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Let \( k \) be a number field, \( \left\lbrack {k : \mathbf{Q}}\right\rbrack = N \) . Let \( {B}_{k} \) be the constant\n\n\[ \n{B}_{k} = \frac{{2}^{{r}_{1}}{\left( 2\pi \right) }^{{r}_{2}}}{{\left| {D}_{k}\right| }^{1/2}}. \]\n\nThen, for \( \mathfrak{c} \) ranging over \( {M}_{k} \) -divisors, the number \( \lambda \... | Proof. We shall first make some remarks concerning \( {M}_{k} \) -divisors. Given an \( {M}_{k} \) -divisor \( \mathfrak{c} \), there exists a fractional ideal \( \mathfrak{a} \) of \( {\mathfrak{o}}_{k} \) such that \( \alpha \in \mathfrak{a} \) if and only if\n\n\[ \n{\left| \alpha \right| }_{\mathfrak{p}} \leqq \mat... | Yes |
Lemma 1. Assume that the associated fractional ideal \( \mathfrak{a} \) is equal to one of the fixed representatives \( {\mathfrak{a}}_{i} \). There exists a unit \( u \) of \( {\mathfrak{o}}_{k} \) such that we have, for all \( v \in {S}_{\infty } \), \[ {c}_{1}\left( k\right) \parallel \mathfrak{c}{\parallel }_{k}^{1... | Proof. Let \( V = \parallel \mathfrak{c}{\parallel }_{k} \) and let \( {\mathfrak{c}}_{v}^{\prime } = {\mathfrak{c}}_{v}{\left( V\mathbf{N}\mathfrak{a}\right) }^{-1/N} \) for all \( v \in {S}_{\infty } \). Then \[ \mathop{\prod }\limits_{{v \in {S}_{\infty }}}{\mathfrak{c}}_{v}^{\prime {N}_{v}} = 1 \] Consider now the ... | Yes |
Lemma 2. Let \( \mathfrak{a} \) be an ideal of the ring of integers of \( k \), and let \( F \) be a fundamental domain of \( \mathfrak{a} \), as lattice in \( {\mathbf{R}}^{N} \) . Then\n\n\[ \operatorname{Vol}\left( F\right) = {2}^{-{r}_{2}}{\left| {D}_{k/\mathbf{Q}}\left( \mathfrak{a}\right) \right| }^{1/2} = {2}^{-... | Proof. The ideal \( \mathfrak{a} \) has a basis \( {\alpha }_{1},\ldots ,{\alpha }_{N} \) over \( \mathbf{Z} \) .\n\nLet \( {\sigma }_{1},\ldots ,{\sigma }_{{r}_{1}} \) be the real embeddings of \( k \) . Let \( {\tau }_{1},\ldots ,{\tau }_{{r}_{2}} \) and their conjugates be the complex ones. Each \( \alpha \) in \( k... | Yes |
Theorem 3. Let \( L \) be a lattice of dimension \( N \) in \( {\mathbf{R}}^{N} \), and let \( C \) be a closed, convex, symmetric subset of \( {\mathbf{R}}^{N} \). If\n\n\[ \mu \left( C\right) \geqq {2}^{N}\mu \left( F\right) \]\n\nwhere \( F \) is a fundamental domain for \( L \), then there exists a lattice point \(... | Proof. We shall first prove the theorem under the assumption \( \mu \left( C\right) > {2}^{N}\mu \left( F\right) \).\n\nUnder this assumption, we contend that there exist two elements in \( \frac{1}{2}C \) whose difference is in \( L \). Indeed, we have\n\n\[ \frac{1}{2}C = \mathop{\bigcup }\limits_{{x \in L}}\left( {\... | Yes |
Theorem 4. In any ideal class, there exists an ideal \( \mathfrak{b} \) such that\n\n\[ \mathrm{N}\mathfrak{b} \leqq {C}_{k}{d}_{k}^{1/2} \]\n\nwhere \( {C}_{k} \) is the Minkowski constant:\n\n\[ {C}_{k} = \frac{N!}{{N}^{N}}{\left( \frac{4}{\pi }\right) }^{{r}_{2}} \] | Proof. We have \( \mathbf{N}\mathfrak{b} \geqq 1 \) whence\n\n\[ {d}_{k} \geqq {\left( \frac{\pi }{4}\right) }^{2{r}_{2}}\frac{{N}^{2N}}{{\left( N!\right) }^{2}} \geqq {\left( \frac{\pi }{4}\right) }^{N}\frac{{N}^{2N}}{{\left( N!\right) }^{2}}.\n\nIf \( N = 2 \), then we obtain at once \( d > \frac{9}{4} > 1 \) . Our a... | Yes |
Theorem 5. If \( k \) is a number field, denote by \( {N}_{k} \) and \( {d}_{k} \) the degree \( \left\lbrack {k : \mathbf{Q}}\right\rbrack \) and absolute value of the discriminant respectively. Then the quotient \( {N}_{k}/\log {d}_{k} \) is bounded for all \( k \neq \mathbf{Q} \) . Furthermore, there exists only a f... | Proof. The first assertion follows from a trivial computation involving the inequality of the Corollary to Theorem 4, and the standard estimate from Stirling's formula\n\n\[ N! = {N}^{N}{e}^{-N}\sqrt{2\pi N}{e}^{\theta /{12N}},\;0 < \theta < 1. \]\n\nWe leave it to the reader. This shows that the degree is bounded when... | Yes |
Theorem 1. The group of \( \mathfrak{c} \) -ideal classes \( I\left( \mathfrak{c}\right) /{P}_{\mathfrak{c}} \) is finite. If \( h \) is the class number of \( k \), and \( {h}_{\mathrm{c}} \) is the order of \( I\left( \mathrm{c}\right) /{P}_{\mathrm{c}} \), and \( s\left( \mathrm{c}\right) \) is the number of real \(... | \[ {h}_{\mathrm{c}} = \frac{{h\varphi }\left( {\mathrm{c}}_{0}\right) {2}^{s\left( \mathrm{c}\right) }}{\left( U : {U}_{\mathrm{c}}\right) } \] It is a reasonable convention to define \[ \varphi \left( \mathfrak{c}\right) = \varphi \left( {\mathfrak{c}}_{0}\right) {2}^{s\left( \mathfrak{c}\right) } \] so as to include ... | Yes |
Theorem 2. Let \( D \) be a subset of \( {\mathbf{R}}^{N} \) and \( L \) a lattice in \( {\mathbf{R}}^{N} \), with fundamental domain \( F \) . Assume that the boundary of \( D \) is \( \left( {N - 1}\right) \) -Lipschitz parametrizable. Let \( \lambda \left( t\right) = \lambda \left( {t, D, L}\right) \) be the number ... | Proof. If a point \( l \in L \) lies in \( {tD} \), then \( {F}_{l} \) intersects \( {tD} \) . Furthermore, either \( {F}_{l} \) is contained in the interior of \( {tD} \), or \( {F}_{l} \) intersects the boundary\n\nof \( {tD} \) . Let:\n\n\[ n\left( t\right) = \text{number of}l \in L\text{such that}l \in {tD}\text{.}... | Yes |
Lemma 2. Let the notation be as in Lemma 1, and let \( L \) be the lattice obtained by translating \( {\mathfrak{{bc}}}_{0} \) by one solution of the two congruences above. Then \( {w}_{\mathrm{c}}j\left( {\widehat{x}, t}\right) \) is equal to the number of elements of \( L \) lying in \[ {\left( \mathbf{N}b \cdot t\ri... | We are therefore in the situation discussed in Chapter V, §2. Observe that the volume of a fundamental domain for \( b{c}_{0} \) in \( {\mathbf{R}}^{N} \) is the same as the volume of a fundamental domain for the translated lattice \( L \) . | No |
Theorem 3. Let \( \mathfrak{c} \) be a cycle of \( k \), and let \( \mathfrak{K} \) be a class of \( I\left( \mathfrak{c}\right) \) modulo \( {P}_{\mathfrak{c}} \) . Then \[ j\left( {\Omega, t}\right) = {\rho }_{\mathrm{c}}t + O\left( {t}^{1 - 1/N}\right) \] where \[ {\rho }_{\mathrm{c}} = \frac{{2}^{{r}_{1}}{\left( 2\... | Proof. In Chapter V, §2, we had computed the volume of a fundamental domain for the lattice of an ideal \( {\mathfrak{{bc}}}_{0} \), and found it equal to \[ {2}^{-{r}_{2}}\mathrm{\;N}b{\mathrm{{Nc}}}_{0}\sqrt{{d}_{k}} \] In view of Lemma 2, there remains only to prove Lemma 1, i.e. construct a suitable fundamental dom... | Yes |
Theorem 1. The additive group \( k \) is embedded as a discrete subgroup of the adeles \( A \) . The multiplicative group \( {k}^{ * } \) is embedded as a discrete subgroup of \( J \) . | Proof. Let \( \alpha \in k \) . To say that \( \alpha \) is close to 0 in the adele topology means that \( {\left| \alpha \right| }_{v} \leqq 1 \) for all but a finite number of \( v \) and \( {\left| \alpha \right| }_{v} \) is very small for a finite set of \( v \) . By the product formula, this implies that \( \alpha... | Yes |
Theorem 2. We have\n\n\[ k + {A}_{{S}_{\infty }} = A\text{.}\]\n\nThe factor group \( A/k \) is compact. | Proof. The first statement means that given any adele \( x \), there exists an element \( \alpha \) of \( k \) such that \( x - \alpha \) has integral components at all valuations \( v \) . This is an easy extension of the Chinese remainder theorem, and can be done for instance as follows. Given \( x \in A \), let \( m... | Yes |
Theorem 3. Let \( {\omega }_{1},\ldots ,{\omega }_{N} \) be a basis for the integers \( {\mathfrak{o}}_{k} \) of \( k \) over \( \mathbf{Z} \) . Let \( {F}_{\infty } \) be the subset of\n\n\[ \mathop{\prod }\limits_{{v \in {S}_{\infty }}}{k}_{v} \]\n\nspanned by the vectors \( \sum {t}_{i}{\omega }_{i} \) with \( 0 \le... | Proof. Given \( x \in A \) we can bring it into \( {A}_{{S}_{\infty }} \) by translation with an element of \( k \), uniquely determined up to an element of \( {\mathfrak{o}}_{k} \) . Restricting the components \( {t}_{i} \) to lie in the half-open interval as above determines this algebraic integer uniquely if we requ... | No |
Theorem 4. The factor group \( {J}^{0}/{k}^{ * } = {C}^{0} \) is compact. So is \( {J}_{S}^{0}/{k}_{S} \) for any finite set \( S \supset {S}_{\infty } \) . | Proof. Let\n\n\[\n\psi : J \rightarrow {\mathbf{R}}^{ + }\n\]\n\nbe the map which to each idele \( a \) associates \( \psi \left( a\right) = \parallel a\parallel \) . Then \( \psi \left( {k}^{ * }\right) = 1 \)\n\nand so \( \psi \) is defined on \( J/{k}^{ * } \) . Its kernel is \( {C}^{0} \) . For any real number \( \... | Yes |
Theorem 5. The image \( \log \left( {k}_{S}\right) \) is a discrete subgroup of rank \( s - 1 \) in \( {H}^{s - 1} \) . | Proof. Note first that \( {H}^{s - 1} \) is generated (over \( \mathbf{R} \) ) by \( \log \left( {J}_{S}^{0}\right) \), because we can pick \( s - 1 \) coordinates in \( S \) arbitrarily, and then adjust the last coordinate (at an archimedean absolute value \( v \) ) so that the sum of the logs is equal to 0 . Let \( W... | Yes |
Theorem 6. The subset \( E \) of \( {J}^{0} \) consisting of\n\n\[ \n{E}^{0}{b}^{\left( 1\right) } \cup \cdots \cup {E}^{0}{b}^{\left( h\right) }\n\]\n\nis a fundamental domain for \( {J}^{0}{\;\operatorname{mod}\;{k}^{ * }} \) . | Proof. Starting with any idele \( b \) in \( {J}^{0} \) we can change it into an idele which represents a principal ideal by dividing it by a uniquely determined \( {b}^{\left( \nu \right) } \) . Multiplication by a field element brings us to an idele representing the unit ideal, and therefore takes the idele into \( {... | Yes |
Theorem 7. Let \( \mathfrak{c} \) be admissible for \( K/k \) , and let \( \mathfrak{f} \) be the smallest admissible cycle for \( K/k \) . Then the inclusion\n\n\[ I\left( \mathrm{c}\right) \rightarrow I\left( \mathrm{f}\right) \]\n\ninduces an isomorphism\n\n\[ \begin{array}{l} I\left( \mathrm{c}\right) /{P}_{\mathrm... | Proof. Let \( \mathfrak{a} \in I\left( \mathfrak{c}\right) \) be such that \( \mathfrak{a} = \left( \alpha \right) {N}_{k}^{K}\mathfrak{b} \) with some \( \mathfrak{b} \) in \( I\left( {\mathfrak{f}, K}\right) \) prime to \( \mathfrak{f} \), and some \( \alpha \in {k}^{ * } \) such that \( \alpha \equiv 1\left( {{\;\op... | Yes |
Theorem 8. Let \( K/k \) be a Galois extension. Let \( \mathfrak{c} \) be admissible for \( K/k \) . Then we have an isomorphism\n\n\[ J/{k}^{ * }{N}_{k}^{K}{J}_{K} \approx I\left( \mathfrak{c}\right) /{P}_{\mathfrak{c}}\mathfrak{N}\left( \mathfrak{c}\right) . \]\n\nThe isomorphism is induced by the isomorphism\n\n\[ {... | Thus in Theorem 8, to each idele \( a \) we first select an idele \( b \) in the same coset \( {\;\operatorname{mod}\;{k}^{ * }} \) such that \( b \in {J}_{\mathrm{c}} \) . We then map \( b \) on its ideal \( \left( b\right) \) . We have \( \left( b\right) \in {P}_{\mathfrak{c}}\mathfrak{N}\left( \mathfrak{c}\right) \)... | Yes |
Theorem 1. If the Dirichlet series \( \sum {a}_{n}/{n}^{s} \) converges for some \( s = {s}_{0} \) , then it converges for any \( s \) with \( \operatorname{Re}\left( s\right) > {\sigma }_{0} = \operatorname{Re}\left( {s}_{0}\right) \), uniformly on any compact subset of this region. | Proof. Write \( {n}^{s} = {n}^{{s}_{0}}{n}^{\left( s - {s}_{0}\right) } \), and sum the following series by parts:\n\n\[ \sum \frac{{a}_{n}}{{n}^{{s}_{0}}}\frac{1}{{n}^{\left( s - {s}_{0}\right) }} \]\n\nIf \( {P}_{n}\left( {s}_{0}\right) = \mathop{\sum }\limits_{{m = 1}}^{n}{a}_{m}/{m}^{{s}_{0}} \), then the tail ends... | Yes |
Theorem 2. Assume that there exists a number \( C \) and \( {\sigma }_{1} > 0 \) such that\n\n\[ \left| {A}_{n}\right| = \left| {{a}_{1} + \cdots + {a}_{n}}\right| \leqq C{n}^{{\sigma }_{1}} \]\n\nfor all \( n \) . Then the abscissa of convergence of \( \sum {a}_{n}/{n}^{s} \) is \( \leqq {\sigma }_{1} \) . | Proof. Summing by parts, we find for \( n \geqq m \) ,\n\n\[ {P}_{n}\left( s\right) - {P}_{m}\left( s\right) = {A}_{n}\frac{1}{{n}^{s}} + \mathop{\sum }\limits_{{k = m + 1}}^{{n - 1}}{A}_{k}\left\lbrack {\frac{1}{{k}^{s}} - \frac{1}{{\left( k + 1\right) }^{s}}}\right\rbrack \]\n\n\[ = {A}_{n}\frac{1}{{n}^{s}} + \mathop... | Yes |
Theorem 4. Let \( \left\{ {a}_{n}\right\} \) be a sequence of complex numbers, with partial sums \( {A}_{n} \) . Let \( 0 \leqq {\sigma }_{1} < 1 \), and assume that there is a complex number \( \rho \) , and \( C > 0 \) such that for all \( n \) we have\n\n\[ \left| {{A}_{n} - {n\rho }}\right| \leqq C{n}^{{\sigma }_{1... | Proof. The proof is obtained by considering \( f\left( s\right) - {\rho \zeta }\left( s\right) \), and applying Theorems 2 and 3 directly. | No |
Theorem 5. Let \( k \) be a number field, \( \left\lbrack {k : \mathbf{Q}}\right\rbrack = N \), and let \( \mathcal{R} \) be an ideal class. Then \( \zeta \left( {s,\mathcal{R}}\right) \) is analytic for \( \operatorname{Re}\left( s\right) > 1 - 1/N \), except for a simple pole at \( s = 1 \), with residue \( \rho \), ... | The same holds for \( {\zeta }_{k}\left( s\right) \), except that the residue is equal to \( {h\rho } \), where \( h \) is the class number. | No |
Theorem 7. The Dirichlet series for \( {L}_{\mathfrak{c}}\left( {s, x}\right) \) is convergent in the half plane \( \operatorname{Re}\left( s\right) > 1 - 1/N \) if \( \chi \neq 1 \), and represents \( {L}_{\mathfrak{c}}\left( {s,\chi }\right) \), which is analytic in that half plane. | Proof. By Theorem 3 loc. cit. we know that the number of ideals \( \mathfrak{a} \) in a given class \( \mathfrak{K} \) such that \( \mathrm{{Na}} \leqq n \) is equal to the same number \( {\rho }_{\mathfrak{c}}n \), with an error term \( O\left( {n}^{1 - 1/N}\right) \) . Using the remark at the beginning of the section... | Yes |
Theorem 9. Let \( K/k \) be a Galois extension and \( E \) a finite extension of \( k \) . Then \( {S}_{K/k} \prec {S}_{E/k} \) if and only if \( E \subset K \) . | Proof. A prime \( \mathfrak{p} \) of \( k \) splits completely in \( E \) if and only if it splits completely in the smallest Galois extension \( L \) of \( k \) containing \( E \), because this condition is equivalent to every conjugate of \( E \) over \( k \) being contained in the completion \( {k}_{\mathfrak{p}} \)... | Yes |
Theorem 10. (Tchebotarev). Let \( K/k \) be Galois with group \( G \) . Let \( \sigma \in G \) . Let \( \left\lbrack {K : k}\right\rbrack = N \), and let \( c \) be the number of elements in the conjugacy class of \( \sigma \) in \( G \) . Then those primes \( \mathfrak{p} \) of \( k \) which are unramified in \( K \) ... | Proof. The simple argument which follows is due to Deuring (Math. Ann. 110,1934). Let \( \sigma \) have order \( f \) . Let \( Z \) be the fixed field of \( \sigma \) . Then \( K/Z \) is cyclic of degree \( f \), and therefore a class field. If \( \mathfrak{c} \) is an admissible cycle for \( K/Z \), then we have the A... | Yes |
Lemma 1. If \( B \) is a subgroup of \( A \) which is mapped into itself by \( f \) and \( g \), so that \( f, g \) may be viewed also as endomorphisms of the factor group \( A/B \), then\n\n\[ Q\left( A\right) = Q\left( B\right) Q\left( {A/B}\right) \]\n\nin the sense that if two of the quotients are defined, so is th... | Proof. One may view the quotient \( Q \) as an Euler-Poincaré characteristic of a complex of length 2 (cf. my book Algebra, Chapter IV), and apply a general result, of an elementary nature, to deduce the multi-plicativity property. We shall reproduce a sketch of the proof below in our special case. First, we give a pro... | Yes |
Lemma 2. The projection \( \pi : A \rightarrow {A}_{1} \) induces an isomorphism\n\n\[ \n{H}^{0}\left( {G, A}\right) \approx {H}^{0}\left( {{G}_{1},{A}_{1}}\right) .\n\] | Proof. We first observe that \( {A}^{G} \) consists of all elements of the form\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{s}{\sigma }_{i}{a}_{1},\;\text{ with }{a}_{1} \in {A}_{1}{}^{{G}_{1}} \n\]\n\nNamely, it is clear that such an element is fixed under \( G \) . On the other hand, if\n\n\[ \na = \mathop{\sum }\limits... | Yes |
Lemma 3. There is an isomorphism (to be described in the proof)\n\n\\[ \n{H}^{-1}\left( {G, A}\right) \approx {H}^{-1}\left( {{G}_{1},{A}_{1}}\right) \n\\] | Proof. Let\n\n\\[ \na = \mathop{\sum }\limits_{{i = 1}}^{s}{\sigma }_{i}{a}_{i}^{\prime }\n\\]\n\n\\[ \n{a}_{i}^{\prime } \in {A}_{1}\n\\]\n\nThen\n\n\\[ \n{\\operatorname{Tr}}_{G}\left( a\\right) = \mathop{\sum }\limits_{{j = 1}}^{s}{\sigma }_{j}{\\operatorname{Tr}}_{{G}_{1}}\left( {{a}_{1}^{\prime } + \\cdots + {a}_{... | Yes |
Lemma 4. Hypotheses being as above, we have\n\n\[ \nQ\left( {G,{K}^{ * }}\right) = \left( {{k}^{ * } : {N}_{k}^{K}{K}^{ * }}\right) = \left\lbrack {K : k}\right\rbrack , \n\]\n\n\[ \n\left( {{U}_{k} : {N}_{k}^{K}{U}_{K}}\right) = e,\;Q\left( {G,{U}_{K}}\right) = 1. \n\] | Proof. We use the \( Q \) -machine. By Hilbert’s Theorem 90 we know that \( {H}^{-1}\left( {G,{K}^{ * }}\right) = 1 \) . Hence\n\n\[ \nQ\left( {K}^{ * }\right) = \left( {{k}^{ * } : {N}_{k}^{K}{K}^{ * }}\right) \n\]\n\nis our norm index. We have \( {K}^{ * }/{U}_{K} \approx \mathbf{Z} \) (with trivial action, because \... | Yes |
Corollary 1. Let \( G \) be cyclic of order \( N \) . Then\n\n\[ Q\left( {G, M}\right) = Q\left( {G,{M}^{\prime }}\right) = \mathop{\prod }\limits_{{v \in S}}{N}_{v} \]\n\nwhere \( {N}_{v} \) is the order of the composition group \( {G}_{w} \) for any \( w \mid v \) . | Proof. We have\n\n\[ Q\left( {G,{M}^{\prime }}\right) = \mathop{\prod }\limits_{{v \in S}}Q\left( {G,{M}_{v}^{\prime }}\right) = \mathop{\prod }\limits_{{v \in S}}Q\left( {{G}_{\bar{v}},\mathbf{Z}{Y}_{\bar{v}}}\right) \]\n\nand \( Q\left( {{G}_{w},\mathbf{Z}}\right) = {N}_{w} \), so that our corollary follows because \... | No |
Corollary 2. Let \( K/k \) be cyclic of order \( N \) and let \( {K}_{S} \) be the \( S \) -units in \( K \) . Then\n\n\[ Q\left( {G,{K}_{S}}\right) = \frac{1}{N}\mathop{\prod }\limits_{{v \in S}}{N}_{v} \] | Proof. The map\n\n\[ L : {K}_{S} \rightarrow {E}^{s} \]\n\ngiven by\n\n\[ L\left( \xi \right) = \mathop{\sum }\limits_{{w \in {S}_{K}}}\log \parallel \xi {\parallel }_{w}{X}_{w} \]\n\n\( \left( {w \in {S}_{K}}\right) \)\nis a \( G \) -homomorphism of \( {K}_{S} \) into \( {E}^{s} \), whose image is a lattice in a hyper... | Yes |
Theorem 1. Let \( K/k \) be abelian. Then the reciprocity law map \( \mathfrak{a} \mapsto \left( {\mathfrak{a}, K/k}\right) \) is surjective, as a map of \( I\left( \mathfrak{c}\right) \) into the Galois group, for any cycle \( \mathfrak{c} \) (divisible by the ramified primes). | Proof. Let \( \mathfrak{c} \) be a fixed cycle of \( k \), divisible by the ramified primes, and let \( H \) be the subgroup of \( G \) which is the image of the reciprocity law mapping. Let \( F \) be the fixed field of \( H \) . We must show that \( F = k \) . Any \( \mathfrak{p} \in I\left( \mathrm{c}\right) \) must... | Yes |
Lemma 1. Let \( a, r \) be integers \( > 1 \) . Let \( q \) be a prime number. Then there exists a prime number \( p \) such that a has order \( {q}^{r}\left( {\;\operatorname{mod}\;p}\right) \) . | Proof. We consider the positive number\n\n\[ T = \frac{{a}^{{q}^{r}} - 1}{{a}^{{q}^{r - 1}} - 1}. \]\n\nThen\n\n\[ T = {\left( {a}^{{q}^{r - 1}} - 1\right) }^{q - 1} + q{\left( {a}^{{q}^{r - 1}} - 1\right) }^{q - 2} + \cdots + q. \]\n\nLet \( p \) be a prime dividing \( T \) . If \( p \) also divides \( {a}^{{q}^{r - 1... | Yes |
Lemma 2. Let\n\n\[ n = {q}_{1}^{{r}_{1}}\cdots {q}_{s}^{{r}_{s}} \]\n\nbe a positive integer factorized into powers of primes \( {q}_{i} \). Let a be an integer \( > 1 \). There exists an integer\n\n\[ m = {p}_{1}\cdots {p}_{s}{p}_{1}^{\prime }\cdots {p}_{s}^{\prime} \]\n\nwith distinct primes \( {p}_{i},{p}_{i}^{\prim... | Proof. By letting \( r \rightarrow \infty \), we see that in Lemma 1, we can find arbitrarily large primes \( p \) such that \( a \) has order \( \left( {\;\operatorname{mod}\;p}\right) \) divisible by a fixed power of \( q \). We therefore first find large distinct primes \( {p}_{1},\ldots ,{p}_{s} \) such that \( a \... | Yes |
Lemma 3. Let \( K \) be an abelian extension of the number field \( k \), and let \( S \) be a finite set of prime numbers. Let \( n = \left\lbrack {K : k}\right\rbrack \) . Let \( \mathfrak{p} \) be a prime of \( k \) which is unramified in \( K \) . Then there exists an integer \( m \) relatively prime to the numbers... | Proof. We apply Lemma 2 with \( a = \mathbf{N}\mathfrak{p} \) . We can take \( m \) divisible only by arbitrarily large primes, so that \( K \cap \mathbf{Q}\left( {\zeta }_{m}\right) = \mathbf{Q} \) and (ii) is satisfied. Let \( \sigma = \left( {\mathfrak{p}, k\left( {\zeta }_{m}\right) /k}\right) \) . Then\n\n\[ \sigm... | Yes |
Theorem 3. Let \( K/k \) be an abelian extension. Let \( \mathfrak{c} \) be any admissible cycle for \( K/k \) . Then the Artin map\n\n\[ I\left( \mathfrak{c}\right) \rightarrow G\left( {K/k}\right) \]\n\nhas kernel equal to \( {P}_{\mathfrak{c}}\mathfrak{N}\left( \mathfrak{c}\right) \), and thus induces an isomorphism... | Proof. Let \( \mathfrak{f} \) be the smallest admissible cycle for \( K/k \) . If \( \mathfrak{c} \) is divisible only by the same \( v \) that divide \( \mathfrak{f} \), then we know from Theorem 7 of Chapter VII, \( §4 \) that \( {P}_{\mathfrak{f}}\mathfrak{N}\left( \mathfrak{f}\right) = {P}_{\mathfrak{c}}\mathfrak{N... | Yes |
Theorem 5. The map \( K \mapsto {N}_{k}^{K}{C}_{K} \) (resp. \( K \mapsto {k}^{ * }{N}_{k}^{K}{J}_{K} \) ) establishes a bijection between finite abelian extensions of \( k \) and open subgroups of \( {C}_{k} \) (resp. of \( {J}_{k} \), containing \( {k}^{ * } \) ). If \( K \) belongs to \( H \) and \( {K}^{\prime } \)... | Proof. Suppose that \( H \) belongs to \( K \) and \( {H}^{\prime } \) belongs to \( {K}^{\prime } \) . The kernel of the Artin map\n\n\[ \n{C}_{k} \rightarrow G\left( {K{K}^{\prime }/k}\right) \n\]\n\nis \( H \cap {H}^{\prime } \), because of the consistency property A2. Hence \( H \cap {H}^{\prime } \) belongs to \( ... | No |
Corollary 1. Let \( K/k \) be class field to \( H \) and let \( {H}_{1} \supset H \) . Then \( {H}_{1} \) has a class field, which is the fixed field of \( \left( {{H}_{1}, K/k}\right) \) . | Proof. Let \( {K}_{1} \) be the fixed field of \( \left( {{H}_{1}, K/k}\right) \) . By the consistency property of the Artin map, we see at once that \( {H}_{1} \) is the kernel of the map \( {C}_{k} \rightarrow G\left( {{K}_{1}/k}\right) \), so \( {K}_{1} \) is the class field belonging to \( {H}_{1} \) . | Yes |
Corollary 2. Let \( K/k \) be an abelian extension and let \( \mathfrak{c} \) be a cycle of \( k \) , admissible for \( K/k \) . Let \( {W}_{\mathrm{c}} \) be the neighborhood of 1 in \( {J}_{k} \) defined in Chapter VII,§4. If \( {k}^{ * }{W}_{\mathfrak{c}} \) belongs to the class field \( {K}_{\mathfrak{c}}/k \), the... | Proof. Let \( H \) belong to \( K \) . As we saw in Chapter VII, \( §4 \), we have the inclusion \( H \supset {k}^{ * }{W}_{\mathrm{c}} \) . Hence our assertion is now obvious. | No |
Corollary 3. Let \( K/\mathbf{Q} \) be an abelian extension of the rationals. Then \( K \) is cyclotomic, i.e. there exists a root of unity \( \zeta \) such that \( K \subset \mathbf{Q}\left( \zeta \right) \) . | Proof. Let \( m{v}_{\infty } \) be an admissible cycle for \( K/\mathbf{Q} \), and let \( H \) be the class group of \( K \) . Then\n\n\[ \n{H}_{m} = {\mathbf{Q}}^{ * }{W}_{m{v}_{\infty }} \subset H \n\]\n\nwhence \( K \subset \mathbf{Q}\left( {\zeta }_{m}\right) \) by Theorem 5 . | No |
Theorem 6. Let \( K/k \) be the class field of \( H \), and let \( E/k \) be finite. Then \( {KE}/E \) is the class field of \( {N}_{E/k}^{-1}\left( H\right) \) . | Proof. The kernel of the Artin map\n\n\[ \n{C}_{E} \rightarrow G\left( {{KE}/E}\right) \n\]\n\nis precisely equal to \( {N}_{E/k}^{-1}\left( H\right) \) because for \( b \in {C}_{E} \) we have\n\n\[ \n{\operatorname{res}}_{K}\left( {b,{KE}/E}\right) = \left( {{N}_{k}^{E}b, K/k}\right) , \n\]\n\nand an automorphism of \... | Yes |
Theorem 7. Let \( E/k \) be a finite extension, and let \( H = {N}_{k}^{E}{C}_{E} \) . Then \( H \) belongs to the maximal abelian subextension of \( E \) . | Proof. Let \( K/k \) be the class field belonging to \( H \) . For any \( b \in {C}_{E} \) we have \( {N}_{k}^{E}b \in H \), and hence\n\n\[ 1 = \left( {{N}_{k}^{E}b, K/k}\right) = {\operatorname{res}}_{K}\left( {b,{KE}/E}\right) . \]\n\nHence \( {KE} \) is the class field to all of \( {C}_{E} \), whence \( {KE} = E \)... | Yes |
Theorem 2. Let \( K/k \) be class field to \( H \) . Let \( v \) be an absolute value on \( k \) . Then \( v \) splits completely in \( K \) if and only if \( {k}_{v}^{ * } \subset H \) . | Proof. If \( v \) splits completely, then \( {k}_{v}^{ * } \subset H \) because every element of \( {k}_{v}^{ * } \) is a local norm. We shall prove the converse. In this section, we prove it only when \( {J}_{k}/H \) has exponent \( n \), and the \( n \) -th roots of unity are contained in \( k \) . We shall complete ... | No |
Theorem 3. Let \( K/k \) be abelian. The Artin map restricted to \( {k}_{v}^{ * } \), namely\n\n\[ a \mapsto \left( {a, K/k}\right) \]\n\n\[ a \in {k}_{v}^{ * } \]\n\nmaps \( {k}_{v}^{ * } \) into the decomposition group \( {G}_{v}\left( { = {G}_{w}}\right. \) for any \( \left. {w \mid v}\right) \), has kernel \( {N}_{... | Proof. We know that the group of local norms is contained in the kernel, since it lies in \( {N}_{k}^{K}{J}_{K} \) . Hence our theorem follows from the local norm index inequality, Chapter IX, §3, once we have proved that the image of \( {k}_{v}^{ * } \) is contained in the decomposition group \( {G}_{v} \), and is equ... | Yes |
Theorem 5. Let \( {k}_{v} \) be a local field. The Artin map defined for abelian extensions \( K \) of \( {k}_{v} \) satisfies the same formalism as the global map, when viewed as defined on the multiplicative group of the local field. The association \[ K \mapsto {N}_{{k}_{v}}^{K}{K}^{ * } \] establishes a bijection b... | Proof. Same as in the global case. We should observe in addition that given any abelian extension \( {k}_{v}^{\prime } \) of \( {k}_{v} \), there exists an abelian extension \( K \) of \( k \) such that \( K{k}_{v} = {k}^{\prime } \) . To see this, note that \( {k}_{v}^{ * } \) is embedded in \( {C}_{k} \) , and hence ... | No |
Theorem 2. If \( K/k \) is abelian, class field to \( H \), and if we identify \( G\left( {K/k}\right) \) with the idele class group \( {C}_{k}/H \) under the Artin map, and \( \chi \) is a simple character, then\n\n\[ L\left( {s, x, K/k}\right) = L\left( {s, x}\right) ,\]\n\ninterpreting the character \( \chi \) on th... | Proof. If \( \mathfrak{p} \) is unramified in \( K \), then the Artin map is given by means of the Frobenius automorphism, which depends only on \( \mathfrak{p} \), and so the p-contribution to the \( L \) -series is the same no matter which \( L \) -series we deal with. Let \( S \) be the kernel of \( \chi \) in \( G ... | Yes |
Lemma 1. We contend that \( \left\{ {{\gamma }_{i,\nu }{\eta }_{i}}\right\} \) is a system of distinct right coset representatives of \( S \) in \( G \) . | Proof. We first prove that they represent distinct cosets. Suppose that\n\n\[ S{\gamma }_{i,\nu }{\eta }_{i} = S{\gamma }_{j,\mu }{\eta }_{j} \]\n\nThen\n\n\[ {\gamma }_{i,\nu }{\eta }_{i}{\eta }_{j}^{-1}{\gamma }_{j,\mu }^{-1} \in S \]\n\nLooking at the effect of this element on \( {\mathfrak{P}}_{j} \), we see that i... | Yes |
Lemma 2. The intersection \( {\sigma }_{i}^{m}{T}_{i} \cap S \) is not empty if and only if \( {f}_{i} \mid m \) . In that case, | Proof. Suppose that \( {\sigma }_{i}^{m}{T}_{i} \) contains an element \( {\sigma }_{i}^{m}\tau \in S \), with \( \tau \in {T}_{i} \) . Then on the residue class field \( {\mathfrak{o}}_{K}/{\mathfrak{P}}_{i} \) the effect of \( {\sigma }_{i}^{m}\tau \) is the same as that of \( {\sigma }_{i}^{m} \), and it also leaves... | Yes |
Consider the function\n\n\[ h\left( x\right) = {e}^{-\pi {x}^{2}} \]\n\nwhere \( {x}^{2} = x \cdot x \) as usual. We contend that \( h \) is self dual, i.e.\n\n\[ \widehat{h} = h\text{.} \]\n | Proof. We differentiate the Fourier transform\n\n\[ \widehat{h}\left( y\right) = \int h\left( x\right) {e}^{-{2\pi ixy}}{dx} \]\n\nunder the integral with respect to \( y \) (we may assume for this proof that we are in the one variable case, since the Fourier transform splits into a product of 1-variable transforms), a... | Yes |
Next, let \( f \) be an arbitrary function in the Schwartz space, and let \( B \) be a non-singular real matrix. Then the function \( {f}_{B} \) defined by \[ {f}_{B}\left( x\right) = f\left( {Bx}\right) \] is also in the Schwartz space, and its Fourier transform is given by \[ {\widehat{f}}_{B}\left( y\right) = \frac{... | This is clear since when we make the change of variables \( z = {Bx} \), we have \( {dz} = \parallel B\parallel {dx} \), and \[ \left\langle {{B}^{-1}z, y}\right\rangle = \left\langle {z,{}^{t}{B}^{-1}y}\right\rangle \] | Yes |
Theorem 1. The polar part of \( F\left( {s,\mathcal{R}}\right) \) at \( s = 0 \) and \( s = 1 \) is given by\n\n\[ \n\frac{{\mu }^{ * }\left( E\right) 2}{w}\left( {\frac{1}{s - 1} - \frac{1}{s}}\right) = \frac{{2}^{{r}_{1}}R}{w}\left( {\frac{1}{s - 1} - \frac{1}{s}}\right) .\n\]\n\nThe zeta function \( \zeta \left( {s,... | Proof. The first statement is obtained by plugging in our value for \( {\mu }^{ * }\left( E\right) \) in the expression (3). The second statement comes from evaluating\n\n\[ \n{A}^{-1}\Gamma {\left( 1/2\right) }^{{r}_{1}}\Gamma {\left( 1\right) }^{{r}_{2}}\n\]\n\nSo Theorem 1 drops out. | Yes |
Theorem 4. Let \( k \) range over a sequence of number fields such that \( {N}_{k}/\log {d}_{k} \rightarrow 0 \) . Assume that for some \( \delta > 0 \), the zeta functions \( {\zeta }_{k}\left( s\right) \) have no zeros on the interval \( \left\lbrack {1 - \delta ,1}\right\rbrack \) . Then for this sequence of fields,... | Proof. We shall consider successively values of \( s > 1 \), and values of \( s < 1 \) to get inequalities for \( {hR} \) . First take \( s \) real \( > 1 \), say \[ s = 1 + \frac{1}{\alpha } \] \( \alpha \geqq 1 \) . For this case, we disregard the sum over \( \mathfrak{b} \) in Theorem 3, except for the fact that it ... | Yes |
Theorem 1. Let \( k \) be a local field. Then the bilinear map\n\n\[ \left( {x, y}\right) \mapsto {e}^{{2\pi i\lambda }\left( {xy}\right) } \]\n\ninduces an identification of the additive group of \( k \) with its own character group. | Proof. It is easily verified that the pairing is continuous, and that the kernels on both sides are trivial, i.e. \( = 0 \) . This induces a natural map of \( k \) into \( k \) which is injective, continuous, and dense. It is in fact bicontinuous, because if the character \( {\lambda }_{x} \) given by\n\n\[ {\lambda }_... | Yes |
Theorem 2. If we define the Fourier transform \( \widehat{f} \) of a function \( f \) in \( {L}_{1}\left( k\right) \) by\n\n\[ \widehat{f}\left( y\right) = \int f\left( x\right) {e}^{-{2\pi i\lambda }\left( {xy}\right) }{dx} \]\n\nthen with our choice of measure, the inversion formula\n\n\[ \widehat{f}\left( x\right) =... | Proof. We need only establish the inversion formula for one non-trivial function, since from abstract Fourier analysis we know it is true save possibly for a constant factor. For \( k \) real, we can take \( f\left( x\right) = {e}^{-\pi {x}^{2}} \) , for \( k \) complex, \( f\left( x\right) = {e}^{-{2\pi }{\left| x\rig... | No |
Proposition 1. The unramified quasi-characters are the maps of the form\n\n\[ c\left( a\right) = \parallel a{\parallel }^{s\log \parallel a\parallel }, \]\n\nwhere \( s \) is any complex number; \( s \) is determined by \( c \) if \( v \) is archimedean, and is determined only \( {\;\operatorname{mod}\;2}{\pi i}/\log \... | Proof. An unramified quasi-character depends only on \( \parallel a\parallel \) . Use the fact that\n\n\[ {k}^{ * } \approx U \times {\mathbf{R}}^{ + }\;\text{ or }\;{k}^{ * } \approx U \times \mathbf{Z} \]\n\naccording as \( v \) is archimedean or not. In the \( \mathfrak{p} \) -adic case, the decomposition is of cour... | Yes |
Proposition 3. A function \( g\left( a\right) \) is in \( {L}_{1}\left( {k}^{ * }\right) \) if and only if \( g\left( x\right) \parallel x{\parallel }^{-1} \) is in \( {L}_{1}\left( {k - 0}\right) \), and for such functions, we have\n\n\[ \n{\int }_{{k}^{ * }}g\left( a\right) {d}_{1}^{ * }a = {\int }_{k - 0}g\left( x\r... | In fact it will be convenient to take a Haar measure on \( {k}^{ * } \) which differs from the above by a constant in the \( \mathfrak{p} \) -adic case, and gives the units measure 1 in general. Thus we take:\n\n\[ \n{d}^{ * }a = \frac{da}{\parallel a\parallel }\;\text{ if }v\text{ is archimedean. } \]\n\n\[ \n{d}^{ * ... | Yes |
Proposition 4. If \( v \) is \( \mathfrak{p} \)-adic, then \( {\int }_{U}{d}^{ * }a = {\left( \mathbf{N}\mathfrak{D}\right) }^{-1/2} \) . | Proof. This is seen immediately, taking into account that \( \parallel a\parallel = 1 \) if \( a \in U \), that the units are the complement of \( \mathfrak{p} \) in \( \mathfrak{o} \), together with the definition of the additive Haar measure in \( §1 \) . | No |
Theorem 3. A zeta function has an analytic continuation to the domain of all quasi-characters given by a functional equation of the type\n\n\\[ \n\\zeta \\left( {f, c}\\right) = \\rho \\left( c\\right) \\zeta \\left( {\\widehat{f},\\widehat{c}}\\right) \n\\]\n\nThe factor \\( \\rho \\left( c\\right) \\), which is indep... | Proofs. For the first, we have\n\n\\[ \n\\zeta \\left( {f, c}\\right) = \\rho \\left( c\\right) \\zeta \\left( {\\widehat{f},\\widehat{c}}\\right) = \\rho \\left( c\\right) \\rho \\left( \\widehat{c}\\right) \\zeta \\left( {\\widehat{\\widehat{f}},\\widehat{\\widehat{c}}}\\right) ,\n\\]\n\nand\n\n\\[ \n\\zeta \\left( {... | Yes |
Theorem 4. With the above notation, we have in all cases, putting \( c\left( a\right) = \chi \left( a\right) \parallel a{\parallel }^{s} : \)\n\n\[ \n{f}_{m}\left( x\right) = {i}^{\left| m\right| }{f}_{-m}\left( x\right) \;\text{ (if }v\text{ real,}{f}_{-m} = {f}_{m}\text{ ) }\n\]\n\n\[ \n\zeta \left( {{f}_{c}, c}\righ... | Proof. If \( v \) is real, our first assertion concerning \( \widehat{f} \) is easy and is left to the reader.\n\nFor the zeta functions, we use repeatedly the definition of the Gamma function:\n\n\[ \n\Gamma \left( s\right) = {\int }_{0}^{\infty }{e}^{-u}{u}^{s - 1}{du} \n\]\n\nThe computations are also quite simple a... | No |
Proposition 5. We have\n\n\[ \n{\\widehat{f}}_{m}\\left( x\\right) = \\left\\{ \\begin{array}{lll} {\\left( \\mathbf{N}\\mathfrak{D}\\right) }^{1/2}\\left( {\\mathbf{N}{\\mathfrak{f}}_{\\chi }}\\right) & x \\equiv 1 & \\left( {\\;\\operatorname{mod}\\;{\\mathfrak{f}}_{\\chi }}\\right) \\\\ 0 & x ≢ 1 & \\left( {\\;\\ope... | Proof. This comes immediately from the fact that the integral of a character over a compact group is \( \\mu \\left( G\\right) \) or 0 according as the character is trivial or not. (The compact group here is \( {\\left( {\\mathfrak{D}}_{\\chi }\\right) }^{-1} \).) | Yes |
Theorem 5. Let \( x \) be an unramified character of \( {k}^{ * } \), and let \( f \) be the characteristic function of an ideal \( {\mathfrak{p}}^{n} \) . Then\n\n\[ \zeta \left( {f,\chi, s}\right) = \frac{{\left( \mathbf{N}\mathfrak{D}\right) }^{-1/2}\chi {\left( \mathfrak{p}\right) }^{n}{\left( \mathbf{N}\mathfrak{p... | Proof. This is an easy computation, using the definition of multiplicative Haar measure in terms of the additive one, and taking the integral as a sum over integrals over the annuli \( {\mathfrak{p}}^{\nu } - {\mathfrak{p}}^{\nu + 1} \) for \( \nu \) ranging from \( n \) to \( \infty \) . On each such annulus, \( \para... | No |
Corollary 1. We have\n\n\[ \zeta \left( {{f}_{0},{x}_{0}, s}\right) = \frac{{\left( \mathbf{N}\mathfrak{D}\right) }^{s - 1/2}}{1 - \frac{1}{\mathbf{N}{\mathfrak{p}}^{s}}} \]\n\nand\n\n\[ \zeta \left( {{\widehat{f}}_{0},{\chi }_{0},1 - s}\right) = \frac{1}{1 - {\mathrm{{Np}}}^{s - 1}}. \] | Proof. Put \( n = - \) ord \( \mathfrak{D} \) in the first case, and \( n = 0 \) in the second. | No |
Theorem 6. Let \( \pi \) be a prime element, \( \chi \) a character of \( {k}^{ * }, m > 0 \) the order of its conductor, and \( \{ \epsilon \} \) a set of unit representatives for \( U/\left( {1 + {\mathfrak{f}}_{x}}\right) \) . Let \( o\left( \chi \right) = \operatorname{ord}\left( {\mathfrak{D}}_{\chi }\right) \), a... | Proof. By definition, \[ \zeta \left( {{f}_{m}, c}\right) = {\int }_{{\mathfrak{D}}_{\chi }^{-1}}\eta \left( a\right) \chi \left( a\right) \parallel a{\parallel }^{s}{d}^{ * }a \] \[ = \mathop{\sum }\limits_{{-o\left( \chi \right) }}^{\infty }{\left( \mathbf{N}\mathfrak{p}\right) }^{-{\nu s}}{\int }_{{A}_{\nu }}\eta \l... | Yes |
Theorem 7. The restricted direct product of the groups \( {\widehat{G}}_{v} \) relative to the subgroups \( {H}_{v}^{ \bot } \) (which are compact by compact-discrete duality) is naturally isomorphic, topologically and algebraically, to the character group \( \widehat{G} \) of \( G \) . | The isomorphism is of course given by the correspondence\n\n\[ \n x = \prod {x}_{v} \n\] | No |
Assume that for each \( v \) we are given a continuous function \( {f}_{v} \in {L}_{1}\left( {G}_{v}\right) \) such that \( {f}_{v} = 1 \) on \( {H}_{v} \) for almost all \( v \) . We define \[ f\left( a\right) = \prod {f}_{v}\left( {a}_{v}\right) \] on \( G \) (actually a finite product). Then \( f \) is continuous. I... | Proof. Immediate. | No |
Theorem 9. Let \( {f}_{v}\left( {a}_{v}\right) \) be in \( {L}_{1}\left( {G}_{v}\right) \), continuous, and assume \( {\widehat{f}}_{v} \) is in \( {L}_{1}\left( {\widehat{G}}_{v}\right) \), i.e. assume that \( {f}_{v} \in \operatorname{Inv}\left( {G}_{v}\right) \) . Assume also that \( {f}_{v} \) is the characteristic... | Proof. By Theorem 8, applied to the function \( f\left( a\right) \overline{c\left( a\right) } = \prod {f}_{v}\left( {a}_{v}\right) \overline{{c}_{v}\left( {a}_{v}\right) } \) we see that the Fourier transform of the product is the product of the Fourier transforms. Since \( {f}_{v} \) is in \( \operatorname{Inv}\left( ... | Yes |
Theorem 11. The additive group \( k \) is its own orthogonal complement in the self-duality of \( A \) . | Proof. We first prove that \( k \) is orthogonal to itself. This amounts to proving that if \( x \in k \), then \( \sum {\lambda }_{v}\left( x\right) = 0 \) . We can verify this at once if \( k = \mathbf{Q} \) is the field of rational numbers (using a partial fraction decomposition of a rational number in terms of rati... | Yes |
Proposition 6. Let \( {F}_{\infty } \) be as in Theorem 3 of Chapter VII,§2. Then with our choice of measure, \( {F}_{\infty } \) has volume \( {d}_{k}^{1/2} \) . | Proof. This is an easy determinant computation. Remember that at complex \( v \), our measure is twice Lebesgue measure. | No |
Proposition 7. Let \( F \) be the subset of \( {A}_{k} \) equal to\n\n\[ \mathop{\prod }\limits_{{v \notin {S}_{\infty }}}{\mathfrak{o}}_{v} \times {F}_{\infty } \]\n\nThen \( F \) has measure \( 1 \) . | Proof. This follows from our choice of measures \( d{x}_{v} \), which insure that \( {\mathfrak{o}}_{v} \) has measure \( {\left( \mathbf{N}{\mathfrak{D}}_{\mathfrak{p}}\right) }^{-1/2} \) for \( v = {v}_{\mathfrak{p}} \) . | No |
Proposition 8. Let the notation be as above, and \( l \) as in Theorem 6, Chapter VII,§3. Then the measure of \( {l}^{-1}\left( P\right) \) is \[ \frac{{2}^{{r}_{1}}{\left( 2\pi \right) }^{{r}_{2}}}{{d}_{k}^{1/2}}R \] | Proof. Let \( Q \) be the unit cube in \( r \) -space. Since \( l \) is a homomorphism, we have \[ \frac{\text{ measure of }{l}^{-1}\left( P\right) }{\text{ measure of }{l}^{-1}\left( Q\right) } = \frac{\text{ volume of }P}{\text{ volume of }Q} = R. \] Thus it suffices to compute the measure of \( {l}^{-1}\left( Q\righ... | No |
Proposition 9. If \( E \) is a fundamental domain for \( {J}^{0}/{k}^{ * } \), then its measure is\n\n\[ \kappa = \frac{{2}^{{r}_{1}}{\left( 2\pi \right) }^{{r}_{2}}{hR}}{w{d}_{k}^{1/2}} \] | Proof. Trivial from Proposition 8. | No |
Theorem 12. By analytic continuation we may extend the definition of any zeta function \( \zeta \left( {f, c}\right) \) to the domain of all quasi-characters of \( J/{k}^{ * } \) . The extended function is single valued and holomorphic, except at \( c\left( a\right) = 1 \) and \( c\left( a\right) = \parallel a\parallel... | Proof. We have\n\n\[ \zeta \left( {f, c}\right) = {\int }_{\parallel a\parallel < 1}f\left( a\right) c\left( a\right) {d}^{ * }a + {\int }_{\parallel a\parallel \geqq 1}f\left( a\right) c\left( a\right) {d}^{ * }a. \]\n\nThe second integral obviously converges for \( \operatorname{Re}\left( c\right) \) equal to any rea... | Yes |
Theorem 13. We have\n\n\[ \zeta \left( {f, c}\right) = \]\n\n\[ {\int }_{\parallel a\parallel \geqq 1}f\left( a\right) c\left( a\right) {d}^{ * }a + {\int }_{\parallel a\parallel \geqq 1}\widehat{f}\left( a\right) \widehat{c}\left( a\right) {d}^{ * }a + {\delta }_{x}\left\lbrack {\frac{\kappa \widehat{f}\left( 0\right)... | The two integrals are convergent for all \( c \), uniformly in every strip\n\n\[ {\sigma }_{0} \leqq \operatorname{Re}\left( c\right) \leqq {\sigma }_{1} \]\n\nThe uniformity of convergence of the integrals in a given strip is clear from the above remarks. Furthermore, let us replace \( \left( {f, c}\right) \) by \( \l... | Yes |
Proposition 10. If \( g \) is a translation of \( f \), i.e. \( g\left( x\right) = f\left( {bx}\right) \), then\n\n\[ \widehat{g}\left( y\right) = \parallel b{\parallel }^{-1}\widehat{f}\left( {y/b}\right) \;\text{ and }\;\xi \left( {g, c}\right) = c{\left( b\right) }^{-1}\xi \left( {f, c}\right) . \] | Proof. Directly from the definitions. The assertion is valid locally or globally, i.e. on \( {k}_{v}^{ * } \) or \( {J}_{k} \) . In the local case, \( \parallel \parallel = \parallel {\parallel }_{v} \) of course, and \( b \) is either in \( {k}_{v}^{ * } \) or an idele. | Yes |
Proposition 11. Let \( \chi = {\chi }_{0} \) be the trivial character. Then \( {g}_{0} \) and \( {\widehat{g}}_{0} \) are \( \geqq 0 \), and \[ {g}_{0}\left( 0\right) = {\widehat{g}}_{0}\left( 0\right) = {d}_{k}^{1/2}{\left( 2\pi \right) }^{-{r}_{2}}. \] | Proof. Immediate from the definitions. | No |
Proposition 12. Let \( b \) be the idele having components \( {b}_{v} = {d}_{k}^{1/N} \) at archimedean \( v \), and \( {b}_{\mathfrak{p}} = {\pi }^{-{\nu }_{\mathfrak{p}}} \), where \( {\nu }_{\mathfrak{p}} = {\operatorname{ord}}_{\mathfrak{p}}\mathfrak{D} \). Then \( \parallel b\parallel = 1 \), and \[ {\widehat{g}}_... | Proof. This is an easy consequence of Proposition 10, together with the explicit determination of \( {g}_{0} \) in terms of \( {f}_{0} \). | No |
Theorem 14. Assume that \( \chi \) is normalized so that\n\n\[ \mathop{\sum }\limits_{{v \in {S}_{\infty }}}{N}_{v}{\varphi }_{v}\left( x\right) = 0 \]\n\nThen:\n\n\[ \zeta \left( {{g}_{\chi },\chi, s}\right) = \Lambda \left( {s,\chi }\right) \mathop{\prod }\limits_{{\mathfrak{p} \in {S}_{\chi }}}{\tau }_{\mathfrak{p}}... | Proof. Just put together the local results of the computations of \( §4 \) , together with Proposition 10, and be careful about all the possible cancellations which take place. | No |
Corollary 1. We have the functional equation\n\n\\[ \nW\\left( x\\right) \\Lambda \\left( {s, x}\\right) = \\Lambda \\left( {1 - s,\\bar{x}}\\right) \n\\]\nwhere \\( W\\left( x\\right) \\) is a constant of absolute value 1, given by\n\n\\[ \nW\\left( \\chi \\right) = {4}^{-{i\\Phi }}{i}^{-M}{\\left( {\\mathbf{{Nf}}}_{\... | Proof. From the local computations of \\( §4 \\), we know that each expression \\( {\\tau }_{\\mathfrak{p}}\\left( \\chi \\right) \\) has absolute value \\( {\\left( {\\mathbf{{Nf}}}_{\\chi ,\\mathfrak{p}}\\right) }^{1/2} \\) which is just enough to cancel out in the equality\n\n\\[ \n\\zeta \\left( {{g}_{\\chi },\\chi... | Yes |
For a fixed character \( \chi \), put \( \Lambda \left( s\right) = \Lambda \left( {s,\widetilde{\chi }}\right) \). Then | Proof. A trivial computation, using the relation \[ \Lambda \left( {\bar{s},\chi }\right) = \overline{\Lambda \left( {s,\bar{\chi }}\right) }.\] | No |
Corollary 3. Let \( {\Lambda }_{0}\left( s\right) = \Lambda \left( {s,{\chi }_{0}}\right) \) . Then | \[ {\Lambda }_{0}\left( s\right) = \zeta \left( {{g}_{0},{\chi }_{0}, s}\right) = {\left( {2}^{-2{r}_{2}}{\pi }^{-N}{d}_{k}\right) }^{s/2}{\Gamma }^{{r}_{1}}\left( {s/2}\right) {\Gamma }^{{r}_{2}}\left( s\right) {\zeta }_{k}\left( s\right) \] and \[ {\Lambda }_{0}\left( s\right) = {\Lambda }_{0}\left( {1 - s}\right) \] | Yes |
Proposition 13. Let\n\n\[ \kappa = \frac{{2}^{{r}_{1}}{\left( 2\pi \right) }^{{r}_{2}}{hR}}{w{d}_{k}^{1/2}} \]\n\nbe the volume of the fundamental domain for \( {J}_{k}^{0}{\;\operatorname{mod}\;{k}^{ * }} \) . Then the residue of \( \zeta \left( {{g}_{0}, s}\right) = \zeta \left( {{g}_{0},{\chi }_{0}, s}\right) \) at ... | Proof. The residue for \( \zeta \left( {{g}_{0},{\chi }_{0}, s}\right) \) comes from the general Theorem 12 and that for the zeta function comes from \( s = 1 \) in Corollary 3 above together with the values \( \Gamma \left( \frac{1}{2}\right) = {\pi }^{1/2} \), and \( \Gamma \left( 1\right) = 1 \) . | Yes |
Theorem 15. We have an integral expression\n\n\[ \zeta \left( {{g}_{0}, s}\right) = {\int }_{\parallel a\parallel \geqq 1}{\widehat{g}}_{0}\left( a\right) \left( {\parallel a{\parallel }^{s} + \parallel a{\parallel }^{1 - s}}\right) {d}^{ * }a + \frac{\lambda }{s\left( {s - 1}\right) }.\] | Proof. If we write down the integral expression of Theorem 13, \( §7 \) and use Proposition 10 together with the fact that \( \parallel b\parallel = 1 \) and that the multiplicative measure is invariant under multiplicative translation, we get what we want. | No |
Theorem 1. Let \( f\\left( s\\right) \) be as above. Let \( g\\left( s\\right) = \\sum {b}_{n}/{n}^{s} \) be a Dirichlet series with complex coefficients \( {b}_{n} \), and assume that there is a number \( C \) such that \( \\left| {b}_{n}\\right| < C{a}_{n} \). Assume that the series for \( g\\left( s\\right) \) conve... | Proof. We naturally set \( \\alpha = 0 \) if there is no pole at \( s = 0 \). Suppose the \( {b}_{n} \) are real. Then the function \( \\left( {{Cf} + g}\\right) /\\left( {C + \\alpha }\\right) \) for large enough constant \( C \) satisfies the same conditions as \( f\\left( s\\right) \). From this our assertion is imm... | Yes |
Proposition 1. Let \( {b}_{n}\left( {n = 2,3,\ldots }\right) \) be complex numbers such that\n\n\[ \Psi \left( N\right) = \mathop{\sum }\limits_{{n = 2}}^{N}{b}_{n} = {\alpha N} + o\left( N\right) \]\n\nfor some complex \( \alpha \) . Then:\n\n\[ \pi \left( N\right) = \mathop{\sum }\limits_{{n = 2}}^{N}\frac{{b}_{n}}{\... | Proof. We have \( {b}_{n} = \Psi \left( n\right) - \Psi \left( {n - 1}\right) \) for \( n \geqq 2 \), putting \( \Psi \left( 1\right) = 0 \) . Hence\n\n\[ \pi \left( N\right) = \mathop{\sum }\limits_{{n = 2}}^{N}\frac{\Psi \left( n\right) - \Psi \left( {n - 1}\right) }{\log n} = \mathop{\sum }\limits_{2}^{N}\frac{\Psi ... | Yes |
Theorem 2. Let \( \chi \) be a Hecke character, \( \chi \neq {\chi }_{0} \) . Then \[ L\left( {1, x}\right) \neq 0. \] | Proof. Assume that \( L\left( {1, x}\right) = 0 \) . We have for \( s \) real \( > 1 \) : \[ L\left( {s,\chi }\right) = \exp \left( {\mathop{\sum }\limits_{{\mathfrak{p}, m}}\frac{\chi \left( {\mathfrak{p}}^{m}\right) }{m\mathbf{N}{\mathfrak{p}}^{ms}}}\right) \] where \( \exp \left( x\right) = {e}^{x} \) . Consider the... | Yes |
Lemma 1. Let \( f\left( s\right) = \sum {a}_{n}{n}^{-s} \) be a function defined by a Dirichlet series with \( {a}_{n} \) real \( \geqq 0 \), such that the series converges for \( \operatorname{Re}\left( s\right) > {\sigma }_{0} \) . Suppose that \( f\left( s\right) \) is holomorphic at \( {\sigma }_{0} \) . Then the s... | Proof. Let \( \delta \) be small \( > 0 \) . We may assume \( {\sigma }_{0} = 0 \) (after a translation). We have for \( 0 < \sigma < \delta \) ,\n\n\[ f\left( \sigma \right) = \mathop{\sum }\limits_{n}{a}_{n}{e}^{-\left( {\sigma - \delta }\right) \log \left( n\right) }{e}^{-\delta \log \left( n\right) }.\]\n\nWe repla... | Yes |
Theorem 3. Let \( x \) be a Hecke character which is non-trivial on \( {J}^{0} \) . Then\n\n\( L\left( {s, x}\right) \) has no zero on the line \( s = 1 + {it} \) . | Proof. This is essentially a corollary of Theorem 2, if we replace \( s \) by \( s + {it} \), and \( x \) by the character\n\n\[ a \mapsto \chi \left( a\right) \parallel a{\parallel }^{-{it}}. \] | Yes |
Theorem 4. Let \( \kappa \) be the residue of the zeta function \( {\zeta }_{k}\left( s\right) \) at 1 . Let \( n\left( {A}_{x}^{S}\right) \) and \( n\left( {P}_{x}\right) \) be the number of elements in \( {A}_{x}^{S} \) and \( {P}_{x} \) respectively. Then these two numbers are asymptotic to\n\n\[ n\left( {A}_{x}^{S}... | Indeed, the residue at 1 of the function obtained from the zeta function \( {\zeta }_{k}\left( s\right) \) by omitting the factors involving the primes in \( S \) has residue \( {\alpha \beta } \) , whence the first assertion. As for the second, we apply the Tauberian theorem to the logarithmic derivative of the zeta f... | Yes |
Theorem 5. Let \( x \) be a Hecke character which is non-trivial on \( {J}^{0} \), and\n\n\( S \) a finite set containing those primes where \( \chi \) ramifies. Then\n\n\[ \mathop{\lim }\limits_{{r \rightarrow \infty }}\frac{1}{n\left( {A}_{r}^{S}\right) }\mathop{\sum }\limits_{{\mathfrak{a} \in {A}_{r}^{S}}}\chi \lef... | Proof. This is immediate, since we know that the L-series is holomorphic at 1, and does not vanish on the line \( 1 + {it} \) . Thus the residue of both the \( L \) -series and its logarithmic derivative is 0, whence our results follow from Theorem 1, Proposition 1, Theorem 3, and the Tauberian theorem. | No |
Let \( k \) be a number field of class number 1 , so that\n\n\[ J = {k}^{ * }{J}_{S} \]\n\nwhere \( S \) is the set of archimedean absolute values. Let \( U \) be the group of units of \( {\mathfrak{o}}_{k} \), viewed as a subgroup of\n\n\[ {k}_{\infty }^{ * } = \mathop{\prod }\limits_{{v \in S}}{k}_{v}^{ * } \]\n\n(i.... | In particular, if \( k \) is a real quadratic field of class number 1, we may take an embedding \( k \rightarrow \mathbf{R} \) of \( k \) into one of its (real) completions, giving rise to an absolute value denoted by \( \left| \right| \), and \( {k}_{\infty }^{ * } = {\mathbf{R}}^{ * } \times {\mathbf{R}}^{ * } \) . W... | Yes |
Lemma 1. There exists an absolute constant \( {c}_{1} \) such that the inequality\n\n\[ \kappa \left( k\right) \leqq {c}_{1}^{N}{\left( 1 + \alpha \right) }^{N}{d}_{k}^{1/{2\alpha }}\;\left( {N = \left\lbrack {k : \mathbf{Q}}\right\rbrack }\right) \]\n\nholds for all number fields \( k \) and all \( \alpha \geqq 1 \) . | Proof. According to Chapter XIV, Theorem 14, Corollary 3 and Theorem 15 together with the fact that the integrals expressing the zeta function are \( \geqq 0 \) for real \( s \), we get for \( s > 1 \) :\n\n\[ {\left( {2}^{-2{r}_{2}}{\pi }^{-N}{d}_{k}\right) }^{s/2}{\Gamma }^{{r}_{1}}\left( \frac{s}{2}\right) {\Gamma }... | Yes |
Lemma 2. There exists a constant \( {c}_{2} \) such that for \( k \neq \mathbf{Q} \) , \[ \log \left( {hR}\right) /\log \left( {d}^{1/2}\right) \leqq {c}_{2} \] If \( k \) ranges over a sequence of fields such that \( N/\log d \) tends to 0, then for this sequence \[ \lim \sup \left\lbrack {\left( {\frac{\log {hR}}{\lo... | Proof. We use the elementary estimate that the number of roots of unity \( w \) in a number field \( k \) is \( \leqq {c}_{3}{N}^{2} \) for some absolute constant \( {c}_{3} \) . (Use the fact that the field of \( n \) -th roots of unity over \( \mathbf{Q} \) has degree \( \varphi \left( n\right) \) , together with an ... | Yes |
Lemma 3. Let \( {s}_{0} \) be real, \( 0 < {s}_{0} < 1 \), and assume that \( \zeta \left( {{g}_{0},{s}_{0}}\right) \leqq 0 \) (or what is the same thing, that \( \left. {{\zeta }_{k}\left( {s}_{0}\right) \leqq 0}\right) \) . Then\n\n\[ \kappa \left( k\right) \geqq {s}_{0}\left( {1 - {s}_{0}}\right) {2}^{-N}{e}^{-{4\pi... | Proof. By Theorem 15 of Chapter XIV, §8, we have\n\n\[ \frac{\kappa {g}_{0}\left( 0\right) }{{s}_{0}\left( {1 - {s}_{0}}\right) } \geqq {\int }_{\parallel a\parallel \geqq 1}{\widehat{g}}_{0}\left( a\right) \parallel a{\parallel }^{{s}_{0}}{d}^{ * }a. \]\n\nWe shrink the domain of integration to a domain \( P = \prod {... | Yes |
Theorem 1. Let \( \epsilon > 0 \) . There exists a number \( {c}_{4}\left( \epsilon \right) \) such that for all fields \( k \) normal over \( \mathbf{Q} \), the inequality holds:\n\n\[ \kappa \left( k\right) \geqq {c}_{4}{\left( \epsilon \right) }^{-N}{d}_{k}^{-\epsilon }.\] | Proof. If we had the Riemann hypothesis, we could dispense with our hypothesis that \( k \) is normal over \( \mathbf{Q} \) . Indeed, our arguments are split into two cases.\n\nCase 1. For all normal fields \( k \) the function \( {\zeta }_{k}\left( s\right) \) does not vanish for real \( s \) with \( 1 - \epsilon /N <... | No |
Lemma 4. Let \( k \) be a number field, \( \psi \neq 1 \) a character of finite period of \( {C}_{k} \), and \( \alpha \geqq 1 \) . Then \[ \left| {L\left( {1,\psi }\right) }\right| \leqq {c}_{6}^{N}{\left( 1 + \alpha \right) }^{N}{d}_{\psi }^{1/{2\alpha }}. \] | Proof. We have, with the same notation as Chapter XIV, §8, \[ \zeta \left( {{g}_{\psi },\psi ,1}\right) = {\int }_{\parallel a\parallel \geqq 1}{g}_{\psi }\left( a\right) \psi \left( a\right) \parallel a\parallel {d}^{ * }a + {\int }_{\parallel a\parallel \geqq 1}{\dot{g}}_{\psi }\left( a\right) \bar{\psi }\left( a\rig... | Yes |
Theorem 4. If \( k \) ranges over a sequence of fields normal over \( \mathbf{Q} \) for which \( N/\log d \) tends to 0, then\n\n\[ \log \left( {Rh}\right) \sim \log {d}^{1/2} \] | It is a simple exercise to estimate the discriminant of the smallest normal extension \( {k}^{\prime } \) containing a given number field \( k \) over \( \mathbf{Q} \) . One finds that\n\n\[ {d}_{{k}^{\prime }} \leqq {d}_{k}^{{N}^{\prime }/2} \]\n\nwhere \( {N}^{\prime } = \left\lbrack {{k}^{\prime } : \mathbf{Q}}\righ... | No |
Theorem 5. If \( k \) ranges over all number fields \( \neq \mathbf{Q} \) and \( {N}^{\prime } \) is the degree over \( \mathbf{Q} \) of the smallest normal field \( {k}^{\prime } \) over \( \mathbf{Q} \) containing \( k \), then the set of values\n\n\[ \left\lbrack {\frac{\log \left( {Rh}\right) }{\log {d}^{1/2}} - 1}... | Proof. Immediate, taking into account that \( {N}^{\prime } \leqq N \) !. | No |
Proposition 1. Let \( G \) be a finite group of order \( g \) . Then\n\n\[ \n{u}_{G} = \frac{1}{g}\sum {\lambda }_{A}^{ * }\n\]\n\nthe sum being taken over all cyclic subgroups of \( G \) . | Proof. Given two functions \( \chi ,\psi \) on \( G \), we have the usual scalar product:\n\n\[ \n\langle \psi ,\chi {\rangle }_{G} = \frac{1}{g}\mathop{\sum }\limits_{{\sigma \in G}}\psi \left( \sigma \right) \overline{\chi \left( \sigma \right) }.\n\]\n\nLet \( \psi \) be any function on \( G \) . Then:\n\n\[ \n\left... | Yes |
Proposition 2. If \( A \neq \{ 1\} \), the function \( {\lambda }_{A} \) is a linear combination of irreducible non-trivial characters of \( A \) with positive integral coefficients. | Proof. If \( A \) is cyclic of prime order, then by Proposition 1, we know that \( {\lambda }_{A} = g{u}_{A} \), and our assertion follows from the standard structure of the regular representation.\n\nIn order to prove the assertion in general, it suffices to prove that the Fourier coefficients of \( {\lambda }_{A} \) ... | Yes |
Lemma 1. Let \( f\left( s\right) \) be holomorphic in the upper part of a strip: \( {\sigma }_{0} \leqq \sigma \leqq {\sigma }_{1} \), and \( t \geqq {t}_{1} > 0 \) . Assume that \( f\left( s\right) \) is \( O\left( {e}^{{t}^{c}}\right) \) for some constant \( c > 0 \), and \( t \rightarrow \infty \) in this strip, and... | This is nothing but the Phragmen-Lindelöf Theorem, proved in Chapter XIII, §5. | Yes |
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