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Corollary 10.4.5. We have \( G\left( {\tau ,0}\right) = - 1/2 \) and \( G\left( {\tau , - k}\right) = 0 \) for all \( k \in \) \( {\mathbb{Z}}_{ \geq 1} \) . | Proof. Immediate from the preceding corollary and left to the reader. | No |
Theorem 10.4.6 (Kronecker's limit formula). Define\n\n\[ \nq = {e}^{2i\pi \tau } = {e}^{-{2\pi y}}{e}^{2i\pi x}\;\text{ and }\;\eta \left( \tau \right) = {e}^{{i\pi \tau }/{12}}\mathop{\prod }\limits_{{n \geq 1}}\left( {1 - {q}^{n}}\right) .\n\]\n\n(1) Around \( s = 1 \) we have\n\n\[ \nG\left( {\tau, s}\right) = \frac... | Proof. We must give the Taylor expansions around \( s = 1 \) up to terms in \( O\left( {s - 1}\right) \) of all the terms occurring in the Fourier expansion of \( G\left( {\tau, s}\right) \) . For notational simplicity set \( G\left( s\right) = {\pi }^{1/2}\Gamma \left( {s - 1/2}\right) \zeta \left( {{2s} - 1}\right) {... | Yes |
Corollary 10.4.8. Let \( Q \) be as above and set \( \tau = \left( {-b + i\sqrt{{4ac} - {b}^{2}}}\right) /\left( {2a}\right) \) , so that \( \Im \left( \tau \right) > 0 \), and let \( D = {b}^{2} - {4ac} \) be the discriminant of \( Q \) . Then\n\n(1) \( {\zeta }_{Q}\left( s\right) = {\left( \left| D\right| /4\right) }... | Proof. By definition we have\n\n\[ \n{\zeta }_{Q}\left( s\right) = \frac{{a}^{-s}}{2}\mathop{\sum }\limits_{{\left( {m, n}\right) \in {\mathbb{Z}}^{2}}}^{\prime }\frac{1}{{\left| m - n\tau \right| }^{2s}} = {\left( a\Im \left( \tau \right) \right) }^{-s}G\left( {\tau, s}\right) = {\left( \left| D\right| /4\right) }^{-s... | Yes |
Theorem 10.5.1. Let \( K \) be a number field of degree \( n \) and signature \( \left( {{r}_{1},{r}_{2}}\right) \) . Denote by \( d\left( K\right), h\left( K\right), R\left( K\right) \), and \( w\left( K\right) \) (standard notation) the discriminant, class number, regulator, and number of roots of unity in \( K \) . ... | Proof. We will not prove this theorem, but we will make a number of remarks on the proof. There are two ways to prove the analytic continuation and functional equation of \( {\zeta }_{K}\left( s\right) \) to the whole plane. One is Hecke’s initial proof: he essentially copies the proof that we have given for the ordina... | No |
Corollary 10.5.2. For \( k \in {\mathbb{Z}}_{ \leq 0} \), the order of the (possible) zero at \( s = k \) of \( {\zeta }_{K}\left( s\right) \) is given by\n\n\[ \left\{ \begin{array}{ll} {r}_{1} + {r}_{2} - 1 & \text{ if }k = 0, \\ {r}_{1} + {r}_{2} & \text{ if }k < 0, k \equiv 0\left( {\;\operatorname{mod}\;2}\right) ... | Proof. Follows immediately from the fact that \( \gamma \left( s\right) \) has simple poles for all \( s \in {\mathbb{Z}}_{ \leq 0} \), and left to the reader. | No |
Corollary 10.5.4. Let \( K \) be a totally real number field of degree \( n \) . For every \( k \geq 1 \) there exists \( {r}_{k} \in {\mathbb{Q}}^{ * } \) such that\n\n\[{\zeta }_{K}\left( {2k}\right) = {r}_{k}\frac{{\pi }^{2kn}}{{\left| d\left( K\right) \right| }^{1/2}}.\] | Proof. Clear by using the functional equation. | No |
Proposition 10.5.5. Let \( K = \mathbb{Q}\left( \sqrt{D}\right) \) be a quadratic field of discriminant D. We have\n\n\[ \n{\zeta }_{K}\left( s\right) = \zeta \left( s\right) L\left( {{\chi }_{D}, s}\right) \n\]\n\nwhere as usual \( {\chi }_{D}\left( n\right) = \left( \frac{D}{n}\right) \) is the Legendre-Kronecker cha... | Proof. Indeed, if \( p \) is a prime number, then we know that \( p \) is inert, splits, or ramifies in \( K/\mathbb{Q} \) according to whether \( \left( \frac{D}{p}\right) = - 1,1 \), or 0 . Thus\n\n\[ \n{\zeta }_{K}\left( s\right) = \mathop{\prod }\limits_{p}\mathop{\prod }\limits_{{\mathfrak{p} \mid p{\mathbb{Z}}_{K... | Yes |
Corollary 10.5.6. Let \( D \) be a nonsquare integer congruent to 0 or 1 modulo 4, and let \( {\chi }_{D} = \left( \frac{D}{ \cdot }\right) \) be the corresponding Legendre-Kronecker symbol. Then \( L\left( {{\chi }_{D},1}\right) > 0 \) | Proof. Write \( D = {D}_{0}{f}^{2} \), where \( {D}_{0} \) is a fundamental discriminant. Since \( L\left( {{\chi }_{D},1}\right) = \mathop{\prod }\limits_{{p \mid f}}\left( {1 - {\chi }_{{D}_{0}}\left( p\right) /p}\right) L\left( {{\chi }_{{D}_{0}},1}\right) \) and \( 1 - {\chi }_{{D}_{0}}\left( p\right) /p > 0 \), we... | Yes |
Proposition 10.5.7. Let \( K \) be an imaginary quadratic field of discriminant \( D \), and denote by \( w\left( D\right) \) the number of roots of unity of \( K \) (so that \( w\left( D\right) = 2 \) if \( D < - 4, w\left( {-4}\right) = 4, w\left( {-3}\right) = 6 \) ).\n\n(1) We have the finite sum decomposition\n\n\... | Proof. (1) is a trivial consequence of the definition of the class group \( {Cl}\left( K\right) \) . | No |
Corollary 10.5.8. Let \( K \) be an imaginary quadratic field of discriminant \( D \), let \( \delta \) be any integer such that \( D \equiv \delta \left( {\;\operatorname{mod}\;2}\right) \), and assume that the class number of \( K \) is equal to 1 . Then \[ {\zeta }_{K}\left( s\right) = \frac{1}{w\left( D\right) }\ma... | Proof. Clear. | No |
Corollary 10.5.9. Let \( K \) be an imaginary quadratic field of discriminant D. The statements of Corollary 10.4.8 (2), (3), and (4) are valid verbatim if we replace \( {\zeta }_{Q}\left( s\right) \) by \( \left( {\omega \left( D\right) /2}\right) {\zeta }_{K}\left( {\mathcal{A}, s}\right) \) ; in other words:\n\n(1) ... | Proof. This is an immediate consequence of the proposition and of Corollary 10.4.8: we note that the quadratic form \( {Q}_{\mathcal{A}}\left( {x, y}\right) \) has determinant \( \left( {4\Re {\left( \tau \right) }^{2} - 4{\left| \tau \right| }^{2}}\right) /\left( {4\Im {\left( \tau \right) }^{2}}\right) = - 1 \), and ... | Yes |
Proposition 10.5.10. Let \( D < 0 \) be a fundamental discriminant, and denote by \( h\left( D\right) \) the class number of \( K = \mathbb{Q}\left( \sqrt{D}\right) \) . Denote by \( \mathcal{Q}\left( D\right) \) the set of equivalence classes of quadratic numbers \( \tau = \left( {-b + \sqrt{D}}\right) /\left( {2a}\ri... | Proof. First note that if \( \left( {1,{\tau }^{\prime }}\right) \) is another \( \mathbb{Z} \) -basis of an ideal \( \mathfrak{b} \) with \( \Im \left( {\tau }^{\prime }\right) > \) 0 we have \( {\tau }^{\prime } = \gamma \left( \tau \right) = \left( {{a\tau } + b}\right) /\left( {{c\tau } + d}\right) \) for some \( \... | Yes |
Proposition 10.5.11 (Lerch, Chowla-Selberg). For any negative fundamental discriminant \( D \) we have the identity\n\n\[ \mathop{\prod }\limits_{{\tau \in \mathcal{Q}\left( D\right) }}\Im \left( \tau \right) {\left| \eta \left( \tau \right) \right| }^{4} = {\left( 4\pi {\left| D\right| }^{1/2}\right) }^{-h\left( D\rig... | Proof. Since \( D < 0 \) we have \( L\left( {{\chi }_{D},0}\right) = {2h}\left( D\right) /w\left( D\right) \), so Proposition 10.3.5 gives\n\n\[ \frac{{L}^{\prime }\left( {{\chi }_{D},0}\right) }{L\left( {{\chi }_{D},0}\right) } = \frac{w\left( D\right) }{{2h}\left( D\right) }\mathop{\sum }\limits_{{r = 1}}^{\left| D\r... | Yes |
Theorem 10.5.13. Let \( D < 0 \) be congruent to 0 or 1 modulo 4, and write \( D = {D}_{0}{f}^{2} \), where \( {D}_{0} \) is a fundamental discriminant, in other words the discriminant of the quadratic field \( \mathbb{Q}\left( \sqrt{D}\right) \). Then\n\n\[ \mathop{\prod }\limits_{{\tau \in \mathcal{Q}\left( D\right) ... | where\n\n\[ e\left( p\right) = \frac{\left( {1 - {p}^{-{v}_{p}\left( f\right) }}\right) \left( {1 - \left( \frac{{D}_{0}}{p}\right) }\right) }{\left( {1 - 1/p}\right) \left( {p - \left( \frac{{D}_{0}}{p}\right) /p}\right) }.\]\n\nNote that we have the following well-known formula, coming directly from Dirichlet's class... | Yes |
Corollary 10.5.14. We have the formulas\n\n\[ \mathop{\prod }\limits_{{n \geq 1}}\left( {1 + {e}^{-{\pi n}}}\right) = {2}^{-1/8}{e}^{\pi /{24}} \]\n\n\[ \mathop{\prod }\limits_{{n \geq 1}}\tanh \left( {{\pi n}/2}\right) = {\left( 2\pi \right) }^{-3/4}\Gamma \left( {1/4}\right) \]\n\n\[ \mathop{\prod }\limits_{{n \geq 1... | Proof. These formulas are simply obtained by replacing the eta function by its infinite product expansion and using the special values. The details are left to the reader (Exercise 59). | No |
Theorem 10.5.15. For each \( \tau \in \mathcal{Q}\left( D\right) \) there exists an algebraic number \( \alpha \left( \tau \right) \) that can be given explicitly such that\n\n\[ \Im \left( \tau \right) {\left| \eta \left( \tau \right) \right| }^{4} = \frac{\alpha \left( \tau \right) }{{4\pi }{\left| D\right| }^{1/2}}{... | Proof. This proof requires a basic knowledge of modular forms and complex multiplication. If \( {\tau }_{1} \) and \( {\tau }_{2} \) are in \( \mathcal{Q}\left( D\right) \) there exists some integral matrix \( \gamma = \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \) with nonzero determinant \( N \) such t... | Yes |
Lemma 10.5.16. Let \( {D}_{1} \) and \( {D}_{2} \) be two coprime fundamental discriminants, and set \( D = {D}_{1}{D}_{2} \) and \( K = \mathbb{Q}\left( \sqrt{D}\right) \) . Then:\n\n(1) \( D \) is a fundamental discriminant.\n\n(2) For any \( \lambda \in {\mathbb{Z}}_{K} \) such that \( \mathcal{N}\left( \lambda \rig... | Proof. (1). If \( {D}_{1} \) and \( {D}_{2} \) are squarefree, hence congruent to 1 modulo 4, then \( {D}_{1}{D}_{2} \) is squarefree and congruent to 1 modulo 4, so is fundamental. Otherwise, by symmetry we may assume that \( {D}_{1} = 4{d}_{1} \) with \( {d}_{1} \equiv 2 \) or 3 modulo 4 and squarefree. It follows th... | Yes |
Corollary 10.5.17. Let \( {D}_{1} \) and \( {D}_{2} \) be two coprime fundamental discriminants, and set \( D = {D}_{1}{D}_{2} \) and \( K = \mathbb{Q}\left( \sqrt{D}\right) \). (1) For any ideal class \( \mathcal{A} \in {Cl}\left( K\right) \) there exists an integral ideal \( \mathfrak{a} \in \mathcal{A} \) such that ... | Proof. (1). Let \( \mathfrak{b} \in \mathcal{A} \) be any integral ideal. By the approximation theorem for Dedekind domains there exists \( \alpha \in K \) such that \( {v}_{\mathfrak{p}}\left( \alpha \right) = - {v}_{\mathfrak{p}}\left( \mathfrak{b}\right) \) for every prime ideal \( \mathfrak{p} \) above a prime numb... | Yes |
Proposition 10.5.19. As above, let \( {D}_{1} \) and \( {D}_{2} \) be two coprime fundamental discriminants, \( D = {D}_{1}{D}_{2}, K = \mathbb{Q}\left( \sqrt{D}\right) \), and let \( {\chi }_{{D}_{1}} \) be the character of \( {Cl}\left( K\right) \) defined in the above corollary. Then\n\n\[ \n{L}_{K}\left( {{\chi }_{... | Proof. It is sufficient to show that the corresponding Euler factors are the same on both sides. Let \( p \) be a prime number. As usual we consider three cases. If \( p \) is inert in \( K \) there is a single prime ideal \( \mathfrak{p} = p{\mathbb{Z}}_{K} \) above \( p \) that is a principal ideal, so that \( {\chi ... | Yes |
Corollary 10.5.20. Let \( {D}_{1} \) and \( {D}_{2} \) be two coprime fundamental discriminants, \( D = {D}_{1}{D}_{2}, K = \mathbb{Q}\left( \sqrt{D}\right) \), and assume that \( {D}_{1} > 0,{D}_{2} < 0 \) hence \( D < 0 \) . Then\n\n\[ L\left( {{\chi }_{{D}_{1}},1}\right) = - \frac{\omega \left( {D}_{2}\right) }{h\le... | Proof. By the proposition we have\n\n\[ {L}_{K}\left( {{\chi }_{{D}_{1}}, s}\right) = \mathop{\sum }\limits_{{\mathcal{A} \in {Cl}\left( K\right) }}{\chi }_{{D}_{1}}\left( \mathcal{A}\right) {\zeta }_{K}\left( {\mathcal{A}, s}\right) = L\left( {{\chi }_{{D}_{1}}, s}\right) L\left( {{\chi }_{{D}_{2}}, s}\right) .\n\nNot... | Yes |
Proposition 10.5.21. Let \( m \) be an integer, and for all primes \( p \nmid m \) denote by \( {f}_{p} \) the order of \( p \) modulo \( m \) (i.e., of the class of \( p \) in \( {\left( \mathbb{Z}/m\mathbb{Z}\right) }^{ * } \) ), and set \( {g}_{p} = \phi \left( m\right) /{f}_{p} \) . Then\n\n\[ \mathop{\prod }\limit... | Proof. Let \( G \) be the group of Dirichlet characters modulo \( m \) and \( {H}_{p} \) the group of \( {f}_{p} \) th roots of unity. If \( \chi \in G \), then \( \chi \left( p\right) \in {H}_{p} \) . The map \( \chi \mapsto \chi \left( p\right) \) is a group homomorphism from \( G \) to \( {H}_{p} \) . I claim that t... | Yes |
Corollary 10.5.23. If \( p \) is a prime then\n\n\[ d\left( {\mathbb{Q}}_{{p}^{k}}\right) = \varepsilon {p}^{k{p}^{k} - \left( {k + 1}\right) {p}^{k - 1}}, \]\n\nwhere \( \varepsilon = - 1 \) if \( {p}^{k} = 4 \) or \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) and \( \varepsilon = 1 \) otherwise. | Proof. The above result is trivially true for \( {p}^{k} = 2 \), so assume \( {p}^{k} > 2 \) . By the above theorem and Proposition 2.1.29, whose notation we keep, we have\n\n\[ \left| {d\left( K\right) }\right| = \mathop{\prod }\limits_{{\chi {\;\operatorname{mod}\;{p}^{k}}}}f\left( \chi \right) = \mathop{\prod }\limi... | Yes |
Proposition 10.5.26. Let \( p \geq 3 \) be a prime number, let \( k \in {\mathbb{Z}}_{ \geq 1} \), and set \( N = \phi \left( {p}^{k}\right) = {p}^{k - 1}\left( {p - 1}\right) \) . We have\n\n\[ \n{h}_{{p}^{k}}^{ - } = \frac{{p}^{k}}{{2}^{N/2 - 1}}\mathop{\prod }\limits_{{\chi \text{ odd }}}L\left( {\chi ,0}\right) = {... | Proof. By Lemma 10.2.1, for all nontrivial characters \( \chi \) we have \( L\left( {{\chi }_{f}, s}\right) = \) \( L\left( {\chi, s}\right) \), so it follows from Theorem 10.5.22 that \( {\zeta }_{\mathbb{Q}\left( {\zeta }_{{p}^{k}}\right) }\left( s\right) = \zeta \left( s\right) \mathop{\prod }\limits_{{\chi \neq {\c... | Yes |
Corollary 10.5.27. Let \( p \geq 3 \) be a prime number, let \( g \in \mathbb{Z} \) be a primitive root modulo \( {p}^{k} \), set \( N = \phi \left( {p}^{k}\right) = {p}^{k - 1}\left( {p - 1}\right) \), and let \( {\zeta }_{N} \in \mathbb{C} \) be a primitive \( N \) th root of unity. Denote by \( P \) the polynomial\n... | Proof. Indeed, note that it is easy to describe all characters modulo \( {p}^{k} \) : such a character is uniquely determined by its value on \( g \) ; hence \( {\chi }_{1}\left( {g}^{j}\right) = {\zeta }_{N}^{j} \) defines a character modulo \( p \) that generates the group of characters modulo \( {p}^{k} \), so that ... | Yes |
Corollary 10.5.28. With the same notation we have\n\n\[ \n{h}_{{p}^{k}}^{ - } = - \frac{1}{{\left( -2{p}^{k}\right) }^{N/2 - 1}}\mathop{\prod }\limits_{{d \mid N, d \nmid N/2}}\operatorname{Res}\left( {P\left( X\right) ,{\Phi }_{d}\left( X\right) }\right) ,\n\]\n\nwhere \( {\Phi }_{d}\left( X\right) \) denotes the dth ... | Proof. This immediately follows from the equality\n\n\[ \n{X}^{N/2} + 1 = \frac{{X}^{N} - 1}{{X}^{N/2} - 1} = \mathop{\prod }\limits_{{d \mid N, d \nmid N/2}}{\Phi }_{d}\left( X\right)\n\]\n\ntogether with the multiplicativity of the resultant. | Yes |
Theorem 10.5.29. For any nontrivial Dirichlet character \( \chi \) we have \( L\left( {\chi ,1}\right) \neq 0 \). | First Proof. By what we have seen in the preceding section, we have\n\n\[ \n{\zeta }_{{\mathbb{Q}}_{m}}\left( s\right) = \zeta \left( s\right) \mathop{\prod }\limits_{\substack{{\chi {\;\operatorname{mod}\;m}} \\ {\chi \neq {\chi }_{0}} }}L\left( {{\chi }_{f}, s}\right) , \n\] \n\nand since both \( \zeta \left( s\right... | Yes |
Theorem 10.5.30 (Dirichlet). Let \( a \) and \( m \) be coprime integers. There exist infinitely many primes \( p \) that are congruent to a modulo \( m \) . More precisely, the set of such primes has an analytic density \( 1/\phi \left( m\right) \), where the analytic density \( d\left( P\right) \) of a set \( P \) of... | Proof. First note that for \( s > 1 \) real,\n\n\[ \log \left( {L\left( {\chi, s}\right) }\right) = - \mathop{\sum }\limits_{p}\log \left( {1 - \chi \left( p\right) {p}^{-s}}\right) = \mathop{\sum }\limits_{p}\frac{\chi \left( p\right) }{{p}^{s}} - S\left( s\right) ,\]\n\nwhere \( S\left( s\right) = \mathop{\sum }\limi... | Yes |
Lemma 10.6.1. Let \( P\left( X\right) \in \mathbb{Z}\left\lbrack X\right\rbrack \) be an irreducible monic polynomial, and \( K = \mathbb{Q}\left( \theta \right) \), where \( \theta \) is a root of \( P\left( X\right) \). If \( p \) is a prime number such that \( {p}^{2} \nmid \operatorname{disc}\left( P\right) \), the... | Proof. If \( {p}^{2} \nmid \operatorname{disc}\left( P\right) \), then a fortiori \( p \) does not divide the index \( \left\lbrack {\mathbb{Z}}_{K}\right. \) : \( \mathbb{Z}\left\lbrack \theta \right\rbrack \rbrack \), so the factorization of \( p{\mathbb{Z}}_{K} \) into prime ideals mimics that of the polynomial \( P... | Yes |
Proposition 10.7.1. For any fixed \( k \geq 0 \) we have \( {\zeta }^{\left( k\right) }\left( s\right) = O\left( {\log {\left( t\right) }^{k + 1}}\right) \) uniformly in the region \( 1 - C/\log \left( t\right) \leq \sigma \leq 2 \), and in particular for \( s = 1 + {it} \) (recall that we also assume \( t \geq {t}_{0}... | Proof. By the Euler-MacLaurin formula for \( n = 1 \) we have for \( \sigma > 0 \) ,\n\n\[ \zeta \left( s\right) = \mathop{\sum }\limits_{{m = 1}}^{N}\frac{1}{{m}^{s}} + \frac{{N}^{1 - s}}{s - 1} - \frac{{N}^{-s}}{2} - s{\int }_{N}^{\infty }\frac{{B}_{1}\left( {\{ t\} }\right) }{{t}^{s + 1}}{dt}. \]\n\nDifferentiating ... | Yes |
Lemma 10.7.2. For all \( \sigma > 1 \) and \( t \in \mathbb{R} \) we have\n\n\[ \zeta {\left( \sigma \right) }^{3}{\left| \zeta \left( \sigma + it\right) \right| }^{4}\left| {\zeta \left( {\sigma + {2it}}\right) }\right| \geq 1 \]\n\nand\n\[ - \Re \left( {3\frac{{\zeta }^{\prime }\left( \sigma \right) }{\zeta \left( \s... | Proof. By expanding the logarithm of the Euler product defining \( \zeta \left( s\right) \) it is clear that \( \zeta \left( s\right) = \exp \left( {\mathop{\sum }\limits_{p}\mathop{\sum }\limits_{{k \geq 1}}{p}^{-{ks}}/k}\right) \), hence that\n\n\[ \log \left( \left| {\zeta \left( s\right) }\right| \right) = \mathop{... | Yes |
Corollary 10.7.3. For \( \sigma > 1 \) we have\n\n\[ \frac{1}{\zeta \left( {\sigma + {it}}\right) } = O\left( \frac{\log {\left( t\right) }^{1/4}}{{\left( \sigma - 1\right) }^{3/4}}\right) . | Proof. Since \( \zeta \left( \sigma \right) = 1/\left( {\sigma - 1}\right) + O\left( 1\right) \), by the lemma and Proposition 10.7.1 we have \( \left| {\zeta {\left( \sigma + it\right) }^{-4}}\right| = O\left( {\zeta {\left( \sigma \right) }^{3}\log \left( t\right) }\right) = O\left( {{\left( \sigma - 1\right) }^{-3}\... | Yes |
Corollary 10.7.4. The function \( \zeta \left( s\right) \) does not vanish in the closed half-plane \( \Re \left( s\right) \geq 1 \), and in particular on the line \( \Re \left( s\right) = 1 \) . | Proof. Since the Euler product is convergent for \( \Re \left( s\right) > 1 \) and none of its terms vanish, we know that \( \zeta \left( s\right) \neq 0 \) for \( \Re \left( s\right) > 1 \) . Now assume by contradiction that \( \zeta \left( {1 + i{t}_{0}}\right) = 0 \) for some \( {t}_{0} \in \mathbb{R} \) . The funct... | Yes |
Lemma 10.7.5. The functions \( \left( {s - 1}\right) \zeta \left( s\right) \) and \( s\left( {1 - s}\right) {\pi }^{-s/2}\Gamma \left( {s/2}\right) \zeta \left( s\right) \) have order 1. | Proof. Indeed, for \( \sigma > 0 \) by the integral representation we have for a suitable constant \( A,\left| {\Gamma \left( {s/2}\right) }\right| \leq \left| {\Gamma \left( {\sigma /2}\right) }\right| = O\left( {e}^{{A\sigma }\log \left( \sigma \right) }\right) \), and on the other hand, Euler-MacLaurin immediately g... | Yes |
Corollary 10.7.7. Set \( b = \log \left( {2\pi }\right) - 1 - \gamma /2 \) . Then for all \( s \in \mathbb{C} \) we have the convergent product\n\n\[ \zeta \left( s\right) = \frac{{e}^{bs}}{s\left( {s - 1}\right) \Gamma \left( {s/2}\right) }\mathop{\prod }\limits_{\rho }\left( {1 - \frac{s}{\rho }}\right) {e}^{s/\rho }... | Proof. We apply Hadamard's theorem to the function\n\n\[ f\left( s\right) = s\left( {1 - s}\right) {\pi }^{-s/2}\Gamma \left( {s/2}\right) \zeta \left( s\right) = 2\left( {1 - s}\right) {\pi }^{-s/2}\Gamma \left( {s/2 + 1}\right) \zeta \left( s\right) .\n\nSince the zeros of \( \zeta \left( s\right) \) for \( s = - {2k... | Yes |
Theorem 10.7.8. There exists a constant \( C > 0 \) such that \( \zeta \left( s\right) \neq 0 \) for \( t \geq {t}_{0} \) in the region\n\n\[ \Re \left( s\right) > 1 - \frac{C}{\log \left( t\right) }.\] | Proof. Here we will use the second inequality of Lemma 10.7.2. Fix some \( \sigma > 1 \) (we will see at the end of the proof how to choose it appropriately). Since \( \zeta \left( \sigma \right) = 1/\left( {\sigma - 1}\right) + O\left( 1\right) \) and \( {\zeta }^{\prime }\left( \sigma \right) = - 1/{\left( \sigma - 1... | Yes |
Lemma 10.7.9. The function \( \Phi \left( s\right) - 1/\left( {s - 1}\right) \) is holomorphic in the closed half-plane \( \Re \left( s\right) \geq 1 \) . | Proof. It is clear that the series for \( \Phi \left( s\right) \) converges absolutely for \( \Re \left( s\right) > 1 \) and normally for \( \Re \left( s\right) \geq 1 + \varepsilon \) for any fixed \( \varepsilon > 0 \), hence defines an analytic function in \( \Re \left( s\right) > 1 \) . For \( \Re \left( s\right) >... | Yes |
Lemma 10.7.10. We have \( \theta \left( x\right) = O\left( x\right) \) . | Proof. For a positive integer \( n \) we have\n\n\[ \n{2}^{2n} = \mathop{\sum }\limits_{{0 \leq k \leq {2n}}}\left( \begin{matrix} {2n} \\ k \end{matrix}\right) \geq \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \geq \mathop{\prod }\limits_{{n < p \leq {2n}}}p = {e}^{\theta \left( {2n}\right) - \theta \left( n\ri... | Yes |
Lemma 10.7.11. The integral\n\n\[ \n{\int }_{1}^{\infty }\frac{\theta \left( x\right) - x}{{x}^{2}}{dx} \n\]\n\nconverges. | Proof. For \( \Re \left( s\right) > 1 \) we have by Stieltjes integration\n\n\[ \n\Phi \left( s\right) = \mathop{\sum }\limits_{p}\frac{\log p}{{p}^{s}} = {\int }_{1}^{\infty }\frac{{d\theta }\left( x\right) }{{x}^{s}} = s{\int }_{1}^{\infty }\frac{\theta \left( x\right) }{{x}^{s + 1}}{dx} = s{\int }_{0}^{\infty }{e}^{... | No |
Lemma 10.7.13. \( \theta \left( x\right) \sim x \) . | Proof. Assume that for some \( \lambda > 1 \) there exist arbitrary large \( x \) such that \( \theta \left( x\right) \geq {\lambda x} \) . Since \( \theta \left( x\right) \) is nondecreasing, we have\n\n\[{\int }_{x}^{\lambda x}\frac{\theta \left( t\right) - t}{{t}^{2}}{dt} \geq {\int }_{x}^{\lambda x}\frac{{\lambda x... | Yes |
Theorem 10.7.14 (Prime number theorem). If \( \pi \left( x\right) \) denotes the number of prime numbers less than or equal to \( x \) we have\n\n\[ \pi \left( x\right) \sim \frac{x}{\log x}. \] | Proof. We have\n\n\[ \theta \left( x\right) = \mathop{\sum }\limits_{{p \leq x}}\log p \leq \mathop{\sum }\limits_{{p \leq x}}\log x = \pi \left( x\right) \log x. \]\n\nOn the other hand, for any \( \varepsilon > 0 \) ,\n\n\[ \theta \left( x\right) \geq \mathop{\sum }\limits_{{{x}^{1 - \varepsilon } \leq p \leq x}}\log... | Yes |
Lemma 10.7.15. For all \( y > 0 \) we have for any \( \sigma > 1 \) ,\n\n\[ \n\max \left( {\log \left( y\right) ,0}\right) = \frac{1}{2i\pi }{\int }_{\Re \left( s\right) = \sigma }\frac{{y}^{s}}{{s}^{2}}{ds} \n\]\n\nthe integral being on the vertical line \( \Re \left( s\right) = \sigma \) . | Proof. Indeed, the given integral is trivially less than \( O\left( {y}^{\sigma }\right) \) . Thus, if \( y < 1 \) it is immediate to check that we can shift the line of integration to the right without changing the value of the integral, and as \( \sigma \) tends to \( + \infty ,{y}^{\sigma } \) tends to 0 . On the ot... | Yes |
Lemma 10.7.16. Let \( \sigma > 1 \) be fixed. There exists \( s \) with \( \Re \left( s\right) = \sigma \) such that\n\n\[ \left| {{F}_{k}\left( x\right) }\right| \leq {x}^{\sigma }\left| {{G}_{k}\left( s\right) }\right| {\left| s\right| }^{-1/2}. \] | Proof. From the above lemma we have\n\n\[ {F}_{k}\left( x\right) = \frac{1}{2i\pi }{\int }_{\Re \left( s\right) = \sigma }\frac{{x}^{s}{G}_{k}\left( s\right) }{{s}^{2}}{ds} \]\n\nand since \( \left| {{G}_{k}\left( s\right) }\right| \) is bounded by the convergent series \( \mathop{\sum }\limits_{{m \geq 1}}{\left( \log... | Yes |
Lemma 10.7.17. For \( \sigma > 1 \) and \( \Re \left( s\right) = \sigma \) we have\n\n\[ \n{G}_{k}\left( s\right) = O\left( {{\left( \sigma - 1\right) }^{-\left( {3/4}\right) \left( {k + 1}\right) }\log {\left( 2\left| s\right| \right) }^{\left( {{9k} + 1}\right) /4}}\right) ,\n\]\n\nwhere the implied constant depends ... | Proof. By definition we have \( {G}_{k}\left( s\right) = {\left( -1\right) }^{k}{\left( 1/\zeta \left( s\right) \right) }^{\left( k\right) } \) . If \( s \) is close to 1, say \( \left| s\right| \leq 2 \) (still with \( \Re \left( s\right) = \sigma > 1 \) ), then \( {\left( 1/\zeta \left( s\right) \right) }^{\left( k\r... | Yes |
Corollary 10.7.18. For \( x \geq 1 \) we have\n\n\[ \n{F}_{k}\left( x\right) = O\left( {x\log {\left( x\right) }^{\left( {3/4}\right) \left( {k + 1}\right) }}\right) .\n\] | Proof. Combining Lemmas 10.7.16 and 10.7.17, and using the fact that any power of \( \log \left( \left| s\right| \right) \) is negligible compared to \( {\left| s\right| }^{1/2} \), we deduce that for all \( \sigma > 1 \) we have \( {F}_{k}\left( x\right) = O\left( {{x}^{\sigma }{\left( \sigma - 1\right) }^{-\left( {3/... | Yes |
Theorem 10.7.19. For any \( A > 0 \) we have\n\n\[ M\left( x\right) = \mathop{\sum }\limits_{{1 \leq m \leq x}}\mu \left( m\right) = O\left( \frac{x}{\log {\left( x\right) }^{A}}\right) . | Proof. We introduce the function\n\n\[ {H}_{k}\left( x\right) = \mathop{\sum }\limits_{{1 \leq m \leq x}}\mu \left( m\right) {\left( \log \left( m\right) \right) }^{k}, \]\n\nwhich is the function \( {F}_{k}\left( x\right) \) from which we have removed the smoothing factor \( \log \left( {x/m}\right) \) . It is easily ... | Yes |
Corollary 10.7.20. For all \( A > 0 \) we have\n\n\[ \psi \left( x\right) - x = O\left( \frac{x}{\log {\left( x\right) }^{A}}\right) ,\]\n\n\[ \theta \left( x\right) - x = O\left( \frac{x}{\log {\left( x\right) }^{A}}\right) ,\text{ and }\]\n\n\[ \pi \left( x\right) - \operatorname{Li}\left( x\right) = O\left( \frac{x}... | Proof. (Sketch). Since this is very standard (in contrast to the above proof due to Iwaniec), we give only a sketch. Let \( d\left( n\right) \) be the number of divisors of \( n \) and\n\n\[ \Delta \left( x\right) = \mathop{\sum }\limits_{{1 \leq n \leq x}}\left( {\log \left( n\right) - d\left( n\right) + {2\gamma }}\r... | No |
Lemma 11.1.7. Let \( \\chi \) be a periodic function defined on \( \\mathbb{Z} \) of period a power of \( p \), and let \( k \\in {\\mathbb{Z}}_{\\geq 0} \). For all \( x \\in {\\mathbb{C}}_{p} \) we have\n\n\[ \n{\\int }_{{\\mathbb{Z}}_{p}}\\chi \\left( t\\right) {\\left( x + t\\right) }^{k}{\\dt} = {B}_{k}\\left( {\\... | Proof. By definition and Corollary 9.4.17 we have\n\n\[ \n{\\int }_{{\\mathbb{Z}}_{p}}\\chi \\left( t\\right) {\\left( x + t\\right) }^{k}{\\dt} = \\mathop{\\lim }\\limits_{{r \\rightarrow \\infty }}\\frac{1}{{p}^{r}}\\mathop{\\sum }\\limits_{{0 \\leq n < {p}^{r}}}\\chi \\left( n\\right) {\\left( n + x\\right) }^{k}\n\... | Yes |
Lemma 11.2.3. We have\n\n\[ \frac{\partial }{\partial x}\langle x{\rangle }^{1 - s} = \left( {1 - s}\right) \frac{\langle x{\rangle }^{1 - s}}{x} = \left( {1 - s}\right) \frac{\langle x{\rangle }^{-s}}{{\omega }_{v}\left( x\right) }.\] | Proof. Trivial and left to the reader. | No |
Proposition 11.2.6. Assume that \( x \in {\mathbb{{CZ}}}_{p} \) .\n\n(1) For any \( k \in \mathbb{Z} \smallsetminus \{ 0\} \) we have\n\n\[ \n{\zeta }_{p}\left( {1 + k, x}\right) = \frac{{\omega }_{v}{\left( x\right) }^{k}}{k}{\int }_{{\mathbb{Z}}_{p}}\frac{dt}{{\left( x + t\right) }^{k}}.\n\]\n\n(2) For \( k \in {\mat... | Proof. By definition of \( {\omega }_{v} \), if \( x \in {\mathrm{{CZ}}}_{p} \) we have \( {\omega }_{v}\left( {1 + n/x}\right) = 1 \), so that \( {\omega }_{v}\left( {n + x}\right) = {\omega }_{v}\left( x\right) \) for all \( n \in \mathbb{Z} \) . It follows that \( \langle n + x{\rangle }^{-k} = {\left( n + x\right) ... | Yes |
Corollary 11.2.7. We have\n\n\[ \frac{\partial {\zeta }_{p}}{\partial x}\left( {s, x}\right) = - \frac{s}{{\omega }_{v}\left( x\right) }{\zeta }_{p}\left( {s + 1, x}\right) . \] | Proof. Formally, this follows from the integral definition and Lemma 11.2.3, but we would need to justify the derivation under the integral sign. Instead, we use the series given by the proposition, since it is normally convergent for \( x \in {\mathrm{{CZ}}}_{p} \) . In that region we can therefore differentiate termw... | Yes |
Proposition 11.2.8. For fixed \( x \in {\mathrm{C\mathbb{Z}}}_{p} \) the function \( {\zeta }_{p}\left( {s, x}\right) \) is a p-adic meromorphic function on \( \left| s\right| < {R}_{p} = {q}_{p}/{p}^{1/\left( {p - 1}\right) } \), which in addition is analytic, except for a simple pole at \( s = 1 \) with residue 1 . | Proof. Since by definition \( \langle x\rangle \equiv 1\left( {{\;\operatorname{mod}\;{q}_{p}}{\mathbb{Z}}_{p}}\right) \), we know that \( \langle x{\rangle }^{1 - s} \) is an analytic function on \( \left| {1 - s}\right| < {R}_{p} \), or equivalently, on \( \left| s\right| < 1 \) since \( {R}_{p} > 1 \) , and in parti... | Yes |
Theorem 11.2.9. Keep the above notation, and let \( x \in {\mathbb{{CZ}}}_{p} \) . (1) For \( k \in {\mathbb{Z}}_{ \geq 1} \) we have \[ {\zeta }_{p}\left( {1 - k, x}\right) = - {\omega }_{v}{\left( x\right) }^{-k}\frac{{B}_{k}\left( x\right) }{k}, \] which is also equal to \( {\omega }_{v}{\left( x\right) }^{-k}\zeta ... | Proof. (1) Although we have already seen this in Proposition 11.2.6, which was in fact the motivation of our definition of \( {\zeta }_{p}\left( {s, x}\right) \), we prove this from the series expansion. Indeed, by definition of the Bernoulli polynomials, for \( k \geq 1 \) we have \[ {\zeta }_{p}\left( {1 - k, x}\righ... | Yes |
Proposition 11.2.10. For \( \\left| s\\right| < {R}_{p} \) and \( x \\in {\\mathbb{{CZ}}}_{p} \) we have\n\n\[ \n{\\int }_{{\\mathbb{Z}}_{p}}{\\zeta }_{p}\\left( {s, x + t}\\right) {dt} = s{\\zeta }_{p}\\left( {s, x}\\right) + \\left( {x - 1}\\right) \\frac{\\partial {\\zeta }_{p}}{\\partial x}\\left( {s, x}\\right) \n... | Proof. The series given by Theorem 11.2.9 (2) being normally convergent for \( u \\in {\\mathbb{Z}}_{p} \) can thus be integrated term by term, so that using Exercise 3 (a) we obtain\n\n\[ \n{\\int }_{{\\mathbb{Z}}_{p}}{\\zeta }_{p}\\left( {s, x + t}\\right) {dt} = \\frac{\\langle x{\\rangle }^{1 - s}}{s - 1}\\mathop{\... | Yes |
Theorem 11.2.11. Let \( x \in {\mathbb{{CZ}}}_{p} \) .\n\n(1) We have \( 1/\left( {s - 1}\right) \mathop{\sum }\limits_{{j \geq 0}}\left( \begin{matrix} 1 - s \\ j \end{matrix}\right) {B}_{j}{x}^{-j} = 1/\left( {s - 1}\right) + \mathop{\sum }\limits_{{j \geq 0}}{c}_{j}{\left( s - 1\right) }^{j} \) with\n\n\[ \n{c}_{0} ... | Proof. For simplicity of notation, set \( v = \left| {{v}_{p}\left( x\right) }\right| = - {v}_{p}\left( x\right) \geq {v}_{p}\left( {q}_{p}\right) \) . By the Clausen-von Staudt theorem and Lemma 4.2.8, for \( j \geq 1 \) we have\n\n\[ \n{v}_{p}\left( {{B}_{j}{x}^{-j}/j!}\right) \geq - j{v}_{p}\left( x\right) - {v}_{p}... | Yes |
Lemma 11.2.13. Let \( \chi \) be a character modulo \( {p}^{v} \), let \( f \) be a function defined for \( {v}_{p}\left( x\right) < - v \) such that for fixed \( x \) the function \( f\left( {x + t}\right) \) is in \( {S}^{1}\left( {\mathbb{Z}}_{p}\right) \), and set \[ g\left( x\right) = {\int }_{{\mathbb{Z}}_{p}}f\l... | Proof. By definition, we have \[ \mathop{\sum }\limits_{{0 \leq j < {p}^{v}}}\chi \left( {x + j}\right) g\left( \frac{x + j}{{p}^{v}}\right) = \mathop{\lim }\limits_{{r \rightarrow \infty }}\frac{1}{{p}^{r}}\mathop{\sum }\limits_{{0 \leq j < {p}^{v}}}\chi \left( {x + j}\right) \mathop{\sum }\limits_{{0 \leq a < {p}^{r}... | Yes |
Corollary 11.2.14. Definition 11.2.12 makes sense for \( x \in {\mathbb{Z}}_{p} \) and \( \left| s\right| < \) \( {R}_{p} \) . More precisely, for any \( M \in {\mathbb{Z}}_{ \geq 1} \) such that \( {p}^{v} \mid M \) we have \[ {\zeta }_{p}\left( {\chi, s, x}\right) = \frac{\langle M{\rangle }^{1 - s}}{M}\mathop{\sum }... | Proof. Applying the lemma to \( f\left( x\right) = \langle x{\rangle }^{1 - s} \) we obtain the given formula for \( M = {p}^{v} \), which also shows the existence of \( {\zeta }_{p}\left( {\chi, s, x}\right) \) . For a general \( M \) , we write \( M = N{p}^{v} \) and \( j = {p}^{v}a + b \) with \( 0 \leq b < {p}^{v} ... | Yes |
Corollary 11.2.15. Let \( \chi \) be a character modulo \( {p}^{v} \) . Then for any \( x \in {\mathbb{Q}}_{p} \) and \( N \in {\mathbb{Z}}_{ \geq 1} \) such that \( {p}^{v} \mid N \) and \( {Nx} \in {\mathbb{Z}}_{p} \) we have\n\n\[ \mathop{\sum }\limits_{{0 \leq j < N}}\chi \left( {{Nx} + j}\right) {\zeta }_{p}\left(... | Proof. Follows from Corollary 11.2.14 applied to \( M = N \) and \( x \) replaced by \( {Nx} \) . | Yes |
Proposition 11.2.16. If \( x \in {\mathbb{Z}}_{p} \) We have\n\n\[ \frac{\partial {\zeta }_{p}}{\partial x}\left( {\chi, s, x}\right) = - s{\zeta }_{p}\left( {\chi {\omega }^{-1}, s + 1, x}\right) . \] | Proof. By Corollary 11.2.7 we have\n\n\[ \frac{\partial {\zeta }_{p}}{\partial x}\left( {\chi, s, x}\right) = - \frac{s}{{p}^{v}}\mathop{\sum }\limits_{{0 \leq j < {p}^{v}}}\frac{\chi \left( {x + j}\right) }{{p}^{v}{\omega }_{v}\left( {\left( {x + j}\right) /{p}^{v}}\right) }{\zeta }_{p}\left( {s + 1,\left( {x + j}\rig... | Yes |
Proposition 11.2.17. For fixed \( x \in {\mathbb{Z}}_{p} \) the function \( {\zeta }_{p}\left( {\chi, s, x}\right) \) is a p-adic meromorphic function on \( \left| s\right| < {R}_{p} \), which is analytic, except when \( \chi = {\chi }_{0} \) , in which case it has a simple pole at \( s = 1 \) with residue \( 1 - 1/p \... | Proof. By Corollary 11.2.14 and Proposition 11.2.8, around \( s = 1 \) we have \( {\zeta }_{p}\left( {\chi, s, x}\right) = {a}_{-1}/\left( {s - 1}\right) + O\left( 1\right) \), where\n\n\[ \n{a}_{-1} = \frac{1}{{p}^{v}}\mathop{\sum }\limits_{{0 \leq j < {p}^{v}}}\chi \left( {x + j}\right) .\n\]\n\nThe result follows si... | Yes |
Corollary 11.2.18. We have\n\n\[ \n\frac{\partial {\zeta }_{p}}{\partial x}\left( {{\chi \omega },0, x}\right) = - \left( {1 - \frac{1}{p}}\right) \delta \left( \chi \right) \n\]\n\nwhere here and elsewhere \( \delta \left( \chi \right) = 0 \) if \( \chi \neq {\chi }_{0} \) and \( \delta \left( \chi \right) = 1 \) if \... | Proof. By analyticity and Proposition 11.2.16 we have\n\n\[ \n\frac{\partial {\zeta }_{p}}{\partial x}\left( {{\chi \omega },0, x}\right) = \mathop{\lim }\limits_{{s \rightarrow 0}}\frac{\partial {\zeta }_{p}}{\partial x}\left( {{\chi \omega }, s, x}\right) = \mathop{\lim }\limits_{{s \rightarrow 0}}\left( {-s{\zeta }_... | Yes |
Proposition 11.2.19. (1) For any \( k \in {\mathbb{Z}}_{ \geq 1} \) and \( x \in {\mathbb{Z}}_{ \geq 0} \) we have\n\n\[{\zeta }_{p}\left( {\chi {\omega }^{k},1 - k, x}\right) = - \frac{{B}_{k}\left( \chi \right) }{k} - \mathop{\sum }\limits_{{0 \leq r < x}}\chi \left( r\right) {r}^{k - 1}.\] | Proof. (1) and (2). By Corollary 11.2.14 and Proposition 11.2.6 we have\n\n\[{\zeta }_{p}\left( {\chi {\omega }^{k},1 - k, x}\right) = \frac{1}{{p}^{v}}\mathop{\sum }\limits_{{0 \leq j < {p}^{v}}}\chi {\omega }^{k}\left( {x + j}\right) {\zeta }_{p}\left( {1 - k,\left( {x + j}\right) /{p}^{v}}\right)\]\n\n\[= - \frac{1}... | Yes |
Proposition 11.2.23. Let \( \chi \) be a character modulo \( {p}^{v} \) for some \( v \geq 1 \) . For \( x \in {p}^{v}{\mathbb{Z}}_{p} \) we have the power series expansion\n\n\[ \n{\zeta }_{p}\left( {\chi, s, x}\right) = \mathop{\sum }\limits_{{k \geq 0}}\left( \begin{matrix} 1 - s \\ k \end{matrix}\right) {L}_{p}\lef... | Proof. Although this result involves \( p \) -adic \( L \) -functions which we will study in much more detail below, taking simply \( {L}_{p}\left( {\chi, s}\right) = {\zeta }_{p}\left( {\chi, s,0}\right) \) as a definition is enough for the proof. Indeed, since \( x \in {p}^{v}{\mathbb{Z}}_{p} \) and \( \chi \) is def... | Yes |
Lemma 11.3.2. Let \( p \) be a prime number and \( \alpha \) an algebraic number. The following conditions are equivalent:\n\n(1) \( \alpha \) is p-integral.\n\n(2) For any embedding \( \sigma \) of \( \overline{\mathbb{Q}} \) into \( {\mathbb{C}}_{p} \) we have \( \left| {\sigma \left( \alpha \right) }\right| \leq 1 \... | Proof. Clear and left to the reader (Exercise 7). | No |
Lemma 11.3.3. If either \( {\chi }_{1}\left( a\right) \neq 0 \) or \( {\chi }_{2}\left( a\right) \neq 0 \) we have \( {\chi }_{1}{\chi }_{2}\left( a\right) = \) \( {\chi }_{1}\left( a\right) {\chi }_{2}\left( a\right) \) . | Proof. If \( {\chi }_{1}\left( a\right) \neq 0 \) and \( {\chi }_{2}\left( a\right) \neq 0 \) we have by definition \( \left( {{\chi }_{1}{\chi }_{2}}\right) \left( a\right) = \) \( {\chi }_{1}\left( a\right) {\chi }_{2}\left( a\right) \) . If exactly one of them is nonzero, say \( {\chi }_{1}\left( a\right) \neq 0 \) ... | Yes |
Lemma 11.3.7. Let \( \chi \) be a nontrivial primitive character of conductor \( f \) , and let \( m \) be a common multiple of \( f \) and \( p \) . Then\n\n\[ \mathop{\sum }\limits_{{0 \leq a < m}}^{\left( p\right) }\chi \left( a\right) = 0. \] | Proof. By multiplicativity we have\n\n\[ \mathop{\sum }\limits_{{0 \leq a < m}}\chi \left( a\right) = \mathop{\sum }\limits_{{0 \leq a < m}}\chi \left( a\right) - \chi \left( p\right) \mathop{\sum }\limits_{{0 \leq b < m/p}}\chi \left( b\right) . \]\n\nSince \( \chi \) is nontrivial and \( f \mid m \) the first sum is ... | Yes |
Proposition 11.3.8. Let \( \chi \) be a primitive character of conductor \( f \), let \( m \in \) \( {\mathbb{Z}}_{ > 0} \) be a multiple of \( f \), and let \( s \in {\mathbb{C}}_{p} \) be such that \( \left| s\right| < {R}_{p} \) and \( s \neq 1 \) .\n\n(1) We have\n\n\[ \n{L}_{p}\left( {\chi, s}\right) = \frac{\lang... | Proof. (1). Writing \( a = {kf} + r \) we have\n\n\[ \n\frac{\langle m{\rangle }^{1 - s}}{m}\mathop{\sum }\limits_{{0 \leq a < m}}{\chi }_{0, m}\left( a\right) \chi \left( a\right) {\zeta }_{p}\left( {s,\frac{a}{m}}\right) \n\]\n\n\[ \n= \frac{\langle m{\rangle }^{1 - s}}{m}\mathop{\sum }\limits_{{0 \leq r < f}}\chi \l... | No |
Proposition 11.3.9. Keep the above assumptions.\n\n(1) The function \( {L}_{p}\left( {\chi, s}\right) \) is a p-adic analytic function for \( \left| s\right| < {R}_{p} \), except when \( \chi = {\chi }_{0} \), in which case the function \( {\zeta }_{p}\left( s\right) = {L}_{p}\left( {{\chi }_{0}, s}\right) \) has a sim... | Proof. (1). By definition \( {L}_{p}\left( {\chi, s}\right) \) is a \( p \) -adic meromorphic function with a possible simple pole at \( s = 1 \) with residue \( \mathop{\sum }\limits_{{0 \leq a < f}}\chi \left( a\right) {\operatorname{Res}}_{s = 1}{\zeta }_{p}\left( {s, a/f}\right) \) , and by Propositions 11.2.8 and ... | Yes |
Proposition 11.3.10. Let \( \chi \) be a primitive character modulo \( f \). (1) For \( k \in \mathbb{Z} \smallsetminus \{ 0\} \) we have \[ {L}_{p}\left( {\chi, k + 1}\right) = \frac{1}{k}\mathop{\lim }\limits_{{r \rightarrow \infty }}\frac{1}{f{p}^{r}}\mathop{\sum }\limits_{{0 \leq n < f{p}^{r}}}^{\left( p\right) }\f... | Proof. (1). Indeed, by what we have just recalled, for \( p \nmid a \) we have \[ {\zeta }_{p}\left( {k + 1,\frac{a}{m}}\right) = \frac{\omega {\left( a\right) }^{k}\langle m{\rangle }^{k}}{k}\mathop{\lim }\limits_{{r \rightarrow \infty }}\frac{1}{{p}^{r}}\mathop{\sum }\limits_{{0 \leq j < {p}^{r}}}\frac{1}{{\left( mj ... | Yes |
Corollary 11.3.11. Let \( k \in \mathbb{Z} \smallsetminus \{ 0\} \) . If \( \chi \) is a primitive character modulo a power of \( p \) we have\n\n\[ \n{L}_{p}\left( {\chi, k + 1}\right) = \frac{1}{k}{\int }_{{\mathbb{Z}}_{p}^{ * }}\frac{\chi {\omega }^{k}\left( t\right) }{{t}^{k}}{dt} \n\]\n\nand\n\[ \n\mathop{\lim }\l... | Proof. Clear. Note that the integrals are over \( {\mathbb{Z}}_{p}^{ * } \) . | No |
Proposition 11.3.12. Let \( \chi \) be a primitive character modulo \( f \) .\n\n(1) For all \( k \in \mathbb{Z} \smallsetminus \{ 0\} \) we have\n\n\[ \n{L}_{p}\left( {\chi, k + 1}\right) = \mathop{\lim }\limits_{{r \rightarrow \infty }}\frac{{B}_{\phi \left( {p}^{r}\right) - k}\left( {\chi {\omega }^{k}}\right) }{k}.... | Proof. Since \( {L}_{p}\left( {\chi, s}\right) \) is a continuous function of \( s \neq 1 \), for all \( k \in \mathbb{Z} \) we\n\nhave\n\n\[ \n{L}_{p}\left( {\chi, k + 1}\right) = \mathop{\lim }\limits_{{r \rightarrow \infty }}{L}_{p}\left( {\chi, k + 1 - \phi \left( {p}^{r}\right) }\right) .\n\]\n\nSince \( {\omega }... | Yes |
Proposition 11.3.14. Assume that the conductor of \( \chi \) is a power of \( p \) (which is true in particular when \( \chi = {\chi }_{0} \) ). Then for \( k \in \mathbb{Z} \smallsetminus \{ 0\} \) we have\n\n\[ \n{B}_{k, p}\left( \chi \right) = \mathop{\lim }\limits_{{r \rightarrow \infty }}\frac{1}{{p}^{r}}\mathop{\... | Proof. This is a restatement of Corollary 11.3.11. | Yes |
Proposition 11.3.15. (1) If \( \chi \left( {-1}\right) = {\left( -1\right) }^{k - 1} \) we have \( {B}_{k, p}\left( \chi \right) = 0 \), and if \( \chi \left( {-1}\right) = - 1 \) we have \( {\gamma }_{p}\left( \chi \right) = 0 \) . | Proof. All the statements except the last two are clear from the definitions and Proposition 11.3.9. By Lemma 9.5.11 we know that \( {v}_{p}\left( {{B}_{k}\left( \chi \right) }\right) \geq - 1 \) , and since \( {B}_{k, p}\left( \chi \right) = \mathop{\lim }\limits_{{r \rightarrow \infty }}{B}_{\phi \left( {p}^{r}\right... | No |
Corollary 11.3.16. Let \( k \geq 2 \) be an even integer.\n\n(1) We have\n\n\[ \n{B}_{-k, p} \equiv \frac{1}{p}\mathop{\sum }\limits_{{1 \leq a \leq p - 1}}\frac{1}{{a}^{k}}\left( {{\;\operatorname{mod}\;{p}^{v}}{\mathbb{Z}}_{p}}\right) , \]\n\nwhere \( v = 1 \) if \( 5 \leq p \leq k + 3 \), and \( v = 2 \) for \( p \g... | Proof. Immediate consequence of the proposition and of the Kummer congruences, and left to the reader; see Exercise 50. | No |
Lemma 11.3.18. Let \( \chi \) be a primitive character modulo \( {p}^{v} \) for some odd prime \( p \) and some \( v \geq {v}_{p}\left( {q}_{p}\right) \), let \( g \) be a primitive root modulo \( {p}^{v} \), and let \( n \in {\mathbb{Z}}_{ \geq 0} \) . Then\n\n\[ \mathop{\sum }\limits_{{0 \leq k < \phi \left( {p}^{v}\... | Proof. The result being trivial if \( v = 1 \), we may assume that \( v \geq 2 \) . To simplify notation set \( \zeta = \chi \left( g\right) \), which is a root of unity of order \( o\left( \chi \right) \), and write \( k = {p}^{v - 1}a + b \) with \( 0 \leq a < p - 1 \) and \( 0 \leq b < {p}^{v - 1} \) . If \( S \) de... | Yes |
Corollary 11.3.20. Let \( \chi \) be a primitive character of conductor \( f \), let \( m \) be the least common multiple of \( f \) and \( {q}_{p} \), and for simplicity of notation set \[ {T}_{n}\left( \chi \right) = \mathop{\sum }\limits_{{0 \leq a < m}}^{\left( p\right) }\chi \left( a\right) \frac{{\log }_{p}{\left... | Proof. (1). As already remarked in the theorem, we have \( {q}_{p}^{n - 1}/n! \in p{\mathbb{Z}}_{p} \) for \( n \geq 2 \), so (1) follows when \( \chi \) is \( p \) -adically tame. If on the contrary \( \chi \) is wild, hence \( p \) odd, the theorem says that the congruence is true modulo \( \left( {{p}^{n - 2}/n!}\ri... | Yes |
Corollary 11.3.22. Let \( \chi \) be an even primitive character modulo \( f \), and assume that \( \chi \) is p-adically tame. Define\n\n\[ \n{M}_{p}\left( {\chi, s}\right) = \left\{ \begin{array}{ll} {L}_{p}\left( {\chi, s}\right) & \text{ if }\chi \text{ is nontrivial,} \\ {L}_{p}\left( {\chi, s}\right) - \frac{1 - ... | Proof. Clear. | No |
Proposition 11.4.1. Let \( \chi \) be an even primitive character of conductor \( f \) and let \( k \in \mathbb{Z} \) be arbitrary, not necessarily positive.\n\n(1) If \( \chi \) is nontrivial and \( p \) -adically tame then \( {L}_{p}\left( {\chi ,1 - k}\right) \) is p-integral and\n\n\[ {L}_{p}\left( {\chi ,1 - k}\ri... | Proof. Since \( {L}_{p}\left( {\chi, s}\right) \) is an analytic function for \( \left| s\right| < {R}_{p} \) and \( {R}_{p} > 1 \), it follows that the radius of convergence of its Taylor series around \( s = 1 \) is greater than or equal to \( {R}_{p} \) . Thus, by Theorem 11.3.21, if \( s \) is \( p \) -integral (an... | Yes |
Corollary 11.4.3. Let \( k \in {\mathbb{Z}}_{ \geq 1} \), let \( D \) be the discriminant of a quadratic field, and assume that \( \operatorname{sign}\left( D\right) = {\left( -1\right) }^{k} \). (1) Assume that \( D \neq - 4, D \neq \pm 8 \), and that either \( D \neq {\left( -1\right) }^{\left( {p - 1}\right) /2}p \)... | Proof. Immediate from the preceding corollary since the absolute value of the discriminant of a quadratic field is not divisible by the square of an odd prime and is a power of 2 only for \( \left| D\right| = 4 \) and \( \left| D\right| = 8 \) ; see Exercise 12. | No |
Proposition 11.4.4. For any \( k \geq 2 \) even, set\n\n\[ \n{z}_{p}\left( k\right) = \left\{ \begin{array}{ll} \left( {{p}^{k - 1} - 1}\right) \frac{{B}_{k}}{k} & \text{ if }\left( {p - 1}\right) \nmid k, \\ \left( {{p}^{k - 1} - 1}\right) \frac{{B}_{k}}{k} + \frac{1 - 1/p}{k} & \text{ if }\left( {p - 1}\right) \mid k... | Proof. Since \( {k}^{\prime } \equiv k\left( {{\;\operatorname{mod}\;p} - 1}\right) \) we have \( {\omega }^{{k}^{\prime }} = {\omega }^{k} \), so by Corollary 11.3.22, since \( p \mid {b}_{i} \) for \( i \geq 1 \) we have\n\n\[ \n{M}_{p}\left( {{\omega }^{k},1 - k}\right) = \mathop{\sum }\limits_{{j \geq 0}}{b}_{j}{\l... | Yes |
Corollary 11.4.5. Let \( k \) and \( {k}^{\prime } \) be even and such that \( \left( {p - 1}\right) \mid k \) and \( \left( {p - 1}\right) \mid \) \( {k}^{\prime } \), and assume that \( \min \left( {k - 2 - {v}_{p}\left( k\right) ,{k}^{\prime } - 2 - {v}_{p}\left( {k}^{\prime }\right) }\right) \geq e \) . Then if \( ... | Proof. Left to the reader (Exercise 25). | No |
Corollary 11.4.6. Let \( k \geq 2 \) be even.\n\n(1) If \( \left( {k, p}\right) \neq \left( {2,2}\right) ,\left( {2,3}\right) \), and \( \left( {4,2}\right) \) we have\n\n\[ \frac{{B}_{k}}{k} \equiv {B}_{1}\left( {\omega }^{k - 1}\right) \left( {\;\operatorname{mod}\;p}\right) \text{ if }\left( {p - 1}\right) \nmid k, ... | Proof. By Proposition 11.3.9 (2) we have\n\n\[ {L}_{p}\left( {{\omega }^{k},0}\right) = - \left( {1 - {\omega }^{k - 1}\left( p\right) }\right) {B}_{1}\left( {\omega }^{k - 1}\right) = - {B}_{1}\left( {\omega }^{k - 1}\right) \]\n\nsince \( k - 1 \) is odd, so \( {\omega }^{k - 1} \) is a nontrivial character (recall t... | Yes |
Corollary 11.4.7. Let \( k \geq 2 \) be an even integer.\n\n(1) We have \( {B}_{k} \equiv k + 1/2\left( {\;\operatorname{mod}\;{2}^{2 + {v}_{2}\left( k\right) }}\right) \), so in particular \( {B}_{k} \equiv k + 1/2 \) \( \left( {\;\operatorname{mod}\;4}\right) \) and \( {B}_{k} \equiv 1/2\left( {\;\operatorname{mod}\;... | Proof. Left to the reader (Exercise 25). | No |
Corollary 11.4.8. If \( k \neq 0 \) is even then \( {B}_{k, p}/k \) is p-integral if \( \left( {p - 1}\right) \nmid k \) , and \( \left( {{B}_{k, p} - \left( {1 - 1/p}\right) }\right) /k \) is p-integral if \( \left( {p - 1}\right) \mid k \) (including for \( p = 3 \) and for \( \left( {k, p}\right) = \left( {2,2}\righ... | Proof. Immediate from the above results and the definitions of \( {B}_{k, p} \) and \( {\gamma }_{p} \), and left to the reader (Exercise 35). | No |
Corollary 11.4.9. Assume that \( p \geq 5 \) . Then \( p \mid {h}_{p}^{ - } \) if and only if \( {v}_{p}\left( {B}_{k}\right) \geq 1 \) for some even \( k \) such that \( 2 \leq k \leq p - 3 \) . | Proof. Indeed, since the odd characters of \( {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{ * } \simeq \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{p}\right) /\mathbb{Q}}\right) \) are the \( {\omega }^{k - 1} \) for \( k \) even with \( 2 \leq k \leq p - 1 \), by Proposition 10.5.26 we have\n\n\[ {v}_{p}\left( {h}... | Yes |
Proposition 11.4.11 (Carlitz). Let \( p \) be a prime number and let \( n \geq 1 \) be an integer. We have the congruence\n\n\[ p + \left( {p - 1}\right) \mathop{\sum }\limits_{{1 \leq j \leq \left( {n - 1}\right) /\left( {p - 1}\right) }}\left( \begin{matrix} n \\ \left( {p - 1}\right) j \end{matrix}\right) \equiv 0\l... | Proof. For \( p = 2 \) the left-hand side is equal to \( {2}^{n} \), and since \( n - 1 \geq {v}_{2}\left( n\right) \) for all \( n \geq 1 \) the result is clear, so we may assume that \( p \geq 3 \) . By the recurrence formula for Bernoulli numbers (Proposition 9.1.3) we have\n\n\[ 1 - \frac{n}{2} + n\mathop{\sum }\li... | Yes |
Theorem 11.4.12. For \( n \geq 0, k \in \mathbb{Z} \smallsetminus \{ 0\} \), and \( h \in \mathbb{Z} \) set \( {b}_{n}\left( {h, k}\right) = \) \( {k}^{n}\left( {{B}_{n}\left( {h/k}\right) - {B}_{n}}\right) \) . We have \( {b}_{0}\left( {h, k}\right) = 0,{b}_{1}\left( {h, k}\right) = h,{b}_{2}\left( {h, k}\right) = h\l... | Proof. As in the proof of Theorem 9.5.29, for \( n > 2 \) we have \[ \frac{{b}_{n}\left( {h, k}\right) }{nk} = \frac{{h}^{n}}{nk} - \frac{{h}^{n - 1}}{2} + \mathop{\sum }\limits_{{2 \leq m \leq n - 1}}\left( \begin{matrix} n - 1 \\ m - 1 \end{matrix}\right) \frac{{B}_{m}}{m}{h}^{n - m}{k}^{m - 1}. \] Fix some prime num... | Yes |
Corollary 11.5.3. For \( x \in {\mathrm{{CZ}}}_{p} \) define \( {\psi }_{p}\left( x\right) = \left( {d/{dx}}\right) \left( {{\operatorname{Log\Gamma }}_{p}\left( x\right) }\right) \) . We have \( {\psi }_{p}\left( {x + 1}\right) = {\psi }_{p}\left( x\right) + 1/x,{\psi }_{p}\left( {1 - x}\right) = {\psi }_{p}\left( x\r... | Proof. The proofs of the proposition and its corollary follow immediately from the corresponding properties of \( {\zeta }_{p}\left( {s, x}\right) \) and are left to the reader (Exercise 20). | No |
Proposition 11.5.4. Let \( \chi \) be a primitive character of conductor \( f \), and let \( N \in {\mathbb{Z}}_{ \geq 1} \) be a common multiple of \( f \) and \( {q}_{p} \) . Recall from Proposition 11.3.9 that \( {L}_{p}\left( {{\chi \omega },0}\right) = - \left( {1 - \chi \left( p\right) }\right) {B}_{1}\left( \chi... | Proof. Recall that by definition of the \( p \) -adic \( L \) -function, for any \( N \) divisible by \( {q}_{p} \) and by \( f \) we have\n\n\[ {L}_{p}\left( {\chi, s}\right) = \mathop{\sum }\limits_{{0 \leq k < N}}^{\left( p\right) }\chi \left( k\right) {\zeta }_{p}\left( {s, k, N}\right) = {\omega }_{v}{\left( N\rig... | Yes |
Corollary 11.5.5. Let \( \chi \) be a primitive character of conductor \( f \), and let \( N \in {\mathbb{Z}}_{ \geq 1} \) be a common multiple of \( f \) and \( {q}_{p} \) . We have\n\n\[ \n{L}_{p}^{\prime }\left( {\chi ,0}\right) = \left( {1 - \chi {\omega }^{-1}\left( p\right) }\right) {B}_{1}\left( {\chi {\omega }^... | Proof. Clear. | No |
Proposition 11.5.6. (1) For all \( k \in {\mathbb{Z}}_{ \geq 1} \) we have\n\n\[{\zeta }_{p}\left( {k + 1, x}\right) = {\left( -1\right) }^{k - 1}{\omega }_{v}{\left( x\right) }^{k}\frac{{\psi }_{p}^{\left( k\right) }\left( x\right) }{k!}.\] | Proof. (1) immediately follows by comparing the expansions\n\n\[{\zeta }_{p}\left( {k + 1, x}\right) = \frac{\langle x{\rangle }^{-k}}{k}\mathop{\sum }\limits_{{j \geq 0}}{\left( -1\right) }^{j}\left( \begin{matrix} k + j - 1 \\ k - 1 \end{matrix}\right) {B}_{j}{x}^{-j}\;\text{ and }\]\n\n\[{\left( -1\right) }^{k - 1}{... | Yes |
Proposition 11.5.7. For \( \left| x\right| > 1 \) we have\n\n\[ \n{\operatorname{log\Gamma }}_{p}\left( x\right) = {\int }_{{\mathbb{Z}}_{p}}\left( {\left( {x + t}\right) {\log }_{p}\left( {x + t}\right) - \left( {x + t}\right) }\right) {dt} \]\n\n\[ \n{\psi }_{p}\left( x\right) = {\int }_{{\mathbb{Z}}_{p}}{\log }_{p}\... | Proof. From the formal power series expansion for \( \log \left( {1 + T}\right) \) we deduce that \( \left( {1 + T}\right) \log \left( {1 + T}\right) - T = \mathop{\sum }\limits_{{j \geq 1}}{\left( -1\right) }^{j - 1}{T}^{j + 1}/\left( {j\left( {j + 1}\right) }\right) \) . Thus, since \( \left| x\right| > 1 \) we have ... | Yes |
Proposition 11.5.8. For \( \left| x\right| > 1 \) we have\n\n\[ \mathop{\lim }\limits_{{n \rightarrow 0}}\frac{{B}_{n}\left( x\right) - {x}^{n}}{n{x}^{n}} = {\psi }_{p}\left( x\right) - {\log }_{p}\left( x\right) ,\]\n\nwhere \( n \) tends \( p \) -adically to 0 in \( {\mathbb{Z}}_{ > 0} \) . | Proof. By definition\n\n\[ \frac{{B}_{n}\left( x\right) - {x}^{n}}{n{x}^{n}} = \frac{1}{n}\mathop{\sum }\limits_{{1 \leq j \leq n}}\left( \begin{array}{l} n \\ j \end{array}\right) {B}_{j}{x}^{-j} = \mathop{\sum }\limits_{{1 \leq j \leq n}}\left( \begin{array}{l} n - 1 \\ j - 1 \end{array}\right) \frac{{B}_{j}}{j}{x}^{... | Yes |
Proposition 11.5.9. If \( x \in {\mathbb{{CZ}}}_{p} \) we have\n\n\[ \n{\int }_{{\mathbb{Z}}_{p}}\log {\Gamma }_{p}\left( {x + t}\right) {dt} = \left( {x - 1}\right) {\psi }_{p}\left( x\right) - x + \frac{1}{2}.\n\] | Proof. It is not difficult to show that we can differentiate with respect to \( s \) under the integral sign in Proposition 11.2.10; hence after setting \( s = 0 \) we obtain\n\n\[ \n{\int }_{{\mathbb{Z}}_{p}}{\omega }_{v}^{-1}\left( {x + t}\right) \log {\Gamma }_{p}\left( {x + t}\right) = \left( {x - 1}\right) {\omega... | Yes |
Proposition 11.5.10. For all \( a \in {\mathbb{Z}}_{p}^{ * } \) we have\n\n\[ \log {\Gamma }_{p}\left( \frac{a}{{q}_{p}}\right) \equiv \left\{ \begin{array}{lll} - \omega \left( a\right) /p & \left( {{\;\operatorname{mod}\;p}{\mathbb{Z}}_{p}}\right) & \text{ if }p \geq 5, \\ {2\omega }\left( a\right) /3 & \left( {{\;\o... | Proof. Assume first that \( p \geq 3 \) . By Proposition 11.5.2 we have\n\n\[ \log {\Gamma }_{p}\left( \frac{a}{p}\right) = \frac{1}{p}\left( {a{\log }_{p}\left( a\right) - a}\right) - \frac{1}{2}{\log }_{p}\left( a\right) + \mathop{\sum }\limits_{{k \geq 1}}\frac{{B}_{2k}}{{2k}\left( {{2k} - 1}\right) }\frac{{p}^{{2k}... | Yes |
Proposition 11.5.12. Let \( M \) be such that \( {p}^{v} \mid M \) . For \( x \in {\mathbb{Z}}_{p} \) we have\n\n\[ \log {\Gamma }_{p}\left( {\chi, x}\right) = \mathop{\sum }\limits_{{0 \leq j < M}}\chi \left( {x + j}\right) \log {\Gamma }_{p}\left( \frac{x + j}{M}\right) \]\n\n\[ + {\log }_{p}\left( M\right) \mathop{\... | Proof. Immediate from the definition and Corollary 11.2.14, since by Proposition 11.2.6 we have \( {\zeta }_{p}\left( {0, x}\right) = - {\omega }_{v}{\left( x\right) }^{-1}\left( {x - 1/2}\right) \) when \( x \in {\mathbb{{CZ}}}_{p} \) . | No |
Proposition 11.5.15. Let \( \chi \) be a character modulo \( {p}^{v} \) .\n\n(1) For all \( k \in {\mathbb{Z}}_{ \geq 1} \) we have\n\n\[ \n{\zeta }_{p}\left( {\chi, k + 1, x}\right) = {\left( -1\right) }^{k - 1}\frac{{\psi }_{p}^{\left( k\right) }\left( {\chi {\omega }^{k}, x}\right) }{k!}.\n\]\n\n(2) Around \( s = 1 ... | Proof. The proofs of these results are immediate from Proposition 11.2.20, Corollary 11.2.15, and Proposition 11.5.6, and left to the reader (Exercise 15). | No |
Corollary 11.5.16. For all \( k \in {\mathbb{Z}}_{ \geq 1} \) we have\n\n\[ \n{\psi }_{p}^{\left( k\right) }\left( {\chi ,0}\right) = {\left( -1\right) }^{k - 1}k!{L}_{p}\left( {\chi {\omega }^{-k}, k + 1}\right) = {\left( -1\right) }^{k - 1}\left( {k - 1}\right) !{B}_{-k, p}\left( \chi \right) ,\n\]\n\nand for \( k = ... | Proof. Clear from the above proposition since \( {L}_{p}\left( {\chi, s}\right) = {\zeta }_{p}\left( {\chi, s,0}\right) \) if \( \chi \) has \( p \) -power conductor. | No |
Proposition 11.5.17. Let \( \chi \) be a character modulo \( {p}^{v} \), let \( x \in {\mathbb{Z}}_{p} \), and let \( N \in {\mathbb{Z}}_{ \geq 1} \) be such that \( {p}^{v} \mid N \). (1) We have the distribution formula \[ \mathop{\sum }\limits_{{0 \leq k < N}}\chi \left( {x + k}\right) {\operatorname{log\Gamma }}_{p... | Proof. By Corollary 11.2.14 we have \[ {\zeta }_{p}\left( {\chi, s, x}\right) = \frac{\langle N{\rangle }^{1 - s}}{N}\mathop{\sum }\limits_{{0 \leq k < N}}\chi \left( {x + k}\right) {\zeta }_{p}\left( \frac{x + k}{N}\right) . \] Differentiating this equality with respect to \( s \), setting \( s = 0 \), replacing \( \c... | Yes |
Proposition 11.5.18. For all \( x \in {\mathbb{Z}}_{p} \) we have the Volkenborn integral representations\n\n\[ \log {\Gamma }_{p}\left( {\chi, x}\right) = {\int }_{{\mathbb{Z}}_{p}}\chi \left( {x + t}\right) \left( {\left( {x + t}\right) {\log }_{p}\left( {x + t}\right) - \left( {x + t}\right) }\right) {dt} \]\n\n\[ {... | Proof. Recall that by definition we have\n\n\[ {\zeta }_{p}\left( {\chi, s, x}\right) = \frac{1}{s - 1}{\int }_{{\mathbb{Z}}_{p}}\chi \left( {x + t}\right) \langle x + t{\rangle }^{1 - s}{dt}. \]\n\nUsing the uniformity estimate given in Proposition 11.2.4 it is easy to show that we can differentiate with respect to \(... | No |
Proposition 11.5.19. For \( x \in {p}^{v}{\mathbb{Z}}_{p} \) we have the convergent power series expansion\n\n\[ \log {\Gamma }_{p}\left( {\chi, x}\right) = {L}_{p}^{\prime }\left( {{\chi \omega },0}\right) - {\gamma }_{p}\left( \chi \right) x + \mathop{\sum }\limits_{{k \geq 2}}{\left( -1\right) }^{k}\frac{{B}_{1 - k,... | Proof. The proof is essentially the same as that of Proposition 11.2.23, and in fact the result can be deduced from that proposition. Nonetheless, I prefer to redo it here. By Proposition 11.5.18 we have\n\n\[ \log {\Gamma }_{p}\left( {\chi, x}\right) = {\int }_{{\mathbb{Z}}_{p}}\chi \left( {x + t}\right) \left( {\left... | Yes |
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