) \) to \( \operatorname{Gal}\left( {L/K}\right) \) . By Proposition 3.5.6 it follows that\n\n\[ \operatorname{Ker}\left( {\operatorname{Art}}_{N/K}\right) \subset \operatorname{Ker}\left( {\operatorname{Art}}_{L/K}\right) = \overline{D}, \]\n\nproving the second inclusion. Furthermore, this also proves that\n\n\[ \operatorname{Gal}\left( {N/L}\right) = {\operatorname{Art}}_{N/K}\left( {\overline{D}/\operatorname{Ker}\left( {\operatorname{Art}}_{N/K}\right) }\right) \]\n\nso \( L = {N}^{\mathcal{H}} \), where, with an evident abuse of notation, we can write indifferently\n\n\[ \mathcal{H} = {\operatorname{Art}}_{N/K}\left( {\overline{D}/\operatorname{Ker}\left( {\operatorname{Art}}_{N/K}\right) }\right) = {\operatorname{Art}}_{N/K}\left( \overline{D}\right) = {\operatorname{Art}}_{N/K}\left( {\overline{D}/C{l}_{{\mathfrak{m}}_{N}}{\left( K\right) }^{n}}\right) \text{,} \]\n\nfinishing the proof of the lemma.
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Yes
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There exists a subgroup \( {G}_{n} \) of \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) such that the extension \( {K}_{z}/K \) is Abelian with Galois group \( {G}_{z} \) given by \( {G}_{z} = \left\{ {{\tau }_{a}/a \in }\right. \) \( \left. {G}_{n}\right\} \), where \( {\tau }_{a} \) is the \( K \) -automorphism of \( {K}_{z} \) sending \( {\zeta }_{n} \) to \( {\zeta }_{n}^{a} \) . In particular, \( \left\lbrack {{K}_{z} : K}\right\rbrack \) divides \( \phi \left( n\right) \) . Conversely, if \( a \notin {G}_{n} \), there does not exist a \( K \) - automorphism \( {\tau }_{a} \) of \( {K}_{z} \) such that \( {\tau }_{a}\left( {\zeta }_{n}\right) = {\zeta }_{n}^{a} \) .
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Proof. By Proposition 5.3.1, we deduce as for Proposition 5.3.2 that \( {K}_{z}/K \) is Abelian with Galois group isomorphic to a subgroup of \( \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\right) \) , hence to a subgroup of \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \), where \( a \in {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) corresponds to \( {\tau }_{a} \in \) \( \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\right) \) by \( {\tau }_{a}\left( {\zeta }_{n}\right) = {\zeta }_{n}^{a} \) . For the converse, we can either note that all the \( K \) -automorphisms of \( {K}_{z} \) are accounted for by the \( {\tau }_{a} \) for \( a \in {G}_{n} \) or, more positively, notice that \( {G}_{n} \) is the set of \( a \in {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) such that \( {\zeta }_{n}^{a} \) is a root of \( f\left( X\right) = 0 \), where \( f\left( X\right) \) is the minimal polynomial of \( {\zeta }_{n} \) in \( K\left\lbrack X\right\rbrack \) (see also Exercise 11).
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Yes
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Proposition 5.5.2. Let \( K \) be a number field, \( {K}_{z} = K\left( {\zeta }_{n}\right) \), let\n\n\[ \n{G}_{z} = \operatorname{Gal}\left( {{K}_{z}/K}\right) = \left\{ {{\tau }_{a}/a \in {G}_{n}}\right\} \n\] \n\nfor some subgroup \( {G}_{n} \) of \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) as in Proposition 5.5.1, let \( L/K \) be a cyclic extension of degree \( n \), and let \( {L}_{z} = L{K}_{z} = L\left( {\zeta }_{n}\right) \) . Then \( {L}_{z}/{K}_{z} \) is a cyclic extension of degree \( m \) dividing \( n \), and if we write \( {L}_{z} = {K}_{z}\left( \theta \right) \) with \( \theta = \sqrt[m]{\alpha } \) for some \( \alpha \in {K}_{z} \), the extension \( {L}_{z}/K \) is Abelian if and only if for each \( a \in {G}_{n} \) there exists \( {\gamma }_{a} \in {K}_{z} \) such that\n\n\[ \n{\tau }_{a}\left( \alpha \right) = {\gamma }_{a}^{m}{\alpha }^{a} \n\] \n\nIf this condition is satisfied, we can choose an extension of \( {\tau }_{a} \) to \( {L}_{z} \) (again denoted by \( {\tau }_{a} \) ) such that \( {\tau }_{a}\left( \theta \right) = {\gamma }_{a}{\theta }^{a} \), and we have\n\n\[ \n\operatorname{Gal}\left( {{L}_{z}/K}\right) = \left\{ {{\sigma }^{j}{\tau }_{a}/0 \leq j < m, a \in {G}_{n}}\right\} \n\] \n\nwhere \( \sigma \) is the \( {K}_{z} \) -automorphism of \( {L}_{z} \) such that \( \sigma \left( \theta \right) = {\zeta }_{m}\theta \) .
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Proof. We have already seen that \( {L}_{z}/{K}_{z} \) is a cyclic extension of degree \( m \) dividing \( n \) . Assume that the extension \( {L}_{z}/K \) is Abelian. Since the extension \( {L}_{z}/K \) is normal we must have \( {\tau }_{a}\left( \theta \right) \in {L}_{z} \) for any extension of \( {\tau }_{a} \) to \( {L}_{z} \), which we denote again by \( {\tau }_{a} \) by abuse of notation. Thus \( {K}_{z}\left( {{\tau }_{a}\left( \theta \right) }\right) \subset {L}_{z} = {K}_{z}\left( \theta \right) \) . Applying this to \( {\tau }_{a}^{-1} = {\tau }_{{a}^{-1}} \), we obtain \( {K}_{z}\left( {{\tau }_{a}^{-1}\left( \theta \right) }\right) \subset {K}_{z}\left( \theta \right) \), hence \( {K}_{z}\left( \theta \right) \subset \) \( {K}_{z}\left( {{\tau }_{a}\left( \theta \right) }\right) \), which finally shows that we have the equality \( {K}_{z}\left( {{\tau }_{a}\left( \theta \right) }\right) = {K}_{z}\left( \theta \right) \) . Since \( {\tau }_{a}{\left( \theta \right) }^{m} = {\tau }_{a}\left( {\theta }^{m}\right) = {\tau }_{a}\left( \alpha \right) \), it follows from the uniqueness theorem of Kummer theory (Corollary 10.2.7 in our case) that there exist \( {\gamma }_{a} \in {K}_{z} \) and \( b \) coprime to \( m \) such that \( {\tau }_{a}\left( \alpha \right) = {\gamma }_{a}^{m}{\alpha }^{b} \), hence we can choose \( {\tau }_{a}\left( \theta \right) = {\gamma }_{a}{\theta }^{b} \) (all the other extensions \( \tau \) of \( {\tau }_{a} \) to \( {L}_{z} \) are obtained from this one by setting \( \tau \left( \theta \right) = {\zeta }_{m}^{s}{\gamma }_{a}{\theta }^{b} \) for \( 0 \leq s < m) \) .\n\nWe now use the fact that \( {L}_{z}/K \) is not only normal but Abelian. Let \( \sigma \) be the generator of the cyclic group \( \operatorname{Gal}\left( {{L}_{z}/{K}_{z}}\right) \), which sends \( \theta \) to \( {\zeta }_{m}\theta \) and leaves \( {K}_{z} \) fixed (note that \( {\zeta }_{m} = {\zeta }_{n}^{n/m} \in {K}_{z} \) ). Then\n\n\[ \n\sigma \left( {{\tau }_{a}\left( \theta \right) }\right) = \sigma \left( {{\gamma }_{a}{\theta }^{b}}\right) = {\gamma }_{a}{\zeta }_{m}^{b}{\theta }^{b}, \n\] \n\nwhile\n\n\[ \n{\tau }_{a}\left( {\sigma \left( \theta \right) }\right) = {\tau }_{a}\left( {{\zeta }_{m}\theta }\right) = {\zeta }_{m}^{a}{\gamma }_{a}{\theta }^{b}. \n\] \n\nComparing, we obtain \( b \equiv a{\;\operatorname{mod}\;m} \) ; in other words, \( b = a \) since \( b \) is only defined modulo \( m \) anyway. This shows that \( {\tau }_{a}\left( \alpha \right) = {\gamma }_{a}^{m}{\alpha }^{a} \) as claimed.\n\nConversely, if this condition is satisfied, then for all \( a \in {G}_{n},{\tau }_{a}
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Yes
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Corollary 5.5.3. Keep the hypotheses and notation of the proposition, and assume that \( {L}_{z} = {K}_{z}\left( \theta \right) \) is an Abelian extension of \( K \), with \( {\theta }^{m} = \alpha \) . Let \( \mathfrak{p} \) be a prime ideal of \( K \) not dividing \( n \) or \( \alpha \), and denote as usual by \( {\sigma }_{\mathfrak{p}} \) the Frobenius automorphism associated to \( \mathfrak{p} \) in the extension \( {L}_{z}/K \) . Then\n\n\[ \n{\sigma }_{\mathfrak{p}}\left( \theta \right) = {\zeta }_{m}^{{u}_{\mathfrak{p}}}{\gamma }_{\mathcal{N}\left( \mathfrak{p}\right) }{\theta }^{\mathcal{N}\left( \mathfrak{p}\right) }\n\]\n\nwhere \( {u}_{\mathrm{p}} \) is the unique integer modulo \( m \) (which exists) such that\n\n\[ \n{\zeta }_{m}^{{u}_{\mathfrak{p}}}{\gamma }_{\mathcal{N}\left( \mathfrak{p}\right) } \equiv 1\left( {{\;\operatorname{mod}\;\mathfrak{p}}{\mathbb{Z}}_{{K}_{z}}}\right)\n\]
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Proof. Since the extension \( {K}_{z}/K \) is ramified only at prime ideals dividing \( n \) and \( {L}_{z}/{K}_{z} \) is ramified only at prime ideals dividing \( n \) and \( \alpha \), it follows that \( \mathfrak{p} \) is unramified in \( {L}_{z}/K \) ; hence \( {\sigma }_{\mathfrak{p}} \) is well-defined.\n\nWe have \( {\sigma }_{\mathfrak{p}}\left( {\zeta }_{m}\right) \equiv {\zeta }_{m}^{\mathcal{N}\left( \mathfrak{p}\right) }\left( {{\;\operatorname{mod}\;\mathfrak{p}}{\mathbb{Z}}_{{K}_{z}}}\right) \), and since \( \mathfrak{p} \nmid n \) we have in fact the equality \( {\sigma }_{\mathfrak{p}}\left( {\zeta }_{m}\right) = {\zeta }_{m}^{\mathcal{N}\left( \mathfrak{p}\right) } \) . It follows that \( {\left. {\sigma }_{\mathfrak{p}}\right| }_{{K}_{z}} = {\tau }_{\mathcal{N}\left( \mathfrak{p}\right) } \), hence by the proposition \( {\sigma }_{\mathfrak{p}}\left( \alpha \right) = {\gamma }_{\mathcal{N}\left( \mathfrak{p}\right) }^{m}{\alpha }^{\mathcal{N}\left( \mathfrak{p}\right) } \), so\n\n\[ \n{\sigma }_{\mathfrak{p}}\left( \theta \right) = {\zeta }_{m}^{u}{\gamma }_{\mathcal{N}\left( \mathfrak{p}\right) }{\theta }^{\mathcal{N}\left( \mathfrak{p}\right) }\n\]\n\nfor some \( u \) . We conclude by using the congruence \( {\sigma }_{\mathfrak{p}}\left( \theta \right) \equiv {\theta }^{N\left( \mathfrak{p}\right) }\left( {{\;\operatorname{mod}\;\mathfrak{p}}{\mathbb{Z}}_{{L}_{z}}}\right) \) and the fact that \( \mathfrak{p} \nmid \alpha \) .
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Yes
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Lemma 5.5.4. Assume that \( \eta \) is such that \( {L}_{z} = K\left( \eta \right) \), and let\n\n\[ \n{P}_{\eta }\left( X\right) = \mathop{\sum }\limits_{{i = 0}}^{d}{\left( -1\right) }^{i}{t}_{i}{X}^{d - i} \n\] \n\nbe the characteristic polynomial of \( \eta \) in \( L\left\lbrack X\right\rbrack \) with \( d = \left\lbrack {{L}_{z} : L}\right\rbrack \) . Then \( L = \) \( K\left( {t}_{i}\right) \) for at least one value of \( i \) (note that \( K\left( {t}_{i}\right) \subset L \) for all \( i \) ).
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Proof. Since \( L/K \) is cyclic of prime power degree \( {\ell }^{r} \), there exists a maximal nontrivial subextension \( {L}_{1}/K \) of degree \( {\ell }^{r - 1} \) . So assume the conclusion of the lemma is false. Then \( {t}_{i} \in {L}_{1} \) for all \( i \), so that \( \left\lbrack {{L}_{z} : {L}_{1}}\right\rbrack \leq d \), which is absurd since \( \left\lbrack {{L}_{z} : {L}_{1}}\right\rbrack = d\ell \) .
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Yes
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Proposition 6.1.2. Let \( K \) be a totally real number field distinct from \( \mathbb{Q} \) , and let \( L/K \) be a finite Abelian extension, where \( L \) is also a totally real field. Assume that \( N \) is a quadratic extension of \( L \) satisfying the following conditions.\n\n(1) \( N/K \) is Abelian.\n\n(2) There exists a unique real embedding \( \tau \) of \( K \) which is unramified in \( N/K \).\n\n(3) Any prime ideal of \( L \) that is above a prime ideal of \( K \) ramified in \( L/K \) is inert or ramified in \( N/L \).\n\nLet \( S \) be the set of infinite places of \( K \) together with the places ramified in \( N/K \), and let \( \varepsilon \in N \) be an \( S \) -unit given by Stark’s Conjecture 6.1.1 for the extension \( N/K \). Then \( \varepsilon \) is in fact a unit and\n\n\[ N = K\left( \varepsilon \right) = \mathbb{Q}\left( \varepsilon \right) \;\text{ and }\;L = K\left( {\varepsilon + {\varepsilon }^{-1}}\right) = \mathbb{Q}\left( {\varepsilon + {\varepsilon }^{-1}}\right) .
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See [Rob1] for the proof. Note that it is not known whether such an extension \( N \) always exists. However, this is not a problem for computational purposes since we can always reduce to cases where \( N \) is known to exist (see Section 6.2.1).
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No
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Lemma 6.1.3. Let \( \\chi \) be an odd character of \( G \) as above, and set \( A\\left( \\chi \\right) = \\) \( \\mathop{\\prod }\\limits_{{\\mathfrak{p} \\mid \\mathfrak{f},\\mathfrak{p} \\nmid \\mathfrak{f}\\left( \\chi \\right) }}\\left( {1 - \\chi \\left( \\mathfrak{p}\\right) }\\right) \) . Then
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\[ {L}_{S}^{\\prime }\\left( {0,\\chi }\\right) = A\\left( \\chi \\right) {L}^{\\prime }\\left( {0,\\chi }\\right) = \\frac{A\\left( \\chi \\right) W\\left( \\chi \\right) }{2}\\frac{\\Lambda \\left( {1,\\bar{\\chi }}\\right) }{{\\pi }^{\\left( {m - 1}\\right) /2}}.\]
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Yes
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Proposition 6.1.5. For \( x > 0 \) and \( s \in \mathbb{C} \), we have\n\n\[ \n{f}_{2}\left( {s, x}\right) = 2\sqrt{\pi }{\left( x/2\right) }^{s}{\int }_{2/x}^{\infty }{t}^{s - 1}{e}^{-t}{dt}.\n\]\n\nIn particular,\n\n\[ \n{f}_{2}\left( {1, x}\right) = x\sqrt{\pi }{e}^{-2/x}\n\]\n\nand\n\[ \n{f}_{2}\left( {0, x}\right) = 2\sqrt{\pi }{\int }_{2/x}^{\infty }\frac{{e}^{-t}}{t}{dt} = 2\sqrt{\pi }{E}_{1}\left( \frac{2}{x}\right) ,\n\]\n\nwhere \( {E}_{1}\left( x\right) \) is the exponential integral function.
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Proof. By the duplication formula for the gamma function we have\n\n\[ \n{\left( \frac{x}{2}\right) }^{-s}{f}_{2}\left( {s, x}\right) = 2\sqrt{\pi }\frac{1}{2i\pi }{\int }_{\delta - i\infty }^{\delta + i\infty }{\left( \frac{x}{2}\right) }^{z - s}\frac{\Gamma \left( z\right) }{z - s}{dz}.\n\]\n\nReplacing \( x \) by \( 2/x \) and differentiating with respect to \( x \), we obtain\n\n\[ \n\frac{d}{dx}\left( {{x}^{s}{f}_{2}\left( {s,\frac{2}{x}}\right) }\right) = - 2\sqrt{\pi }\frac{1}{2i\pi }{\int }_{\delta - i\infty }^{\delta + i\infty }{x}^{-\left( {z - s + 1}\right) }\Gamma \left( z\right) {dz}.\n\]\n\nOn the other hand, by the formula giving the inverse Mellin transform, from \( \Gamma \left( z\right) = {\int }_{0}^{\infty }{x}^{z - 1}{e}^{-x}{dx} \) we deduce that\n\n\[ \n\frac{1}{2i\pi }{\int }_{\delta - i\infty }^{\delta + i\infty }{x}^{-z}\Gamma \left( z\right) {dz} = {e}^{-x},\n\]\n\nfrom which it follows that\n\n\[ \n\frac{d}{dx}\left( {{x}^{s}{f}_{2}\left( {s,\frac{2}{x}}\right) }\right) = - 2\sqrt{\pi }{x}^{s - 1}{e}^{-x},\n\]\n\nhence that\n\[ \n{x}^{s}{f}_{2}\left( {s,\frac{2}{x}}\right) = C + 2\sqrt{\pi }{\int }_{x}^{\infty }{t}^{s - 1}{e}^{-t}{dt}\n\]\n\nfor some constant \( C \) possibly depending on \( s \) but not on \( x \) . Coming back to the definition, we see that\n\n\[ \n{x}^{s}{f}_{2}\left( {s,\frac{2}{x}}\right) = \frac{{2}^{z}}{2i\pi }{\int }_{\delta - i\infty }^{\delta + i\infty }{x}^{-\left( {z - s}\right) }\frac{\Gamma \left( {z/2}\right) \Gamma \left( {\left( {z + 1}\right) /2}\right) }{z - s}{dz}\n\]\n\nand since \( \delta > \operatorname{Re}\left( s\right) \), an easy analytic argument using the fact that the gamma function is bounded in vertical strips shows that the integral tends to 0 when \( x \) tends to infinity. It follows that the constant \( C \) is equal to 0, giving the first formula of the proposition. The other formulas are immediate consequences.
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Yes
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