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Theorem 9.3. If \( p \) is regular then the second case of Fermat’s Last Theorem has no solutions. | Proof. If \( p \) is regular then \( p \nmid {h}^{ + }\left( {\mathbb{Q}\left( {\zeta }_{p}\right) }\right) \), so Assumption I is satisfied.\n\nFrom formulas in the previous section,\n\n\[ \n{\eta }_{a} = \frac{\omega + {\zeta }^{a}\theta }{1 - {\zeta }^{a}}{\rho }_{a}^{-p} \]\n\n\[ \n= \left( {\omega + {\zeta }^{a}\f... | Yes |
Theorem 9.5. Let the notation be as in Proposition 8.18 and Corollary 8.19. If there exists a prime \( l \equiv 1{\;\operatorname{mod}\;p} \) with \( l < {p}^{2} - p \) such that\n\n\[ \n{Q}_{i}^{k} ≢ 1{\;\operatorname{mod}\;l}\;\text{ for all }i \in \left\{ {{i}_{1},\ldots ,{i}_{s}}\right\} ,\n\]\n\nthen the second ca... | Proof. By Corollary \( {8.19}, p \nmid {h}^{ + }\left( {\mathbb{Q}\left( {\zeta }_{p}\right) }\right) \), so Assumption I is satisfied. Suppose that\n\n\[ \n{x}^{p} + {y}^{p} = {z}^{p},\;p \nmid {xy}, p \mid z, z \neq 0,\n\]\n\nwhere \( x, y, z \in \mathbb{Z} \) are relatively prime. Let \( l \) be as in the statement ... | No |
Lemma 9.6. \( l \nmid {xy} \). | Proof. Suppose \( l \mid y \), hence \( l \nmid {xz} \). Since\n\n\[ \mathop{\prod }\limits_{{a = 0}}^{{p - 1}}\left( {y - {\zeta }^{a}z}\right) = - {x}^{p} \]\n\nthe standard argument shows that the numbers\n\n\[ y - {\zeta }^{a}z,\;0 \leq a \leq p - 1, \]\n\nare relatively prime in \( \mathbb{Z}\left\lbrack {\zeta }_... | Yes |
Lemma 9.7. \( l \mid z \) (this is where \( l < {p}^{2} - p \) is used most strongly). | Proof. Write \( l = 1 + {kp} \) with \( k < p - 1 \) . By the equations obtained in Lemma 9.6 (we only needed \( p \nmid x \) ),\n\n\[ \left( {y - {\zeta }^{a}z}\right) {\sigma }_{a}^{-p} = {\gamma }_{a} = {\gamma }_{-a} = \left( {y - {\zeta }^{-a}z}\right) {\sigma }_{-a}^{-p},\;1 \leq a \leq p - 1. \]\n\nLet \( \widet... | Yes |
Lemma 9.9. \( {\eta }_{a}/{\eta }_{b} \) is a pth power. | Proof. Let \( \widetilde{l} \) be a prime of \( \mathbb{Q}\left( {\zeta }_{p}\right) \) lying above \( l \) . Then\n\n\[ \n{\eta }_{a} = \frac{\omega + {\zeta }^{a}\theta }{1 - {\zeta }^{a}}{\rho }_{a}^{-p} = \left( {\omega + {\zeta }^{a}\frac{\omega + \theta }{1 - {\zeta }^{a}}}\right) {\rho }_{a}^{-p}\n\]\n\n\[ \n\eq... | No |
Theorem 10.1. Suppose the extension of number fields \( L/K \) contains no unramified abelian subextensions \( F/K \) with \( F \neq K \) . Then \( {h}_{K} \) divides \( {h}_{L} \) . In fact, the norm map from the class group of \( L \) to the class group of \( K \) is surjective. | Proof. The first statement is Proposition 4.11. The second is proved in the appendix on class field theory. | No |
Theorem 10.2. Let \( n ≢ 2{\;\operatorname{mod}\;4} \) be arbitrary and let \( {h}_{n} = h\left( {\mathbb{Q}\left( {\zeta }_{n}\right) }\right) \) . If \( 2 \mid {h}_{n}^{ + } \) then \( 2 \mid {h}_{n}^{ - } \) . | Proof. \( \mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}{\left( {\zeta }_{n}\right) }^{ + } \) is totally ramified at infinity, so Theorem 10.1 implies that the norm map on the ideal class groups\n\n\[ N : C \rightarrow {C}^{ + } \]\n\nis surjective, so \( {h}_{n}^{ - } = {h}_{n}/{h}_{n}^{ + } = \left| {\ker N}\right... | Yes |
Theorem 10.3. Let \( K \) be a \( {CM} \) -field. The kernel of the map \( {C}^{ + } \rightarrow C \) from the ideal class group of \( {K}^{ + } \) to that of \( K \) has order 1 or 2 . | Proof. Let \( I \) be an ideal of \( {K}^{ + } \) and suppose \( I = \left( \alpha \right) \) in \( K \) . Then \( \left( 1\right) = \bar{I}/I = \) \( \left( {\bar{\alpha }/\alpha }\right) \), so \( \bar{\alpha }/\alpha \) is a unit, hence a root of unity by Lemma 1.6. This root of unity does not depend on the class of... | Yes |
Theorem 10.4. (a) Suppose \( L/K \) is a Galois extension and \( \operatorname{Gal}\left( {L/K}\right) \) is a p-group \( \left( {p = \text{any prime}}\right) \) . Assume there is at most one prime (finite or infinite) which ramifies in \( L/K \) . If \( p \mid {h}_{L} \) then \( p \mid {h}_{K} \) . | Proof. Assume \( p \mid {h}_{L} \) . Let \( H \) be the Hilbert \( p \) -class field of \( L \), so \( H \) is the maximal unramified abelian \( p \) -extension of \( L \) and \( \operatorname{Gal}\left( {H/L}\right) \) is isomorphic to the \( p \) -Sylow subgroup of the ideal class group of \( L \) . Since \( L/K \) i... | Yes |
Corollary 10.5. Let \( n \geq 1 \) . Then \( p\left| {h\left( {\mathbb{Q}\left( {\zeta }_{p}\right) }\right) \Leftrightarrow p}\right| h\left( {\mathbb{Q}\left( {\zeta }_{{p}^{n}}\right) }\right) \) . | Proof. Theorems 10.1 and 10.4. | No |
Theorem 10.8. Let \( L/K \) be cyclic of degree \( n \) . Let \( p \) be prime, \( p \nmid n \), and assume all fields \( E \) with \( K \subseteq E \subsetneqq L \) satisfy \( p \nmid {h}_{E} \) . Let \( A \) be the p-Sylow subgroup of the ideal class group of \( L \), and let \( f \) be the order of \( p{\;\operatorn... | Proof. Let \( V = {A}^{{p}^{a - 1}}/{A}^{{p}^{a}} \), so \( V \) has \( {p}^{{r}_{a}} \) elements. Let \( \sigma \) generate \( \operatorname{Gal}\left( {L/K}\right) \) . Then \( \sigma \) acts on \( V \) . Let \( v \in V, v \neq 0 \), and suppose the orbit of \( v \) under the action of \( \operatorname{Gal}\left( {L/... | Yes |
Theorem 10.11. Let \( p \) be an odd prime. Let \( L \) be a CM-field with \( {\zeta }_{p} \in L \), and let A be the p-Sylow subgroup of the ideal class group of L. Then\n\n\[ \n\text{p-rank}{A}^{ + } \leq 1 + \text{p-rank}{A}^{ - }\text{.} \n\]\n\nLet \( W \) be the roots of unity in \( L \) . If \( L\left( {W}^{1/p}... | Proof. In the notation at the beginning of the section, let \( K = {L}^{ + } \), the maximal real subfield of \( L \) . Then \( G = \operatorname{Gal}\left( {L/K}\right) = \{ 1, J\} \), where \( J = \) complex conjugation, and\n\n\[ \n{A}^{ + } = \frac{1 + J}{2}A,\;{A}^{ - } = \frac{1 - J}{2}A.\n\]\n\nAs in the above t... | Yes |
Proposition 10.12. Let \( L \) be a \( {CM} \) -field and let \( {A}_{L} \) and \( {A}_{{L}^{ + }} \) be the 2-Sylow subgroups of the ideal class groups of \( L \) and \( {L}^{ + } \), respectively. Then\n\n\[ \text{2-rank}{A}_{{L}^{ + }} \leq 1 + 2\text{-rank}{A}_{L}^{ - }\text{.} \] | Proof. Let \( i\left( {A}_{{L}^{ + }}\right) \) denote the image in \( {A}_{L} \) and let \( {\left( {A}_{{L}^{ + }}\right) }_{2} \) and \( {\left( {A}_{L}^{ - }\right) }_{2} \) denote the elements of order 2 in the respective groups. As noted above,\n\n\[ i\left( {\left( {A}_{{L}^{ + }}\right) }_{2}\right) \subseteq {... | Yes |
Let \( A \) be the p-Sylow subgroup of the ideal class group of \( \mathbb{Q}\left( {\zeta }_{p}\right) \) and let\n\n\[ A = {\bigoplus }_{i = 0}^{p - 2}{\varepsilon }_{i}A \]\n\nbe the decomposition according to idempotents. If \( p \nmid h\left( {\mathbb{Q}{\left( {\zeta }_{p}\right) }^{ + }}\right) \) then\n\n\[ {\v... | Proof. By Theorem 10.14, each \( {\varepsilon }_{i}A \) is a cyclic group. By Proposition 6.16, \( {B}_{1,{\omega }^{-i}} \) annihilates \( {\varepsilon }_{i}A \), so\n\n\[ \left| {{\varepsilon }_{i}A}\right| \leq p\text{-part of}{B}_{1,{\omega }^{-i}}\text{.} \]\n\nSince\n\n\[ \mathop{\prod }\limits_{{i = 3}}^{{p - 2}... | Yes |
Corollary 10.17. Suppose \( p \nmid h\left( {\mathbb{Q}{\left( {\zeta }_{p}\right) }^{ + }}\right) \) . Let \( {i}_{1},\ldots ,{i}_{s} \) be the even indices \( i \) such that \( 2 \leq i \leq p - 3 \) and \( p \mid {B}_{i} \) . If\n\n\[ \n{B}_{1,{\omega }^{i - 1}} ≢ 0{\;\operatorname{mod}\;{p}^{2}} \n\] \n\nand \n\n\[... | Proof. Let \( f\left( {T,{\omega }^{i}}\right) = {a}_{0} + {a}_{1}T + \cdots \), with \( {a}_{j} \in {\mathbb{Z}}_{p} \) for all \( p \) . Then, for \( s \in {\mathbb{Z}}_{p} \) , \n\n\[ \n{L}_{p}\left( {s,{\omega }^{i}}\right) = f\left( {{\left( 1 + p\right) }^{s} - 1,{\omega }^{i}}\right) \equiv {a}_{0} + {a}_{1}{sp}... | Yes |
Lemma 11.2. Let \( \gamma = {0.577}\ldots \) be the Euler-Mascheroni constant and let \( A \) be a positive integer. Then\n\n\[ \gamma + \log A \leq \mathop{\sum }\limits_{{a = 1}}^{A}\frac{1}{a} \leq \frac{1}{A} + \gamma + \log A. \] | Proof. Since\n\n\[ \log \left( \frac{A + 1}{A}\right) = \frac{1}{A} - \frac{1}{2{A}^{2}} + \frac{1}{3{A}^{3}}\cdots \]\n\nhas alternating signs and decreasing terms, we have\n\n\[ \frac{1}{A} - \frac{1}{2{A}^{2}} < \log \left( \frac{A + 1}{A}\right) < \frac{1}{A}. \]\n\nTherefore\n\n\[ \frac{1}{A + 1} - \log \left( \fr... | Yes |
Lemma 11.3. Let \( \pi \left( m\right) \) be the number of distinct prime divisors of \( m \) . Then\n\n\[ \mathop{\sum }\limits_{\substack{{n = 1} \\ {\left( {n, m}\right) = 1} }}^{{jm}}\frac{1}{n} = \left( {\gamma + \log \left( {jm}\right) + \mathop{\sum }\limits_{{p \mid m}}\frac{\log p}{p - 1}}\right) \mathop{\prod... | Proof. We shall use induction on \( \pi \left( m\right) \) . By Lemma 11.2, the lemma is true for \( \pi \left( m\right) = 0 \) . Assume it is true for \( \pi \left( m\right) \) and then replace \( m \) by \( {mq} \), with \( q \) prime, \( \left( {q, m}\right) = 1 \) (the cases \( m{q}^{2} \), etc., are obtained by va... | Yes |
Lemma 11.4.\n\n\\[ \n\\mathop{\\sum }\\limits_{{j = 1}}^{N}\\frac{\\log j}{j} < {0.11} + \\frac{1}{2}{\\left( \\log N\\right) }^{2}\\;\\text{ for }N \\geq 1, \n\\]\n\nand\n\n\\[ \n\\mathop{\\sum }\\limits_{{j = 1}}^{N}\\frac{\\log j}{j} < \\frac{1}{2}{\\left( \\log N\\right) }^{2}\\;\\text{ for }N \\geq {21}. \n\\] | Proof. A calculation shows that\n\n\\[ \n\\mathop{\\sum }\\limits_{{j = 1}}^{{21}}\\frac{\\log j}{j} < \\frac{1}{2}{\\left( \\log {21}\\right) }^{2} \n\\]\n\nSince \\( \\left( {\\log x}\\right) /x \\) is decreasing for \\( x > e \\) ,\n\n\\[ \n\\mathop{\\sum }\\limits_{{j = {22}}}^{N}\\frac{\\log j}{j} < {\\int }_{21}^... | Yes |
Lemma 11.8 (Polya-Vinogradov). Let \( \chi \neq 1 \) be primitive with conductor \( f \). Let \( S\left( {u,\chi }\right) = \mathop{\sum }\limits_{{0 \leq n < u}}\chi \left( n\right) \). Then \[ \left| {S\left( {u,\chi }\right) }\right| < {f}^{1/2}\log f \] | Proof. Let \( \zeta = {e}^{{2\pi i}/f} \) and let \( \tau \left( \chi \right) = \mathop{\sum }\limits_{{c = 1}}^{{f - 1}}\chi \left( c\right) {\zeta }^{c} \). By Lemma 4.7, \[ \overline{\chi \left( n\right) }\tau \left( \chi \right) = \mathop{\sum }\limits_{c}\chi \left( c\right) {\zeta }^{cn} \] We may assume \( u \) ... | Yes |
Lemma 11.9. If \( \chi \) is a primitive nontrivial character of conductor \( f \) and \( 1 \geq \) \( \sigma \geq 1 - 1/{4f} \), then\n\n\[ \left| {{L}^{\prime }\left( {\sigma ,\chi }\right) }\right| \leq \left( {1.3}\right) {\left( \log f\right) }^{2}. \] | Proof. As in the proof of Lemma 11.6, we have for \( \sigma = \operatorname{Re}\left( s\right) > 0 \) ,\n\n\[ L\left( {s,\chi }\right) = \mathop{\sum }\limits_{{n = 1}}^{{f - 1}}\chi \left( n\right) {n}^{-s} + s{\int }_{f}^{\infty }S\left( {u,\chi }\right) {u}^{-s - 1}{du}. \]\n\nDifferentiate:\n\n\[ {L}^{\prime }\left... | Yes |
Lemma 11.10. If \( \chi \) is a primitive quadratic character of conductor \( f \), then\n\n\[ L\left( {\sigma ,\chi }\right) \geq 0\;\text{ for }\sigma \geq 1 - \frac{1}{4f}. \] | Proof. By the analytic class number formula (see the discussion following Theorem 4.9),\n\n\[ L\left( {1,\chi }\right) = \left\{ \begin{array}{ll} \frac{{2h}\log \varepsilon }{\sqrt{f}}, & \chi \text{ even,} \\ \frac{2\pi h}{w\sqrt{f}}, & \chi \text{ odd,} \end{array}\right. \]\n\nwhere \( h \) and \( \varepsilon \) ar... | Yes |
Lemma 11.11. If \( \chi \) is a quadratic character \( {\;\operatorname{mod}\;m} \), then\n\n\[ L\left( {\sigma ,{\chi }_{m}}\right) \geq 0\;\text{ for }\sigma \geq 1 - 1/{4m} \] | Proof.\n\n\[ L\left( {\sigma ,{\chi }_{m}}\right) = L\left( {\sigma ,\chi }\right) \mathop{\prod }\limits_{{p \mid m}}\left( {1 - \frac{\chi \left( p\right) }{{p}^{\sigma }}}\right) \geq 0 \]\n\nsince \( 1 - 1/{4m} \geq 1 - 1/{4f} \) . | No |
Lemma 11.12. If \( \phi \left( m\right) \geq {20} \), then\n\n\[ \mathop{\prod }\limits_{{\chi \neq 1}}L\left( {1,{\chi }_{m}}\right) \geq \frac{1}{{16m\phi }{\left( m\right) }^{1/2}} \] | Proof. We shall use Lemma 11.7 with \( f\left( s\right) = \mathop{\prod }\limits_{{\chi \neq 1}}L\left( {s,{\chi }_{m}}\right) \) and \( \alpha = 1 - 1/{4m} \) . \( M \) is given by Lemma 11.6. Clearly (a) and (c) are satisfied. Since \( f\left( s\right) \zeta \left( s\right) \) is the Dedekind zeta function of \( \mat... | Yes |
Lemma 11.13. If \( \phi \left( m\right) \geq {220} \) then\n\n\[ \left| {\mathop{\prod }\limits_{{\chi \text{ odd }}}L\left( {1,{\chi }_{m}}\right) }\right| \geq \frac{1}{{16m\phi }{\left( m\right) }^{1/2}}{\left( {1.7}\right) }^{\left( {2 - \phi \left( m\right) }\right) /4}. \]\n | Proof. Lemmas 11.5 and 11.12. | No |
Lemma 11.14.\n\n\[ \mathop{\prod }\limits_{{\chi \text{ odd }}}\mathop{\prod }\limits_{{p \mid m}}{\left( 1 - \frac{\chi \left( p\right) }{p}\right) }^{-1} \geq {e}^{-{7\phi }\left( m\right) /{24}}. \] | Proof. Write \( m = {m}_{p}{m}_{p}^{\prime } \), where \( {m}_{p} \) is a power of \( p \) and \( p \nmid {m}_{p}^{\prime } \) . Then \( \chi \left( p\right) \neq \) \( 0 \Leftrightarrow {f}_{\chi } \mid {m}_{p}^{\prime } \) . There are at most \( \frac{1}{2}\phi \left( {m}_{p}^{\prime }\right) \) such odd characters. ... | Yes |
Proposition 11.15. If \( \phi \left( m\right) \geq {220} \) then\n\n\[ \log {h}_{m}^{ - } \geq \frac{1}{4}\log {d}_{m} - \left( {1.37}\right) \phi \left( m\right) \]\n\nwhere \( {d}_{m} \) is the absolute value of the discriminant of \( \mathbb{Q}\left( {\zeta }_{m}\right) \) . | Proof. From the class number formula (in particular, see the discussion preceding Theorem 4.17),\n\n\[ \log {h}_{m}^{ - } = \frac{1}{2}\log \left( \frac{{d}_{m}}{{d}_{m}^{ + }}\right) + \log \mathop{\prod }\limits_{{\chi \text{ odd }}}L\left( {1,{\chi }_{m}}\right) + \log w \]\n\n\[ + \log Q - \frac{\phi \left( m\right... | Yes |
Proposition 11.16. Assume \( \phi \left( m\right) \geq {220} \) . If \( m \) is a prime power then\n\n\[ \log {h}_{m}^{ - } \geq \frac{1}{4}\log {d}_{m} - \left( {1.08}\right) \phi \left( m\right) \]\n\nIf \( m \) is arbitrary then\n\n\[ \log {h}_{m}^{ - } \geq \frac{1}{4}\log {d}_{m} - \left( {1.08}\right) \phi \left(... | Proof. In Lemma 11.14, all the factors are 1 if \( m \) is a prime power. If \( m \) is arbitrary, we have, as in the proof of the lemma,\n\n\[ \log \mathop{\prod }\limits_{{\chi \text{ odd }}}\mathop{\prod }\limits_{{p \mid m}}{\left( 1 - \frac{\chi \left( p\right) }{p}\right) }^{-1} \geq - \frac{\phi \left( m\right) ... | Yes |
Corollary 11.17. If \( \phi \left( {p}^{a}\right) \geq {220} \), then \( {h}_{{p}^{a}}^{ - } > 1 \) . | Proof. We have\n\n\[ \frac{\log \left( {d}_{{p}^{a}}\right) }{\phi \left( {p}^{a}\right) } = \log \left( {p}^{a}\right) - \frac{\log p}{p - 1} \geq \log \left( {p}^{a}\right) - \log 2 \]\n\n\[ \geq \log \left( {220}\right) - \log \left( 2\right) \geq {4.7}\text{.} \]\n\nTherefore\n\n\[ \log \left( {h}_{{p}^{a}}^{ - }\r... | Yes |
Corollary 11.18. \( {h}_{{p}^{a}}^{ - } = 1 \) if and only if \( {p}^{a} \) is one of the following: an odd prime \( p \leq {19} \), or \( 4,8,9,{16},{25},{27},{32} \) (one could also include \( {p}^{a} = 1 \) ). | Proof. We know that \( \phi \left( {p}^{a}\right) < {220} \) . The table in the appendix yields the answer. | No |
Lemma 11.22. If \( L/K \) is an extension of degree \( n \) in which no finite primes ramify, then\n\n\[ \n{D}_{L} = {D}_{K}^{n} \n\]\n\nwhere \( {D}_{L} \) and \( {D}_{K} \) are the absolute values of the discriminants of \( L \) and \( K \) , respectively. | Proof. A well-known formula (see Lang [1], pp. 60, 66) states that\n\n\[ \n{D}_{L} = {D}_{K}^{n}N{\mathcal{D}}_{L/K} \n\]\n\nwhere \( N{\mathcal{D}}_{L/K} \) is the norm of the relative different. Since no primes ramify, \( {\mathcal{D}}_{L/K} = \left( 1\right) \) . The result follows. | Yes |
Lemma 11.23. Let \( B\left( n\right) \) be the lower bound for \( {D}^{1/n} \) for totally real fields of degree \( \geq n \) (as given in the table of the previous section). Let \( {d}_{m}^{ + } \) and \( {h}_{m}^{ + } \) be the discriminant and class number of \( \mathbb{Q}{\left( {\zeta }_{m}\right) }^{ + } \) . If\... | Proof. Let \( H \) be the Hilbert class field of \( \mathbb{Q}{\left( {\zeta }_{m}\right) }^{ + } \), so \( H/\mathbb{Q}{\left( {\zeta }_{m}\right) }^{ + } \) is an unramified extension of degree \( {h}_{m}^{ + } \), and\n\n\[{n}_{H} = \left\lbrack {H : \mathbb{Q}}\right\rbrack = \frac{1}{2}\phi \left( m\right) {h}_{m}... | Yes |
If \( \phi \) is a measure, then\n\n\[ \n{\int }_{X}{fd\phi } : \operatorname{Step}\left( X\right) \rightarrow {\mathbb{C}}_{p}\n\]\n\nextends uniquely to a continuous \( {\mathbb{C}}_{p} \) -linear map\n\n\[ \n{\int }_{X}{fd\phi } : C\left( {X,{\mathbb{C}}_{p}}\right) \rightarrow {\mathbb{C}}_{p}\n\] | Proof. Since the step functions are dense, the map must be unique if it exists.\n\nObserve that if \( K \) is the constant used above and \( {\chi }_{i, a} \) is the characteristic function of the previous section,\n\n\[ \n\left| {{\int }_{X}{\chi }_{i, a}{d\phi }}\right| = \left| {{\phi }_{i}\left( a\right) }\right| \... | Yes |
Corollary 12.5. Let \( \phi \) be a measure. Then \( \left( {{\Gamma }_{p}\phi }\right) \left( s\right) \) is an analytic function of \( s \) . | Proof. Let \( \theta = \psi = 1 \) . Clearly any function of the form \( g\left( {{\kappa }_{0}^{s} - 1}\right) \) is analytic. | No |
Corollary 12.6. If \( {\int }_{\Delta \times \left( {1 + p{Z}_{p}}\right) }{\theta \psi d\phi } = 0 \) for all \( \theta \in \widehat{\Delta } \) and all \( \psi \) of the second kind, then \( \phi = 0 \) . (In other words, a measure is determined by its values on characters of finite order. Note that \( \theta \left( ... | Proof. Let \( {g}_{\theta } \) be one of the corresponding power series. Then\n\n\[ {g}_{\theta }\left( {\psi \left( {\kappa }_{0}\right) - 1}\right) = 0\;\text{ for all }\psi \]\n\nhence\n\n\[ {g}_{\theta }\left( {{\zeta }_{{p}^{n}} - 1}\right) = 0\;\text{ for all }n. \]\n\nBy the \( p \) -adic Weierstrass Preparation... | Yes |
Proposition 12.7. Let \( g \in \mathcal{O}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) and let \( d{\widetilde{\phi }}_{a} \) be the corresponding measure on \( \widetilde{X} = {\mathbb{Z}}_{p} \) . For \( k \geq 0 \) , \[ \left( {{D}^{k}g}\right) \left( 0\right) = {\int }_{y \in \widetilde{X}}{y}^{k}d{\wi... | Proof. First let \( g = {\left( 1 + T\right) }^{n} \) . As mentioned above, the left-hand side is \( {n}^{k} \) . The measure \( d\widetilde{\phi } \) is the delta measure \( {\delta }_{n} \), so \[ \int {y}^{k}d\widetilde{\phi } = {n}^{k} \] By linearity, the result holds for polynomials. Let \( g\left( T\right) \in \... | Yes |
Proposition 12.8. Let \( g \in \mathcal{O}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) and let \( d{\phi }_{g} \) and \( d{\widetilde{\phi }}_{g} \) be the corresponding measures on \( X = 1 + p{\mathbb{Z}}_{p} \) and \( \widetilde{X} = {\mathbb{Z}}_{p} \) . Let \( {\chi }_{{\widetilde{X}}^{ \times }}\left... | Proof. By Proposition 12.7, it suffices to prove the first equality. As usual, let \( N \geq 1 \) be large and write\n\n\[ g\left( T\right) = {P}_{N}\left( T\right) {q}_{N}\left( T\right) + {r}_{N}\left( T\right) \]\n\nThen\n\n\[ {Ug} = U\left( {{P}_{N}{q}_{N}}\right) + U{r}_{N} \]\n\nBut\n\n\[ {P}_{N}\left( {\zeta \le... | Yes |
Corollary 12.9. Let \( g \in \mathcal{O}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) . Fix a congruence class \( \alpha {\;\operatorname{mod}\;p} - 1 \) . Then there exists \( h\left( T\right) \in \mathcal{O}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) such that\n\n\[ \left( {{D}^{k}{Ug}}\righ... | Proof. Decompose\n\n\[ {\widetilde{X}}^{ \times } = {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{ \times } \times \left( {1 + p{\mathbb{Z}}_{p}}\right) \]\n\nThen\n\n\[ \left( {{D}^{k}{Ug}}\right) \left( 0\right) = {\int }_{{\widetilde{X}}^{ \times }}{\omega }^{\alpha }\left( y\right) \langle y{\rangle }^{k}d{\widetilde{\p... | Yes |
Proposition 12.11. The universal punctured ordinary distribution \( {A}_{n}^{0} \) on \( \left( {1/n}\right) \mathbb{Z}/\mathbb{Z} \) requires at most \( \phi \left( n\right) + \pi \left( n\right) - 1 \) generators, where \( \pi \left( n\right) \) equals the number of distinct prime factors of \( n \) . | Proof. \( {A}_{n}^{0} \) is generated by\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\nwith relations\n\n\[ g\left( \frac{a}{r}\right) = \mathop{\sum }\limits_{{k = 0}}^{{\left( {n/r}\right) - 1}}g\... | Yes |
Lemma 12.16. Let \( n = m{m}^{\prime } \) with \( {f}_{\chi } \mid m \) and \( \left( {m,{m}^{\prime }}\right) = 1 \) . Then\n\n\[ \mathop{\sum }\limits_{\substack{{a = 1} \\ {\left( {a, n}\right) = 1} }}^{n}\chi \left( a\right) h\left( \frac{a{m}^{\prime }}{n}\right) = \phi \left( {m}^{\prime }\right) \left( {\mathop{... | Proof. See the calculations following the proof of Lemma 8.7. | No |
Proposition 12.17. Let \( n = \mathop{\prod }\limits_{{i = 1}}^{s}{p}_{i}^{{e}_{i}} \) . Let I run through all subsets of \( \{ 1,\ldots, s\} \) , except \( \{ 1,\ldots, s\} \), and let \( {n}_{I} = \mathop{\prod }\limits_{{i \in I}}{p}_{i}^{{e}_{i}} \) . Then\n\n\[ \mathop{\sum }\limits_{I}\mathop{\sum }\limits_{\subs... | Proof. See the end of the proof of Theorem 8.3. This is where we need \( \chi \neq 1 \) (if \( \chi = 1 \), include \( I = \{ 1,\ldots, s\} \) and the result holds). | No |
Theorem 12.18. Let \( n > 2 \) . For some integers \( a, b, c, d \), we have\n\nUniversal punctured \( = {A}_{n}^{0} \simeq {\mathbb{Z}}^{\phi \left( n\right) + \pi \left( n\right) - 1} \) ,\n\nUniversal even punctured \( = {\left( {A}_{n}^{0}\right) }^{ + } \simeq {\mathbb{Z}}^{\left( {1/2}\right) \phi \left( n\right)... | The proof of Bass' theorem is now immediate. Since\n\n\[ \n{\left( {A}_{n}^{0}\right) }^{ + } \rightarrow \text{group generated by}\left\{ {\log \left| {{\zeta }_{n}^{a} - 1}\right| ,0 < a < n}\right\} \n\]\n\nis surjective, and the latter is free abelian of rank (at least, hence exactly) \( \frac{1}{2}\phi \left( n\ri... | Yes |
Proposition 13.1. Let \( {K}_{\infty }/K \) be a \( {\mathbb{Z}}_{p} \) -extension. Then, for each \( n \geq 0 \), there is a unique field \( {K}_{n} \) of degree \( {p}^{n} \) over \( K \), and these \( {K}_{n} \), plus \( {K}_{\infty } \), are the only fields between \( K \) and \( {K}_{\infty } \) . | Proof. The intermediate fields correspond to the closed subgroups of \( {\mathbb{Z}}_{p} \) . Let \( S \neq 0 \) be a closed subgroup and let \( x \in S \) be such that \( {v}_{p}\left( x\right) \) is minimal. Then \( x\mathbb{Z} \), hence \( x{\mathbb{Z}}_{p} \), is in \( S \) . By the choice of \( x \), we must have ... | Yes |
Lemma 13.3. Let \( {K}_{\infty }/K \) be a \( {\mathbb{Z}}_{p} \) -extension. At least one prime ramifies in this extension, and there exists \( n \geq 0 \) such that every prime which ramifies in \( {K}_{\infty }/{K}_{n} \) is totally ramified. | Proof. Since the class number of \( K \) is finite, the maximal abelian unramified extension of \( K \) is finite, so some prime must ramify in \( {K}_{\infty }/K \) . We know that only finitely many primes of \( K \) ramify in \( {K}_{\infty }/K \) by Proposition 13.2. Call them \( {h}_{1},\ldots ,{h}_{s} \), and let ... | Yes |
Theorem 13.4. Suppose the \( {\mathbb{Z}}_{p} \) -rank of \( {\bar{E}}_{1} \) is \( {r}_{1} + {r}_{2} - 1 - \delta \), with \( \delta \geq 0 \) . Then there are \( {r}_{2} + 1 + \delta \) independent \( {\mathbb{Z}}_{p} \) -extensions of \( K \) . In other words, if \( \widetilde{K} \) is the compositum of all \( {\mat... | Proof. Let \( \widetilde{K} \) be as above and \( F \) the maximal abelian extension of \( K \) which is unramified outside \( p \) . Then \( \widetilde{K} \subseteq F \) . Let \( J \) denote the idèles of \( K \) . By class field theory, there is a closed subgroup \( H \) with \[ {K}^{ \times } \subseteq H \subseteq J... | Yes |
Lemma 13.5. \( {U}_{1} \cap H = {U}_{1} \cap \overline{{K}^{ \times }{U}^{\prime \prime }} = \overline{\psi \left( {E}_{1}\right) } \). | Proof. Let \( \varepsilon \in {E}_{1} \). Then \( \psi \left( \varepsilon \right) \in {U}_{1} \). Also\n\n\[ \psi \left( \varepsilon \right) = \left( \varepsilon \right) \left( \frac{\psi \left( \varepsilon \right) }{\varepsilon }\right) \in {K}^{ \times }{U}^{\prime \prime } \]\n\nsince \( \psi \left( \varepsilon \rig... | No |
Corollary 13.6. Let \( H \) be the Hilbert class field of \( K \) and let \( F \) be the maximal abelian extension of \( K \) unramified outside \( p \) . Then \[ \operatorname{Gal}\left( {F/H}\right) \simeq \left( {\mathop{\prod }\limits_{{p \mid p}}{U}_{p}}\right) /\bar{E} \] where \( \bar{E} \) is the closure of \( ... | Proof. \( \operatorname{Gal}\left( {F/K}\right) \simeq {J}^{\prime } \), and the closed subgroup \( {J}^{\prime \prime } \) corresponds to \( H \) . Hence \( \operatorname{Gal}\left( {F/H}\right) \simeq {J}^{\prime \prime } \simeq {U}^{\prime }/{U}^{\prime } \cap H \) . The same proof as for Lemma 13.5 shows that \( {U... | No |
Lemma 13.7. Suppose \( f, g \in \Lambda \) are relatively prime. Then the ideal \( \left( {f, g}\right) \) is of finite index in \( \Lambda \) . | Proof. Let \( h \in \left( {f, g}\right) \) be of minimal degree. Then \( h = {p}^{s}H \) with \( H = 1 \) or \( H \) distinguished. Suppose \( H \neq 1 \) . Since \( f \) and \( g \) are relatively prime, we may assume \( H \) does not divide \( f \) . But\n\n\[ f = {Hq} + r,\;\deg r < \deg H = \deg h, \]\n\nso\n\n\[ ... | Yes |
Lemma 13.8. Suppose \( f, g \in \Lambda \) are relatively prime. Then\n\n(1) the natural map\n\n\[ \Lambda /\left( {fg}\right) \rightarrow \Lambda /\left( f\right) \oplus \Lambda /\left( g\right) \]\n\nis an injection with finite cokernel;\n\n(2) there is an injection\n\n\[ \Lambda /\left( f\right) \oplus \Lambda /\lef... | Proof. (1) Since \( \Lambda \) is a unique factorization domain, the map is an injection. Consider \( \left( {a{\;\operatorname{mod}\;f}, b{\;\operatorname{mod}\;g}}\right) \) . If \( a - b \in \left( {f, g}\right) \), then \( a - b = {fA} + {gB} \), for some \( A, B \) . Let\n\n\[ c = a - {fA} = b + {gB}. \]\n\nThen\n... | Yes |
Proposition 13.9. The prime ideals of \( \Lambda \) are \( 0,\left( {p, T}\right) ,\left( p\right) \), and the ideals \( \left( {P\left( T\right) }\right) \) where \( P\left( T\right) \) is irreducible and distinguished. The ideal \( \left( {p, T}\right) \) is the unique maximal ideal. | Proof. All the above are easily seen to be prime ideals. Let \( h \neq 0 \) be prime. Let \( h \in p \) be of minimal degree. Then \( h = {p}^{s}H \) with \( H = 1 \) or \( H \) distinguished. Since \( p \) is prime, \( p \in p \) or \( H \in p \) . If \( 1 \neq H \in p \) then \( H \) must be irreducible by the minima... | Yes |
Lemma 13.10. Let \( f \in \Lambda \) with \( f \notin {\Lambda }^{ \times } \) . Then \( \Lambda /\left( f\right) \) is infinite. | Proof. We may assume \( f \neq 0 \) . It suffices to consider \( f = p \) and \( f = \) distinguished. If \( f = p,\Lambda /\left( f\right) \simeq \mathbb{Z}/p\mathbb{Z}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) . If \( f \) is distinguished, use the division algorithm. | No |
Lemma 13.11. \( \Lambda \) is a Noetherian ring. | Proof. It is known (Lang’s Algebra) that if \( A \) is Noetherian then so is \( A\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) . One could also use the Hilbert basis theorem \( (A \) Noetherian \( \Rightarrow A\left\lbrack T\right\rbrack \) Noetherian) since the generators of an ideal may always be assumed ... | No |
Lemma 13.16 (Nakayama’s Lemma). Let \( X \) be a compact \( \Lambda \) -module. Then\n\n\[ X\text{is finitely generated over}\Lambda \Leftrightarrow X/\left( {p, T}\right) X\text{is finite.} \]\n\nIf \( {x}_{1},\ldots ,{x}_{n} \) generate \( X/\left( {p, T}\right) X \) over \( \mathbb{Z} \), then they also generate \( ... | Proof. Consider a small neighborhood \( U \) of 0 in \( X \) . Since \( {\left( p, T\right) }^{n} \rightarrow 0 \) in \( \Lambda \) , each \( z \in X \) has a neighborhood \( {U}_{z} \) such that \( {\left( p, T\right) }^{n}{U}_{z} \subseteq U \) for large \( n \) . Since \( X \) is compact, finitely many \( {U}_{z} \)... | Yes |
Lemma 13.20. Assume \( E \) is as in Proposition 13.19, with \( r = 0 \) . Then\n\n\[ m = 0 \Leftrightarrow p - \operatorname{rank}\left( {E/{v}_{n, e}E}\right) \text{is bounded as} n \rightarrow \infty \text{.} \] | Proof. Recall that the \( p \) -rank of a finite abelian group \( A \) is the number of direct summands of \( p \) -power order when \( A \) is decomposed into cyclic groups of prime power order. It is also equal to\n\n\[ {\dim }_{\mathbb{Z}/p\mathbb{Z}}\left( {A/{pA}}\right) \]\n\nRecall that \( {v}_{n, e} \) is disti... | Yes |
Proposition 13.22. Suppose \( {K}_{\infty }/K \) is a \( {\mathbb{Z}}_{p} \) -extension in which exactly one prime is ramified, and assume it is totally ramified. Then\n\n\[ \n{A}_{n} \simeq {X}_{n} \simeq X/\left( {{\left( 1 + T\right) }^{{p}^{n}} - 1}\right) X \]\n\nand\n\n\[ \np \nmid {h}_{0} \Leftrightarrow p \nmid... | Proof. Since \( {K}_{\infty }/K \) satisfies the \ | No |
Proposition 13.23. \( \mu = 0 \Leftrightarrow p - \operatorname{rank}\left( {A}_{n}\right) \) is bounded as \( n \rightarrow \infty \) . | Proof. We have \( {Y}_{e} \sim E \) with \( E \) as in Lemma 13.20. By the lemma, \( \mu = 0 \Leftrightarrow \) \( p \) -rank \( \left( {E/{v}_{n, e}E}\right) \) is bounded. From the proof of Lemma 13.21, we have an exact sequence\n\n\[ 0 \rightarrow {C}_{n} \rightarrow {Y}_{e}/{v}_{n, e}Y \rightarrow E/{v}_{n, e}E \ri... | Yes |
Proposition 13.24. Let \( p \) be prime. Suppose \( K \) is a CM-field with \( {\zeta }_{p} \in K \) and let \( {K}_{\infty }/K \) be the cyclotomic \( {\mathbb{Z}}_{p} \) -extension. Then\n\n\[ \mu = 0 \Leftrightarrow {\mu }^{ - } = 0. \] | Proof. \ | No |
Proposition 13.25. Suppose \( {K}_{\infty }/K \) is a \( {\mathbb{Z}}_{p} \) -extension and assume \( \mu = 0 \) . Then\n\n\[ \nX \simeq \mathop{\lim }\limits_{ \leftarrow }{A}_{n} \simeq {\mathbb{Z}}_{p}^{\lambda } \oplus \left( {\text{ finite }p\text{-group }}\right)\n\]\n\nas \( {\mathbb{Z}}_{p} \) -modules. | Proof. We have\n\n\[ \nX \sim E = {\bigoplus }_{j}\Lambda /\left( {{g}_{j}\left( T\right) }\right)\n\]\n\nwhere each \( {g}_{j} \) is distinguished and \( \sum \deg {g}_{j} = \lambda \) . By the division algorithm,\n\n\[ \n\Lambda /\left( {{g}_{j}\left( T\right) }\right) \simeq {\mathbb{Z}}_{p}^{\deg {g}_{j}}\n\]\n\nTh... | Yes |
Proposition 13.26. Let \( p \) be odd. Suppose \( K \) is a CM-field and \( {K}_{\infty }/K \) is the cyclotomic \( {\mathbb{Z}}_{p} \) -extension of \( K \) . Then the map\n\n\[ \n{A}_{n}^{ - } \rightarrow {A}_{n + 1}^{ - }\n\]\n\nis injective. | Proof. Suppose \( I \) is an ideal in \( {A}_{n} \) which becomes principal in \( {K}_{n + 1} \), so\n\n\[ \nI = \left( \alpha \right) \;\text{ with }\alpha \in {K}_{n + 1}.\n\]\n\nLet \( \sigma \) be a generator for \( \operatorname{Gal}\left( {{K}_{n + 1}/{K}_{n}}\right) \) . Then\n\n\[ \n\left( {\alpha }^{\sigma - 1... | Yes |
Lemma 13.27. If \( {\varepsilon }_{1} \in {W}_{n + 1} \) and \( N{\varepsilon }_{1} = 1 \) then \( {\varepsilon }_{1} = {\varepsilon }_{2}^{\sigma - 1} \) with \( {\varepsilon }_{2} \in {W}_{n + 1} \) (so \( {H}^{1}\left( {\operatorname{Gal}\left( {{K}_{n + 1}/{K}_{n}}\right) ,{W}_{n + 1}}\right) = 0). \) | Proof. Hilbert’s Theorem 90 tells us that \( {\varepsilon }_{1} = {y}^{\sigma - 1} \) with \( y \in {K}_{n + 1} \), but we already know this with \( y = {\alpha }_{1} \) . We want \( y \in {W}_{n + 1} \) . Consider the following two sequences:\n\n\[ 1 \rightarrow {W}_{n} \rightarrow {W}_{n + 1}\xrightarrow[]{\sigma - 1... | Yes |
Proposition 13.28. Let \( p \) be odd, let \( K \) be a \( {CM} \) -field, and let \( {K}_{\infty }/K \) be the cyclotomic \( {\mathbb{Z}}_{p} \) -extension. Then \( {X}^{ - } = \mathop{\lim }\limits_{ \leftarrow }{A}_{n}^{ - } \) contains no finite \( \Lambda \) -submodules. Therefore there is an injection, with finit... | Proof. Suppose \( F \subseteq {X}^{ - } \) is a finite \( \Lambda \) -module. Let \( {\gamma }_{0} \) be a generator of \( \operatorname{Gal}\left( {{K}_{\infty }/K}\right) \) . Since \( F \) is finite, \( {\gamma }_{0}^{{p}^{n}} \) acts trivially on \( F \) for all sufficiently large \( n \) , say \( n \geq {n}_{0} \)... | Yes |
Let \( p \) be odd. Let \( K \) be a CM-field and let \( {K}_{\infty }/K \) be the cyclotomic \( {\mathbb{Z}}_{p} \) -extension. If \( {\mu }^{ - } = 0 \) then\n\n\[ \n{X}^{ - } \simeq {\mathbb{Z}}_{p}^{{\lambda }^{ - }}\n\] | Proof. Proposition 13.28 plus the analogue of Proposition 13.25 for \( {X}^{ - } \) . | No |
Theorem 13.31. \( {\mathcal{X}}_{\infty } \sim {\Lambda }^{{r}_{2}} \oplus \) ( \( \Lambda \) -torsion). | One advantage of using \( {\mathcal{X}}_{\infty } \) rather than \( X \) is that it is easier to describe how \( {M}_{\infty } \) is generated. Since all \( p \) -power roots of unity are in \( {K}_{\infty },{M}_{\infty }/{K}_{\infty } \) is a Kummer extension. There is a subgroup\n\n\[ \nV \subseteq {K}_{\infty }^{ \t... | Yes |
Proposition 13.32. \( {\varepsilon }_{j}{\mathcal{X}}_{\infty }\left( {-1}\right) \simeq {\operatorname{Hom}}_{{\mathbb{Z}}_{p}}\left( {{\varepsilon }_{i}{A}_{\infty },{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}}\right) \) as \( \Lambda \) -modules, where \( i + j \equiv 1{\;\operatorname{mod}\;\left| \Delta \right| } \) and \( ... | Proof. We shall show more generally that\n\n\[ \n{\operatorname{Hom}}_{{\mathbb{Z}}_{p}}\left( {B,{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}}\right) \simeq {\operatorname{Hom}}_{{\mathbb{Z}}_{p}}\left( {B,{W}_{{p}^{\infty }}}\right) { \otimes }_{{\mathbb{Z}}_{p}}{T}^{\left( -1\right) }\n\] \n\nfor any \( \Lambda \) -module \( B... | Yes |
Lemma 13.33. Let \( u \in {U}_{1} \) and let \( 1 \leq n \leq p - 2 \) . Then \( {\phi }_{k}\left( u\right) = 0 \) for \( 1 \leq k \leq \) \( n \Leftrightarrow u \equiv 1{\;\operatorname{mod}\;{\left( {\zeta }_{p} - 1\right) }^{n + 1}}. \) | Proof. If \( u \equiv 1{\;\operatorname{mod}\;{\left( {\zeta }_{p} - 1\right) }^{n + 1}} \) we may take \( f\left( T\right) = 1 + {T}^{n + 1}g\left( T\right) \) with \( g \in \Lambda \) . Since \( {f}^{\prime }\left( T\right) \in {T}^{n}\Lambda ,{\phi }_{k}\left( u\right) = 0 \) for all \( k \leq n \) .\n\nConversely, ... | Yes |
Lemma 13.34. Let \( {\sigma }_{a} \in \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{p}\right) /\mathbb{Q}}\right) \) . Then\n\n\[{\phi }_{k}\left( {{\sigma }_{a}u}\right) = {a}^{k}{\phi }_{k}\left( u\right)\] | Proof. Let \( u = f\left( {{\zeta }_{p} - 1}\right) \) . Define\n\n\[g\left( T\right) = f\left( {{\left( 1 + T\right) }^{a} - 1}\right) .\]\n\nThen \( {\sigma }_{a}u = g\left( {{\zeta }_{p} - 1}\right) \), and\n\n\[\left( {1 + T}\right) \frac{{g}^{\prime }}{g} = a{\left( 1 + T\right) }^{a}\frac{{f}^{\prime }}{f}\left( ... | Yes |
Lemma 13.35. Let \( {\varepsilon }_{i},0 \leq i \leq p - 2 \), be the idempotents of \( {\mathbb{Z}}_{p}\left\lbrack {\operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{p}\right) /\mathbb{Q}}\right. }\right\rbrack \) .\n\nIf \( u \in {U}_{1} \) then\n\n\[ \n{\phi }_{k}\left( {{\varepsilon }_{i}u}\right) = \left\{ \be... | Proof. By Lemma 13.34,\n\n\[ \n{\phi }_{k}\left( {{\varepsilon }_{i}u}\right) = \frac{1}{p - 1}\mathop{\sum }\limits_{{a = 1}}^{{p - 1}}{\omega }^{-i}\left( a\right) {a}^{k}{\phi }_{k}\left( u\right) \equiv \left\{ \begin{array}{ll} 0{\;\operatorname{mod}\;p}, & \text{ if }k \neq i, \\ {\phi }_{k}\left( u\right) {\;\op... | Yes |
If \( i ≢ 1{\;\operatorname{mod}\;p} - 1 \) then \( {\varepsilon }_{i}{U}_{1} \) is cyclic as a \( {\mathbb{Z}}_{p} \) -module. For \( i = 1 \) , \( {\varepsilon }_{1}{U}_{1} \simeq \left\langle {\zeta }_{p}\right\rangle \times \) (cyclic \( {\mathbb{Z}}_{p} \) -module). If \( 2 \leq i \leq p - 2 \) and \( u \in {\vare... | Proof. First, let \( i \geq 1 \) be arbitrary. Since \( \left( {{\zeta }_{p}^{a} - 1}\right) /\left( {{\zeta }_{p} - 1}\right) \equiv a{\;\operatorname{mod}\;\left( {{\zeta }_{p} - 1}\right) } \), \[ {\eta }_{i}\overset{\text{ def }}{ = }{\left( 1 - {\left( {\zeta }_{p} - 1\right) }^{i}\right) }^{{\varepsilon }_{i}} = ... | Yes |
Corollary 13.37. Let \( 2 \leq i \leq p - 2 \) . There exists \( \lambda = {\lambda }_{i} \neq 1 \) with \( {\lambda }^{p - 1} = 1 \) such that\n\n\[{\xi }_{i} = {\varepsilon }_{i}\left( \frac{\lambda - {\zeta }_{p}}{\omega \left( {\lambda - 1}\right) }\right)\]\n\ngenerates \( {\varepsilon }_{i}{U}_{1} \) . (We divide... | Proof. By Lemma 13.36, it suffices to find \( \lambda \) such that \( {\phi }_{i}\left( {\xi }_{i}\right) \neq 0 \), and by Lemma 13.35, we can work with \( \left( {\lambda - {\zeta }_{p}}\right) /\omega \left( {\lambda - 1}\right) \) . Let\n\n\[f\left( T\right) = \frac{\lambda - 1 - T}{\omega \left( {\lambda - 1}\righ... | Yes |
Theorem 13.38. Let \( u = \left( {u}_{n}\right) \in U \) . Then there exists a unique \( {f}_{u} \in \Lambda \) such that\n\n\[ \n{f}_{u}\left( {{\zeta }_{{p}^{n + 1}} - 1}\right) = {u}_{n}\;\text{ for all }n \geq 0.\n\]\n\nThe map\n\n\[ \nU \rightarrow {\Lambda }^{ \times }\n\]\n\n\[ \nu \mapsto {f}_{u}\n\]\n\ngives a... | Proof. Corollary 7.4 implies the uniqueness of \( {f}_{u} \) .\n\nFor simplicity, let \( {v}_{n} = {\zeta }_{{p}^{n + 1}} - 1 \) . Assume for the moment that \( {f}_{u} \) exists. Since\n\n\[ \n{f}_{u}\left( 0\right) \equiv {f}_{u}\left( {{\zeta }_{p} - 1}\right) = {u}_{0} \equiv 1{\;\operatorname{mod}\;\left( {{\zeta ... | Yes |
There exists a unique map \( N : \Lambda \rightarrow \Lambda \) such that \[ \left( {Nf}\right) \left( {{\left( 1 + T\right) }^{p} - 1}\right) = \mathop{\prod }\limits_{{{\zeta }^{p} = 1}}f\left( {\zeta \left( {1 + T}\right) - 1}\right) . \] | Proof. Let \( g\left( T\right) \) be the power series on the right, defined by the product. Observe that for \( {\zeta }^{p} = 1 \) , \[ g\left( {\zeta \left( {1 + T}\right) - 1}\right) = g\left( T\right) . \] Suppose we have \( {a}_{0},\ldots ,{a}_{n - 1} \in {\mathbb{Z}}_{p} \) and \( {g}_{n}\left( T\right) \in \Lamb... | Yes |
Lemma 13.40. Let \( f \in \Lambda \) . Then\n\n\[ \left( {Nf}\right) \left( {v}_{n - 1}\right) = {N}_{n, n - 1}\left( {f\left( {v}_{n}\right) }\right) \] | Proof. \( \left( {Nf}\right) \left( {v}_{n - 1}\right) = \left( {Nf}\right) \left( {{\left( 1 + {v}_{n}\right) }^{p} - 1}\right) \)\n\n\[ = \prod f\left( {\zeta \left( {1 + {v}_{n}}\right) - 1}\right) = {N}_{n, n - 1}\left( {f\left( {v}_{n}\right) }\right) \] \n\nas in a previous calculation. | Yes |
Lemma 13.41. Let \( f \in \Lambda \) and assume \( f\left( {{\left( 1 + T\right) }^{p} - 1}\right) \equiv 1{\;\operatorname{mod}\;{p}^{k}}\Lambda \) . Then \( f\left( T\right) \equiv 1{\;\operatorname{mod}\;{p}^{k}}\Lambda \) | Proof. We may assume \( f \neq 1 \) . Let\n\n\[ f\left( T\right) = 1 + {p}^{\mu }\mathop{\sum }\limits_{{i = 0}}^{\infty }{a}_{i}{T}^{i} \]\n\nfor some \( \mu \geq 0 \), with \( \mu \) maximal. Let \( {a}_{n} \) be the first coefficient such that \( p \nmid {a}_{n} \) . Then\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{\inft... | Yes |
Corollary 13.42. \( N : {\Lambda }^{ \times } \rightarrow {\Lambda }^{ \times } \) is continuous. | Proof. Since \( N \) is a homomorphism it suffices to check continuity at 1 . Lemma 13.41 and the definition of \( N \) yield the result. | No |
Lemma 13.43. Suppose \( f \in {\Lambda }^{ \times } \) . Then\n\n\[ \frac{{N}^{k}f}{f} \equiv 1{\;\operatorname{mod}\;p}\Lambda \]\n\nfor all \( k \geq 0 \) . | Proof. Since\n\n\[ \frac{{N}^{k}f}{f} = \frac{N\left( {{N}^{k - 1}f}\right) }{{N}^{k - 1}f}\cdots \frac{N\left( f\right) }{f} \]\n\nit suffices to consider \( k = 1 \) . We have\n\n\[ \frac{\left( {Nf}\right) \left( {{\left( 1 + T\right) }^{p} - 1}\right) }{f\left( {{\left( 1 + T\right) }^{p} - 1}\right) } = \frac{\pro... | Yes |
Lemma 13.44. Let \( k \geq 1 \) . Then\n\n\[ f \equiv 1{\;\operatorname{mod}\;{p}^{k}}\Lambda \Rightarrow {Nf} \equiv 1{\;\operatorname{mod}\;{p}^{k + 1}}\Lambda . \]\n | Proof. Write \( f\left( T\right) = 1 + {p}^{k}{f}_{1}\left( T\right) \) . Then\n\n\[ f\left( {\zeta \left( {1 + T}\right) - 1}\right) \equiv 1 + {p}^{k}{f}_{1}\left( T\right) {\;\operatorname{mod}\;\left( {{\zeta }_{p} - 1}\right) }{p}^{k}. \]\n\nTherefore\n\n\[ \left( {Nf}\right) \left( {{\left( 1 + T\right) }^{p} - 1... | Yes |
Corollary 13.45. Let \( m \geq k \geq 0 \) and let \( f \in {\Lambda }^{ \times } \) . Then\n\n\[ {N}^{m}f \equiv {N}^{k}f{\;\operatorname{mod}\;{p}^{k + 1}}.\] | Proof. Lemma 13.43 implies \( {N}^{m - k}f/f \equiv 1{\;\operatorname{mod}\;p}\Lambda \) . Lemma 13.44 yields the result. | No |
Corollary 13.46. Let \( f \in {\Lambda }^{ \times } \) . Then \( {N}^{\infty }f = \lim {N}^{k}f \) exists. | Proof. Corollary 13.45, plus the completeness of \( \Lambda \) . | No |
Lemma 13.48. Let \( u \in U \) be associated to \( {f}_{u} \), and let\n\n\[ x = \sum {b}_{a}{\sigma }_{a} \in {\mathbb{Z}}_{p}\left\lbrack {\operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{{p}^{\infty }}\right) /\mathbb{Q}}\right) }\right\rbrack \]\n\nThen\n\n\[ {f}_{{u}^{x}}\left( T\right) = \prod {f}_{u}{\left( ... | It is trivial to check that everything above is defined; for example,\n\n\[ {\left( 1 + T\right) }^{a} = \sum \left( \begin{array}{l} a \\ n \end{array}\right) {T}^{n} \in \Lambda \] | No |
Lemma 13.49. Let \( a \in {\mathbb{Z}}_{p}^{ \times } \) and \( {\sigma }_{a} \in \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{{p}^{\infty }}\right) /\mathbb{Q}}\right) \) . Then, for \( k \geq 1 \) , \[ {\delta }_{k}\left( {{\sigma }_{a}u}\right) = {a}^{k}{\delta }_{k}\left( u\right) \] | Proof. See the proof of Lemma 13.34. | No |
Lemma 13.50. If \( u \in U \) then\n\n\[{\delta }_{k}\left( {{\varepsilon }_{i}u}\right) = \left\{ \begin{array}{ll} 0, & \text{ if }k ≢ i{\;\operatorname{mod}\;p} - 1 \\ {\delta }_{k}\left( u\right) , & \text{ if }k \equiv i{\;\operatorname{mod}\;p} - 1. \end{array}\right.\] | Proof. See the proof of Lemma 13.35. Note that\n\n\[{\varepsilon }_{i} = \frac{1}{p - 1}\mathop{\sum }\limits_{{a = 1}}^{{p - 1}}{\omega }^{-i}\left( a\right) {\sigma }_{\omega \left( a\right) }\]\n\nis the idempotent since \( \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{p}\right) /\mathbb{Q}}\right) \) corresp... | No |
Proposition 13.51. Let \( 2 \leq i \leq p - 1 \) . There exists \( {h}_{i}\left( T\right) \in {\Lambda }^{ \times } \) such that \n\n\[ \n\left( {1 - {p}^{k - 1}}\right) {\delta }_{k}\left( {\xi }_{i}^{\infty }\right) = {h}_{i}\left( {{\kappa }_{0}^{k} - 1}\right) \;\text{ for }k \equiv i{\;\operatorname{mod}\;p} - 1. ... | Proof. By Lemma 13.50 we may compute \n\n\[ \n{\delta }_{k}\left( \frac{\lambda - {\zeta }_{{p}^{n + 1}}}{\omega \left( {\lambda - 1}\right) }\right) \n\] \n\nLet \n\n\[ \nf\left( T\right) = \left( {1 + T}\right) \frac{d}{dT}\log \left( \frac{\lambda - \left( {1 + T}\right) }{\omega \left( {\lambda - 1}\right) }\right)... | Yes |
Let \( h\left( T\right) \in \Lambda \) and \( u \in U \) . Then\n\n\[ {\delta }_{k}\left( {h\left( T\right) u}\right) = h\left( {{\kappa }_{0}^{k} - 1}\right) {\delta }_{k}\left( u\right) \] | Proof. Since both sides are continuous in \( h \), we may assume \( h \) is a polynomial; by linearity, we may assume \( h\left( T\right) = {\left( 1 + T\right) }^{n} \) . Then \( h\left( T\right) u = {\gamma }_{0}^{n}u \) . By Lemma 13.49,\n\n\[ {\delta }_{k}\left( {{\gamma }_{0}^{n}u}\right) = {\kappa }_{0}^{nk}{\del... | Yes |
Lemma 13.53. Let \( {u}_{n} \in {U}_{1}^{\left( n\right) } \). Then\n\n\[ \n{u}_{n} \in {U}_{n}^{\prime } \Leftrightarrow \text{ for all }m \geq n\text{, there exists }{u}_{m} \in {U}_{1}^{\left( m\right) }\text{ with }{N}_{m, n}\left( {u}_{m}\right) = {u}_{n}\text{. }\n\] | Proof. For simplicity, let \( {K}_{m} = {\mathbb{Q}}_{p}\left( {\zeta }_{{p}^{m + 1}}\right) \). An element \( a \in 1 + p{\mathbb{Z}}_{p} \) yields \( {\sigma }_{a} \in \operatorname{Gal}\left( {{K}_{m}/{\mathbb{Q}}_{p}}\right) \), and\n\n\[ \n{\sigma }_{a} = 1 \Leftrightarrow a \equiv 1{\;\operatorname{mod}\;{p}^{m +... | Yes |
Theorem 14.2. If \( K/{\mathbb{Q}}_{p} \) is a finite abelian extension, then\n\n\[ K \subseteq {\mathbb{Q}}_{p}\left( {\zeta }_{n}\right) \]\n\nfor some \( n \) . | Proof. We first show it suffices to prove 14.2.\n\n## 14.2 (for all \( p \) ) \( \Rightarrow \) 14.1 .\n\nAssume \( K/\mathbb{Q} \) is abelian. Let \( p \) be a prime which ramifies in this extension. Let \( {K}_{p} \) be the completion at a prime above \( p \) . Then \( {K}_{p}/{\mathbb{Q}}_{p} \) is abelian, so\n\n\[... | No |
Lemma 14.3. If \( F/\mathbb{Q} \) is an extension in which no finite prime ramifies, then \( F = \mathbb{Q} \) . | Proof. A theorem of Minkowski (see Exercise 2.5) states that every ideal class of \( F \) contains an integral ideal of norm less than or equal to\n\n\[ \frac{n!}{{n}^{n}}{\left( \frac{4}{\pi }\right) }^{{r}_{2}}\sqrt{{d}_{F}} \]\n\nwhere \( n = \left\lbrack {F : \mathbb{Q}}\right\rbrack ,{d}_{F} \) is the absolute val... | Yes |
Lemma 14.4. Let \( K \) and \( L \) be finite extensions of \( {\mathbb{Q}}_{p} \) such that \( K/L \) is unramified. Then\n\n(a) \( K = L\left( {\zeta }_{n}\right) \) for some \( n \) with \( p \nmid n \), and\n\n(b) \( \operatorname{Gal}\left( {K/L}\right) \) is cyclic.\n\nAlso, for fixed \( L \) and for every intege... | We sketch the proof. First consider (a) and (b), and assume \( K/L \) is Galois. Let \( {\mathcal{O}}_{K} \) and \( {h}_{K} \) be the integers and maximal ideal for \( K \) and define \( {\mathcal{O}}_{L} \) and \( {h}_{L} \) similarly. Since \( K/L \) is unramified, there is a canonical isomorphism\n\n\[ \operatorname... | Yes |
Lemma 14.5. Let \( K \) and \( L \) be finite extensions of \( {\mathbb{Q}}_{p} \) and let \( {h}_{L} \) be the maximal ideal of the integers of \( L \) . Suppose \( K/L \) is totally ramified of degree \( e \) with \( p \nmid e \) (i.e., \( K/L \) is tamely ramified). Then there exists \( \pi \in L \) of order 1 at \(... | Proof. Let \( \left| x\right| \) be the absolute value on \( {\mathbb{C}}_{p} \) ( \( = \) completion of the algebraic closure of \( {\mathbb{Q}}_{p} \) ). Let \( {\pi }_{0} \in {\mathcal{M}}_{L} \) be of order 1 . Choose \( \beta \in K \) to be a uniformizing parameter, so that \[ {\left| \beta \right| }^{e} = \left| ... | Yes |
Lemma 14.6. \( {\mathbb{Q}}_{p}\left( {\left( -p\right) }^{1/\left( {p - 1}\right) }\right) = {\mathbb{Q}}_{p}\left( {\zeta }_{p}\right) \). | Proof. Let\n\n\[ g\left( X\right) = \frac{{\left( X + 1\right) }^{p} - 1}{X} \]\n\n\[ = {X}^{p - 1} + p{X}^{p - 2} + \cdots + p. \]\n\nThen\n\n\[ 0 = g\left( {{\zeta }_{p} - 1}\right) \equiv {\left( {\zeta }_{p} - 1\right) }^{p - 1} + p{\;\operatorname{mod}\;{\left( {\zeta }_{p} - 1\right) }^{p}}, \]\n\nso\n\n\[ u = \f... | Yes |
Lemma 14.7. Let \( F \) be a field of characteristic \( \neq p \), let \( M = F\left( {\zeta }_{p}\right) \), and let \( L = M\left( {a}^{1/p}\right) \) for some \( a \in M \) . Define the character \( \omega : \operatorname{Gal}\left( {M/F}\right) \rightarrow {\mathbb{Z}}_{p}^{ \times } \) by \( \sigma {\zeta }_{p} = ... | Proof of Lemma 14.7. Let \( G = \operatorname{Gal}\left( {M/F}\right) \) and \( H = \operatorname{Gal}\left( {L/M}\right) \) . Then \( G \) acts on \( H \) as follows. If \( \sigma \in G \), extend it to an element of \( \operatorname{Gal}\left( {L/F}\right) \) . Then define \( {h}^{\sigma } = {\sigma h}{\sigma }^{-1} ... | Yes |
Lemma 14.9. \( {A}^{2} + {B}^{2} = - 1 \) has no solutions in \( {\mathbb{Q}}_{2} \) . | Proof. We may transform this to\n\n\[ \n{A}_{1}^{2} + {A}_{2}^{2} + {A}_{3}^{2} = 0 \n\] \n\nwith \( {A}_{i} \in {\mathbb{Z}}_{2},1 \leq i \leq 3 \), and \( 2 \nmid {A}_{i} \) for some \( i \) . But there are no nontrivial solutions mod 8 . This completes the proof of the lemma. | Yes |
Theorem 15.2. Let \( F \) be a totally real abelian number field with \( \Delta = \operatorname{Gal}\left( {F/\mathbb{Q}}\right) \) , let \( E \) be the group of units of the ring of integers of \( F \), let \( {C}^{\prime } \) be the group of cyclotomic units defined above, and let \( A \) be the class group of \( F \... | Proof. Choose \( n \) large enough that \( {p}^{n} > \left| A\right| \) and \( {p}^{n} > \left| {E/{C}^{\prime }}\right| \) . Then the Sylow \( p \) -subgroup of \( A \) is isomorphic to \( A/{A}^{{p}^{n}} \) and the Sylow \( p \) -subgroup of \( E/{C}^{\prime } \) is isomorphic to \( E/{E}^{{p}^{n}}{C}^{\prime } \) . ... | Yes |
Lemma 15.3. Let \( \delta \in {C}^{\prime } \) . There exists a unit \( \varepsilon \in L \) such that \( {N}_{L/F}\left( \varepsilon \right) = 1 \) and \( \varepsilon \equiv \delta \left( {{\;\operatorname{mod}\;\sigma }\mathcal{L}}\right) \) for all \( \sigma \) . | Proof. From the definition of \( {C}^{\prime } \), we can write \( \delta = \pm {N}_{\mathbf{Q}\left( {\zeta }_{m}\right) /F}\left( {\mathop{\prod }\limits_{a}{\left( {\zeta }_{m}^{a} - 1\right) }^{{b}_{a}}}\right) \) . Let \( \ell ,\lambda ,\mathcal{L} \), and \( \sigma \) be as above. Note that \( \ell \nmid m \) sin... | Yes |
Lemma 15.5. \( {\bar{\varepsilon }}_{\rho }{\mathbb{F}}_{p}\left\lbrack \Delta \right\rbrack \) is an irreducible \( {\mathbb{F}}_{p}\left\lbrack \Delta \right\rbrack \) -module. | Proof. Suppose \( 0 \neq N \subsetneqq {\bar{\varepsilon }}_{\rho }{\mathbb{F}}_{p}\left\lbrack \Delta \right\rbrack \) . Then \( 0 \neq N \otimes {\overline{\mathbb{F}}}_{p} \subsetneqq {\bar{\varepsilon }}_{\rho }{\overline{\mathbb{F}}}_{p}\left\lbrack \Delta \right\rbrack \) (the inequalities hold because the dimens... | Yes |
Lemma 15.6. Suppose \( \theta \in {\varepsilon }_{\rho }\mathbb{Z}/{p}^{n}\mathbb{Z}\left\lbrack \Delta \right\rbrack \) and \( {p}^{a} \) is the highest power of \( p \) dividing \( \theta \), with \( 0 \leq a < n \) . Then there exists \( {\theta }^{\prime } \in {\varepsilon }_{\rho }\mathbb{Z}/{p}^{n}\mathbb{Z}\left... | Proof. By assumption, \( 0 \neq \overline{{p}^{-a}\theta } \in {\bar{\varepsilon }}_{\rho }{\mathbb{F}}_{p}\left\lbrack \Delta \right\rbrack \) . Since this module is irreducible, \( \overline{{p}^{-a}\theta }{\bar{\varepsilon }}_{\rho }{\mathbb{F}}_{p}\left\lbrack \Delta \right\rbrack = {\bar{\varepsilon }}_{\rho }{\m... | Yes |
Theorem 15.8. Let \( i \) be odd with \( 3 \leq i \leq p - 2 \) . If \( p \) divides the numerator of the Bernoulli number \( {B}_{p - i} \), then \( {\varepsilon }_{i}A \neq 0 \) . | Proof (assuming Theorem 15.7). Let \( {U}_{1} \) be the local units of \( \mathbb{Z}{\left\lbrack {\zeta }_{p}\right\rbrack }^{ + } \) that are congruent to 1 modulo the prime above \( p \) . Let \( {\bar{E}}_{1} \) be the closure of \( E \cap {U}_{1} \) and \( {\bar{C}}_{1} \) the closure of \( C \cap {U}_{1} \) .\n\n... | Yes |
Lemma 15.9. (a) Assume \( \ell \nmid L \) . Then \( {N}_{\ell L/L}\alpha \left( {\ell L}\right) = \alpha {\left( L\right) }^{{\mathrm{{Frob}}}_{\ell } - 1} \), where \( {\mathrm{{Frob}}}_{\ell } \) is the Frobenius for \( \ell \) for the extension \( F\left( L\right) /\mathbb{Q} \) . | Proof. The norm for \( \mathbb{Q}\left( {{\zeta }_{m},{\zeta }_{\ell L}}\right) /\mathbb{Q}\left( {{\zeta }_{m},{\zeta }_{L}}\right) \) of \( 1 - {\zeta }_{m}^{j}{\zeta }_{\ell L} \) is\n\n\[ \mathop{\prod }\limits_{{k = 1}}^{{\ell - 1}}\left( {1 - {\zeta }_{m}^{j}{\zeta }_{L}{\zeta }_{\ell }^{k}}\right) = \frac{1 - {\... | Yes |
Proposition 15.10. There exists \( {\beta }_{L} \in F{\left( L\right) }^{ \times } \) and \( \kappa \left( L\right) \in {F}^{ \times } \) such that\n\n\[ \n{D}_{L}\alpha \left( L\right) = \kappa \left( L\right) {\beta }_{L}^{M} \n\]\n\nand \( {\left( \left( \sigma - 1\right) {D}_{\ell }\alpha \left( L\right) \right) }^... | Proof. We claim that \( {D}_{L}\alpha \left( L\right) \in {\left( F{\left( L\right) }^{ \times }/F{\left( L\right) }^{\times M}\right) }^{G} \) (=the elements fixed by \( G \) ), where \( G = \operatorname{Gal}\left( {F\left( L\right) /F}\right) \) . The proof of the claim is by induction on the number of prime factors... | Yes |
Lemma 15.11. There exists \( \beta \in F{\left( L\right) }^{ \times } \) such that \( c\left( \sigma \right) = {\beta }^{\sigma - 1} \) for all \( \sigma \in G \) . | Proof. By the theorem on linear independence of characters, there exists \( x \in F{\left( L\right) }^{ \times } \) such that \( y = \mathop{\sum }\limits_{{\sigma \in G}}c\left( \sigma \right) \sigma \left( x\right) \neq 0 \) . Let \( \tau \in G \) . The cocycle condition implies that\n\n\[ \n{\tau y} = \mathop{\sum }... | Yes |
Proposition 15.12. Suppose \( \kappa \left( L\right) \equiv {s}^{a}\left( {\;\operatorname{mod}\;\lambda }\right) \) . Then the \( \lambda \) -adic valuation of \( \kappa \left( {\ell L}\right) \) satisfies\n\n\[ \n{v}_{\lambda }\left( {\kappa \left( {\ell L}\right) }\right) \equiv - a\;\left( {\;\operatorname{mod}\;M}... | Proof. From Proposition 15.10 and Lemma 15.5, we have\n\n\[ \n\left( {{\sigma }_{\ell } - 1}\right) {\beta }_{\ell L} = {\left( \left( {\sigma }_{\ell } - 1}\right) {D}_{\ell L}\alpha \left( \ell L\right) \right) }^{1/M} = {\left( \left( \ell - 1 - {N}_{\ell }\right) {D}_{L}\alpha \left( \ell L\right) \right) }^{1/M} \... | Yes |
Lemma 15.13. Let \( \lambda ,\ell \), and \( s \) be as above, with \( \ell \equiv 1\left( {{\;\operatorname{mod}\;m}{ML}}\right) \) . Assume the ideal class \( \mathfrak{C} \) of \( \lambda \) is in \( {\varepsilon }_{x}A \) and also that the classes of the prime ideals of \( F \) dividing \( L \) are in \( {\varepsil... | Proof. Let \( \sigma \in \operatorname{Gal}\left( {F/\mathbb{Q}}\right) \) . Then \( s \) is a primitive root \( {\;\operatorname{mod}\;\sigma }\lambda \) . Let\n\n\[ \kappa \left( L\right) \equiv {s}^{{a}_{\sigma }}\;\left( {{\;\operatorname{mod}\;\sigma }\lambda }\right) \]\n\nWe then have \( {\sigma }^{-1}\kappa \le... | Yes |
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