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Theorem 4.5 (Dirichlet). Let \( \left( {a, n}\right) = 1 \) . Then there are infinitely many primes \( p \equiv a\left( {\;\operatorname{mod}\;n}\right) \) | Proof. Let \( \chi \) be a Dirichlet character. Then for \( \operatorname{Re}\left( s\right) > 1 \) we have\n\n\[ \log L\left( {s,\chi }\right) = - \mathop{\sum }\limits_{p}\log \left( {1 - \chi \left( p\right) {p}^{-s}}\right) \]\n\n\[ = \mathop{\sum }\limits_{p}\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{\chi {\le... | Yes |
Corollary 4.6. \[ \mathop{\prod }\limits_{{\chi \in X}}\tau \left( \chi \right) = \left\{ \begin{array}{ll} \sqrt{\left| d\left( K\right) \right| }, & \text{ if }K\text{ is totally real } \\ {i}^{\deg \left( {K/\mathbb{Q}}\right) /2}\sqrt{\left| d\left( K\right) \right| }, & \text{ if }K\text{ is complex. } \end{array}... | Proof. It follows from the above proof that \( \mathop{\prod }\limits_{{x \in X}}{W}_{x} = 1 \) . The result follows immediately. | No |
Lemma 4.7. For every integer \( b \) , \[ \mathop{\sum }\limits_{{a = 1}}^{f}\bar{\chi }\left( a\right) {e}^{{2\pi iab}/f} = \chi \left( b\right) \tau \left( \bar{\chi }\right) \] | Proof. If \( \left( {b, f}\right) = 1 \), then change variables: \( c \equiv {ab}\left( {\;\operatorname{mod}\;f}\right) \) . Since everything depends only on residue classes \( {\;\operatorname{mod}\;f} \), the result follows in this case. If \( \left( {b, f}\right) = d > 1 \) then the result is still true, since both... | Yes |
Lemma 4.8. \( \left| {\tau \left( \chi \right) }\right| = \sqrt{{f}_{\chi }} \) | Proof.\n\n\[ \phi \left( f\right) {\left| \tau \left( \chi \right) \right| }^{2} = \mathop{\sum }\limits_{{b = 1}}^{f}{\left| \chi \left( b\right) \tau \left( \chi \right) \right| }^{2}\;\text{ (note only }\phi \left( f\right) \text{ terms are non-zero) } \]\n\n\[ = \mathop{\sum }\limits_{{b = 1}}^{f}\mathop{\sum }\lim... | Yes |
Theorem 4.9.\n\n\[ L\left( {1,\chi }\right) = {\pi i}\frac{\tau \left( \chi \right) }{f}{B}_{1,\bar{\chi }} = {\pi i}\frac{\tau \left( \chi \right) }{f}\frac{1}{f}\mathop{\sum }\limits_{{a = 1}}^{f}\bar{\chi }\left( a\right) a\;\text{ if }\chi \left( {-1}\right) = - 1. \]\n\n\[ L\left( {1,\chi }\right) = - \frac{\tau \... | Note that the theorem implies that \( {B}_{1,\chi } \neq 0 \) if \( \chi \) is odd. There is no elementary proof known for this fact. | No |
Theorem 4.10. Let \( K \) be a \( {CM} \) -field, \( {K}^{ + } \) its maximal real subfield, and let \( h \) and \( {h}^{ + } \) be the respective class numbers. Then \( {h}^{ + } \) divides \( h \) . | Proof. We need the following result from class field theory. | No |
Proposition 4.11. Let \( K/L \) be an extension of number fields such that there is no nontrivial unramified (at all primes, including archimedean ones) subextension \( F/L \) with \( \operatorname{Gal}\left( {F/L}\right) \) abelian. Then the class number of \( L \) divides the class number of \( K \) . | Proof. Let \( H \) be the maximal unramified (at all primes) abelian extension of \( L \). By class field theory, \( \operatorname{Gal}\left( {H/L}\right) \) is isomorphic to the ideal class group of \( L \). The assumptions on \( K/L \) imply that \( H \cap K = L \). Therefore \( \left\lbrack {{KH} : K}\right\rbrack =... | Yes |
Theorem 4.12. Let \( K \) be a CM-field and let \( E \) be its unit group. Let \( {E}^{ + } \) be the unit group of \( {K}^{ + } \) and let \( W \) be the group of roots of unity in \( K \) . Then\n\n\[ Q\overset{\text{ def }}{ = }\left\lbrack {E : W{E}^{ + }}\right\rbrack = 1\text{ or }2. \] | Proof. Define \( \phi : E \rightarrow W \) by \( \phi \left( \varepsilon \right) = \varepsilon /\bar{\varepsilon } \) . Since \( \overline{{\varepsilon }^{\sigma }} = {\left( \bar{\varepsilon }\right) }^{\sigma } \) for all embeddings \( \sigma (K \) is \( {CM} \) ), we have \( \left| {\phi {\left( \varepsilon \right) ... | Yes |
Corollary 4.13. Let \( K = \mathbb{Q}\left( {\zeta }_{n}\right) \). Then \( Q = 1 \) if \( n \) is a prime power and \( Q = 2 \) if \( n \) is not a prime power. | Proof. The proof when \( n \) is an odd prime power is exactly the same as that given in Proposition 1.5. For \( p = 2 \) we must argue a little differently. Suppose \( \varepsilon \) is a unit in \( Q\left( {\zeta }_{{2}^{m}}\right) \) such that \( \varepsilon /\bar{\varepsilon } \notin {W}^{2} \). Then \( \varepsilon... | Yes |
Theorem 4.14. Let \( C \) be the ideal class group of \( \mathbb{Q}\left( {\zeta }_{n}\right) \) and \( {C}^{ + } \) the ideal class group of the real subfield \( \mathbb{Q}{\left( {\zeta }_{n}\right) }^{ + } \) . Then the natural map \( {C}^{ + } \rightarrow C \) is an injection. | Proof. Suppose \( I \) is an ideal of \( \mathbb{Q}{\left( {\zeta }_{n}\right) }^{ + } \) which becomes principal when lifted to \( \mathbb{Q}\left( {\zeta }_{n}\right) \) . We must show \( I \) was principal to begin with. Let \( I = \left( \alpha \right) \) with \( \alpha \in \mathbb{Q}\left( {\zeta }_{n}\right) \) .... | Yes |
Lemma 4.15. Let \( {\varepsilon }_{1},\ldots ,{\varepsilon }_{r} \) be independent units of a number field \( K \) which generate a subgroup \( A \) of the units of \( K \) modulo roots of unity, and let \( {\eta }_{1},\ldots \) , \( {\eta }_{r} \) generate a subgroup \( B \) . If \( A \subseteq B \) is of finite index... | Proof. We may write\n\n\[ {\varepsilon }_{i} = \left( {\mathop{\prod }\limits_{l}{\eta }_{l}^{{a}_{il}}}\right) \cdot \text{ (root of unity),}\;\text{ with }{a}_{il} \in \mathbb{Z}. \]\n\nTherefore\n\n\[ {\delta }_{j}\log \left| {\varepsilon }_{i}^{{\sigma }_{j}}\right| = \mathop{\sum }\limits_{l}{a}_{il}{\delta }_{j}\... | Yes |
Lemma 4.18. \( \log {d}_{n} = \phi \left( n\right) \log n + o\left( {\phi \left( n\right) \log n}\right) \) | Proof. From Proposition 2.7, we have\n\n\[ \log {d}_{n} = \phi \left( n\right) \log n - \phi \left( n\right) \mathop{\sum }\limits_{{p \mid n}}\frac{\log p}{p - 1}. \]\n\nLet \( m = \log n/\log 2 \). Since \( {2}^{m} = n \), it follows that \( n \) has at most \( m \) prime factors. Clearly\n\n\[ \mathop{\sum }\limits_... | Yes |
Lemma 4.19. If \( n \) is not a prime power then \( {d}_{n} = {\left( {d}_{n}^{ + }\right) }^{2} \) . If \( n = {p}^{a} \) then \( {d}_{n} = \) \( p{\left( {d}_{n}^{ + }\right) }^{2} \) if \( p \neq 2,{d}_{n} = 4{\left( {d}_{n}^{ + }\right) }^{2} \) if \( p = 2 \) . In all cases we have\n\n\[ \log {d}_{n}^{ + } = \frac... | Proof. Recall the formula (see Lang [1], pp. 60, 66, or Long [1] p. 82)\n\n\[ \left| {d\left( L\right) }\right| = \left( {N{\mathcal{D}}_{L/K}}\right) {\left| d\left( K\right) \right| }^{\deg \left( {L/K}\right) } \]\n\nwhere \( L/K \) is any extension of number fields, \( {\mathcal{D}}_{L/K} \) is the relative differe... | Yes |
Proposition 5.1. \( {\overline{\mathbb{Q}}}_{p} \) is not complete. | Proof. Let\n\n\[ \alpha = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\zeta }_{{n}^{\prime }}{p}^{n} \]\n\nwhere \( {n}^{\prime } = n \) if \( \left( {n, p}\right) = 1 \) and \( {n}^{\prime } = 1 \) otherwise. If \( {\overline{\mathbb{Q}}}_{p} \) were complete, then the series would converge to \( \alpha \in {\overline{\... | Yes |
Proposition 5.2. \( {\mathbb{C}}_{p} \) is algebraically closed. | Proof. We need the following lemma, due to Krasner.\n\nLemma 5.3. Suppose \( K | No |
Lemma 5.3. Suppose \( K \) is a complete field with respect to a non-archimedean valuation. Let \( \alpha ,\beta \in \bar{K} \), the algebraic closure of \( K \), with \( \alpha \) separable over \( K\left( \beta \right) \) . Finally, suppose that for all conjugates \( {\alpha }_{i} \neq \alpha \) of \( \alpha \) we ha... | Proof. Consider the extension \( K\left( {\alpha ,\beta }\right) /K\left( \beta \right) \) and let \( L/K\left( \beta \right) \) be the Galois closure. Let \( \sigma \in \operatorname{Gal}\left( {L/K\left( \beta \right) }\right) \) . Then \( \sigma \left( {\beta - \alpha }\right) = \beta - \sigma \left( \alpha \right) ... | Yes |
Proposition 5.4. There exists a unique extension of \( {\log }_{p} \) to all of \( {\mathbb{C}}_{p}^{ \times } \) such that \( {\log }_{p}\left( p\right) = 0 \) and \( {\log }_{p}\left( {xy}\right) = {\log }_{p}x + {\log }_{p}y \) for all \( x, y \in {\mathbb{C}}_{p}^{ \times } \) . | Proof. We need to investigate the multiplicative structure of \( {\mathbb{C}}_{p}^{ \times } \) . For each rational number \( r \) choose a power \( {p}^{r} \) of \( p \) in such a way that \( {p}^{r}{p}^{s} = {p}^{r + s} \) (one way: let \( {p}^{r} \) be the positive real \( r \) th power of \( p \) in \( \overline{\m... | Yes |
Lemma 5.5. If \( \left| x\right| < {p}^{-1/\left( {p - 1}\right) } \) then \( \left| {{\log }_{p}\left( {1 + x}\right) }\right| = \left| x\right| \) and if \( \left| x\right| \leq {p}^{-1/\left( {p - 1}\right) } \) then \( \left| {{\log }_{p}\left( {1 + x}\right) }\right| \leq \left| x\right| \). | Proof. If \( n < p \) then \( \left| n\right| = 1 \), and in general \( \left| n\right| \geq 1/n \) . Therefore, if \( \left| x\right| < \) \( {p}^{-1/\left( {p - 1}\right) } \) we have\n\n\[ \left| \frac{{x}^{n}}{n}\right| = {\left| x\right| }^{n - 1} \cdot \left| x\right| < \left| x\right| \;\text{ if }2 \leq n < p \... | Yes |
Proposition 5.6. \( {\log }_{p}x = 0 \Leftrightarrow x \) is a rational power of \( p \) times a root of unity (of arbitrary order). | Proof. Clearly such \( x \) satisfy \( {\log }_{p}x = 0 \) . Conversely, suppose \( {\log }_{p}x = 0 \) . Since \( {\mathbb{C}}_{p}^{ \times } = {p}^{\mathbb{Q}} \times W \times {U}_{1} \), we may assume \( x = 1 + y \) with \( \left| y\right| < 1 \) . Let \( N \) be large enough that \( \left| {y}^{{p}^{N}}\right| < {... | Yes |
Proposition 5.7. If \( \left| x\right| < {p}^{-1/\left( {p - 1}\right) } \) then\n\n\[{\log }_{p}\exp \left( x\right) = x\]\n\nand\n\n\[\exp {\log }_{p}\left( {1 + x}\right) = 1 + x.\] | Proof. Both are formal power series identities, so we need only check convergence. Since \( \left| {{x}^{n}/n!}\right| < 1 \) for \( n \geq 1 \) (because \( {v}_{p}\left( {n!}\right) < n/\left( {p - 1}\right) \) ), we have \( \left| {\exp \left( x\right) - 1}\right| < 1 \) for all \( x \) with \( \left| x\right| < {p}^... | Yes |
Proposition 5.8. Suppose \( r < {p}^{-1/\left( {p - 1}\right) } < 1 \) and\n\n\[ f\left( X\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}\left( \begin{matrix} X \\ n \end{matrix}\right) \]\n\nwith \( \left| {a}_{n}\right| \leq M{r}^{n} \) for some \( M \) . Then \( f\left( X\right) \) may be expressed as a p... | Proof of Proposition 5.8. Let\n\n\[ {P}_{i}\left( X\right) = \mathop{\sum }\limits_{{n \leq i}}{a}_{n}\left( \begin{matrix} X \\ n \end{matrix}\right) = \mathop{\sum }\limits_{{n \leq i}}{a}_{n, i}{X}^{n},\;i = 1,2,3,\ldots \]\n\nThen\n\n\[ {a}_{n, i} = {a}_{n}\frac{\text{ integer }}{n!} + {a}_{n + 1}\frac{\text{ integ... | Yes |
Theorem 5.9. Suppose \( q \mid F \) and \( p \nmid a \) . Then there exists a p-adic meromorphic function \( {H}_{p}\left( {s, a, F}\right) \) on\n\n\[ \left\{ {s \in {\mathbb{C}}_{p}\left| \right| s \mid < q{p}^{-1/\left( {p - 1}\right) } > 1}\right\} \]\n\nsuch that\n\n\[ {H}_{p}\left( {1 - n, a, F}\right) = {\omega ... | Proof. Let\n\n\[ {H}_{p}\left( {s, a, F}\right) = \frac{1}{s - 1}\frac{1}{F}\langle a{\rangle }^{1 - s}\mathop{\sum }\limits_{{j = 0}}^{\infty }\left( \begin{matrix} 1 - s \\ j \end{matrix}\right) \left( {B}_{j}\right) {\left( \frac{F}{a}\right) }^{j}. \]\n\nAssume convergence for the moment. Then\n\n\[ {H}_{p}\left( {... | Yes |
Theorem 5.10 (von Staudt-Clausen). Let \( n \) be even and positive. Then\n\n\[ \n{B}_{n} + \mathop{\sum }\limits_{{\left( {p - 1}\right) \mid n}}\frac{1}{p} \in \mathbb{Z} \n\]\n\nwhere the sum is over those primes \( p \) such that \( p - 1 \) divides \( n \) (in particular,2 and 3 appear in the denominator of each B... | Proof. We shall show that for each prime \( p \) we have \( {B}_{n} = - 1/p \) or \( 0{\;\operatorname{mod}\;{\mathbb{Z}}_{p}} \) , depending on whether \( p - 1 \) does or does not divide \( n \) . Assume by induction that this is true for \( m < n \) . In particular, \( p{B}_{m} \in {\mathbb{Z}}_{p} \) for \( m < n \... | Yes |
Theorem 5.11. Let \( \chi \) be a Dirichlet character of conductor \( f \) and let \( F \) be any multiple of \( q \) and \( f \) . Then there exists a p-adic meromorphic (analytic if \( \chi \neq 1 \) ) function \( {L}_{p}\left( {s,\chi }\right) \) on \( \left\{ {s \in {\mathbb{C}}_{p}\left| \right| s \mid < q{p}^{-1/... | Proof of Theorem 5.11. We show that the formula gives the desired function. Since \[ {L}_{p}\left( {s,\chi }\right) = \mathop{\sum }\limits_{\substack{{a = 1} \\ {p \nmid a} }}^{F}\chi \left( a\right) {H}_{p}\left( {s, a, F}\right) \] the analyticity properties follow at once. At \( s = 1,{L}_{p}\left( {s,\chi }\right)... | Yes |
Corollary 5.13. Suppose \( \chi \neq 1,{pq}\rangle f \) . Let \( m, n \in \mathbb{Z} \) . Then\n\n\[ \n{L}_{p}\left( {m,\chi }\right) \equiv {L}_{p}\left( {n,\chi }\right) \left( {\;\operatorname{mod}\;p}\right) \n\]\n\nand both numbers are p-integral. | Proof. Both sides are congruent to \( {a}_{0} \) in the notation of the theorem. | No |
Corollary 5.14 (Kummer’s Congruences). Suppose \( m \equiv n ≢ 0\left( {{\;\operatorname{mod}\;p} - 1}\right) \) are positive even integers. Then\n\n\[ \n\frac{{B}_{m}}{m} \equiv \frac{{B}_{n}}{n}\left( {\;\operatorname{mod}\;p}\right) \n\]\n\nMore generally, if \( m \) and \( n \) are positive even integers with \( m ... | Proof. Consider \( {L}_{p}\left( {s,{\omega }^{m}}\right) = {L}_{p}\left( {s,{\omega }^{n}}\right) \) . Then\n\n\[ \n{L}_{p}\left( {1 - m,{\omega }^{m}}\right) = - \left( {1 - {p}^{m - 1}}\right) \left( {{B}_{m}/m}\right) \n\]\n\nand similarly for \( n \) . Also\n\n\[ \n{L}_{p}\left( {1 - m,{\omega }^{m}}\right) = {a}_... | Yes |
Corollary 5.15. Suppose \( n \) is odd, \( n ≢ - 1\left( {{\;\operatorname{mod}\;p} - 1}\right) \) . Then\n\n\[ \n{B}_{1,{\omega }^{n}} \equiv \frac{{B}_{n + 1}}{n + 1}\left( {\;\operatorname{mod}\;p}\right) \n\]\n\nand both sides are p-integral. | Proof. Since \( n ≢ - 1,{\omega }^{n + 1} \neq 1 \) . Also \( {\omega }^{n}\left( p\right) = 0 \) since \( {\omega }^{n} \neq 1 \) . Therefore, by Corollary 5.13,\n\n\[ \n{B}_{1,{\omega }^{n}} = \left( {1 - {\omega }^{n}\left( p\right) }\right) {B}_{1,{\omega }^{n}} = - {L}_{p}\left( {0,{\omega }^{n + 1}}\right) \n\]\n... | Yes |
Let \( p \) be an odd prime and let \( {h}_{p}^{ - } \) be the relative class number of \( \mathbb{Q}\left( {\zeta }_{p}\right) \) . Then \( p \mid {h}_{p}^{ - } \Leftrightarrow p \) divides the numerator of \( {B}_{j} \) for some \( j = 2,4,\ldots, p - 3 \) . | The odd characters corresponding to \( \mathbb{Q}\left( {\zeta }_{p}\right) \) are \( \omega ,{\omega }^{3},\ldots ,{\omega }^{p - 2} \) . Therefore, by Theorem 4.17\n\n\[ \n{h}_{p}^{ - } = {2p}\mathop{\prod }\limits_{\substack{{j = 1} \\ {j\text{ odd }} }}^{{p - 2}}\left( {-\frac{1}{2}{B}_{1,{\omega }^{j}}}\right)\n\]... | Yes |
Theorem 5.17. There are infinitely many irregular primes. | Proof. Suppose \( {p}_{1},\ldots ,{p}_{r} \) are all the irregular primes and let \( m = N\left( {{p}_{1} - 1}\right) \) \( \cdots \left( {{p}_{r} - 1}\right) \), where \( N \) will be chosen later. It follows from Exercise 4.3 that \( \left| {{B}_{n}/n}\right| \rightarrow \infty \) as \( n \rightarrow \infty \) , \( n... | Yes |
Theorem 5.18. Let \( \chi \) be an even nontrivial Dirichlet character of conductor \( f \) , let \( \bar{\chi } = {\chi }^{-1} \), let \( \zeta \) be a primitive \( f \) th root of unity, and let \( \tau \left( \chi \right) = \mathop{\sum }\limits_{{a = 1}}^{f}\chi \left( a\right) {\zeta }^{a} \) be a Gauss sum. Then ... | Proof. (The proof is not especially enlightening. The reader could possibly omit it without seriously impairing the understanding of subsequent results.) We shall consider the cases \( f = p \) and \( f \neq p \) separately. I. \( f = p \) (the argument in this case is essentially due to Kummer). Then \( \chi = {\omega... | No |
Lemma 5.19.\n\n\[ \mathop{\sum }\limits_{{j = 0}}^{i}\left( \begin{array}{l} i \\ j \end{array}\right) {\left( -1\right) }^{j}{j}^{m} = 0\;\text{ for }i > m \] | Proof. The left-hand side is the coefficient of \( {t}^{m}/m \) ! in the Taylor expansion of \( {\left( 1 - {e}^{t}\right) }^{i} = {t}^{i} + \) higher terms. The result follows immediately. | No |
Lemma 5.20. Let \( \chi \neq 1 \) be a Dirichlet character of conductor \( f \) and let \( \zeta \) be a primitive \( f \) th root of unity. Then for \( n \geq 1 \)\n\n\[ \n\frac{{B}_{n,\chi }}{n} = - \frac{\tau \left( \chi \right) }{f}\mathop{\sum }\limits_{{a = 1}}^{{f - 1}}\mathop{\sum }\limits_{{i = 1}}^{n}\frac{\b... | Proof. Let\n\n\[ \n{f}_{a}\left( t\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }\mathop{\sum }\limits_{{i = 1}}^{\infty }\frac{1}{i{\left( {\zeta }^{a} - 1\right) }^{i}}\mathop{\sum }\limits_{{j = 0}}^{i}\left( \begin{array}{l} i \\ j \end{array}\right) {\left( -1\right) }^{i - j}{j}^{n}\frac{{t}^{n}}{n!} \n\]\n\n... | Yes |
Lemma 5.21.\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{i}\left( \begin{array}{l} i \\ j \end{array}\right) {\left( -1\right) }^{i - j}{j}^{m}\;\text{ and }\;\mathop{\sum }\limits_{\substack{{j = 1} \\ {p/j} }}^{i}\left( \begin{array}{l} i \\ j \end{array}\right) {\left( -1\right) }^{i - j} \] are both divisible by i! for \... | Proof. Write the monomial \( {X}^{m} \) as\n\n\[ {X}^{m} = \mathop{\sum }\limits_{{i = 0}}^{\infty }{a}_{i}\left( \begin{matrix} X \\ i \end{matrix}\right) \]\n\nThen the \( {a}_{i} \) are uniquely determined and\n\n\[ {a}_{i} = \mathop{\sum }\limits_{{j = 1}}^{i}\left( \begin{array}{l} i \\ j \end{array}\right) {\left... | Yes |
Proposition 5.22. Suppose \( \chi \neq 1, f \neq p,\zeta = {\zeta }_{f} \) . Then for \( s \in {\mathbb{Z}}_{p} \) we have\n\n\[ \n{L}_{p}\left( {s,\chi }\right) = - \frac{\tau \left( \chi \right) }{f}\mathop{\sum }\limits_{{a = 1}}^{{f - 1}}\bar{\chi }\left( a\right) \mathop{\sum }\limits_{{i = 1}}^{\infty }\frac{1}{i... | Proof. \( {L}_{p}\left( {s,\chi }\right) = \lim {L}_{p}\left( {1 - n,\chi }\right) \), where \( n = \left( {p - 1}\right) m \rightarrow \infty \), and \( m \rightarrow \left( {1 - s}\right) / \) \( \left( {p - 1}\right) p \) -adically. So\n\n\[ \n{L}_{p}\left( {s,\chi }\right) = \lim - \left( {1 - \chi \left( p\right) ... | No |
Proposition 5.23. If \( p \) is a regular prime and \( k \) is an even integer with \( k ≢ 0 \) \( \left( {{\;\operatorname{mod}\;p} - 1}\right) \), then \( {L}_{p}\left( {1,{\omega }^{k}}\right) ≢ 0\left( {\;\operatorname{mod}\;p}\right) \) . In particular, \( {L}_{p}\left( {1,{\omega }^{k}}\right) \neq 0 \) . | Proof. We know from Corollary 5.13 that \( {L}_{p}\left( {1,{\omega }^{k}}\right) \equiv {L}_{p}\left( {1 - k,{\omega }^{k}}\right) = \) \( - \left( {1 - {p}^{k - 1}}\right) \left( {{B}_{k}/k}\right) ≢ 0\left( {\;\operatorname{mod}\;p}\right) \), since \( p \nmid {B}_{k} \) . | Yes |
Theorem 5.25. If \( K/\mathbb{Q} \) is abelian then \( {R}_{p}\left( K\right) \neq 0 \) . | Proof. We shall need several preparatory results. | No |
Lemma 5.26. Let \( G \) be a finite abelian group and let \( f \) be a function on \( G \) with values in some field of characteristic 0 . Then\n\n(a)\n\n\[ \det {\left( f\left( \sigma {\tau }^{-1}\right) \right) }_{\sigma ,\tau \in G} = \mathop{\prod }\limits_{{\chi \in \widehat{G}}}\mathop{\sum }\limits_{{\sigma \in ... | Proof. (a) We may assume \( f \) takes values in an algebraically closed field \( F \) . Let \( V \) be the finite-dimensional vector space of all \( F \) -valued functions \( h\left( X\right) \) on \( G \) . Then \( G \) acts on \( V \) by translation: \( {\sigma h}\left( X\right) = h\left( {\sigma X}\right) \) . Defi... | Yes |
Lemma 5.27. Let \( K/\mathbb{Q} \) be a finite Galois extension. If \( K \) is real then let \( {\sigma }_{1},\ldots ,{\sigma }_{r + 1} \) be the elements of \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) . If \( K \) is complex then let \( {\sigma }_{1},\ldots ,{\sigma }_{r + 1} \) , \( {\bar{\sigma }}_{1},\ldot... | Proof. We shall find a unit \( \varepsilon \) such that \( \left| {\varepsilon }^{{\sigma }_{1}}\right| > 1 \) but \( \left| {\varepsilon }^{{\sigma }_{i}}\right| < 1 \) for \( i \neq 1 \) (the absolute value is the complex absolute value corresponding to a fixed embedding of \( K \) into \( \mathbb{C} \) ). The existe... | Yes |
Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be algebraic over \( \mathbb{Q} \) and suppose \( {\log }_{p}{\alpha }_{1},\ldots ,{\log }_{p}{\alpha }_{n} \) are linearly independent over \( \mathbb{Q} \) . Then they are linearly independent over \( \overline{\mathbb{Q}} = \) the algebraic closure of \( \mathbb{Q} \) in... | for a proof, see Brumer [1] | No |
Corollary 5.30. Let \( \chi \neq 1 \) be an even Dirichlet character. Then \( {L}_{p}\left( {1,\chi }\right) \neq 0 \) . | By Theorem 5.18, we know that \( {L}_{p}\left( {1,\chi }\right) \) is essentially a linear form in logarithms. So why did we not apply Theorem 5.29 directly? The problem is that the logarithms in question are generally not independent over \( \mathbb{Q} \) . In certain cases we know all relations. For example, the only... | No |
Theorem 5.31. Let \( K \) be totally real. Then \( {R}_{p}\left( K\right) \neq 0 \Leftrightarrow \) the \( {\mathbb{Z}}_{p} \) -rank of \( {\bar{E}}_{1} \) is \( {r}_{1} - 1 \) . | Proof of Theorem 5.31. Suppose the \( {\mathbb{Z}}_{p} \) -rank of \( {\bar{E}}_{1} \) is less than \( r = {r}_{1} - 1 \) . Let \( {\varepsilon }_{1},\ldots ,{\varepsilon }_{r} \) be a \( \mathbb{Z} \) -basis for \( {E}_{1} \) modulo roots of unity. Then \( {\varepsilon }_{1},\ldots ,{\varepsilon }_{r} \) must be \( {\... | Yes |
Corollary 5.32. If \( K/\mathbb{Q} \) is abelian then the \( {\mathbb{Z}}_{p} \) -rank of \( {\bar{E}}_{1} \) is \( {r}_{1} + {r}_{2} - 1 \) . | Proof. If \( K \) is real, use Theorems 5.25 and 5.31. If \( K \) is complex, then the corollary is true for \( {K}^{ + } \) . Since \( {r}_{1} + {r}_{2} - 1 \) is the same for both fields, the result follows easily. | No |
Theorem 5.34. If \( p \mid {h}^{ + }\left( {\mathbb{Q}\left( {\zeta }_{p}\right) }\right) \) then \( p \mid {h}^{ - }\left( {\mathbb{Q}\left( {\zeta }_{p}\right) }\right) \) . Therefore \( p \mid h\left( {\mathbb{Q}\left( {\zeta }_{p}\right) }\right) \Leftrightarrow p \) divides \( {B}_{j} \) for some \( j = 2,4,\ldots... | Proof of Theorem 5.34. The characters corresponding to \( \mathbb{Q}{\left( {\zeta }_{p}\right) }^{ + } \) are \( 1,{\omega }^{2} \) , \( \ldots ,{\omega }^{p - 3} \) . Let \( n = \frac{1}{2}\left( {p - 1}\right) \) . Then\n\n\[ \frac{{2}^{n - 1}{h}^{ + }{R}_{p}^{ + }}{\sqrt{{d}^{ + }}} = \mathop{\prod }\limits_{\subst... | Yes |
Lemma 5.35. Let \( K/\mathbb{Q} \) be an extension of degree \( n \), with \( n \leq p - 1 \) . Assume that \( p \) is totally ramified: \( \left( p\right) = {p\not{} }^{n} \) . Suppose \( \varepsilon \) is a unit of \( K \) which is congruent to a rational integer modulo \( {k}^{c}\left( {c > 0}\right) \) . Then \( {\... | Proof. Let \( \pi \) generate \( {p\not{} } \) in \( {\mathcal{O}}_{p\not{} } \), so \( \varepsilon = a + b{\pi }^{c} + \cdots = a\left( {1 + \left( {b/a}\right) {\pi }^{c} + \cdots }\right) \) with \( a, b,\ldots \in \mathbb{Z} \) . Since\n\n\[ \left| {\pi }^{c}\right| = {p}^{-c/n} \leq {p}^{-1/\left( {p - 1}\right) }... | Yes |
Theorem 5.37 (Ankeny-Artin-Chowla). Let \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) and let \( h \) and \( \varepsilon = \) \( \left( {t + u\sqrt{p}}\right) /2 > 1 \) be the class number and fundamental unit for \( \mathbb{Q}\left( \sqrt{p}\right) \) . Then \[ \frac{u}{t}h \equiv {B}_{\left( {p - 1}\right) ... | Proof. Until now we have been able to ignore the ambiguity in sign for the \( p \) -adic regulator. But now we are forced to choose signs. From the classical class number formula for \( \mathbb{Q}\left( \sqrt{p}\right) \), we have (Exercise 4.6) \[ {\varepsilon }^{-{2h}} = \mathop{\prod }\limits_{{a = 1}}^{{p - 1}}{\le... | Yes |
Proposition 5.38. Let \( m \geq 1 \) be squarefree and assume 3 does not split completely in \( \mathbb{Q}\left( \sqrt{-m}\right) \) . If 3 divides the class number of \( \mathbb{Q}\left( \sqrt{3m}\right) \) then 3 divides the class number of \( \mathbb{Q}\left( \sqrt{-m}\right) \) (we allow \( 3 \mid m \), in which ca... | Proof of Proposition 5.38. We may assume \( m > 3 \) since the proposition is vacuously true for \( m \leq 3 \) . Let \( \chi \) be the character for \( \mathbb{Q}\left( \sqrt{-m}\right) \) . Then \( {\chi \omega } = \) \( \chi {\omega }_{3} \) is the character for \( \mathbb{Q}\left( \sqrt{3m}\right) \) . Let \( \vare... | Yes |
Lemma 6.1. (a) \( g\left( \bar{\chi }\right) = \chi \left( {-1}\right) \overline{g\left( \chi \right) } \); (b) if \( \chi \neq 1, g\left( \chi \right) g\left( \bar{\chi }\right) = \chi \left( {-1}\right) q \); (c) if \( \chi \neq 1, g\left( \chi \right) \overline{g\left( \chi \right) } = q \). | Proof. (a) is straightforward. (b) follows from (a) and (c). For (c), \[ g\left( \chi \right) \overline{g\left( \chi \right) } = \mathop{\sum }\limits_{{a, b \neq 0}}\chi \left( {a{b}^{-1}}\right) \psi \left( {a - b}\right) \] \[ = \mathop{\sum }\limits_{{b, c \neq 0}}\chi \left( c\right) \psi \left( {{bc} - b}\right) ... | Yes |
Lemma 6.2. (a) \( J\left( {1,1}\right) = 2 - q \) ; (b) \( J\left( {1,\chi }\right) = J\left( {\chi ,1}\right) = 1 \) if \( \chi \neq 1 \) ; (c) \( J\left( {\chi ,\bar{\chi }}\right) = \chi \left( {-1}\right) \) if \( \chi \neq 1 \) ; (d) \( J\left( {{\chi }_{1},{\chi }_{2}}\right) = g\left( {\chi }_{1}\right) g\left( ... | Proof. (a) and (b) are easy. To prove (c) and (d), we compute \[ g\left( {\chi }_{1}\right) g\left( {\chi }_{2}\right) = \mathop{\sum }\limits_{{a, b}}{\chi }_{1}\left( a\right) {\chi }_{2}\left( b\right) \psi \left( {a + b}\right) \] \[ = \mathop{\sum }\limits_{{a, b}}{\chi }_{1}\left( a\right) {\chi }_{2}\left( {b - ... | No |
Corollary 6.3. If \( {\chi }_{1},{\chi }_{2} \) are characters of orders dividing \( m \), then\n\n\[ \frac{g\left( {\chi }_{1}\right) g\left( {\chi }_{2}\right) }{g\left( {{\chi }_{1}{\chi }_{2}}\right) } \]\n\nis an algebraic integer in \( \mathbb{Q}\left( {\zeta }_{m}\right) \) . | Proof. If \( {\chi }_{1}{\chi }_{2} \) is nontrivial, use the above result. The remaining cases are quickly checked individually. | No |
Lemma 6.4. Assume \( {\chi }^{m} \) is trivial. Then\n\n\[ \frac{g{\left( \chi \right) }^{b}}{g{\left( \chi \right) }^{{\sigma }_{b}}} = g{\left( \chi \right) }^{b - {\sigma }_{b}} \in \mathbb{Q}\left( {\zeta }_{m}\right) \]\n\nand \( g{\left( \chi \right) }^{m} \in \mathbb{Q}\left( {\zeta }_{m}\right) \) . | Proof. The second follows from the first if we let \( b = 1 + m \) . For the first, we have\n\n\[ g{\left( \chi \right) }^{{\sigma }_{b}} = - \sum \chi {\left( a\right) }^{b}\psi \left( a\right) = g\left( {\chi }^{b}\right) . \]\n\nLet \( \tau \in \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{mp}\right) /\mathbb... | Yes |
Lemma 6.5. \( g\left( {\chi }^{p}\right) = g\left( \chi \right) \) . | Proof. Since \( a \mapsto {a}^{p} \) is an automorphism over \( \mathbb{Z}/p\mathbb{Z}, T\left( a\right) = T\left( {a}^{p}\right) \), and \( {a}^{p} \) yields a permutation of \( \mathbb{F} \) . Therefore\n\n\[ g\left( {\chi }^{p}\right) = - \sum \chi \left( {a}^{p}\right) {\zeta }_{p}^{T\left( a\right) } \]\n\n\[ = - ... | Yes |
Corollary 6.7. For any given \( d,{X}^{d} + {Y}^{d} \equiv 1\\left( {\\operatorname{mod}p}\\right) \) has solutions with \( {XY} ≢ 0\\left( {\\operatorname{mod}p}\\right) \), for all sufficiently large \( p \) . | Proof. From the above, the number of points at infinity is \( {N}_{d}\\left( {-1}\\right) \) (or \( {N}_{e}\\left( {-1}\\right) \) ), which is at most \( d \) . The number of solutions with \( X \equiv 0 \) or \( Y \equiv 0 \) is at most \( {2d} \) . Therefore we have a nontrivial solution as soon as \( \\bar{N} > {3d}... | Yes |
Lemma 6.9. Suppose \( M = \mathbb{Q}\left( {\zeta }_{m}\right) \) . Let \( {I}^{\prime } \) be the ideal of \( \mathbb{Z}\left\lbrack G\right\rbrack \) generated by elements of the form \( c - {\sigma }_{c} \), with \( \left( {c, m}\right) = 1 \) . Let \( \beta \in \mathbb{Z}\left\lbrack G\right\rbrack \) . Then\n\n\[ ... | Proof. Since\n\n\[ \n\left( {c - {\sigma }_{c}}\right) \theta = \mathop{\sum }\limits_{a}\left( {c\left\{ \frac{a}{m}\right\} - \left\{ \frac{ac}{m}\right\} }\right) {\sigma }_{a}^{-1} \in \mathbb{Z}\left\lbrack G\right\rbrack , \n\]\n\nwe have \ | Yes |
Lemma 6.11. (a) \( s\left( 0\right) = 0 \) ;\n\n(b) \( 0 \leq s\left( {\alpha + \beta }\right) \leq s\left( \alpha \right) + s\left( \beta \right) \) ;\n\n(c) \( s\left( {\alpha + \beta }\right) \equiv s\left( \alpha \right) + s\left( \beta \right) \left( {{\;\operatorname{mod}\;p} - 1}\right) \) ;\n\n(d) \( s\left( {p... | Proof. (a) is obvious; (b) and (d) follow from Corollary 6.3 and Lemma 6.5, respectively. Since \( {\widetilde{\mathcal{P}}}^{p - 1} = {p\not{} } \), the values of \( {v}_{\widetilde{\mathcal{P}}} \) on \( \mathbb{Q}\left( {\zeta }_{q - 1}\right) \) are divisible by \( p - 1 \) . Therefore (c) also follows from Corolla... | Yes |
Lemma 6.12. \( s\left( \alpha \right) > 0 \) if \( \alpha ≢ 0\left( {{\;\operatorname{mod}\;q} - 1}\right) \), and \( s\left( 1\right) = 1 \) . | Proof. Since \( \pi = {\zeta }_{p} - 1 \in \widetilde{\mathcal{P}} \) ,\n\n\[ g\left( {\omega }^{-\alpha }\right) = - \sum {\omega }^{-\alpha }\left( a\right) {\zeta }_{p}^{T\left( a\right) } \equiv - \sum {\omega }^{-\alpha }\left( a\right) \equiv 0\left( {\;\operatorname{mod}\;\mathcal{P}}\right) . \]\n\nTherefore \(... | Yes |
Proposition 6.13. Let \( 0 \leq \alpha < q - 1 \) and let \( \alpha = {a}_{0} + {a}_{1}p + \cdots + {a}_{f - 1}{p}^{f - 1} \) , \( 0 \leq {a}_{i} \leq p - 1 \), be the standard p-adic expansion of \( \alpha \) . Then\n\n\[ s\left( \alpha \right) = {a}_{0} + {a}_{1} + \cdots + {a}_{f - 1}. \] | Proof. From Lemma 6.11(a), (b), (c) and Lemma 6.12 we immediately have \( s\left( \alpha \right) = \alpha \) for \( 0 \leq \alpha \leq p - 2 \) . If \( q = p \), we are done. Otherwise, \( s\left( {p - 1}\right) > 0 \) and we similarly obtain \( s\left( {p - 1}\right) = p - 1 \) . Lemma 6.11(b) and (d) imply that \( s\... | Yes |
Lemma 6.14. Let \( 0 \leq h < q - 1 \) . Then\n\n\[ s\left( h\right) = \left( {p - 1}\right) \mathop{\sum }\limits_{{i = 0}}^{{f - 1}}\left\{ \frac{{p}^{i}h}{q - 1}\right\} .\n\] | Proof. Let \( h = {a}_{0} + {a}_{1}p + \cdots + {a}_{f - 1}{p}^{f - 1} \) . Then\n\n\[ {p}^{i}h \equiv {a}_{0}{p}^{i} + {a}_{1}{p}^{i + 1} + \cdots + {a}_{f - 1}{p}^{i - 1}\left( {{\;\operatorname{mod}\;q} - 1}\right) .\n\]\n\nIt follows that\n\n\[ \left\{ \frac{{p}^{i}h}{q - 1}\right\} = \frac{1}{q - 1}\left( {{a}_{0}... | Yes |
Lemma 6.15. If \( \mathbb{Q}\left( {\zeta }_{m}\right) \subseteq K \subseteq \mathbb{Q}\left( {\zeta }_{n}\right) \) and \( K/\mathbb{Q}\left( {\zeta }_{m}\right) \) is unramified at all primes, then \( K = \mathbb{Q}\left( {\zeta }_{m}\right) \) . | Proof. Suppose \( K \neq \mathbb{Q}\left( {\zeta }_{m}\right) \) . Then there is a character \( \chi \) for \( K \) of conductor not dividing \( m \) . By Theorem \( {3.5}, K/\mathbb{Q}\left( {\zeta }_{m}\right) \) must be ramified at some prime. Contradiction. For another proof, see Lemma 15.48. | Yes |
Proposition 6.16. \( {A}_{0} = {A}_{1} = 0 \) . For \( i = 3,5,\ldots, p - 2,{B}_{1,{\omega }^{-i}} \) annihilates \( {A}_{i} \) . | Suppose \( {A}_{i} \neq 0 \) . Then we must have \( {B}_{1,{\omega }^{-i}} \equiv 0\left( {\;\operatorname{mod}\;p}\right) \) . But \( {B}_{1,{\omega }^{-i}} \equiv \) \( {B}_{p - i}/\left( {p - i}\right) \left( {\;\operatorname{mod}\;p}\right) \) by Corollary 5.15. We have proved the following. | No |
Theorem 6.18 (Ribet). Let \( i \) be odd, \( 3 \leq i \leq p - 2 \) . If \( p \mid {B}_{p - i} \) then \( {A}_{i} \neq 0 \) . | This will be proved in Chapter 15 by elementary means. Ribet's original proof used delicate techniques from algebraic geometry to construct an abelian unramified extension of degree \( p \) which corresponds by class field theory to \( {A}_{i} \) . | No |
Theorem 6.19 (Iwasawa). \( \left\lbrack {{R}^{ - } : {I}^{ - }}\right\rbrack = {h}^{ - }\left( {\mathbb{Q}\left( {\zeta }_{{p}^{n}}\right) }\right) \) . | Proof of Theorem 6.19. The proof will proceed by considering completions, since we can then work with one prime at a time, which is slightly easier. Let \( q \) be a prime, \( {R}_{q} = {\mathbb{Z}}_{q}\left\lbrack G\right\rbrack ,{I}_{q} = {R}_{q}I \) . Clearly \( I \) is dense in \( {I}_{q} \) in the natural \( q \) ... | No |
Theorem 6.21. \( \left\lbrack {{R}_{q}^{ - } : {I}_{q}^{ - }}\right\rbrack = q \) -part of \( {h}^{ - }\left( {\mathbb{Q}\left( {\zeta }_{{p}^{n}}\right) }\right) \) . | Proof. We first consider \( q \neq 2, p \) . Then \( \left( {1 \pm J}\right) /2 \in {R}_{q} \) ; and we get \( {R}_{q} = {R}_{q}^{ + } \oplus {R}_{q}^{ - } \) and \( {I}_{q} = {I}_{q}^{ + } \oplus {I}_{q}^{ - } \) from the relation \( 1 = \left( {1 + J}\right) /2 + \left( {1 - J}\right) /2 \) . Therefore\n\n\[ \n{I}_{q... | Yes |
Lemma 6.22. (a) \( {I}_{2}^{ - } \subseteq {R}_{2}\widetilde{\theta } \) ;\n\n(b) \( \left\lbrack {{R}_{2}\widetilde{\theta } : {I}_{2}^{ - }}\right\rbrack = 2 \) . | Proof. For the first statement, suppose \( x \in {R}_{2} \) and \( {x\theta } \in {I}_{2}^{ - } = {R}_{2}\theta \cap {R}_{2}^{ - } \) . Then \( {x\theta } = \left\lbrack {\left( {1 - J}\right) /2}\right\rbrack {x\theta } = x\left\lbrack {\left( {1 - J}\right) /2}\right\rbrack \left( {\widetilde{\theta } + \frac{1}{2}N}... | Yes |
Theorem 6.23. Suppose \( p \) is prime and suppose the index of irregularity of \( p \) (= the number of Bernoulli numbers divisible by \( p \) ) satisfies \( i\left( p\right) < \sqrt{p} - 2 \) . Then\n\n\[ \n{X}^{p} + {Y}^{p} = {Z}^{p},\;\left( {{XYZ}, p}\right) = 1, \n\]\n\nhas no integer solutions. | Proof of Theorem 6.23. Let \( \zeta = {\zeta }_{p} \) . As in Chapter 1, we assume we have a solution and obtain\n\n\[ \n\left( {x + {\zeta }^{i}y}\right) = {C}_{i}^{p},\;i = 0,\ldots, p - 1, \n\]\n\nwhere \( {C}_{i} \) is an ideal of \( \mathbb{Q}\left( {\zeta }_{p}\right) \) . Let \( C \) be the subgroup of the ideal... | Yes |
Proposition 7.2. Let \( f, g \in \mathcal{O}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) and assume \( f = {a}_{0} + {a}_{1}T + \cdots \), with \( {a}_{i} \in p \) for \( 0 \leq i \leq n - 1 \), but \( {a}_{n} \in {\mathcal{O}}^{ \times } \) . Then we may uniquely write\n\n\[ g = {qf} + r \]\n\nwhere \( q ... | Proof. We first prove uniqueness, which reduces to considering \( {qf} + r = 0 \) . If \( q, r \neq 0 \), we may assume that either \( \pi /r \) or \( \pi /q \) . Reduction mod \( \pi \) shows that \( \pi \left| {r\text{, so}\pi }\right| {qf} \) . An easy argument shows that since \( \pi \left| {f\text{we must have}\pi... | No |
Theorem 7.3 ( \( p \) -adic Weierstrass Preparation Theorem). Let\n\n\[ f\left( T\right) = \mathop{\sum }\limits_{{i = 0}}^{\infty }{a}_{i}{T}^{i} \in \mathcal{O}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \]\n\nand assume for some \( n \) we have \( {a}_{i} \in h,0 \leq i \leq n - 1 \), but \( {a}_{n} \noti... | Proof. The second part clearly follows from the first part if we factor as large a power of \( \pi \) as possible from the coefficients of \( f\left( T\right) \) .\n\nTo prove the first part, let \( g\left( T\right) = {T}^{n} \) in Proposition 7.2. Then\n\n\[ {T}^{n} = q\left( T\right) f\left( T\right) + r\left( T\righ... | Yes |
Corollary 7.4. Let \( f\left( T\right) \in \mathcal{O}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) be nonzero. Then there are only finitely many \( x \in {\mathbb{C}}_{p},\left| x\right| < 1 \), with \( f\left( x\right) = 0 \) . | Proof. Assume \( f\left( x\right) = 0 \) . Write \( f\left( T\right) = {\pi }^{\mu }P\left( T\right) U\left( T\right) \), as above. Since \( U\left( T\right) \) is invertible, \( U\left( x\right) \neq 0 \) . Therefore \( P\left( x\right) = 0 \) . The result follows. | No |
Lemma 7.5. Suppose \( P\left( T\right) \in \mathcal{O}\left\lbrack T\right\rbrack \) is a distinguished polynomial, and let \( g\left( T\right) \in \mathcal{O}\left\lbrack T\right\rbrack \) be arbitrary. If \( g\left( T\right) /P\left( T\right) \in \mathcal{O}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) th... | Proof. Suppose \( g\left( T\right) = f\left( T\right) P\left( T\right) \) for some \( f\left( T\right) \in \mathcal{O}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) . Let \( x \in {\mathbb{C}}_{p} \) be a zero of \( P\left( T\right) \) . Then\n\n\[ 0 = P\left( x\right) = {x}^{n} + \left( {\text{ multiple of ... | Yes |
Proposition 7.6. (a) \( \frac{1}{2}{\eta }_{n}\left( \theta \right) \in {\mathcal{O}}_{\theta }\left\lbrack {\Gamma }_{n}\right\rbrack \) ; | Proof. (a) is obvious except when \( p = 2 \) . To take care of this case, note that \( {\gamma }_{n}\left( a\right) = {\gamma }_{n}\left( {{q}_{n} - a}\right) \) but \( \theta {\omega }^{-1}\left( {{q}_{n} - a}\right) = - \theta {\omega }^{-1}\left( a\right) \) . The terms for \( a \) and \( {q}_{n} - a \) in \( \frac... | Yes |
Lemma 7.7. \( {\gamma }_{n}\left( a\right) = {\gamma }_{n}\left( b\right) \Leftrightarrow \langle a\rangle \equiv \langle b\rangle {\;\operatorname{mod}\;q}{p}^{n} \) . | Proof. The decomposition \( {\sigma }_{a} = \delta \left( a\right) {\gamma }_{n}\left( a\right) \) corresponds to \( \mathbb{Q}\left( {\zeta }_{{q}_{n}}\right) = \mathbb{Q}\left( {\zeta }_{{q}_{0}}\right) \cdot {\mathbb{B}}_{n} \) , where \( {\mathbb{B}}_{n} \) is the subfield of \( \mathbb{Q}\left( {\zeta }_{q{p}^{n}}... | Yes |
Lemma 7.8. Suppose \( s, t \in \mathbb{Z},\left( {t, d}\right) = 1 \) . Then\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{{d - 1}}\theta {\omega }^{-1}\left( {s + {itq}}\right) = 0 \] | Proof. If \( p \nmid {f}_{\theta {\omega }^{-1}} \), then \( {f}_{\theta {\omega }^{-1}} = d \) . Since \( s + {itq} \) runs through a complete set of residue classes \( {\;\operatorname{mod}\;d} \), the result follows. Now suppose \( p \mid {f}_{\theta {\omega }^{-1}} \), so the conductor is \( {qd} \) . If \( p \mid ... | Yes |
\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{{q}_{n}}\mathop{\sum }\limits_{\substack{{0 < a < {q}_{n}} \\ {\left( {a,{q}_{0}}\right) = 1} }}\chi {\omega }^{-m}\left( a\right) {a}^{m} = \left( {1 - \chi {\omega }^{-m}\left( p\right) {p}^{m - 1}}\right) {B}_{m,\chi {\omega }^{-m}}. \] | Proof. From Proposition 4.1 (recall \( {B}_{m}\left( X\right) = \sum \left( \begin{matrix} m \\ i \end{matrix}\right) {B}_{i}{X}^{m - i} \) ),\n\n\[ {B}_{m,\chi {\omega }^{-m}} = \frac{1}{{q}_{n}}\mathop{\sum }\limits_{{j = 1}}^{{q}_{n}}\chi {\omega }^{-m}\left( j\right) {q}_{n}^{m}{B}_{m}\left( \frac{j}{{q}_{n}}\right... | Yes |
Lemma 7.12. If \( \theta = 1 \), then \( \frac{1}{2}g\left( {T,\theta }\right) \) is a unit of \( {\mathbb{Z}}_{p}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) . | Proof. By Theorem 7.10,\n\n\[ f\left( {0,1}\right) = - {B}_{1,{\omega }^{-1}} = - \frac{1}{q}\mathop{\sum }\limits_{\substack{{a = 1} \\ {n \nmid a} }}^{q}{\omega }^{-1}\left( a\right) a \equiv \frac{1}{p}{\;\operatorname{mod}\;{\mathbb{Z}}_{p}}, \]\n\nsince \( \omega \left( a\right) \equiv a\left( {\;\operatorname{mod... | Yes |
Theorem 7.14. Let \( {p}^{{e}_{n}^{ - }} \) be the exact power of \( p \) dividing \( {h}_{n}^{ - } \), in the notation of the previous theorem. There exist integers \( \lambda ,\mu \), and \( v \), independent of \( n \), with \( \lambda \geq 0,\mu \geq 0 \), such that\n\n\[ \n{e}_{n}^{ - } = {\lambda n} + \mu {p}^{n}... | Proof. In the notation of the previous theorem, let\n\n\[ \nA\left( T\right) = \mathop{\prod }\limits_{{\theta \neq 1}}\frac{1}{2}f\left( {T,\theta }\right) \in {\mathbb{Z}}_{p}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \n\]\n\nThen\n\n\[ \n\frac{{h}_{n}^{ - }}{{h}_{0}^{ - }} = \mathop{\prod }\limits_{\subs... | Yes |
Theorem 7.15. Let \( K \) be an abelian extension of \( \mathbb{Q} \), let \( p \) be any prime, and let \( {K}_{\infty }/K \) be the cyclotomic \( {\mathbb{Z}}_{p} \) -extension of \( K \). Then \( \mu = 0 \). | Proof of Theorem 7.15. Before starting the main part of the proof, we state some facts, some of which will be proved in later chapters but which are needed now.\n\nI. \( K \) is contained in a cyclotomic field (Kronecker-Weber theorem).\n\nII. If \( K \subseteq {K}^{\prime } \) then \( \mu \leq {\mu }^{\prime } \).\n\n... | No |
Proposition 7.16. Let \( R \) be the set of \( \left( {p - 1}\right) \) st roots of unity in \( {\mathbb{Z}}_{p}(R = \{ \pm 1\} \) if \( p = 2 \) ) and let \( {R}^{\prime } \) be a set of representatives for \( R \) modulo \( \{ \pm 1\} \) . (a) Suppose \( \theta = {\omega }^{k}, k ≢ 0\left( {{\;\operatorname{mod}\;p} ... | Proof. (a) Since \( \theta \neq 1 \) is even, we must have \( p \geq 5 \), so we have the luxury of ignoring 2 and letting \( p = q \) . In the notation of Proposition 7.9, we see that if \( \frac{1}{2}{\xi }_{n}\left( \theta \right) \) is expressed as a polynomial in \( 1 + T \), modulo \( {\left( 1 + T\right) }^{{p}^... | Yes |
Lemma 7.19. Suppose \( {\gamma }_{1},\ldots ,{\gamma }_{r} \in {\mathbb{Z}}_{p} \) are linearly independent over \( \mathbb{Q} \) . For almost all \( \beta \in {\mathbb{Z}}_{p} \) the sequence of vectors\n\n\[ \n{X}_{n} = {X}_{n}\left( \beta \right) = \left( {{x}_{n}\left( {\beta {\gamma }_{1}}\right) ,\ldots ,{x}_{n}\... | Proof. Let \( z = \left( {{z}_{1},\ldots ,{z}_{r}}\right) \in {\mathbb{Z}}^{r}, z \neq 0 \), let \( \beta \in {\mathbb{Z}}_{p} \), and let \( {y}_{n} = {X}_{n} \cdot z = \) \( {q}^{-1}{p}^{-n}\mathop{\sum }\limits_{i}{z}_{i}{s}_{n}\left( {\beta {\gamma }_{i}}\right) \) . Since\n\n\[ \n\mathop{\sum }\limits_{i}{z}_{i}{s... | Yes |
Lemma 8.1. Let \( p \) be prime and \( m \geq 1 \). (a) The cyclotomic units of \( \mathbb{Q}{\left( {\zeta }_{{p}^{m}}\right) }^{ + } \) are generated by -1 and the units \[ {\xi }_{a} = {\zeta }_{{p}^{m}}^{\left( {1 - a}\right) /2}\frac{1 - {\zeta }_{{p}^{m}}^{a}}{1 - {\zeta }_{{p}^{m}}},\;1 < a < \frac{1}{2}{p}^{m},... | Proof. Let \( \zeta = {\zeta }_{{p}^{m}} \) . The definition of the cyclotomic units involves \( 1 - {\zeta }^{a} \) for all \( a ≢ 0{\;\operatorname{mod}\;{p}^{m}} \) . If \( k < m \) and \( \left( {b, p}\right) = 1 \), then, using the relation \( 1 - {X}^{pk} = \) \( \prod \left( {1 - {\zeta }^{j{p}^{m - k}}X}\right)... | Yes |
Theorem 8.3. Let \( n ≢ 2{\;\operatorname{mod}\;4} \), and let \( n = \mathop{\prod }\limits_{{i = 1}}^{s}{p}_{i}^{{e}_{i}} \) be its prime factorization. Let \( I \) run through all subsets of \( \{ 1,\ldots, s\} \), except \( \{ 1,\ldots, s\} \), and let \( {n}_{I} = \) \( \mathop{\prod }\limits_{{i \in I}}{p}_{i}^{{... | Proof of Theorem 8.3. The proof will be similar in many ways to that of Theorem 8.2, but will be more technical. As in the proof of that theorem, we have\n\n\[ \nR\left( \left\{ {\xi }_{a}\right\} \right) = \pm \mathop{\prod }\limits_{{\chi \neq 1}}\frac{1}{2}\mathop{\sum }\limits_{\substack{{a = 1} \\ {\left( {a, n}\r... | Yes |
Lemma 8.4. If \( {f}_{\chi }/\left( {n/m}\right) \) then\n\n\[ \mathop{\sum }\limits_{\substack{{a = 1} \\ {\left( {a, n}\right) = 1} }}^{n}\chi \left( a\right) \log \left| {1 - {\zeta }_{n}^{am}}\right| = 0. \] | Proof. We claim that there exists \( b \equiv 1{\;\operatorname{mod}\;\left( {n/m}\right) } \) such that \( \left( {b, n}\right) = 1 \) and \( \chi \left( b\right) \neq 1 \) . If not, then \( \chi : {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \rightarrow {\mathbb{C}}^{ \times } \) may be factored through \( {\le... | Yes |
Lemma 8.5. Let \( n = m{m}^{\prime } \) with \( \left( {m,{m}^{\prime }}\right) = 1 \), and suppose \( {f}_{\chi } \mid m \) . Then\n\n\[ \mathop{\sum }\limits_{\substack{{a = 1} \\ {\left( {a, n}\right) = 1} }}^{n}\chi \left( a\right) \log \left| {1 - {\zeta }_{n}^{a{m}^{\prime }}}\right| = \phi \left( {m}^{\prime }\r... | Proof. Write \( a = b + {cm} \) with \( 1 \leq b < m,0 \leq c < {m}^{\prime } \) . If \( \left( {a, n}\right) = 1 \) then \( \left( {b, m}\right) = 1 \) . Conversely, for each \( b \) with \( \left( {b, m}\right) = 1 \) there are \( \phi \left( {m}^{\prime }\right) \) choices of \( c \) such that \( \left( {b + {cm},{m... | Yes |
Lemma 8.6. Suppose \( F, g, t \) are positive integers with \( {f}_{x}\left| {F\text{and}g}\right| F \) . Then\n\n\[ \mathop{\sum }\limits_{\substack{{a = 1} \\ {\left( {a, g}\right) = 1} }}^{{Ft}}\chi \left( a\right) \log \left| {1 - {\zeta }_{Ft}^{a}}\right| = \mathop{\sum }\limits_{\substack{{b = 1} \\ {\left( {b, g... | Proof. Write \( a = b + {cF},1 \leq b \leq F,0 \leq c < t \) . Then \( \left( {a, g}\right) = 1 \Leftrightarrow \left( {b, g}\right) = 1 \) . Since\n\n\[ \mathop{\prod }\limits_{{c = 0}}^{{t - 1}}\left( {1 - {\zeta }_{Ft}^{b + {cF}}}\right) = 1 - {\zeta }_{F}^{b} \]\n\nand since \( \chi \left( a\right) \) depends only ... | Yes |
Lemma 8.7. Assume \( {f}_{\chi } \mid m \) . Then\n\n\[ \mathop{\sum }\limits_{\substack{{b = 1} \\ {\left( {b, m}\right) = 1} }}^{m}\chi \left( b\right) \log \left| {1 - {\zeta }_{m}^{b}}\right| = \left\lbrack {\mathop{\prod }\limits_{{p \mid m}}\left( {1 - \chi \left( p\right) }\right) }\right\rbrack \mathop{\sum }\l... | Proof. Let \( p, q,\ldots \) represent the primes dividing \( m \) . We only need to consider those which do not divide \( {f}_{x} \) . The right-hand side equals\n\n\[ \mathop{\sum }\limits_{{b = 1}}^{m}\chi \left( b\right) \log \left| {1 - {\zeta }_{m}^{b}}\right| - \mathop{\sum }\limits_{p}\chi \left( p\right) \math... | Yes |
Corollary 8.8. Let \( {C}_{n}^{\prime \prime } \) be the group generated by -1 and the units of the form\n\n\[ \n{\zeta }_{n}^{\left( {1 - a}\right) /2}\frac{1 - {\zeta }_{n}^{a}}{1 - {\zeta }_{n}},\;1 < a < \frac{1}{2}n,\left( {a, n}\right) = 1.\n\]\n\nThen\n\n\[ \n\left\lbrack {{E}_{n}^{ + } : {C}_{n}^{\prime \prime ... | Proof. The regulator of \( {C}_{n}^{\prime \prime } \) is\n\n\[ \n\pm \mathop{\prod }\limits_{{\chi \neq 1}}\frac{1}{2}\mathop{\sum }\limits_{\substack{{a = 1} \\ {\left( {a, n}\right) = 1} }}^{n}\chi \left( a\right) \log \left| {1 - {\zeta }_{n}^{a}}\right|\n\]\n\nBy the above calculations (plus Lemmas 8.7 and 8.6), w... | Yes |
Theorem 8.9. Let \( n ≢ 2{\;\operatorname{mod}\;4} \) and let \( {\left( {A}_{n}^{0}\right) }^{ + } \) be the additive abelian group with generators\n\n\[ \n\left\{ {\left. {g\left( \frac{a}{n}\right) }\right| \;\begin{array}{l} a \\ n \end{array} \in \frac{1}{n}\mathbb{Z}/\mathbb{Z},\frac{a}{n} \neq 0}\right\} \n\]\n\... | Proof. The proof uses distributions, hence will be postponed until Chapter 12. | No |
Proposition 8.11. Let \( g \) be a primitive root \( {\;\operatorname{mod}\;{p}^{n}} \) . Then\n\n\[ \n{\zeta }_{{p}^{n}}^{\left( {1 - g}\right) /2}\frac{1 - {\zeta }_{{p}^{n}}^{g}}{1 - {\zeta }_{{p}^{n}}}\n\]\n\ngenerates \( {C}_{{p}^{n}}^{ + }/\{ \pm 1\} \) as a module over \( \mathbb{Z}\left\lbrack {\operatorname{Ga... | Proof. During this proof, let \( \zeta = {\zeta }_{{p}^{n}} \) . Let \( \left( {a, p}\right) = 1 \) . Then \( a \equiv {g}^{r}{\;\operatorname{mod}\;{p}^{n}} \) for some \( r > 0 \), so\n\n\[ \n{\zeta }^{\left( {1 - a}\right) /2}\frac{1 - {\zeta }^{a}}{1 - \zeta } = {\zeta }^{\left( {1 - {g}^{r}}\right) /2}\frac{1 - {\... | Yes |
Proposition 8.13. Let \( N \geq 1 \) and let \( i \) be even, \( 2 \leq i \leq p - 3 \) . Then \[ {\varepsilon }_{i}{E}_{{p}^{N}}^{ + } \simeq \mathbb{Z}/{p}^{N}\mathbb{Z} \] | Proof. Since \( {L}_{p}\left( {1,{\omega }^{i}}\right) \neq 0,{E}_{i}^{\left( N\right) } \neq \pm 1 \), hence \( {\varepsilon }_{i}{E}_{{p}^{N}}^{ + } \neq 0 \), for large \( N \) . Since \( {E}_{{p}^{N}}^{ + } \simeq {\left( \mathbb{Z}/{p}^{N}\mathbb{Z}\right) }^{\left( {p - 3}\right) /2} \), and \( {\varepsilon }_{i}... | Yes |
Theorem 8.16. \( {E}_{i} \) is a pth power \( \Rightarrow p \mid {B}_{i} \) . | Proof. We may replace \( {E}_{i} \) with \( {E}_{i}^{\left( N\right) } \) for \( N \) sufficiently large. If \( {E}_{i}^{\left( N\right) } = {\eta }^{p} \) then \( {\log }_{p}{E}_{i}^{\left( N\right) } = p{\log }_{p}\eta \) . Since \( {\log }_{p}\eta \in {\mathbb{Z}}_{p}\left\lbrack {\zeta }_{p}\right\rbrack \) (cf. Ex... | Yes |
Corollary 8.17. \( p \mid {h}^{ + }\left( {\mathbb{Q}\left( {\zeta }_{p}\right) }\right) \Rightarrow p \mid {h}^{ - }\left( {\mathbb{Q}\left( {\zeta }_{p}\right) }\right) \) (this is the same as Theorem 5.34). | Proof. Theorems 8.14, 8.16, and 5.16. | No |
Proposition 8.18. Let \( i \) be even, \( 2 \leq i \leq p - 3 \). Let \( l \) be a prime with \( l \equiv 1 \) \( {\;\operatorname{mod}\;p}, \) say \( l = {kp} + 1 \), and let \( t \) be an integer satisfying \( \left( {t, l}\right) = 1 \) and \( {t}^{k} ≢ 1 \) \( {\;\operatorname{mod}\;l} \). Define\n\n\[ d = {d}_{i} ... | Proof. Let \( {R}_{i} = \mathop{\prod }\limits_{{a = 1}}^{{p - 1}}{\left( {\zeta }^{a/2} - {\zeta }^{-a/2}\right) }^{{a}^{p - 1 - i}} \). Recall that \( g \) is a primitive root \( {\;\operatorname{mod}\;p} \). Changing \( a \) to \( {ag} \), we find that\n\n\[ {R}_{i} = \mathop{\prod }\limits_{{a = 1}}^{{p - 1}}{\left... | Yes |
Corollary 8.19. Let \( p \) be an irregular prime and let \( {i}_{1},\ldots ,{i}_{s} \) be the even indices \( 2 \leq i \leq p - 3 \) such that \( p \mid {B}_{i} \) . Suppose there exists a prime \( l \equiv 1{\;\operatorname{mod}\;p} \) and an integer \( t \), as in Proposition 8.18, such that \( {Q}_{i}^{k} ≢ 1{\;\op... | Proof. Theorems 8.14 and 8.16, and Proposition 8.18. | No |
Proposition 8.20. Let \( i \) be even, \( 2 \leq i \leq p - 3 \) . If \( N > {2c}/\left( {p - 1}\right) \) and\n\n\[ \n\eta = a + b{\lambda }^{c} + \cdots \in {\varepsilon }_{i}{E}_{{p}^{N}}^{ + }\n\]\n\nthen\n\n\[ \nc \equiv \frac{i}{2}{\;\operatorname{mod}\;\frac{p - 1}{2}}.\n\] | Proof. Let \( \left( {\alpha, p}\right) = 1 \) . Since \( {\pi }^{{\sigma }_{\alpha }} = {\zeta }^{\alpha } - 1 = {\left( \pi + 1\right) }^{\alpha } - 1 = {\alpha \pi } + \cdots \), it follows that\n\n\[ \n{\sigma }_{\alpha }\left( \lambda \right) \equiv {\alpha }^{2}\lambda {\;\operatorname{mod}\;{\lambda }^{2}}\;\tex... | Yes |
Lemma 8.21. Let \( i \) be even, \( 2 \leq i \leq p - 3 \), and let \( {\widetilde{\eta }}_{i} \) be a generator for \( {\varepsilon }_{i}{E}_{{p}^{N}}^{ + } \) . If \( N \geq 1 + {v}_{p}\left( {{L}_{p}\left( {1,{\omega }^{i}}\right) }\right) \) then\n\n\[ \n{\widetilde{\eta }}_{i} \equiv {a}_{i}^{\prime } + {b}_{i}^{\... | Proof. We have \( {E}_{i}^{\left( N\right) } = {\widetilde{\eta }}_{i}^{{d}_{i}}{\gamma }^{{p}^{N}} \) for some \( {d}_{i} \in \mathbb{Z},\gamma \in {E}^{ + } \) . Therefore\n\n\[ \n{\log }_{p}{E}_{i}^{\left( N\right) } \equiv {d}_{i}{\log }_{p}{\widetilde{\eta }}_{i}{\;\operatorname{mod}\;{p}^{N}}. \n\]\n\nBy Proposit... | Yes |
Theorem 8.22. Let \( M = \mathop{\max }\limits_{i}{v}_{p}\left( {{L}_{p}\left( {1,{\omega }^{i}}\right) }\right) \), where \( i \) is even, \( 2 \leq i \leq p - 3 \) . If \( \eta \) is a unit of \( \mathbb{Z}\left\lbrack {\zeta }_{p}\right\rbrack \) which is congruent to a rational integer \( {\;\operatorname{mod}\;{p}... | Proof. As in the proof of Theorem 5.36, we may assume \( \eta \) is real. Write\n\n\[ \eta = a + b{\lambda }^{c} + \cdots \]\n\nThen, by hypothesis,\n\n\[ c \geq \frac{p - 1}{2}\left( {M + 1}\right) ,\;\text{ so }{v}_{p}\left( {{\log }_{p}\eta }\right) = \frac{2c}{p - 1} \geq M + 1. \]\n\nLet \( N \geq M + 1 \) and let... | Yes |
Corollary 8.23. Suppose \( {p}^{3} \nmid {B}_{pi} \) for all even \( i,2 \leq i \leq p - 3 \) . If \( \eta \) is a unit of \( \mathbb{Z}\left\lbrack {\zeta }_{p}\right\rbrack \) which is congruent to a rational integer \( {\;\operatorname{mod}\;{p}^{2}} \) then \( \eta \) is a pth power. | Proof. By Theorem 5.12,\n\n\[ \n{L}_{p}\left( {s,{\omega }^{i}}\right) = {a}_{0} + {a}_{1}\left( {s - 1}\right) + \cdots \n\] \n\nwith \( {a}_{i} \in {\mathbb{Z}}_{p} \) for all \( i \), and \( p \mid {a}_{i} \) for \( i \geq 1 \) . Therefore \n\n\[ \n- \frac{{B}_{pi}}{pi} \equiv - \left( {1 - {p}^{{pi} - 1}}\right) \f... | No |
Lemma 8.24. Let \( i \) be even, \( 2 \leq i \leq p - 3 \) . Then \( {\varepsilon }_{i}{U}_{1} \) is a cyclic \( {\mathbb{Z}}_{p} \) -module with\n\n\[ \n{\xi }_{i} = {\left( 1 + {\lambda }^{i/2}\right) }^{\left( {p - 1}\right) {\varepsilon }_{i}} \n\]\n\nas a generator. | Proof. As in the proof of Proposition 8.20,\n\n\[ \n{\sigma }_{\alpha }\left( {\lambda }^{i/2}\right) \equiv {\alpha }^{i}{\lambda }^{i/2}{\;\operatorname{mod}\;{\lambda }^{i/2 + 1}} \n\]\n\nTherefore\n\n\[ \n{\left( 1 + {\lambda }^{i/2}\right) }^{\left( {p - 1}\right) {\varepsilon }_{i}} = 1 + \left( {\mathop{\sum }\l... | Yes |
Theorem 8.25. Let \( i \) be even, \( 2 \leq i \leq p - 3 \) . Then\n\n\[ \left\lbrack {{\varepsilon }_{i}{U}_{1}^{\prime } : {\varepsilon }_{i}{\bar{C}}_{1}^{ + }}\right\rbrack = {p}^{{v}_{p}\left( {{L}_{p}\left( {1,{\omega }^{i}}\right) }\right) }.\] | Proof. Let \( d \) be the index, which must be a power of \( p \) since we are working with \( {\mathbb{Z}}_{p} \)-modules. Let \( {\xi }_{i} \) be as in Lemma 8.24 and let \( {\left( {E}_{i}^{\left( N\right) }\right) }^{p - 1} \) be as above. Then\n\n\[ {\xi }_{i}^{d} = {\left( {E}_{i}^{\left( N\right) }\right) }^{\le... | Yes |
Corollary 8.26. \( {\operatorname{Res}}_{s = 1}{\zeta }_{\mathbb{Q}{\left( {\zeta }_{p}\right) }^{ + }, p}\left( s\right) = \left( {1 - 1/p}\right) \left\lbrack {{U}_{1}^{\prime } : {\bar{C}}_{1}^{ + }}\right\rbrack \cdot u \), where \( u \) is a p-adic unit. | Proof. We know from Chapter 5 that the residue is\n\n\[ \left( {1 - \frac{1}{p}}\right) \prod {L}_{p}\left( {1,{\omega }^{i}}\right) \]\n\nThe result now follows easily from the theorem. | No |
Lemma 9.1. If \( \alpha \in \mathbb{Z}\left\lbrack {\zeta }_{p}\right\rbrack \) satisfies \( \alpha \equiv 1{\;\operatorname{mod}\;{\left( 1 - \zeta \right) }^{p}} \) then\n\n\[ \mathbb{Q}\left( {{\zeta }_{p},{\alpha }^{1/p}}\right) /\mathbb{Q}\left( {\zeta }_{p}\right) \]\n\n is unramified at \( \left( {1 - \zeta }\ri... | Proof. Let\n\n\[ f\left( X\right) = \frac{{\left( \left( 1 - \zeta \right) X + 1\right) }^{p} - \alpha }{{\left( 1 - \zeta \right) }^{p}}.\]\n\nClearly \( f \) is monic, and since \( p \) divides the binomial coefficients \( \left( \begin{matrix} p \\ j \end{matrix}\right) ,1 \leq j \leq \) \( p - 1 \), it follows that... | Yes |
Lemma 9.2. Assume \( p \nmid {h}^{ + }\left( {\mathbb{Q}\left( {\zeta }_{p}\right) }\right) \) . Suppose \( \alpha \in \mathbb{Q}\left( {\zeta }_{p}\right) \) satisfies \( \bar{\alpha } = {\alpha }^{-1} \) and suppose the extension \( \mathbb{Q}\left( {{\zeta }_{p},{\alpha }^{1/p}}\right) /\mathbb{Q}\left( {\zeta }_{p}... | Proof. Assume the extension is nontrivial, hence of degree \( p \) . Let\n\n\[ \sigma : {\alpha }^{1/p} \mapsto {\zeta }_{p}{\alpha }^{1/p} \]\n\ngenerate the Galois group and let \( J \) denote complex conjugation, extended so that \( J\left( {\alpha }^{1/p}\right) = {\left( J\alpha \right) }^{1/p} \) . Since \( {J\al... | Yes |
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