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Theorem 1.1. Let \( {f}^{ - } \) be the function such that \( {f}^{ - }\left( x\right) = f\left( {-x}\right) \). Then \( {T}^{2}f = q{f}^{ - } \), that is\n\n\[ \n{T}^{2}f\left( z\right) = {qf}\left( {-z}\right) \n\]
Proof. We have\n\n\[ \n{T}^{2}f\left( z\right) = \mathop{\sum }\limits_{y}\mathop{\sum }\limits_{x}f\left( x\right) \lambda \left( {-{yx}}\right) \lambda \left( {-{zy}}\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{x}f\left( {x - z}\right) \mathop{\sum }\limits_{y}\lambda \left( {-{yx}}\right) \n\]\n\nIf \( x \neq 0 \) ...
Yes
Theorem 1.2. For functions \( f, g \) on \( F \) we have\n\n\[ T\left( {f * g}\right) = \left( {Tf}\right) \left( {Tg}\right) \]\n\n\[ T\left( {fg}\right) = \frac{1}{q}{Tf} * {Tg} \]
Proof. For the first formula we have\n\n\[ T\left( {f * g}\right) \left( z\right) = \mathop{\sum }\limits_{y}\left( {f * g}\right) \left( y\right) \lambda \left( {-{zy}}\right) = \mathop{\sum }\limits_{y}\mathop{\sum }\limits_{x}f\left( x\right) g\left( {y - x}\right) \lambda \left( {-{zy}}\right) .\n\nWe change the or...
Yes
Theorem 1.3. Assume that \( \chi \) has order \( m \) .\n\n(i) \( S{\left( \chi \right) }^{m} \) lies in \( \mathbf{Q}\left( {\mu }_{m}\right) \) .\n\n(ii) Let \( b \) be an integer prime to \( m \), and let \( {\sigma }_{b} = {\sigma }_{b,1} \) . Then \( S{\left( \chi \right) }^{b - {\sigma }_{b}} \) lies in \( \mathb...
Proof. In each case we operate on the given expression by an automorphism \( {\sigma }_{1, v} \) with an integer \( v \) prime to \( {pm} \) . Using GS 5, it is then obvious that the given expression is fixed under such an automorphism, and hence lies in \( \mathbf{Q}\left( {\mu }_{m}\right) \) .
No
Theorem 2.2. We have the factorization\n\n\[ S\left( {\omega }^{-k}\right) \sim {\mathfrak{P}}^{\left( {p - 1}\right) \theta \left( {k,\mathfrak{p}}\right) } \sim {\mathfrak{p}}^{\theta \left( {k,\mathfrak{p}}\right) }.\]
Proof. We have\n\n\[ \operatorname{ord}{\sigma }_{c}^{-1}{PS}\left( {\omega }^{-k}\right) = {\operatorname{ord}}_{\mathfrak{P}}{\sigma }_{c}S\left( {\omega }^{-k}\right) \]\n\n\[ = {\operatorname{ord}}_{\mathfrak{P}}S\left( {\omega }^{-{kc}}\right) \]\n\n\[ = s\left( {kc}\right) \]\n\nby Theorem 2.1. On the other hand,...
No
For any integer \( k \) we have\n\n\[ s\left( k\right) = \left( {p - 1}\right) \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}\left\langle \frac{k{p}^{i}}{q - 1}\right\rangle . \]
Proof. We may assume that \( 1 \leq k < q - 1 \) since both sides are \( \left( {q - 1}\right) \) - periodic in \( k \), and the relation is obvious for \( k = 0 \) . Since \( {p}^{n} \equiv 1\left( {{\;\operatorname{mod}\;q} - 1}\right) \) we find:\n\n\[ k = {k}_{0} + {k}_{1}p + \cdots + {k}_{n - 1}{p}^{n - 1} \]\n\n\...
Yes
Lemma 2. We have \( {I\theta } = {R\theta } \cap R \) .
Proof. Note that \( m \in I \) because\n\n\[ m = - \left( {{\sigma }_{1 + m} - \left( {1 + m}\right) }\right) . \]\n\nSuppose that an element of \( {R\theta } \) lies in \( R \), that is\n\n\[ \sum z\left( b\right) {\sigma }_{b}\theta \in R \]\n\nwith \( z\left( b\right) \in \mathbf{Z} \) . Then\n\n\[ \sum z\left( b\ri...
Yes
Theorem 2.4. The Stickelberger ideal annihilates the ideal class group of \( \\mathbf{Q}\\left( {\\mu }_{m}\\right) \) .
Proof. Let\n\n\[ \n\\alpha = \\mathop{\\sum }\\limits_{r}z\\left( r\\right) {\\theta }_{r}\\left( m\\right) \\in R \n\] \n\nbe an element of the Stickelberger ideal, with \( z\\left( r\\right) \\in \\mathbf{Z} \), and the sum taken with only a finite number of coefficients \( \\neq 0 \) . Then\n\n\[ \n\\mathop{\\sum }\...
Yes
Lemma 1. (i) If \( \chi = \omega \) is the Teichmuller character, then \( {I}_{\chi } = \left( p\right) \). (ii) If \( \chi \) is non-trivial and not equal to the Teichmuller character, then \( {I}_{\chi } = \left( 1\right) \).
Proof. For (i), we can take an integer \( b \) of the form \[ b = \zeta + {pu} \] where \( u \) is a \( p \) -adic unit, and \( \zeta = \omega \left( b\right) \) is a \( \left( {p - 1}\right) \) th root of unity. This makes (i) clear, and (ii) is obvious, from the definitions.
No
Corollary 1. Assume that \( m = p \) is prime \( \geq 3 \) . If \( \chi \) is not equal to the Teichmuller character and is non-trivial, then\n\n\[ \text{ord}{B}_{1,\bar{\chi }}{I}_{\chi } = \text{ord}{B}_{1,\bar{\chi }}\text{.} \]
Proof. Immediate from the lemma and the theorem.
No
Corollary 2. If \( \chi \) is equal to the Teichmuller character then \( {B}_{1,\bar{\chi }}{I}_{\chi } = \left( 1\right) \) , and \( {\mathcal{C}}^{\left( p\right) }\left( \chi \right) = 0 \) .
Proof. Mod \( {\mathbf{Z}}_{p} \), we have the congruence\n\n\[ \n{B}_{1,{\omega }^{-1}} = \frac{1}{p}\mathop{\sum }\limits_{{c = 1}}^{{p - 1}}{c\omega }{\left( c\right) }^{-1} \equiv \frac{1}{p}\mathop{\sum }\limits_{{c = 1}}^{{p - 1}}1 \equiv \frac{p - 1}{p}\;\left( {\;\operatorname{mod}\;{\mathbf{Z}}_{p}}\right) .\n...
No
Corollary 3 (Herbrand’s theorem). Assume again that \( m = p \) . Let \( \chi = {\omega }^{1 - k} \) , with \( 2 \leq k \leq p - 2 \) . If \( {\mathcal{C}}^{\left( p\right) }\left( \chi \right) \neq 0 \), then \( p \mid {B}_{k} \), where \( {B}_{k} \) is the \( k \) th Bernoulli number.
Proof. In the next chapter Theorem 2.5, we shall prove the congruence\n\n\[ \frac{1}{n}{B}_{n,{\omega }^{k - n}} \equiv \frac{1}{k}{B}_{k}\left( {\;\operatorname{mod}\;p}\right) \]\n\nfor \( k \) in the given range, and any positive integer \( n \) . By Corollary 1, we know that \( {B}_{1,\bar{\chi }} \) annihilates \(...
No
Theorem 4.1. The algebraic number \( w\left( {a,\alpha }\right) \) is a root of unity.
Proof. As (α) ranges over all principal fractional ideals, the numbers \( w\left( {a,\alpha }\right) \) form a group. It will therefore suffice to prove that these numbers have absolute value 1 , for then their conjugates also have absolute value 1 , and these numbers form a finite group. In case \( a \) is special the...
Yes
Theorem 4.2. If \( \alpha \) is an algebraic integer in \( \mathbf{Q}\left( {\mu }_{m}\right) \), and \( \alpha \equiv 1\left( {\;\operatorname{mod}\;{m}^{2}}\right) \)\nthen for all a we have \( w\left( {a,\alpha }\right) = 1 \), that is,\n\n\[ J\left( {a,\left( \alpha \right) }\right) = {\alpha }^{\theta \left\lbrack...
Proof. We fix \( \alpha \) and view \( J, w \) as functions of \( a \), omitting \( \alpha \) from the notation. In the Fourier inversion relation, we know that the Fourier coefficients \( \widehat{J}\left( b\right) \) are integers. But \( \alpha \equiv 1\left( {\;\operatorname{mod}\;{m}^{2}}\right) \) implies that\n\n...
Yes
Theorem 6.1. Let \( N \) be the number of points of \( V\left( d\right) \) (in affine space) in the field \( F \) . Then\n\n\[ N = {q}^{2} - \left( {q - 1}\right) \sum {\chi }^{a + b}\left( {-1}\right) J\left( {{\chi }^{a},{\chi }^{b}}\right) .\n\]\nThe sum is taken over integers \( a, b \) satisfying \( 0 < a < d \) a...
Proof. We have\n\n\[ N = \mathop{\sum }\limits_{{a, b, c}}\mathop{\sum }\limits_{{L\left( {u, v, w}\right) = 0}}{\chi }^{a}\left( u\right) {\chi }^{b}\left( v\right) {\chi }^{c}\left( w\right) \]\n\nwhere the sum over \( u, v, w \) is taken over triples of elements of \( F \) lying on the line\n\n\[ u + v + w = 0. \]\n...
Yes
Lemma 1. We have \( {R}^{ - } = 2{\varepsilon }^{ - }R = \left( {1 - {\sigma }_{-1}}\right) R \) and\n\n\[ \left( {{\varepsilon }^{ - }R : {R}^{ - }}\right) = {2}^{M}\text{.} \]
Proof. The inclusion \( \left( {1 - {\sigma }_{-1}}\right) R \subset {R}^{ - } \) is clear. Conversely, let \( P \) be a set of representatives in \( \mathbf{Z}{\left( m\right) }^{ * } \) for \( \mathbf{Z}{\left( m\right) }^{ * }/ \pm 1 \) . Let\n\n\[ \alpha = \sum z\left( c\right) {\sigma }_{c}^{-1} \in {R}^{ - } \]\n...
Yes
Lemma 2. \[ {R\theta } \cap R = {\left( R{\theta }^{\prime } \cap R\right) }^{ - }. \]
Proof. Let \( T = R{\theta }^{\prime } \cap R \) . Clearly \[ {T}^{ - } \subset {\varepsilon }^{ - }{R\theta } = {R\theta }\text{ and }{T}^{ - } \subset R, \] so the inclusion \( \supset \) is obvious. Conversely, let \( \alpha \in {R\theta } \cap R \) . It will suffice to prove that \( \alpha \in R{\theta }^{\prime } ...
Yes
Lemma 3. \[ \left( {{R\theta } : {R\theta } \cap R}\right) = w. \]
Proof. We define a homomorphism \[ T : {R\theta } \rightarrow \frac{1}{w}\mathbb{Z}/\mathbb{Z} \] by mapping an element of the group algebra on its first coefficient \( {\;\operatorname{mod}\;\mathbf{Z}} \) . In other words, if \[ \alpha = \sum a\left( c\right) {\sigma }_{c} \] we let \( {T\alpha } = a\left( 1\right) \...
Yes
Lemma 4. \[ \left( {{R\theta } : {Rm\theta }}\right) = {m}^{M}. \]
Proof. This is obvious if one can show that \( {R\theta } \) is a free abelian group of rank \( M \) . When \( m \) is a prime power, this results from the fact that for odd \( \chi \) we have \[ \chi \left( \theta \right) = {B}_{1,\chi } \neq 0. \]
No
Lemma 5.\n\[ \n\\left( {{\\varepsilon }^{ - }R : {\\varepsilon }^{ - }{Rm\\theta }}\\right) = \\pm {m}^{M}\\mathop{\\prod }\\limits_{{\\chi \\text{ odd }}}{B}_{1,\\chi }.\n\]
Proof. First observe that the sign is whatever is needed to make the righthand side positive. Multiplication by \( {\\varepsilon }^{ - }{m\\theta } \) is an endomorphism of \( \\mathbf{Q}{R}^{ - } \) , which is a semisimple algebra, decomposing into a product of 1-dimensional algebras corresponding to the odd character...
Yes
Theorem 2.1. (i) The values of \( {E}_{k, c}^{\left( N\right) } \) are \( N \) -integral.\n\n(ii) We have the congruence for every prime \( p \) dividing \( N \) :\n\n\[ {E}_{k, c}^{\left( N\right) }\left( x\right) \equiv {x}^{k - 1}{E}_{1, c}^{\left( N\right) }\left( x\right) {\;\operatorname{mod}\;N}{\mathbf{Z}}_{p}....
Proof. For large integer \( v \) the values \( {N}^{v}/{kD}\left( k\right) \) are \( N \) -integral. Let \( M = {N}^{v} \) . The distribution relation yields\n\n\[ {E}_{k, c}^{\left( N\right) }\left( x\right) = \mathop{\sum }\limits_{y}{E}_{k, c}^{\left( M\right) }\left( y\right) \]\n\nwhere the sum is taken over those...
Yes
Theorem 2.3. Let \( c \in {\mathbf{Z}}_{p}^{ * } \) and let \( k \) be an integer \( \geq 1 \) such that \( {c}^{k} \neq 1 \) . Then\n\n\[ \frac{1}{k}{B}_{k} = \frac{1}{1 - {c}^{k}}{\int }_{{\mathbf{z}}_{p}}{x}^{k - 1}d{E}_{1, c}\left( x\right) \]
Proof. By definition,\n\n\[ \frac{1}{k}{B}_{k} = {\int }_{{\mathbf{Z}}_{p}}d{E}_{k} = {\int }_{{\mathbf{Z}}_{p}}d{E}_{k, c} + {\int }_{{\mathbf{Z}}_{p}}{c}^{k}d{E}_{k}\left( {{c}^{-1}x}\right) . \]\n\nOn the last integral to the right, we make the change of variable\n\n\[ x \mapsto {cx} \]\nwhich gives\n\n\[ {\int }_{{...
No
Corollary 1 (Kummer Congruence). Let \( \alpha \) be a residue class \( {\;\operatorname{mod}\;p} - 1 \) and \( \alpha \neq 0 \) . Then for even positive integers \( k \equiv \alpha {\;\operatorname{mod}\;p} - 1 \), the values \( \left( {1/k}\right) {B}_{k} \) are all congruent \( {\;\operatorname{mod}\;p} \), and are ...
Proof. Select \( c \) to be a primitive root \( {\;\operatorname{mod}\;p} \) so that\n\n\[ {c}^{k} ≢ 1{\;\operatorname{mod}\;p}\text{.} \]\n\nThen \( 1 - {c}^{k} \) is a unit at \( p \) . The values \( 1 - {c}^{k} \) and \( {x}^{k - 1}{\;\operatorname{mod}\;p} \) are independent of the choice of \( k \) in the residue ...
No
Corollary 2 (Von Staudt Congruence). Let \( k \equiv 0{\;\operatorname{mod}\;p} - 1 \), and \( k \) even.\n\nThen\n\n\[ \n{B}_{k} \equiv - \frac{1}{p}{\;\operatorname{mod}\;{\mathbf{Z}}_{p}}.\n\]
Proof. Suppose \( p \) odd for simplicity. Let \( c = 1 + p \) . An easy induction shows that\n\n\[ \n{c}^{k} \equiv 1 + {pk}{\;\operatorname{mod}\;{p}^{2}}k{\mathbf{Z}}_{p}\n\]\n\nHence\n\n\[ \n\frac{1}{1 - {c}^{k}} = - \frac{1}{pk}\left( {1 + O\left( p\right) }\right)\n\]\n\nand so\n\n\[ \n{B}_{k} \equiv - \frac{1}{p...
No
Theorem 2.4. Let \( \psi \) be a character of finite order on \( {\mathbf{Z}}_{p}^{ * } \) . Then \[ \frac{1}{n}{B}_{n,\psi } = \frac{1}{1 - \psi \left( c\right) {c}^{n}}{\int }_{{\mathbf{Z}}_{p}^{ * }}\psi \left( a\right) {a}^{n - 1}d{E}_{1, c}\left( a\right) . \]
Proof. We write \( d{E}_{n} = d{E}_{n, c} + {c}^{n}d{E}_{n} \circ {c}^{-1} \), or in other words \[ \frac{1}{n}{B}_{n,\psi } = \int {\psi d}{E}_{n, c} + \int \psi \left( x\right) {c}^{n}d{E}_{n}\left( {{c}^{-1}x}\right) . \] Integrals are taken over \( {\mathbf{Z}}_{p}^{ * } \) . We let \( x \mapsto {cx} \) in the seco...
Yes
Theorem 2.5. Let \( 2 \leq k \leq p - 2 \) . Let \( \omega : \mathbf{Z}{\left( p\right) }^{ * } \rightarrow {\mathbf{Z}}_{p}^{ * } \) be the Teichmuller character such that\n\n\[ \omega \left( a\right) \equiv a\left( {\;\operatorname{mod}\;p}\right) \]\n\nFor any integer \( n \geq 1 \) we have\n\n\[ \frac{1}{n}{B}_{n,{...
Proof. Let \( \psi = {\omega }^{k - n} \) . Choose \( c \) to be a primitive root \( {\;\operatorname{mod}\;p} \), so that \( {c}^{k} ≢ 1{\;\operatorname{mod}\;p} \) . By Theorem 2.3 we get\n\n\[ \frac{1}{k}{B}_{k} \equiv \frac{1}{1 - {c}^{k}}{\int }_{{\mathbf{Z}}_{p}^{ * }}{x}^{k - 1}d{E}_{1, c}\left( x\right) \left( ...
Yes
Theorem 3.1. (i) We have\n\n\[ R{\theta }_{k}^{\prime } \cap R = {I}^{\left( k\right) }{\theta }_{k}^{\prime } \]\n\nIn fact, if an element \( \xi \in R \) is such that \( \xi {\theta }^{\prime } \in R \), then \( \xi \in {I}^{\left( k\right) } \).
Proof. First we prove that for any prime \( \geq 2 \), we have\n\n\[ I{\theta }^{\prime } \subset R,\text{ and }{I}_{p}\theta \subset {R}_{p}. \]\n\nA similar property is due to Mazur and Coates-Sinnott, as mentioned before. Indeed, we have\n\n\[ {\sigma }_{c}^{-1}\left( {{\sigma }_{c} - {c}^{k}}\right) {\theta }_{k} =...
No
Lemma 1. The polynomial \( \left( {1/k}\right) \left( {{\mathbf{B}}_{k}\left( X\right) - {\mathbf{B}}_{k}\left( 0\right) }\right) \) maps \( \mathbf{Z} \) into \( \mathbf{Z} \) and maps \( {\mathbf{Z}}_{l} \) into \( {\mathbf{Z}}_{l} \) for every prime \( l \) .
Proof. A standard property of Bernoulli polynomials states that\n\n\[ \n\frac{1}{k}\left( {{\mathbf{B}}_{k}\left( {X + 1}\right) - {\mathbf{B}}_{k}\left( X\right) }\right) = {X}^{k - 1}.\n\]\n\nHence for any integer \( m \) we see recursively that the first assertion of the lemma is true. The second, concerning \( l \)...
No
Lemma 2. (i) Let \( \xi \in R \) and suppose that \( \xi {\theta }^{\prime } \in {\mathbf{Z}}_{p}\left\lbrack G\right\rbrack = {R}_{p} \) . Then \( \xi \in J \) .
Proof. Write \( \xi = \sum z\left( b\right) {\sigma }_{b} \) with integral coefficients \( z\left( b\right) \) . Then\n\n\[ \xi {\theta }^{\prime } = {N}^{k - 1}\mathop{\sum }\limits_{c}\mathop{\sum }\limits_{b}z\left( b\right) \frac{1}{k}{\mathbf{B}}_{k}^{\prime }\left( \left\langle \frac{bc}{N}\right\rangle \right) {...
Yes
Lemma 3. Let \( {p}^{s} \) be the smallest power of \( p \) such that \( {p}^{s}{\theta }_{k}^{\prime } \) is p-integral. Then\n\n\[ s = n + {\operatorname{ord}}_{p}k. \]\n\nWe have \( {I}^{\left( k\right) } \cap \mathbb{Z} = \left( {p}^{s}\right) \).
Proof. The argument uses the same expression for the Bernoulli polynomial as in the previous lemma. We see that\n\n\[ {p}^{s}\sum \frac{{N}^{k - 1}}{k}\left( \begin{array}{l} k \\ i \end{array}\right) {B}_{i}{\left( \frac{1}{N}\right) }^{k - i}\text{is}p\text{-integral.} \]\n\nThe leading term is \( {p}^{s}/{kN} \) . T...
Yes
Lemma 4. We have \( J = I + \mathbf{Z}N \), and \( \left( {J : I}\right) = {p}^{s - n} = {p}^{\text{ord }k} \) .
Proof. It is clear that \( N \in J \) . Conversely, write an element of \( J \) in the form\n\n\[ \sum m\left( c\right) \left( {{\sigma }_{c} - {c}^{k}}\right) + \sum m\left( c\right) {c}^{k} \]\n\nThe first term is in \( I \), and the second term is an integral multiple of \( N \) . This proves the lemma.
No
Theorem 5.1. \[ \left( {{R}_{0} : {R}_{0} \cap {R\theta }}\right) = N{p}^{\operatorname{ord}k - t}\mathop{\prod }\limits_{{\chi \neq 1}} \pm \frac{1}{k}{B}_{k,\chi }. \]
First observe that since \( \deg \theta \) and \( \deg {\theta }^{\prime } \neq 0 \) we have \[ {R}_{0} \cap {R\theta } = {R}_{0} \cap R{\theta }^{\prime }. \] By Theorem 2.1, we conclude that \[ R{\theta }^{\prime } \cap R = I{\theta }^{\prime },\text{ and hence }R{\theta }^{\prime } \cap {R}_{0} = {I}_{0}{\theta }^{\...
Yes
Lemma 1. We have \( {R}^{ - } = 2{\varepsilon }^{ - }R \) and \( \left( {{\varepsilon }^{ - }R : {R}^{ - }}\right) = {2}^{\phi \left( N\right) /2} \) .
Proof. This is the same as Lemma 1 of \( §1 \) .
No
Lemma 2. \( \left( {{R}^{ - } : 2{\varepsilon }^{ - }I}\right) = {p}^{s} \) where \( s = n + {\operatorname{ord}}_{p}k \) .
Proof. The group \( 2{\varepsilon }^{ - }I \) is generated by elements of the form\n\n\[ \left( {{\sigma }_{c} - {\sigma }_{-c}}\right) - {c}^{k}\left( {{\sigma }_{1} - {\sigma }_{-1}}\right) \]\n\nAn element \( \xi \in {R}^{ - } \) lies in \( \mathbf{Z}\left( {{\sigma }_{1} - {\sigma }_{-1}}\right) {\;\operatorname{mo...
No
Theorem 8.1. The function \( g : \mathbf{Q}/\mathbf{Z} \rightarrow \lim F\left\lbrack {G\left( N\right) }\right\rbrack \) is an ordinary distribution.
Proof. Immediate from the definitions.
No
Theorem 8.2. The dimension of \( {A}_{N} \) is equal to the cardinality of \( {\widehat{G}}_{h}\left( N\right) \) .
Proof. The space generated by the elements \( {g}_{N}\left( r\right) \) with \( r \in {Z}_{N} \) is clearly a \( G\left( N\right) \) -module since\n\n\[ \n{\sigma }_{b}{g}_{N}\left( r\right) = {g}_{N}\left( {rb}\right) ,\;\text{ for }b \in G\left( N\right) .\n\] \n\nWe let the idempotent associated with \( \chi \) be t...
Yes
Theorem 9.1. (i) The elements \( g\left( {T}_{N}\right) \) generate the abelian group generated by \( g\left( {Z}_{N}\right) \) .
Proof. The first statement is obvious from the preceding remarks.
No
Theorem 10.1. (Davenport-Hasse) We have\n\n\[ \mathop{\prod }\limits_{{{\chi }^{m} = 1}}\tau \left( {\chi \psi }\right) = \tau \left( {\psi }^{m}\right) C\left( {\psi, m}\right) \]\n\nwhere \( C\left( {\psi, m}\right) = \psi \left( {m}^{-m}\right) \mathop{\prod }\limits_{{{\chi }^{m} = 1}}\tau \left( \chi \right) \) .
Proof. Let \( {u}_{m}\left( \psi \right) \) be the quotient of the left-hand side by the right-hand side, that is\n\n\[ {u}_{m}\left( \psi \right) = \frac{\prod \tau \left( {\chi \psi }\right) }{\tau \left( {\psi }^{m}\right) C\left( {\psi, m}\right) }.\]\n\nWe have to show \( {u}_{m}\left( \psi \right) = 1 \) . First ...
Yes
Lemma 1. Let \( 0 \leq k < q - 1 \) . Then\n\n\[ k! \equiv {\left( -p\right) }^{\frac{k - s\left( k\right) }{p - 1}}\gamma \left( k\right) {\;\operatorname{mod}\; * }p. \]
Proof. By induction. Suppose first that \( p \nmid k \) . Then \( {k}_{0} \geq 1 \), and\n\n\[ s\left( k\right) = s\left( {k - 1}\right) + 1,\;\gamma \left( k\right) = \gamma \left( {k - 1}\right) {k}_{0}. \]\n\nThe assertion is then obvious from the inductive step for \( k - 1 \) . Next suppose \( p \mid k \), so \( k...
Yes
Theorem 1.1. If \( \chi \) is primitive and \( d \) is not prime to \( m \), then\n\n\[ S\left( {\chi ,\lambda \circ d}\right) = 0. \]
Proof. Using the prime power decomposition, we may assume without loss of generality that \( m = {p}^{n} \) is a prime power. Abbreviate\n\n\[ A = \mathbb{Z}\left( {p}^{n}\right) \]\n\nAlso without loss of generality, we may assume \( d = {p}^{r} \) for some integer \( r \geq 1 \), and \( r < n \) . Form a coset decomp...
Yes
Theorem 1.2. (i) We have \( {T}^{2}f = m{f}^{ - } \) .\n\n(ii) If \( \chi \) is primitive, then\n\n\[ \n{T\chi } = \chi \left( {-1}\right) S\left( \chi \right) {\chi }^{-1}.\n\]\n\n(iii) Again if \( \chi \) is primitive, then\n\n\[ \nS\left( \chi \right) \overline{S\left( \chi \right) } = m \n\]
Proof. Part (i) is proved as for the finite field case. For (ii), if \( y \) is not prime to \( m \), then \( {T\chi }\left( y\right) = 0 \) by Theorem 1.1. If \( y \) is prime to \( m \) then we can make the usual change of variables to get the right answer. Part (iii) is then proved as in the finite field case.
No
Theorem 2.1. Assume that \( \chi \) is a primitive character \( {\;\operatorname{mod}\;m} \) . Then\n\n\[ L\left( {s,\chi }\right) = \frac{1}{m}S\left( \chi \right) \mathop{\sum }\limits_{{b \in \mathbf{Z}{\left( m\right) }^{ * }}}\bar{\chi }\left( b\right) \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\zeta }^{-{nb}...
Proof. If \( x \) is not prime to \( m \) then the Gauss sum is 0 by Theorem 1.1. If \( b \) is prime to \( m \), we can make the change of variables which yields the desired expression.
No
Theorem 2.2. If \( \chi \) is a primitive character, then\n\n\[ L\left( {1,\chi }\right) = - \frac{S\left( \chi \right) }{m}\mathop{\sum }\limits_{{b \in \mathbf{Z}{\left( m\right) }^{ * }}}\bar{\chi }\left( b\right) \log \left( {1 - {\zeta }^{-b}}\right) . \]
Case 1. \( \chi \) is even.\n\nIn this case, adding the sum with \( b \) and \( - b \) yields\n\n\[ 2\sum \bar{\chi }\left( b\right) \log \left( {1 - {\zeta }^{-b}}\right) = \sum \bar{\chi }\left( b\right) \left\lbrack {\log \left( {1 - {\zeta }^{b}}\right) + \log \left( {1 - {\zeta }^{-b}}\right) }\right\rbrack . \]\n...
Yes
Theorem 3.1. We have product expressions:\n\n(i)\n\n\\[ \n\\mathop{\\prod }\\limits_{{\\chi \\neq 1}}m\\left( \\chi \\right) = d \n\\]\n\n(ii)\n\n\\[ \n\\mathop{\\prod }\\limits_{{\\chi \\neq 1}}S\\left( \\chi \\right) = \\left\\{ \\begin{array}{ll} {d}^{1/2} & \\text{ if }K\\text{ is real } \\\\ {i}^{{r}_{2}}{d}^{1/2}...
Proof. It is possible to give essentially algebraic proofs for these facts (although the sign of the Gauss sums is always a little delicate, involving something about the complex numbers). The best way to see the theorem, however, is probably as in Hasse [Ha 1], using the functional equations of the zeta function and \...
Yes
Theorem 3.2. For imaginary \( K \) , \[ h = {h}^{ + }{Qw}{2}^{-N/2}\mathop{\prod }\limits_{{\chi \text{ odd }}} - {B}_{1,\chi } \]
In the next section, we shall analyze more closely the decomposition \[ h = {h}^{ + }{h}^{ - } \] where \( {h}^{ - } \) is defined as \( h/{h}^{ + } \), and we shall see that \( {h}^{ - } \) is an integer. In any case, we have the class number formula: \( {\mathrm{{CNF}}}^{ - } \). \[ {h}^{ - } = {Qw}\mathop{\prod }\li...
No
Theorem 3.3. If \( m \) is a prime power, \( K = \mathbf{Q}\left( {\mu }_{m}\right), G = \operatorname{Gal}\left( {K/\mathbf{Q}}\right) \), and \( \mathcal{S} \) is the Stickelberger ideal, then\n\n\[ \n{h}^{ - } = \left( {\mathbf{Z}{\left\lbrack G\right\rbrack }^{ - } : \mathcal{S}}\right) \n\]
Let \( p \) be a prime number. If \( A \) is an abelian group, we denote by \( {A}^{\left( p\right) } \) its \( p \) -primary part. As Iwasawa observed [Iw 7], knowing the index immediately shows that:\n\nThe group \( {C}_{K}^{-\left( p\right) } \) is generated by one element over \( \mathbf{Z}\left\lbrack G\right\rbra...
No
Theorem 4.1. Let \( K = \mathbf{Q}\left( {\mu }_{m}\right) \) . Then \( {Q}_{K} = 1 \) if \( m \) is a prime power, and 2 if \( m \) is not a prime power.
Proof. Let \( E = {E}_{K} \) be the unit group in \( K \) . For each unit \( u \) in \( E \), the quotient \( \bar{u}/u \) is a unit, of absolute value 1, and for any automorphism \( \sigma \) of \( K \) over \( \mathbf{Q} \), we have\n\n\[ \sigma \left( {\bar{u}/u}\right) = \overline{\sigma u}/{\sigma u} \]\n\nbecause...
Yes
Theorem 4.2. Let \( K = \mathbf{Q}\left( {\mu }_{m}\right) \) . The natural map\n\n\[ \n{C}_{K} + \rightarrow {C}_{K} \n\]\n\nof ideal classes in \( {K}^{ + } \) into the ideal class group of \( K \) is injective.
Proof. Let \( \mathfrak{a} \) be an ideal of \( {K}^{ + } \) and suppose \( \mathfrak{a} = \left( \alpha \right) \) with \( \alpha \) in \( K \) . Then \( \bar{\alpha }/\alpha \) is a unit, and in fact a root of unity as one sees by an argument similar to that in Theorem 4.1. Suppose that \( m \) is composite. By Theor...
Yes
Theorem 4.3. Let \( K \) be an imaginary abelian extension of \( \mathbf{Q} \). Then the norm map\n\n\[ \n{N}_{K/{K}^{ + }} : {C}_{K} \rightarrow {C}_{{K}^{ + }} \n\]\n\non the ideal class group is surjective.
Proof. We have to use class field theory, which gives the more general statement:\n\nLemma. Let \( K \) be an abelian extension of a number field \( F \). Let \( H \) be the\n\nHilbert class field of \( F \) (maximal abelian unramified extension of \( F \)). If\n\n\( K \cap H = F \) then the norm map \( {N}_{K/F} : {C}...
Yes
Theorem 4.4. Let \( K = \mathbf{Q}\left( {\mu }_{m}\right) \) . Then the sequence\n\n\[ 1 \rightarrow {C}_{K}^{ - } \rightarrow {C}_{K}\xrightarrow[]{\text{ norm }}{C}_{{K}^{ + }} \rightarrow 1 \]\n\nis exact.
Proof. We consider the norm map followed by the injection,\n\n\[ {C}_{K}\xrightarrow[]{\text{ norm }}{C}_{{K}^{ + }}\xrightarrow[]{\text{ inj }}{C}_{K \]\n\nThe kernel of this composite map is \( {C}_{K}^{ - } \) by definition, so the theorem is obvious by what had already been proved.
No
Theorem 5.1. Let \( K = \mathbf{Q}\left( {\mu }_{m}\right) \) and \( h = {h}_{K} \) . Assume \( m = {p}^{n} \) is a prime power.\n\nThen\n\n\[ \n{h}^{ + } = \left( {{E}^{ + } : {\mathcal{E}}^{ + }}\right) = \left( {E : \mathcal{E}}\right) .\n\]
Proof. Let \( G \) be any finite abelian group. Then we have the Frobenius determinant formula for any function \( f \) on \( G \) :\n\n\[ \n\mathop{\prod }\limits_{{\chi \neq 1}}\mathop{\sum }\limits_{{a \in G}}\chi \left( a\right) f\left( {a}^{-1}\right) = {\det }_{a, b \neq 1}\left\lbrack {f\left( {a{b}^{-1}}\right)...
No
Lemma 1. We have for \( G = \mathbf{Z}{\left( m\right) }^{ * }/ \pm 1 \) :\n\n\[ \pm {\det }_{a, b \neq 1}\log \left| {{\sigma }_{a}{g}_{b}}\right| = \mathop{\prod }\limits_{{\chi \neq 1}}\mathop{\sum }\limits_{{b \in G}}\chi \left( b\right) \log \left| {1 - {\zeta }^{b}}\right| \]\n\n\[ = \mathop{\prod }\limits_{{\chi...
Proof. The first expression comes from the Frobenius determinant formula (Theorem 6.2), and the second comes from the fact that for non-trivial \( \chi \) ,\n\n\[ \sum \chi \left( b\right) \log \left| {1 - \zeta }\right| = 0 \]
Yes
Lemma 2. Let \( {G}_{\chi } = \mathbf{Z}{\left( m\left( \chi \right) \right) }^{ * }/ \pm 1 \) . For prime power \( m = {p}^{n} \), we have\n\n\[ \mathop{\sum }\limits_{{b \in {G}_{\chi }}}\chi \left( b\right) \log \left| {1 - {\zeta }_{m\left( \chi \right) }^{b}}\right| = \mathop{\sum }\limits_{{b \in G}}\chi \left( b...
Proof. Let \( m\left( \chi \right) = {p}^{s} \) . We write residue classes in \( \mathbf{Z}{\left( {p}^{n}\right) }^{ * } \) in the form\n\n\[ y = b + {p}^{s}c,\;\text{ with }0 \leq c < {p}^{n - s}, \]\n\nand \( b \) ranges over a fixed set of representatives for residue classes of \( \mathbf{Z}{\left( {p}^{s}\right) }...
Yes
Theorem 5.3.\n\[ \n{h}_{F} = \left( {{E}_{F} : {\mathcal{E}}_{F}}\right) \n\]
Proof. Let\n\n\[ \n\alpha = \mathop{\prod }\limits_{{\chi \left( a\right) = 1}}\left( {1 - {\zeta }^{a}}\right) \;\text{ and }\;{\alpha }^{\prime } = \mathop{\prod }\limits_{{\chi \left( a\right) = - 1}}\left( {1 - {\zeta }^{a}}\right) \n\]\n\nwhere \( \zeta \) is a fixed primitive \( D \) th root of unity. Note that t...
Yes
Theorem 6.1. Let \( f \) be any (complex valued) function on \( G \) . Then\n\n\[ \mathop{\prod }\limits_{{\chi \in G}}\mathop{\sum }\limits_{{a \in G}}\chi \left( a\right) f\left( {a}^{-1}\right) = \mathop{\det }\limits_{{a, b}}f\left( {{a}^{-1}b}\right) \]
Proof. Let \( F \) be the space of functions on \( G \) . It is a finite dimensional vector space whose dimension is the order of \( G \) . It has two natural bases. First, the characters \( \{ \chi \} \), and second the functions \( \left\{ {\delta }_{b}\right\}, b \in G \), where\n\n\[ {\delta }_{b}\left( x\right) = ...
Yes
Theorem 6.2. The determinant in Theorem 4.1 splits into\n\n\\[ \n\\mathop{\\det }\\limits_{{a, b}}f\\left( {a{b}^{-1}}\\right) = \\left\\lbrack {\\mathop{\\sum }\\limits_{{a \\in G}}f\\left( a\\right) }\\right\\rbrack \\mathop{\\det }\\limits_{{a, b \\neq 1}}\\left\\lbrack {f\\left( {a{b}^{-1}}\\right) - f\\left( a\\ri...
Proof. Let \\( {a}_{1} = 1,\\ldots ,{a}_{n} \\) be the elements of \\( G \\) . In the determinant\n\n\\[ \n\\det f\\left( {{a}_{i}{a}_{j}^{-1}}\\right) = \\left| \\begin{matrix} f\\left( {{a}_{1}{a}_{1}^{-1}}\\right) & f\\left( {{a}_{1}{a}_{2}^{-1}}\\right) \\cdots f\\left( {{a}_{1}{a}_{n}^{-1}}\\right) \\\\ \\vdots & ...
Yes
Theorem 7.1.\n\[ \pm {D}_{p} = {\left( 2p\right) }^{\left( {p - 3}\right) /2}{h}_{p}^{ - } \]
where\n\[ {D}_{p} = \det \left\lbrack {R\left( {\zeta }^{i + j}\right) - R\left( {-{\zeta }^{i + j}}\right) }\right\rbrack .\n\]\nObserve that each entry in the determinant \( {D}_{p} \) is an integer of absolute value \( \leq p - 1 \) .\n\nThe absolute value of the determinant is the volume of the fundamental domain o...
Yes
Theorem 1.1. Let \( f\left( X\right) = \sum {c}_{k}{X}^{k} \in \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) . Then
\[ {c}_{k} = {\int }_{{\mathbb{Z}}_{p}}\left( \begin{array}{l} x \\ k \end{array}\right) d{\mu }_{f}\left( x\right) \]
No
Theorem 1.2. The power series \( {P\mu } \) is the unique power series \( f \) such that for \( z \) in the maximal ideal of \( \mathfrak{o} \), we have\n\n\[ \n{\int }_{{\mathbf{Z}}_{p}}{\left( 1 + z\right) }^{x}{d\mu }\left( x\right) = f\left( z\right) \n\]
Proof. We have\n\n\[ \n{\int }_{{\mathbf{Z}}_{p}}{\left( 1 + z\right) }^{x}d{\mu }_{f}\left( x\right) = {\int }_{{\mathbf{Z}}_{p}}\mathop{\sum }\limits_{{k = 0}}^{\infty }\left( \begin{array}{l} x \\ k \end{array}\right) {z}^{k}d{\mu }_{f}\left( x\right) . \n\]\n\nWe can interchange the sum and integral, apply Theorem ...
No
Let \( v \) be a measure on \( {\mathbf{Z}}_{p} \) whose support lies in the open closed subset \( 1 + p{\mathbf{Z}}_{p} \). Let \( \gamma \) be a topological generator of \( 1 + p{\mathbf{Z}}_{p} \), for instance \( \gamma = 1 + p \). There is an isomorphism\n\n\[ \n{\mathbf{Z}}_{p} \rightarrow 1 + p{\mathbf{Z}}_{p} \...
By Theorem 1.2, writing \( {\gamma }^{s} = 1 + z \), we get\n\n\[ \n{\int }_{1 + p{\mathbf{Z}}_{p}}{u}^{s}{dv}\left( u\right) = f\left( {{\gamma }^{s} - 1}\right) \n\]
Yes
Theorem 1.4. We have \( \parallel f\parallel = \begin{Vmatrix}{\mu }_{f}\end{Vmatrix} \) .
Proof. Since\n\n\[ \n{c}_{n} = \int \left( \begin{array}{l} x \\ n \end{array}\right) d{\mu }_{f}\left( x\right) \n\] \n\nwe get trivially \( \parallel f\parallel \leq \begin{Vmatrix}{\mu }_{f}\end{Vmatrix} \) . Conversely, given a level \( {p}^{n} \), let \( {x}_{0} \in \mathbf{Z}\left( {p}^{n}\right) \) and let \( \v...
Yes
Theorem 3.1. Let \( g \in \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) be such that \( \mathbf{U}g = g \), and let \( h \) be a power series such that \( {Dh} = g \) . Then \( \mathbf{U}h \in \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) and\n\n\[{\Gamma }_{p}\mathbf{U}h...
Proof. This is an immediate application of Meas 7, after integrating the function \( \langle a{\rangle }^{s} \) .
No
Theorem 3.2. The value of \( {L}_{p}\left( {1 - s,\chi }\right) \) is independent of the choice of \( c \) , and for any positive integer \( k \) ,\n\n\[ \n{L}_{p}\left( {1 - k,\chi }\right) = - \frac{1}{k}{B}_{k,\chi {\omega }^{-k}} \n\] \n\nIn particular, if \( k \equiv 0{\;\operatorname{mod}\;p} - 1 \), and \( p \) ...
Proof. Since the set of sufficiently large integers \( k \equiv 0{\;\operatorname{mod}\;p} - 1 \) is dense \nin \( {\mathbf{Z}}_{p} \), we see that the first assertion follows from the explicit values given at integers of the form \( 1 - k \) as described. For these, we have: \n\n\[ \n{\mathbf{M}}_{p}\left( {\chi {E}_{...
Yes
Theorem 3.3. Let \( g = \\mathbf{U}g \) and let \( h \) be the power series such that\n\n\[ \n{Dh} = g\\text{ and }h\\left( 0\\right) = 0.\n\]\n\nThen\n\n\[ \n{\\mathbf{M}}_{p}{\\mu }_{g}\\left( 0\\right) = - \\frac{1}{p}\\mathop{\\sum }\\limits_{{{\\xi }^{p} = 1}}h\\left( {\\xi - 1}\\right)\n\]
Proof. By Meas 7 we have\n\n\[ \n{\\mathbf{M}}_{p}{\\mu }_{g}\\left( 0\\right) = \\int {a}^{-1}d{\\mu }_{g}\\left( a\\right) = \\int d{\\mu }_{\\mathbf{U}h}\\left( a\\right) = \\mathbf{U}h\\left( 0\\right) .\n\]\n\nThe formula is then clear from the definition of \( \\mathbf{U} \) .
No
Proposition 3.4. Let \( c \in {\mathbf{Z}}_{p}^{ * } \) . The power series associated with the measure \( {E}_{1, c} \) is \[ {f}_{1, c} = \frac{1}{T - 1} - \frac{c}{{T}^{c} - 1},\;\text{ with }T = 1 + X. \]
Proof. It is immediate to verify that as power series in \( X \) the expression on the right-hand side is holomorphic at \( X = 0 \), and that its coefficients are \( p \) -integral because \( c \) is a \( p \) -unit. Let \[ f\left( T\right) = \frac{\log T}{T - 1} - \frac{c\log T}{{T}^{c} - 1} \] Putting \( T = {e}^{Z}...
Yes
Proposition 3.5. Let \( \\chi \) be a non-trivial character on \( {\\mathbf{Z}}_{p}^{ * } \) with conductor \( N \). The power series associated with \( \\chi {E}_{1, c} \) is \[ {g}_{\\chi, c} = {G}_{\\chi }\\left( T\\right) - {c\\chi }\\left( c\\right) {G}_{\\chi }\\left( {T}^{c}\\right) \] where \[ {G}_{\\chi }\\lef...
Proof. Immediate from Meas 5.
No
Theorem 3.6. Let \( \chi \) be a primitive Dirichlet character with conductor \( N \) equal to a power of p. Then\n\n\[ \n{L}_{p}\left( {1,\chi }\right) = - \frac{S\left( {\chi ,\zeta }\right) }{N}\mathop{\sum }\limits_{{a \in \mathbf{Z}{\left( N\right) }^{ * }}}\bar{\chi }\left( a\right) \log \left( {1 - {\zeta }^{a}}...
Proof. By Theorem 3.3, Proposition 3.5, and the definition of \( {L}_{p}\left( {1,\chi }\right) \), we find:\n\n\[ \n\left( {1 - \chi \left( c\right) }\right) {L}_{p}\left( {1,\chi }\right) = \frac{1}{p}\frac{S\left( {\chi ,\zeta }\right) }{N}\mathop{\sum }\limits_{\xi }\mathop{\sum }\limits_{a}\mathop{\sum }\limits_{{...
No
Let \( K \) be the real subfield of \( \mathbf{Q}\left( {\mu }_{m}\right) \) . Then\n\n\[ \n{R}_{p}\left( \mathcal{E}\right) = \left( {E : \mathcal{E}}\right) {R}_{p}\left( E\right) = \left( {E : \mathcal{E}}\right) {R}_{p}.\n\]
We know from Theorem 5.1 of the preceding chapter that\n\n\[ \n{h}^{ + } = \left( {E : \mathcal{E}}\right) \n\]\n\nLet \( {g}_{a} \) ( \( a \) prime to \( m \) ) be the cyclotomic units, and \( {g}_{a}^{ + } \) the corresponding real cyclotomic units. From our definition of the \( p \) -adic log, we know that for any e...
Yes
Theorem 4.2 (Brumer). We have \( {R}_{p} \neq 0 \) for the real cyclotomic field \( \mathbf{Q}{\left( {\mu }_{m}\right) }^{ + } \) .
Proof. The cyclotomic units are algebraic, and it is a known theorem from the theory of transcendental numbers that the logs ( \( p \) -adic or otherwise) of multiplicatively independent algebraic numbers are linearly independent over the algebraic numbers. The proof is the \( p \) -adic analogue of Baker’s proof for t...
Yes
Theorem 4.3 (Leopoldt p-adic Class Number-regulator Formula). Let \( m = {p}^{n} \) be a prime power, and \( {K}^{ + } = \mathbf{Q}{\left( {\mu }_{m}\right) }^{ + } \) . Then\n\n\[ \mathop{\prod }\limits_{\substack{{\chi \neq 1} \\ {\chi \text{ even }} }}\frac{1}{2}{L}_{p}\left( {1,\chi }\right) = \frac{{h}^{ + }}{\sqr...
Proof. From Theorem 4.1 and the complexly derived index\n\n\[ {h}^{ + } = \left( {E : \mathcal{E}}\right) \]\n\nof Theorem 5.1 in the preceding chapter, we find:\n\n\[ \pm {h}^{ + }{R}_{p}\left( E\right) = \pm {R}_{p}\left( \mathcal{E}\right) = \mathop{\prod }\limits_{\substack{{\chi \neq 1} \\ {\chi \text{ even }} }}\...
Yes
Theorem 6.1. If \( f \in \mathcal{L} \) then \( f \) converges on the disc of elements\n\n\[ x \in {\mathbb{C}}_{p}\;\text{ and }\;\left| x\right| \leq {\left| p\right| }^{1/\left( {p - 1}\right) }.\]\n\nFor such \( x \) we have\n\n\[ \left| {f\left( x\right) }\right| \leq \parallel f{\parallel }_{\mathcal{L}} \]
Proof. Obvious, because\n\n\[ {\left| p\right| }^{n/\left( {p - 1}\right) } \leq \left| {n!}\right| \text{ and so }\left| {{a}_{n}\frac{{x}^{n}}{n!}}\right| \leq \left| {a}_{n}\right| . \]
No
Theorem 6.2. Let \( \alpha \) be a residue class \( {\;\operatorname{mod}\;p} - 1 \) . There exists a unique continuous linear map\n\n\[ \n{\Gamma }_{\alpha } : {\mathcal{L}}_{K} \rightarrow C\left( {{\mathbb{Z}}_{p}, K}\right) \n\] \n\nsatisfying any one of the following three equivalent conditions:\n\n\( {\Gamma }_{\...
Proof. Any continuous linear map on the space of polynomials (with Leopoldt norm) extends uniquely by continuity to the Leopoldt Banach algebra. We shall prove that the linear map\n\n\[ \n{\Gamma }_{\alpha } : K\left\lbrack X\right\rbrack \rightarrow C\left( {{\mathbf{Z}}_{p}, K}\right) \n\] \n\nwith values\n\n\[ \n{\G...
Yes
Theorem 6.3. Let \( m \) be an integer \( \geq 0 \), and \( m \equiv \alpha {\;\operatorname{mod}\;p} - 1 \) . Then \[ {\Gamma }_{\alpha }f\left( m\right) = \Gamma \mathrm{U}f\left( m\right) \]
Proof. The two maps \[ f \mapsto {\Gamma }_{\alpha }f\text{ and }f \mapsto \Gamma \mathbb{U}f \] of \( K\left\lbrack X\right\rbrack \rightarrow C\left( {{\mathbf{Z}}_{p}, K}\right) \) are equal on the polynomials \( {\left( 1 + X\right) }^{v} \) . For a fixed \( m \) the maps \[ f \mapsto {\Gamma }_{\alpha }f\left( m\r...
No
Theorem 6.4. For \( f \in {\mathcal{L}}_{K} \) we have\n\n\[{\Gamma }_{0}f\left( 0\right) = \mathbf{U}f\left( 0\right) = f\left( 0\right) - \frac{1}{p}\mathop{\sum }\limits_{{{\zeta p} = 1}}f\left( {\zeta - 1}\right) .\]
Proof. The power series for \( \mathbf{U}f \) in terms of \( X \) or \( Z \) have the same constant term. Hence\n\n\[ \Gamma \mathrm{U}f\left( 0\right) = \mathrm{U}f\left( 0\right) \]\n\nTaking \( \alpha = 0 \), the theorem is obvious from Theorem 6.3, and the fact that \( \zeta - 1 \) lies in the domain of convergence...
No
Theorem 6.5. For \( s \in {\mathbf{Z}}_{p} \) and \( f \in \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) we have\n\n\[{\int }_{{\mathbf{Z}}_{p}^{ * }}\langle a{\rangle }^{s}d{\mu }_{f}\left( a\right) = {\Gamma }_{0}f\left( s\right)\]
Proof. By continuity in \( s \), it suffices to prove the theorem when \( s = k \) is an integer \( \geq 1 \) and \( k \equiv 0{\;\operatorname{mod}\;p} - 1 \) . Let \( \varphi \) be the characteristic function of \( {\mathbf{Z}}_{p}^{ * } \) . Then\n\n\[{\int }_{{\mathbf{Z}}_{p}^{ * }}\langle a{\rangle }^{k}d{\mu }_{f...
Yes
Theorem 1.1. The homomorphism \( \varepsilon \) is an isomorphism.
Proof. A trivial induction shows that\n\n\[ \n{h}_{n} = {\left( 1 + X\right) }^{{p}^{n}} - 1 \in {\left( p, X\right) }^{n + 1} \n\] \n\nwhere \( \left( {p, X}\right) \) denotes the maximal ideal of \( {\mathbf{Z}}_{p}\left\lbrack X\right\rbrack \), generated by \( p \) and \( X \) . It follows that the intersection of ...
Yes
Theorem 1.2. (i) If \( V = \Lambda /{p}^{m} \) then \( {e}_{n} = m{p}^{n} \) .
Proof. In case (i) we have\n\n\[ \mathbf{Z}\left( {p}^{m}\right) \left\lbrack \left\lbrack X\right\rbrack \right\rbrack /\left( {{\left( X + 1\right) }^{{p}^{n}} - 1}\right) \approx \mathbf{Z}\left( {p}^{m}\right) \left\lbrack X\right\rbrack /\left( {{\left( X + 1\right) }^{{p}^{n}} - 1}\right) ,\]\n\nand this is just ...
Yes
Theorem 1.3. Assume that \( V \) is of Iwasawa type. Then the conclusions of Theorem 1.2(i), (ii), (iii) remain valid, except that in 1.2(ii) we have to write the exponent\n\n\[ \n{e}_{n} = {dn} + {c}_{0} \n\]\n\nwith some constant \( {c}_{0} \) .
Proof. Note that Case (i) is unchanged, only Case (ii) is now slightly different, but the proof runs along entirely similar lines as follows. In this case, \( V \) is \( {\mathbf{Z}}_{p} \) -free of rank \( d \) . An argument similar to that of Theorem 1.2(ii) shows that\n\n\[ \n{g}_{n}V = {p}^{n - {n}_{0}}{g}_{{n}_{0}...
Yes
Theorem 2.1. Let \( \mathfrak{o} \) be a complete local ring with maximal ideal \( \mathfrak{m} \) . Let\n\n\[ f\left( X\right) = \mathop{\sum }\limits_{{i = 0}}^{\infty }{a}_{i}{X}^{i} \]\n\nbe a power series in \( \mathfrak{o}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \), such that not all \( {a}_{i} \) l...
Proof. Let \( \alpha \) and \( \tau \) be the projections on the beginning and tail end of the series, given by\n\n\[ \alpha : \sum {b}_{i}{X}^{i} \mapsto \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}{b}_{i}{X}^{i} = {b}_{0} + {b}_{1} + \cdots + {b}_{n - 1}{X}^{n - 1} \]\n\n\[ \tau : \sum {b}_{i}{X}^{i} \mapsto \mathop{\su...
Yes
Theorem 2.2 (Weierstrass Preparation). The power series \( f \) in the previous theorem can be written in the form\n\n\[ f\left( X\right) = \left( {{X}^{n} + {b}_{n - 1}{X}^{n - 1} + \cdots + {b}_{0}}\right) u, \]\n\nwhere \( {b}_{i} \in \mathfrak{m} \), and \( u \) is a unit in \( \mathfrak{o}\left\lbrack \left\lbrack...
Proof. Write\n\n\[ {X}^{n} = {qf} + r \]\n\nby the Euclidean algorithm. Then \( q \) is invertible because\n\n\[ q = {c}_{0} + {c}_{1}X + \cdots \]\n\n\[ f = \cdots + {a}_{n}{X}^{n} + \cdots \]\n\nso that\n\n\[ 1 \equiv {c}_{0}{a}_{n}\left( {\;\operatorname{mod}\;\mathfrak{m}}\right) \]\n\nand \( {c}_{0} \) is a unit i...
Yes
Theorem 3.2. If \( R \) is a matrix of relations, we can transform \( R \) with a finite number of admissible operations into a matrix \( {R}^{\prime } \) of the form\n\n\[ \left( \begin{matrix} {\lambda }_{11} & 0 & \cdots & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & & \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & {\lam...
Proof. By the lemma, we can replace \( R \) by a matrix \( {R}^{\prime } \) of the form\n\n\[ \left( \begin{matrix} {\lambda }_{11} & 0 & \cdots & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & & \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & {\lambda }_{rr} & 0 & \cdots & 0 \\ * & * & \cdots & * & 0 & \cdots & 0 \end{matrix}...
Yes
Theorem 4.1. Assume first that IW is satisfied with \( s = 1 \) . Let \( I \) be the inertia group of any prime above \( \mathfrak{p} \) in \( G \) . Then:\n\n(i) \( G = I{G}_{C} \) is a semidirect product, and the restriction of \( I \) to \( {K}_{\infty } \) gives an isomorphism of \( I \) and \( \Gamma \) .\n\n(ii) ...
Proof. We have an exact sequence\n\n\[ 1 \rightarrow {G}_{c} \rightarrow G \rightarrow \Gamma \rightarrow 1 \]\n\nThe image of \( I \) in \( \Gamma \) by restriction to \( {K}_{\infty } \) is surjective because \( {K}_{\infty } \) is totally ramified over \( {K}_{0} \) . It is injective because \( {M}_{\infty } \) is u...
Yes
Theorem 4.2. Assume that IW is satisfied, with primes \( {\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{s} \) . Let \( {I}_{j} \) be the inertia group of \( {\mathfrak{p}}_{j} \) in \( G \) . Then:\n\n(i) There is a semidirect product decomposition\n\n\[ G = {I}_{1}{G}_{C} \]\n\nand \( {G}^{\prime } = {G}_{C}^{\gamma - 1} ...
Proof. Identical with that of Theorem 4.1, except that in the present more general situation, we have to look at the smallest subgroup of \( G \) containing the commutator group \( {G}^{\prime } \) and all the inertia groups \( {I}_{j} \) instead of a single inertia group \( I \) .
No
Theorem 4.3. Assume that IW is satisfied with one prime. If \( {C}_{0} = \{ 1\} \), then \( {C}_{n} = \{ 1\} \) for all \( n \) .
Proof. If \( {C}_{0} = 1 \), then Theorem 4.1 shows that \( C = {C}^{\gamma - 1} \) . Viewing \( C \) as module over \( {\mathbf{Z}}_{p}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \), this means that \( C = {XC} \) . But \( X \) is contained in the maximal ideal of \( {\mathbf{Z}}_{p}\left\lbrack \left\lbrac...
Yes
Theorem 4.4. For any \( {\mathbf{Z}}_{p} \) -extension the module \( C \) over \( {\mathbf{Z}}_{p}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) is a finitely generated torsion module.
Proof. Suppose first for simplicity that condition IW is satisfied with only one prime. Then \( C/\mathfrak{m}C \) is a factor group of \( C/{C}^{\gamma - 1} \), which is none other than \( {C}_{0} \) by Theorem 4.1, and is therefore finite. That \( C \) is finitely generated is a special case of Nakayama's lemma.\n\nI...
Yes
Theorem 5.1. Let \( H \) be the p-Hilbert class field of \( K \) . Then we have an isomorphism\n\n\[ \text{Gal}\left( {{M}_{p}\left( K\right) /H}\right) \approx p\text{-part of}{U}_{p}/\overline{{\sigma }_{p}E} \]\n\n\[ = {U}_{p}^{\left( 1\right) }/\left( {{U}_{p}^{\left( 1\right) } \cap \overline{{\sigma }_{p}E}}\righ...
Again, as \( p \) is fixed, we write simply \( {U}_{p}/\bar{E} \) . By a quasi-isomorphism, we shall mean a homomorphism with finite kernel and cokernel. We denote a quasi-isomorphism by a single \( \sim \) . The theorem yields a quasi-isomorphism\n\n\[ {G}_{p}^{\mathrm{{ab}}}\left( K\right) \sim {U}_{p}/\bar{E} \]\n\n...
No
Theorem 5.2. Assume the Leopoldt conjecture for \( K \) . Then we have a quasi-isomorphism\n\n\[ \n{G}_{p}^{\mathrm{{ab}}}\left( K\right) \sim {\mathbb{Z}}_{p}^{{r}_{2} + 1} \approx \operatorname{Gal}\left( {{Z}_{p}\left( K\right) /K}\right) \n\]
Proof. The first statement comes from the definitions and\n\n\[ \n\left\lbrack {K : \mathbf{Q}}\right\rbrack = {r}_{1} + 2{r}_{2} \n\]\n\nFor the second statement, we note that the composite of all \( {\mathbf{Z}}_{p} \) -extensions of\n\n\( K \) has a Galois group embedded in the product of \( {\mathbf{Z}}_{p} \) with...
Yes
Theorem 6.1. Let \( \Omega \) be the maximal p-abelian p-ramified extension of \( {K}_{\infty } \) . Then:\n\n(i) \( G = \operatorname{Gal}\left( {\Omega /{K}_{\infty }}\right) \) is finitely generated over the Iwasawa algebra, and in fact\n\n\[ G/{G}^{\gamma - 1} \sim {\mathbb{Z}}_{p}^{\rho }\;\text{ where }\rho = \le...
Proof. By definition,\n\n\[ {\Omega }_{0} = {M}_{p}\left( {K}_{0}\right) \]\n\nand the rank over \( {\mathbf{Z}}_{p} \) of a subgroup of finite index in its Galois group was determined to be \( \left\lbrack {{K}_{0} : \mathbf{Q}}\right\rbrack - {r}_{p} \) in Theorem 5.2. Taking into account \( \Gamma \) itself shows th...
Yes
Theorem 6.2. Assume that \( {K}_{0} \) is totally imaginary, and that each \( {K}_{n} \) satisfies the Leopoldt conjecture (namely\n\n\[ \n\\left. {{r}_{p}\left( {E}_{n}\right) = {r}_{2}\left( {K}_{n}\right) }\\right) \n\]\n\nThen there is a quasi-isomorphism\n\n\[ \nG \\sim {\\Lambda }^{{r}_{2}} \\times {G}_{\\mathrm{...
Proof. From the structure theorem, we know that\n\n\[ \nG \\sim {\\Lambda }^{t} \\times {G}_{\\text{tor }} \n\]\n\nOn the other hand,\n\n\[ \n{r}_{2}\left( {K}_{n}\right) = {r}_{2}{p}^{n} \n\]\n\nBy Theorem 5.2 we know that\n\n\[ \n\\operatorname{Gal}\left( {{\\Omega }_{n}/{K}_{n}}\\right) \\sim {\\mathbb{Z}}_{p}^{{r}_...
Yes
Lemma 1. \( \Omega = {\Omega }_{A}{\Omega }_{B} \) .
Proof. By Kummer theory, \( \Omega \) is a composite of cyclic extensions. Let \( {K}_{\infty }\left( {\alpha }^{1/{p}^{m}}\right) \subset \Omega \) for some \( \alpha \in {K}_{\infty } \) . Then \( \alpha \in {K}_{n} \) for some \( n \) . We take \( n \geq m \) and also such that\n\n\[ \n{K}_{n}\left( {\alpha }^{1/{p}...
Yes
Theorem 2.1. The Galois groups \( \operatorname{Gal}\left( {{\Omega }_{A}/{\Omega }_{E}}\right) \) and \( \operatorname{Gal}\left( {{\Omega }_{B}/{\Omega }_{E}}\right) \) are \( \Lambda \) -torsion modules. So \( \operatorname{Gal}\left( {\Omega /{\Omega }_{E}}\right) \) is a \( \Lambda \) -torsion module.
Proof. We shall analyze each Galois group separately, and get a closer view of its structure.\n\nThe extension \( {\Omega }_{A}/{\Omega }_{E} \).\n\nLet \( {G}_{A/E} = \operatorname{Gal}\left( {{\Omega }_{A}/{\Omega }_{E}}\right) \). For now abbreviate \( {G}_{A/E} = G \), and let\n\n\[ {G}_{n} = \operatorname{Gal}\lef...
Yes
Theorem 2.3. Assume that there is only one prime in \( {K}_{\infty } \) lying above \( p \) . Then\n\n\[ \n{\Omega }_{E} = {\Omega }_{{E}_{p}} = {\Omega }_{B} \n\] \n\nwhere \( {E}_{p} \) is the group of p-units in \( {K}_{\infty } \), and \n\n\[ \n{\Omega }_{{E}_{p}} = \Omega \left( {E}_{p}^{1/{p}^{\infty }}\right) \n...
Proof. We consider the diagram of fields:\n\n![ddc7d6d6-2c43-4de1-8a63-a0f4832b6652_166_0.jpg](images/ddc7d6d6-2c43-4de1-8a63-a0f4832b6652_166_0.jpg)\n\nThe ideal above \( p \) in \( {\mathbf{Q}}_{n} \) is principal, say generated by the element \( {\lambda }_{n} = \) \( 1 - {\zeta }_{n} \) . The degree \( \left\lbrack...
Yes
Theorem 3.1. The association of \( {\left( \mathbf{Q}/\mathbf{Z}\right) }^{\left( p\right) } \rightarrow V \) given by\n\n\[ a \mapsto {g}_{a}\text{ (mod roots of unity) } \]\n\nsatisfies the distribution relations except at 0 .
The theorem means that for \( a \neq 0 \) we have\n\n\[ \mathop{\prod }\limits_{{{pb} = a}}{g}_{b} = {g}_{a} \]\n\nand is obvious in the light of \( \mathbf{{CU}}2 \) .
No
Theorem 3.2. The group \( {\mathcal{G}}_{n}^{ + } \) operates simply transitively on the primitive elements of \( {V}_{n} \), and the induced homomorphism \[ \mathbb{Z}\left\lbrack {\mathcal{G}}_{n}^{ + }\right\rbrack \rightarrow {V}_{n}\text{such that}{\sigma }_{c} \mapsto {g}_{c/{p}^{n}} \] is an isomorphism.
Proof. The homomorphism is obviously surjective. It is injective because \( \mathbf{Z}\left\lbrack {\mathcal{G}}_{n}^{ + }\right\rbrack \) is torsion free, and the ranks of the two groups are equal. This proves the theorem.
Yes
Theorem 3.3. The factor group \( {V}_{m}/{V}_{n} \) for \( m \geq n \) has no torsion.
Proof. The embedding of \( {V}_{n} \) into \( {V}_{m} \) corresponds to the embedding of group rings\n\n\[ \mathbf{Z}\left\lbrack {\mathcal{G}}_{n}^{ + }\right\rbrack \rightarrow \mathbf{Z}\left\lbrack {\mathcal{G}}_{m}^{ + }\right\rbrack \]\n\nwhich sends an element \( {\sigma }_{c} \) on the element \( \sum {\sigma }...
Yes
Theorem 3.4. Let \( p \) be odd, and let \( c \) be a primitive root \( {\;\operatorname{mod}\;{p}^{2}} \). Then \( {E}_{n} \) is generated over \( \mathbf{Z}{\left\lbrack {\mathcal{G}}_{n}\right\rbrack }_{0} \) by the element \[ v = {\sigma }_{c}\pi /\pi = \frac{{\zeta }^{c} - 1}{\zeta - 1} \]
Proof. We write an element \( \alpha \) of degree 0 in the form \[ \alpha = \sum k\left( b\right) \left( {{\sigma }_{b} - 1}\right) \] and observe that \( {\sigma }_{b} - 1 \) is divisible in the integral group ring by \( {\sigma }_{c} - 1 \) because \( {\sigma }_{c} \) is a generator of the cyclic group \( {\mathcal{G...
No
Theorem 3.5. Let \( c \) be a generator of \( 1 + 4{\mathbf{Z}}_{2} \) if \( p = 2 \), and a primitive root \( {\;\operatorname{mod}\;{p}^{2}} \) if \( p > 2 \) . Let\n\n\[ \n{v}_{n} = \text{ image of }\frac{{\zeta }^{c} - 1}{\zeta - 1}\text{ in }{V}_{n} \]\n\nand let \( {V}_{n}^{0} \) be the subgroup of \( {V}_{n} \) ...
Proof. Clear.
No
Theorem 3.6. This distribution is the universal even ordinary distribution with value 0 at 0, and values into abelian groups on which multiplication by 2 is invertible.
Proof. On \( \left( {1/N}\right) \mathbf{Z}/\mathbf{Z} \) the group generated by the image of \( h \) has rank\n\n\[ \frac{1}{2}\left| {\mathbf{Z}{\left( N\right) }^{ * }}\right| - 1 \]\n\nwhich according to Kubert's Theorem 9.1(iii) of Chapter 2 is the maximal possible rank (the value 0 at 0 gives rise to the -1 ). Th...
No
Theorem 4.1. Assuming the Vandiver conjecture, we have \( G = {G}^{ - } \), and \( G \) is a 1-dimensional free module over \( {R}^{ - } \) .
Proof. By the Vandiver conjecture and Theorem 3.3 we have\n\n\[ \n{E}_{p, n} \cap {K}_{\infty }^{*{p}^{n}} = {E}_{p, n}^{{p}^{n}}{\mu }^{\left( p\right) }.\n\] \n\nLet us abbreviate for simplicity\n\n\[ \n{V}_{n}^{1/{p}^{n}} = {E}_{p, n}^{1/{p}^{n}}/{E}_{p, n} \cap {K}_{\infty }^{ * }.\n\] \n\nThen the Kummer theory pa...
Yes
Theorem 4.2. Let \( {C}_{n} = C{l}^{\left( p\right) }\left( {K}_{n}\right) \) be the p-primary part of the ideal class group of \( {K}_{n} \) and let \( C = \) projective limit of the \( {C}_{n} \) under the norm map. Under the Vandiver conjecture, we have \( C = {C}^{ - } \), and \( {C}^{ - } \) is cyclic as a \( \Lam...
Proof. The field diagram (once the theorem is proved) is as follows. \[ G\left\{ \begin{array}{l} \Omega \\ \mid \\ {\Omega }^{\mathrm{{nr}}} \\ \mid \\ {K}_{\infty } \end{array}\right\} {G}_{C} \] What we have to do is to show that the maximal \( p \) -abelian unramified extension of \( {K}_{\infty } \) is in fact con...
Yes
Theorem 4.3. (i) For \( m \geq n \) we have an injection\n\n\[ \operatorname{Ker}\left( {{C}_{n} \rightarrow {C}_{m}}\right) \rightarrow {H}^{1}\left( {{\Gamma }_{m, n},{E}_{m}}\right) \]\n\n(ii) Under the Vandiver conjecture we have \( {H}^{1}\left( {{\Gamma }_{m, n},{E}_{m}}\right) = 0 \), so\n\n\[ {C}_{n} \rightarro...
Proof. Let \( \mathfrak{a} \) be an ideal representing an element of \( {C}_{n} \), becoming principal in \( {K}_{m} \), say \( \mathfrak{a} = \left( \alpha \right) \) with \( \alpha \in {K}_{m} \) . For any element \( \sigma \in {\Gamma }_{m, n} \) we have \( \sigma \mathfrak{a} = \mathfrak{a} \) . Hence \( {\sigma \a...
Yes