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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 \( \Omeg... | 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( ... | Yes |
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}}\... | 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 \... | Yes |
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 hav... | 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 \( 2{R}_{2, i} \) ... | Yes |
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) }... | 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 ... | Yes |
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... | 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 t... | Yes |
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}^{-... | 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}... | Yes |
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... | Proof. The proof is straightforward and left to the reader (see Exercises 24 and 25). | No |
Proposition 2.2.12. (1) If \( I \) and \( J \) are two integral ideals of \( {\mathbb{Z}}_{L} \), we have\n\n\[ \n{\mathcal{N}}_{L/K}\left( {IJ}\right) = {\mathcal{N}}_{L/K}\left( I\right) {\mathcal{N}}_{L/K}\left( J\right) .\n\] | 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). | No |
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... | 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) ... | Yes |
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\\] | 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 \... | Yes |
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 ... | 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... | Yes |
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 ... | 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}... | Yes |
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 \(... | Proof. The proof follows immediately from Proposition 2.3.1 and is left to the reader (Exercise 28). | No |
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}}\r... | 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}}_{... | Yes |
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 l... | \[ \mathfrak{P} = \left( {\left( {1,\mathfrak{p}}\right) ,\left( {\alpha ,\mathfrak{a}}\right) }\right) = \mathfrak{p}{\mathbb{Z}}_{L} + \alpha \mathfrak{a}{\mathbb{Z}}_{L} \] if and only if \( {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( {\alpha \mathfrak{a}}\right) }\right) = f \) or \( {v}_{\mathfrak{p}}\left... | Yes |
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 \( \beta \mathfrak{b} ⊄ \mathfrak{p}{\mathbb{Z}}_{L} \) . | 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{... | Yes |
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}}_... | 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 \... | Yes |
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 \( \mathfrak{D}\left( {L/k}\right) = \mathfrak{D}\left( {L/K}\right) \mathfrak{D}\left( {K/k}\right) \) . | 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 \ma... | Yes |
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 \( \left( {M,{\mathfrak{c}}_{i}}\right) \)... | 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 \... | Yes |
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) ... | 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 \( \gcd \left( {a, b}\right) \operatorname{lcm}\left( {a, b}\right) = {ab} \) ).... | Yes |
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}_{... | 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, he... | Yes |
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 ... | 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... | Yes |
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 \( {\mathbb{Z}}_{K}/\mathfrak{p} \) . | 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}}}\ove... | Yes |
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 \... | ## 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 generalizin... | No |
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... | 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... | Yes |
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} ... | 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{... | Yes |
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 gi... | 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}_{... | Yes |
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... | 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) }... | Yes |
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}... | 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\\] | Yes |
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 }{ \righta... | 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( \bet... | Yes |
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\le... | Proof. The proof is clear from the proposition. | No |
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 ... | 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) \g... | Yes |
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(... | 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 ... | Yes |
Proposition 3.3.4. (1) The relation \( \sim \) defined above between congruence subgroups is an equivalence relation. | 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 \( \left( {{\mathfrak{m}}_{2},{C}_{2}}\right) \sim \) \( \left( {{\mathfrak{m}}_{3},{C}_{3}}\right) \) ; ... | Yes |
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... | 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 \( {P}_{{\\mathfrak{m}}_{2}} \\subset {C}_{2} \),... | Yes |
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} \sub... | Proof. Applying Corollary 3.3.2, we have \[ \frac{{I}_{{\mathrm{m}}_{2}}}{{C}_{2}} \simeq \frac{{I}_{{\mathrm{m}}_{1}}{C}_{2}}{{C}_{2}} \simeq \frac{{I}_{{\mathrm{m}}_{1}}}{{I}_{{\mathrm{m}}_{1}} \cap {C}_{2}} \simeq \frac{{I}_{{\mathrm{m}}_{1}}/{C}_{1}}{\left( {{I}_{{\mathrm{m}}_{1}} \cap {C}_{2}}\right) /{C}_{1}}, \]... | Yes |
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}_{{\mathfr... | 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... | Yes |
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},{... | 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... | Yes |
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 \( \l... | 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. | Yes |
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 \( \mathfrak{f} \) . | 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}} ... | Yes |
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 par... | Proof. This is an immediate consequence of Proposition 3.3.6 and the above proposition. | Yes |
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... | 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 \( {... | Yes |
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 take... | 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 su... | Yes |
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{... | 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) \... | Yes |
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. | 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). | No |
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... | Proof. Once again the proof is left to the reader (Exercise 9). | No |
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 \( {... | 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... | Yes |
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 ... | 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 i... | No |
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}... | 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 \( g = n/f \) . | Yes |
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 \( {h}_{\mathfrak{m}/\mathfrak{p}, C} < {h}_{\mathfrak{m}, C} \). | 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. | No |
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|... | 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 unr... | Yes |
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 \rig... | 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}\ri... | Yes |
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 ... | 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}\ri... | Yes |
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. | 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 \( \\math... | No |
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 no... | 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 \( {b}_{i} \) . B... | Yes |
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 num... | 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) = \\l... | Yes |
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} ... | 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}... | Yes |
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}}... | 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 injec... | Yes |
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} =... | Proof. The proof of this proposition is easy and is left to the reader (Exercise 11). | No |
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 {... | 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) - ... | Yes |
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}_{\mathf... | 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 {\... | Yes |
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 a... | 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... | Yes |
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 \... | 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}... | Yes |
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 \( {\mathbb{Z}}_{K}/\mathfrak{a} \) . | 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}_{... | Yes |
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... | 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... | Yes |
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}\r... | 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... | Yes |
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}}... | 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 (otherwis... | No |
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\\ma... | 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) = {\\big... | Yes |
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 t... | 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 h... | Yes |
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... | 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{{\mathfr... | Yes |
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) \( {... | 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)... | Yes |
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 ... | 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 ... | Yes |
(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 \( 0 \leq j \leq {r}_{u} \) . | 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... | Yes |
Lemma 5.2.6. We have \( U\left( K\right) \cap {K}^{*\ell } = U{\left( K\right) }^{\ell } \) . | 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( {\;\operato... | No |
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 }{... | Proof. The proof is straightforward and is left to the reader (Exercise 3). | No |
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 \( ... | 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}}}\matho... | No |
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} ... | 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 \( ... | Yes |
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 \... | 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 \( {\mathcal{G}}_{i} = \operatorname{Gal}\left( {N/{L}_{i}}\right) \) s... | Yes |
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 aut... | 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... | Yes |
Corollary 5.3.4. With the above notation, the eigenspace \( {W}_{k} \) is equal to \( {e}_{k}W = \left\{ {{x}^{{e}_{k}}/x \in W}\right\} . \) | 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\nSi... | Yes |
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 primitiv... | 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... | Yes |
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( \fra... | 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 ... | Yes |
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 \( \tau \left( \theta \right) = {\theta }^{g}{\beta }^{\mu } \) . | 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 ... | Yes |
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}... | 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{... | Yes |
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 }}}\rig... | 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\[ =... | Yes |
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\[ {\mathfra... | 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}\rig... | Yes |
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 \(... | Proof. It follows from Theorem 5.3.5 and Proposition 5.3.6 that, up to Kummer-equivalence, we ca | No |
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... | 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 }\rig... | Yes |
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... | 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 \le... | Yes |
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 \( ... | 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( \alph... | Yes |
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... | 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 satis... | Yes |
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 \( \operato... | 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} = \ma... | Yes |
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}}_{... | 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}\lef... | Yes |
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 ... | 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 ... | Yes |
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... | 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 ... | Yes |
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 \( {\... | 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 hav... | Yes |
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... | 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\rbr... | Yes |
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) The... | 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). | No |
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 | \[ {L}_{S}^{\\prime }\\left( {0,\\chi }\\right) = A\\left( \\chi \\right) {L}^{\\prime }\\left( {0,\\chi }\\right) = \\frac{A\\left( \\chi \\right) W\\left( \\chi \\right) }{2}\\frac{\\Lambda \\left( {1,\\bar{\\chi }}\\right) }{{\\pi }^{\\left( {m - 1}\\right) /2}}.\] | Yes |
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)... | 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 \(... | Yes |
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