Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Proposition 6.1.6. Let \( \chi \) be a character of \( G \) and let \( n \geq 1 \) be an integer. Set \( \chi \left( \mathfrak{p}\right) = 0 \) if \( \mathfrak{p} \) divides the conductor of \( \chi \) . Then \( {a}_{1}\left( \chi \right) = 1 \), and for \( n > 1 \) , if \( n = {p}_{1}^{{m}_{1}}\ldots {p}_{k}^{{m}_{k}}... | Proof. The first assertion is a translation of the fact that the map \( n \mapsto \) \( {a}_{n}\left( \chi \right) \) is multiplicative. The formulas for \( n = {p}^{m} \) are clear when \( p \) is inert or ramified and are easily proved by induction when \( p \) is split. | No |
Proposition 6.2.5. Keep the above notation. If \( \gamma = a + {b\omega } \in {\mathbb{Z}}_{K} \) satisfies \( \left| {{\tau }_{2}\left( \gamma \right) - \beta }\right| \leq \varepsilon \) and \( \left| {{\tau }_{1}\left( \gamma \right) }\right| \leq B \), then \( q\left( {a, b,1}\right) \leq 3{B}^{2} \) . Conversely, ... | Proof. The first statement is clear. Conversely, assume that \( q\left( {x, y, z}\right) \leq \) \( 3{B}^{2} \) and that \( \left( {x, y, z}\right) \neq \left( {0,0,0}\right) \) . It is clear that \( \left| z\right| \leq 1 \), and Exercise 4 (which is an excellent exercise on the properties of continued fractions) show... | No |
Proposition 6.3.1. Let \( \mathfrak{a} \) and \( \mathfrak{c} \) be fractional ideals of \( K \), and let \( \operatorname{Art}\left( \mathfrak{c}\right) \) be the element of \( \operatorname{Gal}\left( {K\left( 1\right) /K}\right) \) corresponding to the ideal \( \mathfrak{c} \) by the Artin reciprocity map (since \( ... | Since we know that \( K\left( 1\right) = K\left( {j\left( {\mathbb{Z}}_{K}\right) }\right) \), it follows from this proposition that \[ \alpha = {\operatorname{Tr}}_{K\left( 1\right) /K{\left( 1\right) }^{C}}\left( {j\left( {\mathbb{Z}}_{K}\right) }\right) = \mathop{\sum }\limits_{{\overline{\mathfrak{c}} \in \overline... | No |
Proposition 6.3.3. Let \( {\Gamma }^{0}\left( {pq}\right) \) be the group of matrices \( \gamma = \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \in \) \( {\mathrm{{PSL}}}_{2}\left( \mathbb{Z}\right) \) such that \( {pq} \mid b \) .\n\n(1) For any \( \gamma \in {\Gamma }^{0}\left( {pq}\right) \), we have \(... | Proof. Left to the reader (Exercise 10). | No |
Proposition 6.3.4. Let \( \mathfrak{p} \) and \( \mathfrak{q} \) be two primitive ideals of respective norms \( p \) and \( q \) such that the product \( \mathfrak{{pq}} \) is also primitive. If\n\n\[ \mathfrak{p}\mathfrak{q} = {pq}\mathbb{Z} \oplus \frac{-w + \sqrt{D}}{2}\mathbb{Z} \]\n\nthen \( \mathfrak{p} = p\mathb... | Proof. Since \( \mathfrak{{pq}} \subset \mathfrak{p} \), we have \( w \equiv u\left( {{\;\operatorname{mod}\;2}p}\right) \) and similarly for \( \mathfrak{q} \), proving the proposition. | No |
Corollary 6.3.5. Let \( \mathfrak{a},\mathfrak{p},\mathfrak{q} \) be three primitive ideals such that apq is a primitive ideal. Let \( a = \mathcal{N}\left( \mathfrak{a}\right), p = \mathcal{N}\left( \mathfrak{p}\right), q = \mathcal{N}\left( \mathfrak{q}\right) \), and assume that \( e \) is a positive integer such th... | Proof. For (1), we write apq \( = \operatorname{apq}\mathbb{Z} \oplus \left( {\left( {-w + \sqrt{D}}\right) /2}\right) \mathbb{Z} \) . It follows from the proposition that \( {\omega }_{1} = \left( {-w + \sqrt{D}}\right) /2 \) and \( {\omega }_{2} = a \) is a suitable basis. For (2), we note that \( \left( {{\omega }_{... | Yes |
Theorem 6.3.7 (Schertz). Let \( {\left( {\mathfrak{a}}_{i}\right) }_{1 \leq i \leq h\left( K\right) } \) be a system of representatives of the ideal classes of \( K = \mathbb{Q}\left( \sqrt{D}\right) \), chosen to be primitive. Let \( \mathfrak{p} \) and \( \mathfrak{q} \) be ideals of \( K \) of norm \( p \) and \( q ... | We refer to [Sch1] for the proof. The statement (and the proof) given by Schertz is slightly incorrect because of the omission of condition (2) above on classes of order 2 . An example is \( K = \mathbb{Q}\left( \sqrt{-{30}}\right) \), which has class group isomorphic to \( {\left( \mathbb{Z}/2\mathbb{Z}\right) }^{2} \... | Yes |
Proposition 6.3.11. Let \( L \) be a complex lattice.\n\n(1) There exists a unique meromorphic function \( \zeta \left( {z, L}\right) \), called the Weierstrass \( \zeta \) -function, such that \( {\zeta }^{\prime }\left( {z, L}\right) = - \wp \left( {z, L}\right) \) and such that \( \zeta \left( {z, L}\right) \) is an... | Proof. Since the proofs are easy, we leave some details to the reader (Exercise 13).\n\nIt is clear that the series defining \( \zeta \left( {z, L}\right) \) converges uniformly on any compact subset not containing points of \( L \) (this is the case as soon as the general term goes to zero faster than \( 1/{\left| \om... | No |
Corollary 6.3.12. For any \( \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \in {\operatorname{PSL}}_{2}\left( \mathbb{Z}\right) \) and \( \tau \in \mathcal{H} \), we have\n\n\[ \n{E}_{2}\left( \frac{{a\tau } + b}{{c\tau } + d}\right) = {\left( c\tau + d\right) }^{2}{E}_{2}\left( \tau \right) + \frac{{12c}\... | Proof. Set \( {\omega }_{1}^{\prime } = a{\omega }_{1} + b{\omega }_{2},{\omega }_{2}^{\prime } = c{\omega }_{1} + d{\omega }_{2} \), and \( {\tau }^{\prime } = {\omega }_{1}^{\prime }/{\omega }_{2}^{\prime } \) . By assumption, \( \left( {{\omega }_{1}^{\prime },{\omega }_{2}^{\prime }}\right) \) is still an oriented ... | Yes |
Proposition 6.3.14. Let \( L \) be a complex lattice.\n\n(1) There exists a unique holomorphic function \( \sigma \left( {z, L}\right) \), called the Weierstrass \( \sigma \) -function, such that \( {\sigma }^{\prime }\left( {z, L}\right) /\sigma \left( {z, L}\right) = \zeta \left( {z, L}\right) \) and \( \mathop{\lim ... | Proof. Once again, we leave some details to the reader (Exercise 16). The general term of the given product expansion tends to 1 as \( {\left| \omega \right| }^{-3} \), hence the product converges uniformly on any compact subset of \( \mathbb{C} \) and so defines a holomorphic function. By definition, its logarithmic d... | No |
Corollary 6.3.16.\n\n\[ \eta \left( \tau \right) = {q}^{1/{24}}\left( {1 + \mathop{\sum }\limits_{{k \geq 1}}{\left( -1\right) }^{k}\left( {{q}^{k\left( {{3k} - 1}\right) /2} + {q}^{k\left( {{3k} + 1}\right) /2}}\right) }\right) . \] | Proof. This follows from the Jacobi identity by replacing \( \left( {u, q}\right) \) by \( \left( {q,{q}^{3}}\right) \) and rearranging terms. The details are left to the reader (Exercise 17). Note that this is the identity we used in Algorithm 6.3.2. | No |
Corollary 6.3.17.\n\n\[ \n{\eta }^{3}\left( \tau \right) = {q}^{1/8}\mathop{\sum }\limits_{{k \geq 0}}{\left( -1\right) }^{k}\left( {{2k} + 1}\right) {q}^{k\left( {k + 1}\right) /2}.\n\] | Proof. This follows from the Jacobi identity by making \( u \rightarrow 1 \) . | No |
Corollary 6.3.18. With the same notation as above, we have the formula\n\n\\[ \n\\sigma \\left( {z, L}\\right) = \\frac{{\\omega }_{2}}{2i\\pi }{e}^{{\\eta }_{2}{z}^{2}/\\left( {2{\\omega }_{2}}\\right) }\\frac{\\mathop{\\sum }\\limits_{{k \\geq 0}}{\\left( -1\\right) }^{k}\\left( {{u}^{k + 1/2} - {u}^{-\\left( {k + 1/... | ## Proof. Clear from Proposition 6.3.14 (4). | No |
Proposition 6.3.20. Let \( L \) be a complex lattice and \( a \notin L \) . Then\n\n\[ \wp \left( {z, L}\right) - \wp \left( {a, L}\right) = - \frac{\sigma \left( {z - a, L}\right) \sigma \left( {z + a, L}\right) }{\sigma {\left( a, L\right) }^{2}\sigma {\left( z, L\right) }^{2}}. \] | Proof. Using Proposition 6.3.14, it is easy to check that the ratio of the left- and right-hand sides is an elliptic function with no zero or poles, hence it is constant by Liouville’s theorem. Making \( z \rightarrow 0 \) and using the expansions of \( \wp \) and \( \sigma \) around 0 gives the result. | No |
Proposition 6.3.21. Let \( \omega = m{\omega }_{1} + n{\omega }_{2} \in L \) with \( m \) and \( n \) in \( \mathbb{Z} \), and let \( z = {x}_{1}{\omega }_{1} + {x}_{2}{\omega }_{2} \) with \( {x}_{1} \) and \( {x}_{2} \) in \( \mathbb{R} \), as above. Then\n\n\[ \n{\phi }^{ * }\left( {z + \omega, L}\right) = s\left( \... | Proof. The proof of the transformation formula follows immediately from the corresponding formula for the \( \sigma \) -function seen above. It follows in particular that \( \left| {{\phi }^{ * }\left( {z + \omega, L}\right) }\right| = \left| {{\phi }^{ * }\left( {z, L}\right) }\right| \), and since \( {\phi }^{ * } \)... | Yes |
Theorem 6.3.24. Let \( K \) be an imaginary quadratic field of discriminant \( D \) , let \( h \) be its class number, and let \( \mathfrak{f} \) be a conductor of \( K \) different from \( {\mathbb{Z}}_{K} \) (see Proposition 3.3.20 for all the possible conductors). Denote as usual by \( {\zeta }_{m} \) a primitive mt... | Proof. The special cases \( D = - 3,\mathfrak{f} = {\mathfrak{p}}_{3}^{3} \) and \( D = - 4,\mathfrak{f} = 4{\mathbb{Z}}_{K} \) are easily treated directly, so we exclude these cases. We first prove a lemma.\n\nLemma 6.3.25. For any integer \( m \), denote by \( \mathfrak{f}\left( m\right) \) the conductor of the Abeli... | No |
Lemma 6.3.25. For any integer \( m \), denote by \( \mathfrak{f}\left( m\right) \) the conductor of the Abelian extension \( K\left( {\zeta }_{m}\right) /K \). (1) We have \( \mathfrak{f}\left( m\right) \mid m{\mathbb{Z}}_{K} \). (2) If \( K = \mathbb{Q}\left( \sqrt{D}\right) \) with \( D \equiv 8\left( {\;\operatornam... | Proof. (1) is nothing else but Proposition 3.5.5. For (2), we can, for example, use Hecke's Theorem 10.2.9 from which we borrow the notation. Indeed, since \( \left\lbrack {K\left( i\right) : K}\right\rbrack = 2 \), we have \( \mathfrak{f}\left( 4\right) = \mathfrak{d}\left( {K\left( i\right) /K}\right) \). Denote by \... | Yes |
Lemma 6.3.26. Let \( f \) be an integer such that \( f \nmid {12} \) .\n\n(1) If \( D \) is any integer, there exists \( t \) such that \( \left( {{t}^{2} - D,{2f}}\right) = 1 \) and \( f \nmid {2t} \) . | Proof. Let \( {E}_{f} \) be the set of \( t \) such that \( 0 \leq t < {2f} \) and \( \left( {{t}^{2} - D,{2f}}\right) = 1 \) . This condition means that \( t ≢ D\left( {\;\operatorname{mod}\;2}\right) \) and that for each prime \( p > 2 \) dividing \( f, t \) must not be congruent to the square roots of \( D \) modulo... | Yes |
Theorem 7.1.5. Let \( L/K \) be a relative extension of degree \( n \), and let \( h\left( K\right) \) be the class number of \( K \). (1) If \( \left( {n, h\left( K\right) }\right) = 1 \) (for example, if \( K \) has class number 1), the natural map \( {\psi }_{N, i} \) from \( C{l}_{N}\left( {L/K}\right) \) to \( C{l... | Proof. It is clear that statements (3) and (4) imply statement (2), which implies statement (1). Furthermore, the formulas \( {\mathcal{N}}_{L/K}\left( {{i}_{L/K}\left( \overline{\mathfrak{a}}\right) }\right) = {\overline{\mathfrak{a}}}^{n} \) and \( {\overline{\mathfrak{a}}}^{h\left( K\right) } = \overline{1} \) in \(... | Yes |
Proposition 7.2.3. (1) We have the following exact sequence\n\n\[ 1 \rightarrow U\left( K\right) \overset{{i}_{L/K}}{ \rightarrow }U\left( L\right) \overset{j}{ \rightarrow }{U}_{i}\left( {L/K}\right) \overset{\pi }{ \rightarrow }C{l}_{i}\left( K\right) \rightarrow 1 \] | Proof. (1). If \( \alpha \in U\left( L\right) \), then \( \alpha {\mathbb{Z}}_{L} = {\mathbb{Z}}_{L} \), so \( \alpha {\mathbb{Z}}_{K}{\mathbb{Z}}_{L} = {\mathbb{Z}}_{L} \) . Hence \( j\left( \alpha \right) \in \) \( {U}_{i}\left( {L/K}\right) \) is well-defined and is a group homomorphism. If \( \overline{\left( \math... | Yes |
Corollary 7.2.4. Assume that \( \\left( {n, h\\left( K\\right) }\\right) = 1 \) . Then the group \( {U}_{i}\\left( {L/K}\\right) \) is isomorphic to the quotient group \( U\\left( L\\right) /{i}_{L/K}\\left( {U\\left( K\\right) }\\right) \) . | Proof. By Theorem 7.1.5 (1), when \( \\left( {n, h\\left( K\\right) }\\right) = 1 \) we have \( C{l}_{i}\\left( K\\right) = \\{ 1\\} \) , and so the corollary follows immediately from Proposition 7.2.3. | Yes |
Corollary 7.2.5. There exists a six-term exact sequence \[ 1 \rightarrow U\left( K\right) \overset{{i}_{L/K}}{ \rightarrow }U\left( L\right) \overset{j}{ \rightarrow }{U}_{i}\left( {L/K}\right) \] \[ \rightarrow {Cl}\left( K\right) \overset{{i}_{L/K}}{ \rightarrow }{Cl}\left( L\right) \rightarrow C{l}_{i}\left( {L/K}\r... | Proof. This is a trivial consequence of exact sequences (1) and (4). | No |
Proposition 7.2.6. Let \( A \) be an \( m \times n \) integer matrix of rank \( n \), hence with \( m \geq n \). There exist two unimodular matrices \( U \) and \( V \) of size \( n \times n \) and \( m \times m \), respectively, such that \[ B = {VAU} = \left( \begin{matrix} {d}_{1} & 0 & \ldots & 0 \\ 0 & {d}_{2} & \... | Proof. As usual, denote by \( {A}^{t} \) the transpose of \( A \), which is an \( n \times m \) matrix of rank \( n \). By the theorem on the HNF ([Coh0, Theorem 2.4.4]), there exists a unimodular \( m \times m \) matrix \( {V}_{1} \) and an \( n \times n \) matrix \( H \) in HNF such that \( {A}^{t}{V}_{1} = \left( {0... | Yes |
Corollary 7.2.8. Keep the notation of the above proposition, denote as usual by \( \mu \left( L\right) \) the group of roots of unity of \( L \), and set \( w\left( {L/K}\right) = w\left( L\right) /w\left( K\right) \) . Then:\n\n(1)\n\n\[ \frac{U\left( L\right) }{\mu \left( L\right) \cdot {i}_{L/K}\left( {U\left( K\rig... | Proof. Statement (1) follows immediately from Proposition 7.2.7. Note that because of the presence of the integers \( {e}_{i} \), we cannot give such a clean formula for \( U\left( L\right) /{i}_{L/K}\left( {U\left( K\right) }\right) \) .\n\nBy (1), \( \operatorname{diag}\left( {d}_{i}\right) \) is the Smith normal for... | Yes |
Proposition 7.2.11. (1) We have the following exact sequences:\n\n\[ 1 \rightarrow {\\operatorname{Cl}}_{N}\\left( K\\right) \rightarrow {U}_{N}\\left( {L/K}\\right) \rightarrow U\\left( L\\right) \\overset{{\\mathcal{N}}_{L/K}}{ \\rightarrow }\\frac{U\\left( K\\right) }{\\mu \\left( K\\right) } \rightarrow {U}_{N}\\le... | Proof. Note that the map from \( C{l}_{N}\\left( K\\right) \) to \( {U}_{N}\\left( {L/K}\\right) \) is the map sending \( \\overline{\\mathfrak{a}} \) to \( \\left( {\\overline{\\mathfrak{a}},1}\\right) \), and the map from \( {U}_{N}\\left( {L/K}\\right) \) to \( U\\left( L\\right) \) is the map sending \( \\left( {\\... | Yes |
Proposition 7.4.4. We have a canonical isomorphism\n\n\[ \nC{l}_{S}\left( K\right) \simeq {Cl}\left( K\right) / < \overline{{\mathfrak{p}}_{i}}{ > }_{{\mathfrak{p}}_{i} \in S}, \n\] \n\nwhere \( < \overline{{\mathfrak{p}}_{i}} > \) denotes the subgroup of \( {Cl}\left( K\right) \) generated by ideal classes of the prim... | Proof. Let \( \bar{I} \) be an ideal class in \( C{l}_{S}\left( K\right) \), and define \( f\left( \bar{I}\right) = \overline{I \cap {\mathbb{Z}}_{K}} \) in \( {Cl}\left( K\right) / < \overline{{\mathfrak{p}}_{i}} > \) . The map \( f \) is well-defined and is a group homomorphism. Assume that \( f\left( \bar{I}\right) ... | Yes |
Corollary 7.4.5. There exists \( {S}_{1} \) such that for any \( S \supset {S}_{1} \), the ring \( {\mathbb{Z}}_{K, S} \) is a principal ideal domain. | Proof. Let \( \left( \overline{{\mathfrak{a}}_{i}}\right) \) be generators of \( {Cl}\left( K\right) \), and let \( {S}_{1} \) be a set containing all Archimedean places and all prime ideals \( \mathfrak{p} \) such that \( {v}_{\mathfrak{p}}\left( {\mathfrak{a}}_{i}\right) \neq 0 \) for some \( i \) . This set is finit... | Yes |
Proposition 7.4.7. Let \( {Cl}\left( K\right) = \left( {B,{D}_{B}}\right) \) with \( B = \left( \overline{{\mathfrak{b}}_{i}}\right) \) be the \( {SNF} \) of \( {Cl}\left( K\right) \), where \( {\mathfrak{b}}_{i} \) are ideals and \( {D}_{B} = \operatorname{diag}\left( {b}_{i}\right) \). Let\n\n\[ U\left( K\right) = \l... | Proof. Let \( {S}_{0} \) be the row vector of the prime ideals belonging to \( S \) and let \( {B}^{\prime } = \left\lbrack {{\mathfrak{b}}_{1},\ldots ,{\mathfrak{b}}_{m}}\right\rbrack \) be the row vector of the ideals \( {\mathfrak{b}}_{i} \). We use again matrix notation as we did in Chapter 4. I first claim that\n\... | No |
Proposition 7.5.1. The exponent of the quotient group\n\n\[ \n\\left( {{\\mathcal{N}}_{L/K}\\left( {L}^{ * }\\right) \\cap {U}_{S}\\left( K\\right) }\\right) /{\\mathcal{N}}_{L/K}\\left( {{U}_{S}\\left( L\\right) }\\right) \n\]\n\ndivides \( \\left\\lbrack {L : K}\\right\\rbrack \) . | Proof. Indeed, if \( a \\in {U}_{S}\\left( K\\right) \\subset {K}^{ * } \), then\n\n\[ \n{a}^{\\left\\lbrack L : K\\right\\rbrack } = {\\mathcal{N}}_{L/K}\\left( a\\right) \\in {\\mathcal{N}}_{L/K}\\left( {{U}_{S}\\left( L\\right) }\\right) \n\] | Yes |
Proposition 7.5.3. Assume that \( {S}_{0} \) is suitable for the extension \( L/K \), let \( a \in {K}^{ * } \), and call \( {S}_{a} \) the set of prime ideals \( \mathfrak{p} \) of \( K \) such that \( {v}_{\mathfrak{p}}\left( a\right) \neq 0 \) . Then the equation \( {\mathcal{N}}_{L/K}\left( x\right) = a \) is solub... | Proof. Indeed, if \( S = {S}_{0} \cup {S}_{a} \), then \( a \in {U}_{S}\left( K\right) \) by definition, and since \( {S}_{0} \) is suitable and \( S \supset {S}_{0} \), if the equation \( {\mathcal{N}}_{L/K}\left( x\right) = a \) is soluble, we have \( a \in {\mathcal{N}}_{L/K}\left( {{U}_{S}\left( L\right) }\right) .... | Yes |
Theorem 7.5.4. Let \( L/K \) be a Galois extension, and let \( {S}_{0} \) be a set of prime ideals of \( K \) such that \( C{l}_{i}\left( {L/K}\right) \) can be generated by the classes of ideals divisible only by prime ideals of \( L \) above the ideals of \( {S}_{0} \) . Then \( {S}_{0} \) is suitable for the extensi... | Proof of Theorem 7.5.4. The inclusions \( {\mathcal{N}}_{L/K}\left( {{U}_{S}\left( L\right) }\right) \subset {\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap \) \( {U}_{S}\left( K\right) \) and \( {\mathcal{N}}_{L/K}\left( {\mathbb{Z}}_{L, S}\right) \subset {\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap {\mathbb{Z}}_{K... | Yes |
Lemma 7.5.5. Let \( L/K \) be a Galois extension of number fields, let \( S \) be a finite set of prime ideals of \( K \), and let \( I \) and \( J \) be \( S \)-integral ideals of \( L \) such that \( {\mathcal{N}}_{L/K}\left( I\right) = \mathfrak{a}{\mathcal{N}}_{L/K}\left( J\right) \) for some \( S \)-integral ideal... | Proof. We prove the lemma by induction on \( k \), the case \( k = 0 \) being trivial. Assume first \( k = 1 \), and let \( \mathfrak{P} \) be a prime ideal dividing \( J \). Since \( {\mathcal{N}}_{L/K}\left( \mathfrak{P}\right) \) divides \( {\mathcal{N}}_{L/K}\left( J\right) \), it also divides \( {\mathcal{N}}_{L/K... | Yes |
Proposition 7.5.6. Let \( L/K \) be a Galois extension, and let \( r \) be an integer such that \( \operatorname{Gal}\left( {L/K}\right) \) can be generated by \( r \) elements. In addition, for any finite set \( S \) of prime ideals of \( K \), denote by \( C{l}_{i, S}\left( {L/K}\right) \) the quotient of \( C{l}_{i}... | I refer to [Sim1] for the proof. | No |
Corollary 7.5.7. Let \( L/K \) be a Galois extension, and let \( {S}_{0} \) be a set of prime ideals such that \( \left| {C{l}_{i,{S}_{0}}\left( {L/K}\right) }\right| \) is coprime to \( \left\lbrack {L : K}\right\rbrack \) . Then \( {S}_{0} \) is suitable for the extension \( L/K \) . | Proof. This follows immediately from the above proposition and Proposition 7.5.1. This corollary is a slight strengthening of part of Theorem 7.5.4, which asserts only that \( {S}_{0} \) is suitable if \( \left| {C{l}_{i,{S}_{0}}\left( {L/K}\right) }\right| = 1 \) . | Yes |
Proposition 7.5.8. Let \( S \) be a finite set of primes of \( K \) . The exponent of \( \left( {{\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap {U}_{S}\left( K\right) }\right) /{\mathcal{N}}_{L/K}\left( {{U}_{S}\left( L\right) }\right) \) divides the GCD \( \left( {n,\left| H\right| \cdot \left| {C{l}_{i, S}\left( {N/... | Proof. By Proposition 7.5.1, we already know that this exponent divides \( n \) . Set \( h = \left| {C{l}_{i, S}\left( {N/K}\right) }\right| \), and let \( a \in {\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap {U}_{S}\left( K\right) \), so that we may write \( a = {\mathcal{N}}_{L/K}\left( x\right) \) for some \( x \in... | Yes |
Theorem 7.5.10. Keep the above notation, and let \( {S}_{0} \) be a finite set of primes of \( K \) containing all the prime ideals of \( K \) ramified in \( L/K \) . Assume that \( \left| {C{l}_{i,{S}_{0}}\left( {N/K}\right) }\right| \) is coprime to \( n \) and that for all cyclic subgroups \( C \) of \( G = \operato... | We refer to [Sim1] for the (quite technical) proof. Please note the condition that \( {S}_{0} \) must contain all the ramified prime ideals, which did not occur in the previous results. It is not difficult to give examples showing that this condition (or a similar one) is necessary (see [Sim1]). Note also that, at leas... | No |
Proposition 8.1.1. Let \( \gamma = \left( \begin{array}{ll} A & B \\ C & D \end{array}\right) \) . Then\n\n\[ \operatorname{disc}\left( {F \circ \gamma }\right) = {\left( AD - BC\right) }^{n\left( {n - 1}\right) }\operatorname{disc}\left( F\right) . \] | Proof. Let \( \left( {{\alpha }_{i} : {\beta }_{i}}\right) \) be the roots of \( F \) chosen as above so that\n\n\[ F\left( {X, Y}\right) = \mathop{\prod }\limits_{{1 \leq i \leq n}}\left( {{\beta }_{i}x - {\alpha }_{i}y}\right) . \]\n\nThen\n\n\[ F \circ \gamma \left( {X, Y}\right) = \mathop{\prod }\limits_{{1 \leq i ... | Yes |
Proposition 8.1.3. Let \( {f}_{1} \) and \( {f}_{2} \) be two covariants on \( {\Phi }_{n} \) of degree \( {d}_{1} \) , \( {d}_{2} \), weight \( {w}_{1},{w}_{2} \), and with values in \( {\Phi }_{{m}_{1}} \) and \( {\Phi }_{{m}_{2}} \), respectively. For any nonnegative integer \( h \) and \( F \in {\Phi }_{n} \), set\... | Proof. To simplify notation, write \( {\partial }_{X} \) for \( \partial /\partial X \) and \( {\partial }_{Y} \) for \( \partial /\partial Y \) . The only operators that occur are \( {\partial }_{X} \) and \( {\partial }_{Y} \), which commute, and the multiplication operator, which does not commute with \( {\partial }... | Yes |
Proposition 8.1.5. Let \( F \) be an integral form and \( \gamma \in {\mathrm{{GL}}}_{2}\left( \mathbb{Z}\right) \) . Then \( F \circ \gamma \) is irreducible if and only if \( F \) is irreducible, and \( F \circ \gamma \) is primitive if and only \( F \) is primitive. | Proof. This immediately follows from the fact that the action of \( {\mathrm{{GL}}}_{2}\left( \mathbb{Z}\right) \) is reversible, that is, \( F = \left( {F \circ \gamma }\right) \circ {\gamma }^{-1} \) . | Yes |
Proposition 8.2.1. Let \( \mathcal{B} = \left( {1,\alpha ,\beta }\right) \) be an integral basis of a cubic number field \( K \) as above. For \( x \) and \( y \) elements of \( \mathbb{Q} \), set\n\n\[ \n{F}_{\mathcal{B}}\left( {x, y}\right) = \frac{\mathop{\prod }\limits_{{1 \leq i < j \leq 3}}\left( {\left( {{\beta ... | Proof. (1) follows directly from the definitions, after noting that \( d\left( K\right) \) is the square of the determinant of the matrix\n\n\[ \n\left( \begin{matrix} 1 & {\alpha }_{1} & {\beta }_{1} \\ 1 & {\alpha }_{2} & {\beta }_{2} \\ 1 & {\alpha }_{3} & {\beta }_{3} \end{matrix}\right) \;. \n\]\n\nIf we take one ... | Yes |
Proposition 8.2.2. We have \( {\phi }_{\Phi \mathcal{C}} \circ {\phi }_{\mathcal{C}\Phi } = 1 \) ; hence \( {\phi }_{\mathcal{C}\Phi } \) is injective and \( {\phi }_{\Phi \mathcal{C}} \) is surjective. | Proof. Let \( K \) be a cubic field and let \( \left( {1,\alpha ,\beta }\right) \) be an integral basis. We have (for example) \( {K}_{{F}_{K}} = \mathbb{Q}\left( {\left( {{\alpha }_{2} - {\alpha }_{3}}\right) /\left( {{\beta }_{2} - {\beta }_{3}}\right) }\right) \subset {K}^{g} \) . If \( K \) is a cyclic cubic field,... | No |
(1) \( p \mid \operatorname{disc}\left( F\right) \) if and only if \( F \) has at least a double root in \( \overline{{\mathbb{F}}_{p}} \), and if this is the case, all the roots of \( F \) are in fact in \( {\mathbb{F}}_{p} \) itself. | Proof. (1). Assume that \( p \mid \operatorname{disc}\left( F\right) \) . We know that any nonzero polynomial in one variable over \( {\mathbb{F}}_{p} \) can be written as \( \mathop{\prod }\limits_{{i > 1}}{A}_{i}^{i} \), where \( {A}_{i} \in {\mathbb{F}}_{p}\left\lbrack X\right\rbrack \) are pairwise coprime and squa... | Yes |
Corollary 8.3.4. Let \( F = \left( {a, b, c, d}\right) \) be a primitive form, and let \( {H}_{F} = \) \( \left( {P, Q, R}\right) \) be its Hessian. Then \( F \notin {U}_{2} \) if and only if \( \operatorname{disc}\left( F\right) \equiv 0\left( {\;\operatorname{mod}\;{16}}\right) \) or \( \operatorname{disc}\left( F\ri... | Proof. The proof is trivial and is left to the reader (Exercise 7). | No |
Corollary 8.3.6. Let \( F \) be a primitive cubic form and \( p \) be a prime. Then \( F \notin {U}_{p} \) if and only if \( F \) has at least a double root \( \left( {\gamma : \delta }\right) \) modulo \( p \), and \( F\left( {\gamma ,\delta }\right) \equiv 0\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) . | Proof. If \( F \notin {U}_{p} \), then in particular \( F \notin {V}_{p} \), hence \( p \mid f \), and so by Proposition 8.3.5, either \( F\left( {\gamma ,\delta }\right) \equiv 0\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) if \( F \) has a double root, or \( \left( {F, p}\right) = \left( {1}^{3}\right) \) , but th... | Yes |
Lemma 8.4.3. Assume that \( F \) is a primitive cubic form and \( p \) a prime such that \( \left( {F, p}\right) = \left( {1}^{3}\right) \) . Then \( F \in {U}_{p} \) if and only if there exists \( \left( {u, v}\right) \in {\mathbb{Z}}^{2} \) such that \( F\left( {u, v}\right) = {ep} \) with \( p \nmid e \) . | Let us prove the lemma. By lifting the condition \( \left( {F, p}\right) = \left( {1}^{3}\right) \) to \( \mathbb{Z} \), we can write\n\n\[ F\left( {x, y}\right) = \lambda {\left( \delta x - \gamma y\right) }^{3} + {pG}\left( {x, y}\right) ,\]\n\nwith \( \lambda ,\gamma ,\delta \) in \( \mathbb{Z} \) and \( G \) an int... | Yes |
Lemma 8.4.5. Let \( F \) be any form in \( \Phi \) (in other words, primitive and irreducible). Then there exists a number field \( K \) such that \( F \) is rationally equivalent to \( {F}_{K} \) . | Proof. In some algebraic closure of \( \mathbb{Q} \), write \( F = a\left( {x - {\lambda y}}\right) \left( {x - {\lambda }^{\prime }y}\right) (x - \) \( \left. {{\lambda }^{\prime \prime }y}\right) \) . Since \( F \) is irreducible, \( \lambda \) is a cubic irrationality, and we will take \( K = \) \( \mathbb{Q}\left( ... | Yes |
Proposition 8.4.8. Let \( K \) be a cubic number field, and as before write \( d\left( K\right) = {d}_{k}{f}^{2} \), where \( {d}_{k} \) is a fundamental discriminant. Let \( {F}_{K} \) be the cubic form associated to \( K \) by the Davenport-Heilbronn map, and let \( {H}_{K} = \) \( \left( {P, Q, R}\right) \) be its H... | Proof. By Proposition 8.3.3 (2), we have \( p \mid {f}_{H} \) if and only if \( \left( {F, p}\right) = \left( {1}^{3}\right) \) , and by Proposition 8.2.3 (3) this is true if and only if \( p \) is totally ramified, hence by Proposition 8.4.1 (1), if and only if \( p \mid f \) . Hence \( f \) and \( {f}_{H} \) have the... | Yes |
Lemma 8.5.2. Let \( H = \left( {P, Q, R}\right) \) and \( {H}^{\prime } = \left( {{P}^{\prime },{Q}^{\prime },{R}^{\prime }}\right) \) be two reduced, definite, integral, binary quadratic forms such that there exists \( M \in {\mathrm{{GL}}}_{2}\left( \mathbb{Z}\right) \) with \( {H}^{\prime } = H \circ M \) . Then, ei... | Proof. Since \( H \) and \( {H}^{\prime } \) are equivalent, they have the same discriminant and represent the same integers. To say that \( H \) is reduced implies that \( P \) is the minimum of \( H \) on \( {\mathbb{Z}}^{2} - \{ \left( {0,0}\right) \} \) and that \( R \) is the second minimum; hence \( P = {P}^{\pri... | Yes |
Proposition 8.5.5. Let \( F = \left( {a, b, c, d}\right) \) be a reduced form such that \( 0 < \) \( \operatorname{disc}\left( F\right) \leq X \) . We have the following inequalities.\n\n(1)\n\n\[ 1 \leq a \leq \frac{2}{3\sqrt{3}}{X}^{1/4}. \]\n\n(2) If \( a \leq {X}^{1/4}/3 \), we have\n\n\[ 0 \leq b \leq \frac{3a}{2}... | Proof. Let \( H = \left( {P, Q, R}\right) \) be the Hessian of \( F \), and set \( \Delta = \operatorname{disc}\left( F\right) \) so that \( {4PR} - {Q}^{2} = {3\Delta } \) . Since \( F \) is reduced, we have\n\n\[ {3X} \geq {3\Delta } \geq {4PR} - {P}^{2} \geq 3{P}^{2}. \]\n\nHowever, it is easily checked that\n\n\[ P... | Yes |
Proposition 8.5.7. Let \( K \) be a totally real cubic number field, \( {F}_{K} \) the unique reduced form associated to \( K \), and \( {H}_{K} \) its Hessian. Then\n\n(1) \( K \) is cyclic (that is, \( d\left( K\right) = {f}^{2} \) ) if and only if \( {H}_{K} = {f}_{H}\left( {1, \pm 1,1}\right) \).\n\n(2) \( d\left( ... | Proof. Using Proposition 8.4.8, proving this is just a matter of listing reduced quadratic forms. We leave it to the reader (see Exercise 8). | No |
Lemma 8.6.2. Let \( F = \left( {a, b, c, d}\right) \) be a complex cubic form. Then \( F \) is reduced if and only if\n\n\[ \n{d}^{2} - {a}^{2} + {ac} - {bd} > 0 \n\]\n\n\[ \n- {\left( a - b\right) }^{2} - {ac} < {ad} - {bc} < {\left( a + b\right) }^{2} + {ac} \n\]\n\n\[ \na > 0,\;b \geq 0,\;d \neq 0,\text{ and }\;d > ... | Proof. The condition \( B < A \) is equivalent to \( {a\theta } + b < a \) and hence to \( \theta < \left( {a - b}\right) /a \) since \( a > 0 \) . Since \( F \) has only one real root (and again \( a > 0 \) ), this is equivalent to \( F\left( {a - b, a}\right) > 0 \), which gives \( - {\left( a - d\right) }^{2} - {ac}... | Yes |
Proposition 8.6.3. (1) Two equivalent, irreducible, reduced, complex cubic forms are equal. | Proof. (1). Let \( {F}^{\prime } = F \circ M \), where \( F \) and \( {F}^{\prime } \) are reduced and \( M \in \) \( {\mathrm{{GL}}}_{2}\left( \mathbb{Z}\right) \) . Then by the formula proved above, there exists \( \lambda > 0 \) such that \( {H}_{{F}^{\prime }} = \lambda {H}_{F} \circ M \) . As before, we deduce fro... | Yes |
Proposition 8.6.4. Let \( F = \left( {a, b, c, d}\right) \) be a reduced form such that \( - X \leq \) \( \operatorname{disc}\left( F\right) < 0 \) . We have the following inequalities:\n\n(1)\n\n\[ 1 \leq a \leq \frac{2{X}^{1/4}}{{3}^{3/4}} \]\n\n(2)\n\n\[ 0 \leq b \leq \frac{3a}{2} + \sqrt{{\left( \frac{X}{3}\right) ... | Proof. Set \( \Delta = \left| {\operatorname{disc}\left( F\right) }\right| \) and \( {3D} = {4AC} - {B}^{2} \) . The inequalities \( \left| B\right| < \) \( A < C \) imply as usual \( {A}^{2} < D \) or, equivalently, \( {a}^{2} < D \) . In addition, by a computation made above, we have \( \Delta = {3D}\left( {A{\theta ... | Yes |
Lemma 9.2.1. Let \( L/K \) be an Abelian extension of number fields of conductor \( \mathfrak{m} \), let \( {n}_{L} = \left\lbrack {L : \mathbb{Q}}\right\rbrack \) and \( {n}_{K} = \left\lbrack {K : \mathbb{Q}}\right\rbrack \), and assume that the root discriminant satisfies \( {\left| d\left( L\right) \right| }^{1/{n}... | Proof. By Theorem 3.5.11, we have\n\n\[ {\left| d\left( L\right) \right| }^{1/{h}_{\mathfrak{m}, C}} = \left| {d\left( K\right) }\right| \frac{\mathcal{N}\left( \mathfrak{m}\right) }{\mathop{\prod }\limits_{{\mathfrak{p} \mid \mathfrak{m}}}\mathcal{N}{\left( \mathfrak{p}\right) }^{\mathop{\sum }\limits_{{1 \leq k \leq ... | Yes |
Lemma 9.2.2. As usual, denote by \( {V}_{2}\left( K\right) \) the group of 2-virtual units of \( K \) (see Definition 5.2.4), and let \( {I}_{s} \) be the group of squarefree integral ideals of \( K \) whose ideal class is a square. Then\n\n(1) The map \( \phi \), which sends the class modulo \( {V}_{2}\left( K\right) ... | Proof. (1). We first show that the map \( \phi \) is well-defined. Indeed, we can write in a unique way \( \alpha {\mathbb{Z}}_{K} = \mathfrak{a}{\mathfrak{q}}^{2} \) with \( \mathfrak{a} \) integral and squarefree, and the ideal class of \( \mathfrak{a} \) is equal to that of \( {\mathfrak{q}}^{-2} \), so is a square,... | Yes |
Theorem 9.2.6. Let \( K \) be a number field, \( L \) an extension of \( K \) of degree \( n \) , and let \( {L}_{2} \) be the Galois closure of \( L/K \) in some algebraic closure of \( K \) . Assume that \( \operatorname{Gal}\left( {{L}_{2}/K}\right) \) is isomorphic to the dihedral group \( {D}_{n} \), and that \( n... | See Section 10.1.5 for a proof of this theorem in the case \( n \) prime. | No |
Lemma 9.3.6. (1) If the \( {x}_{j} \) for \( 1 \leq j \leq n \) are nonnegative real numbers and \( k \) is a real number such that \( k \geq 1 \), then \[ \mathop{\sum }\limits_{{1 \leq j \leq n}}{x}_{j}^{k} \leq {\left( \mathop{\sum }\limits_{{1 \leq j \leq n}}{x}_{j}\right) }^{k} \] | Proof. For (1), we could say that for all \( p \geq 1 \), the \( {L}^{p} \) norm is less than or equal to the \( {L}^{1} \) norm. For a simpler proof, let us show the statement by induction on \( n \) . It is trivially true for \( n \leq 1 \) . Set \[ f\left( {x}_{n}\right) = {\left( \mathop{\sum }\limits_{{1 \leq j \l... | Yes |
Proposition 9.3.7. If \( {r}_{2} = 0 \), then we have the following inequalities:\n\n(1) \( \left| {a}_{n}\right| < {s}_{2}^{n/2}/{n}^{n/2} \) ;\n\n(2) \( {s}_{2} \geq n + 1 \) or, equivalently, \( {a}_{2} \leq \left( {{a}_{1}^{2} - 1 - n}\right) /2 \) ;\n\n(3) for \( k \) even, \( {s}_{k} > n{\left| {a}_{n}\right| }^{... | Proof. Since \( {T}_{2} = {s}_{2} \), the above remark and the arithmetic-geometric mean inequality used above gives the bound \( \left| {a}_{n}\right| < {s}_{2}^{n/2}/{n}^{n/2} \), the inequality being strict since the arithmetic-geometric mean inequality becomes an equality only if all components are equal, which is ... | Yes |
Proposition 9.3.8. If \( {r}_{2} = 0 \), then for \( 1 \leq k \leq n - 1 \) we have the inequality\n\n\[ \n{a}_{k - 1}{a}_{k + 1} \leq \frac{k\left( {n - k}\right) }{\left( {k + 1}\right) \left( {n - k + 1}\right) }{a}_{k}^{2}.\n\] | Proof. Write as usual \( P\left( X\right) = \mathop{\prod }\limits_{j}\left( {X - {x}_{j}}\right) \) with the \( {x}_{j} \) real by assumption, and let \( x \) be a real number different from the \( {x}_{j} \) . Then\n\n\[ \n\frac{{P}^{\prime }\left( x\right) }{P\left( x\right) } = \mathop{\sum }\limits_{j}\frac{1}{x -... | Yes |
Proposition 9.3.9. Let \( {g}_{1},\ldots ,{g}_{m} \) and \( f \) be \( {C}^{1} \) functions in \( {\mathbb{R}}^{n} \), let \( A \) be the subset of \( {\mathbb{R}}^{n} \) defined by the equations \( {g}_{k}\left( \mathbf{x}\right) = 0 \) for \( 1 \leq k \leq m \), and let \( \mathbf{x} \) be a local extremum of the fun... | Thus, we may apply this proposition to \( A = {G}_{3} \) and \( A = {G}_{4} \), but not to \( G \) itself since there is an inequality in the definition of \( G \), and this is the reason for which we have had to introduce the auxiliary sets \( {G}_{3} \) and \( {G}_{4} \) . Note also that, as in the one variable case,... | No |
Lemma 9.3.11. Let \( {R}_{m, k}\left( x\right) \) be the above function with \( 1 \leq m \leq n - 1 \) .\n\n(1) For \( x > 0 \), the function \( {R}_{m, k}\left( x\right) \) has a unique minimum at 1, it decreases for \( x < 1 \), and it increases for \( x > 1 \) . | The proof of the lemma is straightforward. We have\n\n\[ {R}_{m, k}^{\prime }\left( x\right) = \frac{{km}\left( {n - m}\right) }{2}\left( {{x}^{{km}/2 - 1} - {x}^{k\left( {m - n}\right) /2 - 1}}\right) ,\]\n\nso \( {R}_{m, k}^{\prime }\left( x\right) = 0 \) if and only if \( x = 1 \), and it is negative for \( x < 1 \)... | Yes |
Proposition 9.4.2. Assume that \( P\left( X\right) \) has signature \( \left( {0,2}\right) \), keep the notation of Section 9.3.4, and let \( \mathbf{x} = \left( {{x}_{1},{x}_{2},{x}_{3},{x}_{4}}\right) \) be a local extremum on \( {G}_{3} \) or \( {G}_{4} \) of the function\n\n\[ \n{s}_{3}\left( \mathbf{x}\right) = 2\... | Proof. With the notation of Section 9.3.4, we have \( {g}_{1}\left( \mathbf{x}\right) = 2\left( {{x}_{1} + {x}_{2}}\right) - {a}_{1} \) , \( {g}_{2}\left( \mathbf{x}\right) = 2\left( {{x}_{1}^{2} + {x}_{2}^{2} - {x}_{3}^{2} - {x}_{4}^{2}}\right) - {s}_{2},{g}_{3}\left( \mathbf{x}\right) = \left( {{x}_{1}^{2} + {x}_{2}^... | Yes |
(1) If \( \mathbf{x} \in {G}_{3} \smallsetminus {G}_{4} \), then either \( {x}_{1} = {x}_{2} \) or \( {x}_{4} = 0 \) . | Proof. With the notation of Section 9.3.4, we have \( {g}_{1}\left( \mathbf{x}\right) = {x}_{1} + {x}_{2} + 2{x}_{3} - {a}_{1},{g}_{2}\left( \mathbf{x}\right) = {x}_{1}^{2} + {x}_{2}^{2} + 2{x}_{3}^{2} - 2{x}_{4}^{2} - {s}_{2},{g}_{3}\left( \mathbf{x}\right) = {x}_{1}{x}_{2}\left( {{x}_{3}^{2} + {x}_{4}^{2}}\right) - {... | Yes |
Proposition 9.4.4. Assume that \( P\left( X\right) \) has signature \( \left( {4,0}\right) \), keep the notation of Section 9.3.4, set\n\n\[ Q\left( X\right) = 3{X}^{4} - 2{a}_{1}{X}^{3} + {a}_{2}{X}^{2} - {a}_{4} = X{P}^{\prime }\left( X\right) - P\left( X\right) ,\]\n\n\[ S\left( X\right) = {12}{X}^{3} - 9{a}_{1}{X}^... | Proof. This signature is much simpler than the other two. Indeed, we have \( {g}_{1}\left( \mathbf{x}\right) = {x}_{1} + {x}_{2} + {x}_{3} + {x}_{4} - {a}_{1},{g}_{2}\left( \mathbf{x}\right) = {x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2} - {s}_{2} \) , \( {g}_{3}\left( \mathbf{x}\right) = {x}_{1}{x}_{2}{x}_{3}... | Yes |
Lemma 10.1.1. Let \( R \) be a Dedekind domain, \( M \) a finitely generated, torsion-free \( R \) -module, \( N \) a submodule of \( M \) of finite index, and \( \mathfrak{a} = \) Ann \( \left( {M, N}\right) \) and \( \mathfrak{b} = \left\lbrack {M : N}\right\rbrack \) be as above. Finally, let \( \mathfrak{p} \) be a... | Proof. Since \( \mathfrak{a}\left| \mathfrak{b}\right| {\mathfrak{a}}^{n} \), as already stated \( \mathfrak{a} \) and \( \mathfrak{b} \) have the same prime ideal divisors, so the equivalence of (1) and (2) is trivial.\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) By \( \left( 2\right) \), we have \( \mathfrak{... | Yes |
Lemma 10.1.2. Let \( L/K \) be a finite extension of number fields of degree \( n \) , and let \( \mathfrak{p} \) be a prime ideal of \( K \) that is totally ramified in \( L \), so that \( \mathfrak{p}{\mathbb{Z}}_{L} = {\mathfrak{P}}^{n} \) . Let \( \pi \in \mathfrak{P} \smallsetminus {\mathfrak{P}}^{2} \) . Then \( ... | Proof. Since \( \mathfrak{P} \) is totally ramified, \( {\mathbb{Z}}_{L}/{\mathfrak{P}}^{k} \) is a \( {\mathbb{Z}}_{K}/\mathfrak{p} \) -vector space (of dimension \( k \) ) for all \( k \leq n \) . We will show by induction that for all \( k \leq n \), the classes modulo \( {\mathfrak{P}}^{k} \) of the \( {\pi }^{i} \... | Yes |
Proposition 10.1.3. Let \( L/K \) be a normal extension of number fields as above.\n\n(1) The ideals \( {\mathfrak{P}}_{i} \) are permuted transitively by the Galois group \( G \) : in other words, for every pair \( \left( {i, j}\right) \) there exists a (not necessarily unique) \( {\sigma }_{i, j} \in G \) such that \... | Proof. (1) Fix \( {\mathfrak{P}}_{i} \), and assume that for some \( j,{\mathfrak{P}}_{j} \) is not of the form \( \sigma \left( {\mathfrak{P}}_{i}\right) \) for \( \sigma \in G \) . By the weak approximation theorem (Proposition 1.2.3), we can find \( x \in {\mathfrak{P}}_{j} \) such that \( x \notin \sigma \left( {\m... | Yes |
(1) The map \( s \) defined above is a surjective homomorphism from \( D\left( {\mathfrak{P}/\mathfrak{p}}\right) \) to \( \operatorname{Gal}\left( {\left( {{\mathbb{Z}}_{L}/\mathfrak{P}}\right) /\left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) }\right) \) whose kernel is equal to \( I\left( {\mathfrak{P}/\mathfrak{p}}\ri... | Proof. (1) The finite field extension \( \left( {{\mathbb{Z}}_{L}/\mathfrak{P}}\right) /\left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) \) is Galois (that is, normal and separable, and even cyclic). Hence, in particular, the primitive element theorem applies, so we can find \( \bar{\theta } \in {\mathbb{Z}}_{L}/\mathfrak... | Yes |
Corollary 10.1.6. We have\n\n\\[ \n\\left| {D\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) }\\right| = e\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) f\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) \\;\\text{ and }\\;\\left| {I\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) }\\right| = e\\left( {\\mathfrak{P}/\\mathfrak{p... | Proof. Since there are \( g \) ideals \( {\\mathfrak{P}}_{i} \) above \( \\mathfrak{p} \) and the \( D\\left( {{\\mathfrak{P}}_{i}/\\mathfrak{p}}\\right) \) are conjugate groups, and hence have the same cardinality, we have \( D\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) = n/g = \) \( e\\left( {\\mathfrak{P}/\\mathfr... | Yes |
Corollary 10.1.7. If \( L/K \) is a normal extension of number fields which is not cyclic, no prime ideal of \( K \) is inert in \( L/K \) . | Proof. Indeed, if \( \mathfrak{p} \) is inert, then \( D\left( {\mathfrak{P}/\mathfrak{p}}\right) = \operatorname{Gal}\left( {L/K}\right) \) and \( I\left( {\mathfrak{P}/\mathfrak{p}}\right) = \{ 1\} \) , hence Proposition 10.1.5 shows that \( \operatorname{Gal}\left( {L/K}\right) = D\left( {\mathfrak{P}/\mathfrak{p}}\... | Yes |
Lemma 10.1.8. Recall that \( D = {G}_{-1} \) and \( I = {G}_{0} \). (1) We have the inclusions \[ G \supset {G}_{-1} \supset {G}_{0} \supset {G}_{1}\cdots \] (2) There exists \( n \) such that \( {G}_{k} = \left\{ {1}_{G}\right\} \) for all \( k \geq n \) . | Proof. (1) is trivial. Furthermore, if \( \sigma \in {G}_{k} \) for all \( k \), then for all \( x \in {\mathbb{Z}}_{L} \) and for all \( k,\sigma \left( x\right) \equiv x\left( {\;\operatorname{mod}\;{\mathfrak{P}}^{k + 1}}\right) \), so \( \sigma \left( x\right) = x \) and hence \( \sigma = {1}_{G} \). Since \( G \) ... | Yes |
Lemma 10.1.9. For all \( k \geq 0,{G}_{k} \) is a normal subgroup of \( D = {G}_{-1} \) (not of \( G \) itself, however). | Proof. Let \( \sigma \in {G}_{k} \) and \( \tau \in D \) . For all \( x \in {\mathbb{Z}}_{L} \), we have\n\n\[ \sigma \left( {{\tau }^{-1}\left( x\right) }\right) \equiv {\tau }^{-1}\left( x\right) \left( {\;\operatorname{mod}\;{\mathfrak{P}}^{k + 1}}\right) ; \]\n\nhence\n\n\[ \tau \left( {\sigma \left( {{\tau }^{-1}\... | Yes |
Lemma 10.1.10. Keep the above notation. Then for all \( k \geq 0 \) we have \( {G}_{k}\left( {\mathfrak{P}/{\mathfrak{P}}_{I}}\right) = {G}_{k}\left( {\mathfrak{P}/\mathfrak{p}}\right) \) and we also have \( {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left... | Proof. Since \( I = {G}_{0} \), it is clear that for \( k \geq 0,\sigma \in {G}_{k}\left( {\mathfrak{P}/\mathfrak{p}}\right) \) if and only if \( \sigma \in {G}_{k}\left( {\mathfrak{P}/{\mathfrak{P}}_{I}}\right) \), so the groups are equal. To prove the second statement, we note that the extension \( {L}^{I}/K \) is un... | Yes |
Lemma 10.1.11. Assume that \( \mathfrak{p} = {\mathfrak{P}}^{e} \) is totally ramified in \( L/K \) and let \( \pi \in \mathfrak{P} \smallsetminus {\mathfrak{P}}^{2} \) be a uniformizer at \( \mathfrak{P} \). If \( \sigma \in G = \operatorname{Gal}\left( {L/K}\right) \), then \( \sigma \in {G}_{k} \) if and only if \( ... | Proof. The condition is evidently necessary. Conversely, if it is satisfied, then \( \sigma \left( x\right) \equiv x\left( {\;\operatorname{mod}\;{\mathfrak{P}}^{k + 1}}\right) \) for all \( x \in {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack \). Let \( y \) be an arbitrary element of \( {\mathbb{Z}}_{L} \). By Lemma ... | Yes |
Proposition 10.1.13. Let \( \mathfrak{p} \) be a prime ideal of \( K \) which is not necessarily totally ramified in \( L/K \), let \( \pi \) be a uniformizer of \( \mathfrak{P} \), and let \( \sigma \in {G}_{0} = I \).\n\n(1) We have \( {v}_{\mathfrak{P}}\left( {\sigma \left( \pi \right) /\pi }\right) = 0 \).\n\n(2) F... | Proof. Since \( \pi \) is a uniformizer of \( \mathfrak{P} \), we have\n\n\[ {v}_{\mathfrak{P}}\left( {\sigma \left( \pi \right) /\pi }\right) = {v}_{{\sigma }^{-1}\left( \mathfrak{P}\right) }\left( \pi \right) - {v}_{\mathfrak{P}}\left( \pi \right) = 0 \]\n\nsince \( \sigma \) is in the inertia group, hence a fortiori... | Yes |
Proposition 10.1.14. As above, let \( \pi \) be a uniformizer of \( \mathfrak{P} \) . (1) The map that sends \( \sigma \in {G}_{0} \) to \( \sigma \left( \pi \right) /\pi \) induces an injection \( {\theta }_{0} \) from \( {G}_{0}/{G}_{1} \) to \( {\left( {\mathbb{Z}}_{L}/\mathfrak{P}\right) }^{ * } \) . (2) For \( k \... | Proof. First note that if \( {v}_{\mathfrak{P}}\left( \alpha \right) \geq k \), by Lemma 1.2.31 we may write \( \alpha = x/d \) with \( {v}_{\mathfrak{P}}\left( d\right) = 0 \), hence with \( {v}_{\mathfrak{P}}\left( x\right) \geq k \) . It follows that we can send \( \alpha \) to \( {\mathfrak{P}}^{k}/{\mathfrak{P}... | No |
Corollary 10.1.15. Let \( p \) be the prime number below \( \mathfrak{p} \) and \( \mathfrak{P} \) or, equivalently, the characteristic of the residue fields \( {\mathbb{Z}}_{K}/\mathfrak{p} \) and \( {\mathbb{Z}}_{L}/\mathfrak{P} \) . Then we have the following.\n\n(1) The group \( {G}_{0}/{G}_{1} \) is a cyclic group... | Proof. Since \( {\left( {\mathbb{Z}}_{L}/\mathfrak{P}\right) }^{ * } \) is the multiplicative group of a finite field with \( {p}^{f\left( {\mathfrak{P}/p}\right) } \) elements, it is a cyclic group of order \( {p}^{f\left( {\mathfrak{P}/p}\right) } - 1 \) . Proposition 10.1.14 thus implies that \( {G}_{0}/{G}_{1} \) i... | Yes |
Proposition 10.1.16. Let \( \sigma \in {G}_{0} \) and let \( \tau \in {G}_{k} \) for some \( k \geq 1 \) . Then in \( {\mathfrak{P}}^{k}/{\mathfrak{P}}^{k + 1} \) we have\n\n\[{\theta }_{k}\left( {{\sigma \tau }{\sigma }^{-1}}\right) = {\theta }_{0}{\left( \sigma \right) }^{k}{\theta }_{k}\left( \tau \right) .\] | Proof. Note first that this formula makes sense, since \( {\theta }_{0}{\left( \sigma \right) }^{k} \in {\left( {\mathbb{Z}}_{L}/\mathfrak{P}\right) }^{ * } \) , which operates multiplicatively on the \( {\mathbb{Z}}_{L}/\mathfrak{P} \) -vector space \( {\mathfrak{P}}^{k}/{\mathfrak{P}}^{k + 1} \) .\n\nSet \( {\pi }_{1... | Yes |
Corollary 10.1.17. Let \( \sigma \in {G}_{0} \) and let \( \tau \in {G}_{k} \) for some \( k \geq 1 \) . Then \( {\sigma \tau }{\sigma }^{-1}{\tau }^{-1} \in {G}_{k + 1} \) if and only if either \( \tau \in {G}_{k + 1} \) or \( {\sigma }^{k} \in {G}_{1} \) . | Indeed, by the above proposition\n\n\[ \n{\sigma \tau }{\sigma }^{-1}{\tau }^{-1} \in {G}_{k + 1} \Leftrightarrow {\sigma \tau }{\sigma }^{-1} \in \tau {G}_{k + 1} \n\]\n\n\[ \n\Leftrightarrow {\theta }_{k}\left( {{\sigma \tau }{\sigma }^{-1}}\right) = {\theta }_{k}\left( \tau \right) \n\]\n\n\[ \n\Leftrightarrow {\the... | Yes |
Corollary 10.1.18. Let \( {e}_{0} = e/\left( {{p}^{\infty }, e}\right) \) be the order of \( {G}_{0}/{G}_{1} \) . (1) If \( {G}_{0} \) is an Abelian group, then for \( k \geq 0,{G}_{k} \neq {G}_{k + 1} \) implies \( {e}_{0} \mid k \) . | Proof. Let \( \sigma \in {G}_{0} \) be such that the image of \( \sigma \) in \( {G}_{0}/{G}_{1} \) generates the cyclic group \( {G}_{0}/{G}_{1} \) . (1). Assume that \( {e}_{0} \nmid k \), and in particular \( k \geq 1 \) . This means that \( {\sigma }^{k} \notin {G}_{1} \) . It follows that for any \( \tau \in {G}_{... | Yes |
Lemma 10.1.20. Let \( \pi \) be as above, and let \( f \) be the minimal polynomial of \( \pi \) in \( {\mathbb{Z}}_{K}\left\lbrack X\right\rbrack \) . Then \[ {\mathbb{Z}}_{K}{\left\lbrack \pi \right\rbrack }^{ * } = \frac{1}{{f}^{\prime }\left( \pi \right) }{\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack \] | Proof. Set \( {a}_{i, j} = {\operatorname{Tr}}_{L/K}\left( {{\pi }^{i}{\pi }^{j}/{f}^{\prime }\left( \pi \right) }\right) \), and let \( A \) be the \( n \times n \) matrix whose entries are the \( {a}_{i, j} \). For any \( x \in L \) we can write \( x = \mathop{\sum }\limits_{i}{x}_{i}{\pi }^{i}/{f}^{\prime }\left( \p... | Yes |
Corollary 10.1.21. With the above notation,\n\n\[ \n{v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = {v}_{\mathfrak{P}}\left( {{f}^{\prime }\left( \pi \right) }\right) \n\] | Proof. As in Section 10.1.1, set\n\n\[ \n\mathfrak{a} = \operatorname{Ann}\left( {{\mathbb{Z}}_{L},{\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack }\right) = \left\{ {x \in {\mathbb{Z}}_{K}/x{\mathbb{Z}}_{L} \subset {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack }\right\} \n\]\n\nSince \( {\mathbb{Z}}_{K}\left\lbrack \p... | Yes |
Theorem 10.1.22. Let \( L/K \) be a normal extension of number fields, set \( G = \operatorname{Gal}\left( {L/K}\right) \), let \( \mathfrak{P} \) be a prime ideal of \( L \), and let \( \mathfrak{p} \) be the prime ideal of \( K \) below \( \mathfrak{P} \) . The valuation at \( \mathfrak{P} \) of the relative differen... | Proof. By Lemma 10.1.10, if we set \( {K}^{\prime } = {L}^{I} \), the ideal \( \mathfrak{P} \) has the same ramification groups (for \( k \geq 0 \) ) in the extension \( L/{K}^{\prime } \) as in the extension \( L/K \), and \( {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/{K}^{\prime }}\right) }\right) = {v}_{\mathfr... | Yes |
Proposition 10.1.23. Let \( L/K \) be a cyclic extension of number fields of prime degree \( \ell \), let \( \mathfrak{p} \) be a prime ideal of \( K \) above \( \ell \), and denote as usual by \( {G}_{k} \) the ramification groups of \( \mathfrak{P}/\mathfrak{p} \) for any prime ideal \( \mathfrak{P} \) of \( L \) abo... | Proof. If \( \mathfrak{p} \) is unramified in \( L/K \), we have \( {G}_{0} = \{ 1\} \), so the inequality is trivial. Thus, since \( \ell \) is prime, we may assume that \( \mathfrak{p} \) is totally ramified in \( L/K \), so that \( \mathfrak{p}{\mathbb{Z}}_{L} = {\mathfrak{P}}^{\ell } \) . For simplicity set \( e = ... | Yes |
Corollary 10.1.24. Let \( L/K \) be a cyclic extension of number fields of prime degree \( \ell \) and let \( \mathfrak{p} \) be a prime ideal of \( K \) above \( \ell \) . If \( \mathfrak{f} \) denotes the conductor of \( L/K \), then \( {v}_{\mathfrak{p}}\left( \mathfrak{f}\right) \leq \lfloor \ell e\left( {\mathfrak... | Proof. Indeed, we may assume that \( \mathfrak{p} \mid \mathfrak{f} \), so \( \mathfrak{p} \) is ramified, hence totally ramified in \( L/K \) . Thus, using the same notation as that of the proposition, we have \( {v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {L/K}\right) }\right) = {v}_{\mathfrak{P}}\left( {\mathfrak{D... | No |
Proposition 10.1.26. Keep the above hypotheses and notation, and let \( \mathfrak{p} \) be a prime ideal of \( K \) . (1) The prime ideal \( \mathfrak{p} \) cannot be inert in \( N/K \) . | Proof. Note first that if \( g \) is the number of prime ideals of \( N \) above \( \mathfrak{p} \) and if \( {\mathfrak{P}}_{N} \) is one of them, we have \( e\left( {{\mathfrak{P}}_{N}/\mathfrak{p}}\right) f\left( {{\mathfrak{P}}_{N}/\mathfrak{p}}\right) g = \left\lbrack {N : K}\right\rbrack = 2\ell \), hence the pos... | Yes |
Lemma 10.1.27. Keep the above hypotheses and notation. We have\n\n\[ \n{\mathcal{N}}_{L/K}\left( {\mathfrak{d}\left( {N/L}\right) }\right) = \mathfrak{d}\left( {{K}_{2}/K}\right) \n\] | Proof. We are going to show that for every prime ideal \( \mathfrak{p} \) of \( K \), the \( \mathfrak{p} \) - adic valuations of both sides are equal. Thus, let \( \mathfrak{p} \) be a prime ideal of \( K \) , and assume first that \( \mathfrak{p} \nmid 2 \) . If \( \mathfrak{p} \mid \mathfrak{d}\left( {{K}_{2}/K}\rig... | Yes |
Proposition 10.1.28. Keep the above hypotheses and notation. In particular, recall that \( \mathfrak{f} \) is an ideal of \( K \) such that \( \mathfrak{f}{\mathbb{Z}}_{{K}_{2}} \) is the conductor of \( N/{K}_{2} \) . Let \( \mathfrak{p} \) be a prime ideal of \( K \) .\n\n(1) We have\n\n\[ \mathfrak{d}\left( {L/K}\ri... | Proof. (1). By Theorem 2.5.1 we know that\n\n\[ \mathfrak{d}\left( {N/K}\right) = \mathfrak{d}{\left( L/K\right) }^{2}{\mathcal{N}}_{L/K}\left( {\mathfrak{d}\left( {N/L}\right) }\right) = \mathfrak{d}{\left( {K}_{2}/K\right) }^{\ell }{\mathcal{N}}_{{K}_{2}/K}\left( {\mathfrak{d}\left( {N/{K}_{2}}\right) }\right) \]\n\n... | Yes |
Lemma 10.2.1 (Dirichlet's Character Independence Theorem). Let \( G \) be a group, let \( L \) be a field, and let \( {\chi }_{1},\ldots ,{\chi }_{m} \) be distinct characters of \( G \) with values in \( {L}^{ * } \) . The characters \( {\chi }_{i} \) are \( L \) -linearly independent, in other words a relation \( \ma... | Proof. Assume that the characters are \( L \) -linearly dependent. Choose a dependence relation of minimal length, so that, up to reordering of the \( {\chi }_{i} \) ,\n\n\[ \forall h \in G\;\mathop{\sum }\limits_{{1 \leq i \leq n}}{a}_{i}{\chi }_{i}\left( h\right) = 0 \]\n\n(1)\n\nwith \( n \) minimal. For any \( g \i... | Yes |
Corollary 10.2.2. Let \( K \) and \( L \) be fields, and let \( {\sigma }_{1},\ldots ,{\sigma }_{m} \) be distinct homomorphisms from \( K \) to \( L \) . Then the \( {\sigma }_{i} \) are \( L \) -linearly independent. | Proof. Simply apply the preceding lemma to \( G = {K}^{ * } \) and to \( {\chi }_{i} \) equal to the restriction of \( {\sigma }_{i} \) to \( {K}^{ * } \) . | No |
Lemma 10.2.3 (Noether’s Theorem). Let \( L/K \) be a normal extension with Galois group \( G \), and let \( \phi \) be a map from \( G \) to \( {L}^{ * } \) . We will say that \( \phi \) satisfies the cocycle condition if for all \( g, h \) in \( G \) we have\n\n\[ \phi \left( {gh}\right) = \phi \left( g\right) \cdot g... | Proof. If \( \phi \left( g\right) = \alpha /g\left( \alpha \right) \), we have\n\n\[ \phi \left( g\right) \cdot g\left( {\phi \left( h\right) }\right) = \frac{\alpha }{g\left( \alpha \right) }g\left( \frac{\alpha }{h\left( \alpha \right) }\right) = \frac{\alpha }{g\left( {h\left( \alpha \right) }\right) } = \phi \left(... | Yes |
Lemma 10.2.4 (Hilbert’s Theorem 90). Let \( L/K \) be a cyclic extension with Galois group \( G \) generated by an element \( \sigma \) . Then \( \alpha \in L \) is an element of relative norm equal to 1 if and only if there exists \( \beta \in L \) such that \( \alpha = \beta /\sigma \left( \beta \right) \) . | Proof. Clearly \( {\mathcal{N}}_{L/K}\left( {\sigma \left( \beta \right) }\right) = {\mathcal{N}}_{L/K}\left( \beta \right) \), hence the relative norm of \( \beta /\sigma \left( \beta \right) \) is equal to 1 . Conversely, assume that \( {\mathcal{N}}_{L/K}\left( \alpha \right) = 1 \) . Let \( n = \left| G\right| \) ,... | Yes |
Lemma 10.2.6. Any homomorphism from \( G \) to \( {\mu }_{n} \) is of the form\n\n\[ \sigma \mapsto < \sigma ,\bar{b} > \]\n\nfor some \( \bar{b} \in B \) . | To prove the lemma, let \( \phi \) be a homomorphism from \( G \) to \( {\mathbf{\mu }}_{n} \) . Recall that \( {\mathbf{\mu }}_{n} \subset K \), hence that any element of \( G = \operatorname{Gal}\left( {L/K}\right) \) fixes \( {\mathbf{\mu }}_{n} \) pointwise. Thus, for all \( \sigma \) and \( \tau \) in \( G \) we h... | Yes |
Corollary 10.2.7. Let \( K \) be a number field and \( n \geq 1 \) be an integer such that \( {\zeta }_{n} \in K \) .\n\n(1) An extension \( L/K \) is a cyclic extension of degree \( n \) if and only if there exists \( \alpha \in {K}^{ * } \) such that \( \bar{\alpha } \) is exactly of order \( n \) in \( {K}^{ * }/{K}... | Proof. (1) Let \( L/K \) be a cyclic extension of degree \( n \) . By Theorem 10.2.5, there exists a subgroup \( B \) of \( {K}^{ * }/{K}^{*n} \) such that \( L = {K}_{B} \) and \( B \simeq \) \( \operatorname{Gal}\left( {L/K}\right) \simeq \mathbb{Z}/n\mathbb{Z} \) . If \( \bar{\alpha } \) is a generator of \( B \), i... | Yes |
Theorem 10.2.9. Let \( K \) be a number field, \( \ell \) a prime number such that \( {\zeta }_{\ell } \in K \), and \( L = K\left( \sqrt[\ell ]{\alpha }\right) \), where \( \alpha \in {K}^{ * } \smallsetminus {K}^{*\ell } \) . If \( \mathfrak{p} \) is a prime ideal of \( {\mathbb{Z}}_{K} \) , we set\n\n\[ e\left( {\ma... | Proof. To simplify notation, write \( \zeta \) for \( {\zeta }_{\ell } \) . For (1), assume that \( \ell \nmid {v}_{\mathfrak{p}}\left( \alpha \right) \) . Let \( \pi \) be a uniformizer of \( \mathfrak{p} \) in \( K \) . There exist integers \( x \) and \( y \) such that \( x\ell + y{v}_{\mathfrak{p}}\left( \alpha \ri... | Yes |
Lemma 10.2.10. Assume that \( {v}_{\mathfrak{p}}\left( \alpha \right) = 0 \) and \( \mathfrak{p} \mid \ell \) . The congruence \( {x}^{\ell } \equiv \alpha \) \( \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k}}\right) \) has a solution in \( K \) for \( k = z\left( {\mathfrak{p},\ell }\right) \) if and only if it has ... | Proof. Assume that the congruence has a solution \( x \) for some \( k \geq z\left( {\mathfrak{p},\ell }\right) \) . We apply a Newton-Hensel iteration: in other words, we set\n\n\[ \n{x}_{1} = x + y\;\text{ with }\;y = - \frac{{x}^{\ell } - \alpha }{\ell {x}^{\ell - 1}}.\n\]\n\nThen \( {v}_{\mathfrak{p}}\left( y\right... | Yes |
Lemma 10.2.11. With this notation, we have \( a \geq 1 \) and \( \ell \nmid a \) . | Proof. This follows immediately from Proposition 10.2.13, which we will prove in the next section. | No |
Corollary 10.2.12. Let \( K \) be a number field, \( \ell \) a prime number such that \( {\zeta }_{\ell } \in K \), and \( L = K\left( \sqrt[\ell ]{\alpha }\right) \), where \( \alpha \in {K}^{ * } \smallsetminus {K}^{*\ell } \) . Let \( \mathfrak{p} \) be a prime ideal of \( {\mathbb{Z}}_{K} \) . Then we have the foll... | Proof. This is an immediate consequence of Theorem 10.2.9. | No |
Proposition 10.2.13. Keep the above notation and hypotheses, and let \( \pi \) be a uniformizer of \( \mathfrak{p} \) . Let \( k \) be an integer such that \( 1 \leq k \leq z\left( {\mathfrak{p},\ell }\right) - 1 = \) \( \ell e\left( {\mathfrak{p}/\ell }\right) /\left( {\ell - 1}\right) \), and let \( {x}_{k - 1} \in {... | Proof. Assume first that the congruence modulo \( {\mathfrak{p}}^{k} \) is satisfied. Since \( {\mathbb{Z}}_{K}/\mathfrak{p} \) is a perfect field of characteristic \( \ell \), the map \( x \mapsto {x}^{\ell } \) is a bijection (hence an injection) from \( {\mathbb{Z}}_{K}/\mathfrak{p} \) to itself, so if we write \( {... | Yes |
(1) The function \( \gamma \left( s\right) \) is a meromorphic function of \( s \) . | Proof. It is clear from the definition that \( \gamma \left( s\right) \) is a meromorphic function whose poles are the complex numbers \( s \) such that \( {a}_{i}s + {b}_{i} = - k \) for some nonnegative integer \( k \) -in other words, the numbers \( s = - \left( {{b}_{i} + k}\right) /{a}_{i} \) . Since the \( {a}_{i... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.