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Proposition 9.21. Again maintaining the notations and hypotheses of Proposition 9.19, let \( \chi \) be an irreducible character of \( G = \operatorname{Gal}\left( {L/K}\right) \) . Then, \( L\left( {s,\chi }\right) = {Z}_{K}\left( {\chi \left( {\phi }_{q}\right) u}\right) \) where, as usual, \( u = {q}^{-s} \), and \(... | Proof. Since \( G \) is abelian, \( \chi \) is a linear character, i.e., a homomorphism from \( G \) to \( {\mathbb{C}}^{ * } \) . From the definitions, and Proposition 9.19,\n\n\[ L\left( {s,\chi }\right) = \mathop{\prod }\limits_{{P \in {\mathcal{S}}_{K}}}{\left( 1 - \chi \left( \left( P, L/K\right) \right) N{P}^{-s}... | Yes |
Theorem 9.23. (E. Artin) Let \( L/K \) be a finite abelian extension of global function fields. Let \( S \) be the set of primes’ of \( K \) which are ramified in \( L \) . Then the Artin map, \( \left( {*, L/K}\right) \), takes \( D\left( S\right) \) onto \( \operatorname{Gal}\left( {L/K}\right) \) and there is an eff... | As we have already mentioned, this is a very deep result whose proof is long and involved. We have proved a portion of the Theorem in Proposition 9.18. The Artin map is onto and the norms of divisors are contained in the kernel. The exisitence of a divisor \( \mathcal{F} \) such that \( {\mathcal{P}}^{\mathcal{F}}{N}_{... | No |
Theorem 9.25. Let \( L/K \) be a finite, abelian extension of global function fields, \( G = \operatorname{Gal}\left( {L/K}\right) \), and \( \chi \) a linear character of \( G \) . Then, \( L\left( {s,\chi }\right) \) has an analytic continuation to an entire function in the whole complex plane if and only if \( {K}_{... | Proof. From the discussion preceding the statement of the theorem, \( L\left( {s,\chi }\right) = L\left( {s,\lambda }\right) \) where \( \lambda = \chi \circ \rho \) . Here, \( \rho : C{l}_{{\mathcal{F}}_{\chi }} \rightarrow {G}_{\chi } \) and \( \chi : {G}_{\chi } \rightarrow {\mathbb{C}}^{ * } \) . Let \( {G}_{\chi }... | Yes |
Lemma 10.1. Let \( P \) be a prime of \( K \) not containing \( a \in {K}^{ * } \) . Then, \( a \) is a primitive root modulo \( P \) if and only if there is no prime \( l \in \mathbb{Z} \) satisfying both of the following conditions:\n\n\[ \n\text{i)}{NP} \equiv 1\;\left( {\;\operatorname{mod}\;l}\right) \;\text{and}\... | Proof. If there is a prime \( l \) satisfying both conditions, then the order of \( a \) modulo \( P \) divides \( \left( {{NP} - 1}\right) /l \), so that \( a \) cannot be a primitive root. So, if \( a \) is a primitive root, there is no prime \( l \) for which both conditions hold.\n\nNow, suppose there is no prime \... | Yes |
Proposition 10.2. Let \( L = K\left( {\zeta }_{l}\right) \). Then \( \left\lbrack {L : K}\right\rbrack = f\left( l\right) \), the smallest positive integer \( f \) such that \( {q}^{f} \equiv 1\left( {\;\operatorname{mod}\;l}\right) \). A prime \( P \in {\mathcal{S}}_{K} \) splits completely in \( L \) if and only if \... | Proof. Since \( L = K\mathbb{F}\left( {\zeta }_{l}\right) \) it is a constant field extension and \( \left\lbrack {L : K}\right\rbrack = \) \( \left\lbrack {\mathbb{F}\left( {\zeta }_{l}\right) : \mathbb{F}}\right\rbrack \) by Proposition 8.1. Now, \( \operatorname{Gal}\left( {\mathbb{F}\left( {\zeta }_{l}\right) : \ma... | Yes |
Proposition 10.3. Let \( K \) be a function field over a constant field \( F \) of characteristic \( p \) . Let \( l \) be a prime number not equal to \( p \) and \( a \in {K}^{ * } \), not an \( l \) -th power in \( {K}^{ * } \) . Let \( \alpha \) be a root of \( {X}^{l} - a = 0 \) and \( L = K\left( \alpha \right) \)... | Proof. Note to begin with that since \( l \neq p,{X}^{l} - a \) is a separable polynomial and so \( L \) is a separable extension of \( K \) . Also, since \( a \) is not an \( l \) -th power, \( {x}^{l} - a \) is irreducible (see Lang [4]) and so \( \left\lbrack {L : K}\right\rbrack = l \) .\n\nSuppose first that \( l ... | Yes |
Proposition 10.4. Let \( K/F \) be a function field, with constant field \( F \) of characteristic \( p \) (possibly, \( p = 0 \) ). Let \( l \neq p \) be a prime and \( L = K\left( \alpha \right) \) where \( {\alpha }^{l} = a \in {K}^{ * } \) . Assume that \( a \) is geometric at \( l \) and that \( a \) is not an \( ... | Proof. This is an application of the Riemann-Hurwitz Theorem, Theorem 7.16, which asserts that\n\n\[ 2{g}_{L} - 2 = \left\lbrack {L : K}\right\rbrack \left( {2{g}_{K} - 2}\right) + {\deg }_{L}{\mathfrak{D}}_{L/K}. \]\n\nHere, \( L/K \) is presumed to be a finite, separable, geometric extension of function fields and \(... | Yes |
Proposition 10.5. Let \( K \) be a global function field over a constant field \( \mathbb{F} \) with \( q \) elements. Let \( l \) be a prime different from the characteristic of \( \mathbb{F} \) . Let \( a \in {K}^{ * } \) . Assume that \( K \) contains a primitive \( l \) -th root of unity, \( {\zeta }_{l} \) . Suppo... | Proof. Since \( {\zeta }_{l} \in K \), the extension \( L/K \) is a cyclic Galois extension. Also, we must have \( {\zeta }_{l} \in {\mathbb{F}}^{ * } \), which implies \( q \equiv 1\left( {\;\operatorname{mod}\;l}\right) \) and so \( {NP} \equiv 1\left( {\;\operatorname{mod}\;l}\right) \) for all primes \( P \) of \( ... | Yes |
Proposition 10.6. Let \( K \) be a global function field with constant field \( \mathbb{F} \). Let \( l \) be a prime different from the characteristic of \( \mathbb{F} \). Let \( a \in {K}^{ * } \). Let \( {E}_{l} = K\left( {{\zeta }_{l},\sqrt[l]{a}}\right) \). Let \( P \) be a prime of \( K \) such that \( {\operator... | Proof. Consider the tower of fields \( K \subseteq K\left( {\zeta }_{l}\right) \subseteq {E}_{l} \). A prime of \( K \) splits completely in \( L \) if and only if it splits completely in \( K\left( {\zeta }_{l}\right) \) and every prime above it in \( K\left( {\zeta }_{l}\right) \) splits completely in \( {E}_{l} \).\... | Yes |
Proposition 10.7. Assume a is geometric and not an \( l \) -th power in \( K \) . Let \( m \) be a square-free integer prime to \( q \) and \( {E}_{m} \) the compositum of the fields \( {E}_{l} \) for all \( l \mid m \) . Let \( f\left( m\right) \) denote the order of \( q \) modulo \( m \) . Then, \( \left\lbrack {{E}... | Proof. The field \( {E}_{m} \) contains \( {\zeta }_{m} \), a primitive \( m \) -th root of unity.\n\nLet \( l \mid m \) . We claim \( a \) is an \( l \) -th power in \( K \) if and only if it is an \( l \) -th power in \( K\left( {\zeta }_{m}\right) \) . One way is obvious, so suppose \( a \) is an \( l \) -th power i... | Yes |
Theorem 10.9. (Bilharz) Assume a is geometric element of \( {K}^{ * } \) . Then, with the above notations, the Dirichlet density of \( {\mathcal{M}}_{a} \) exists and is given by \[ \delta \left( {\mathcal{M}}_{a}\right) = \mathop{\sum }\limits_{\substack{{m = 1} \\ {\left( {m, p}\right) = 1} }}^{\infty }\frac{\mu \lef... | The sum is easily seen to be absolutely convergent using Romanoff's theorem. | No |
Proposition 10.10. With the same notations as Theorem 10.9,\n\n\[ \mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{\mu \left( m\right) }{{m}_{a}f\left( m\right) } \geq \mathop{\prod }\limits_{{l \neq p}}\left( {1 - \frac{1}{{l}_{a}f\left( l\right) }}\right) . \] | Let \( S \) denote the set of primes \( l \neq p \) for which \( a \) is an \( l \) -th power. \( S \) is a finite set, possibly empty. We can rewrite the right-hand side of the equation in Proposition 10.10 as follows:\n\n\[ \mathop{\prod }\limits_{{l \notin S}}\left( {1 - \frac{1}{{lf}\left( l\right) }}\right) \matho... | Yes |
Theorem 10.11. (Bilharz) Let a be a geometric element of \( {K}^{ * } \) . Then there are infinitely many primes \( P \in {\mathcal{S}}_{K} \) for which a is a primitive root provided that there is no prime divisor \( l \) of \( q - 1 \) for which \( a \) is an \( l \) -th power. If there is such a prime divisor, then ... | Proof. We have just shown on the basis of Theorem 10.9 and Proposition 10.10 that if \( a \) is not an \( l \) -th power for some prime \( l \mid q - 1 \), then \( \delta \left( {\mathcal{M}}_{a}\right) \neq 0 \) . Since we pointed out earlier that \( {\mathcal{M}}_{a} \) differs from the set of primes for which \( a \... | Yes |
Lemma 10.13. The sum\n\n\[ \mathop{\sum }\limits_{{l \neq p}}\frac{1}{f\left( l\right) {q}^{\frac{f\left( l\right) }{2}}} \]\n\nis convergent. | Proof. We will break the sum up into two subsums, the first over all primes\n\n\( l \) such that \( l \leq {q}^{f\left( l\right) /2} \) and the second over all primes such that \( l > {q}^{f\left( l\right) /2} \) .\n\nFor the first subsum we have\n\n\[ \mathop{\sum }\limits_{\substack{{l \neq p} \\ {l \leq {q}^{f\left(... | Yes |
Lemma 10.14. Let \( \gamma \) denote Euler’s constant.\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\frac{\mu {\left( d\right) }^{2}}{d} \leq \frac{6}{{\pi }^{2}}{e}^{\gamma }\log \log \left( n\right) + O\left( 1\right) . \] | Proof. First of all, note that\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\frac{\mu {\left( d\right) }^{2}}{d} = \mathop{\prod }\limits_{{p \mid n}}\left( {1 + \frac{1}{p}}\right) \]\n\nThe product is over all prime divisors \( p \) of \( n \) . We break this product up into two parts - first the product over prime diviso... | Yes |
Lemma 10.15.\n\n\\[ d\\left( {T}_{n}\\right) \\geq \\mathop{\\prod }\\limits_{{i = 1}}^{n}\\left( {1 - \\frac{1}{{a}_{i}}}\\right) \\] | Proof. The proof is by induction on \\( n \\) . If \\( n = 1 \\) the assertion reads \\( d\\left( {T}_{1}\\right) = \\) \\( 1 - 1/{a}_{1} \\) which we have already noted. So, we assume the result is true for \\( n \\) and prove it for \\( n + 1 \\) .\n\nNote that \\( {T}_{n} = {T}_{n + 1} \\cup S \\) where \\( S \\) is... | Yes |
Lemma 10.16. Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be real numbers with \( 0 \leq {x}_{i} \leq 1 \) for each i. As above, let \( \left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right\} \) be a finite set of positive integers. Then\n\n\[ 1 - \mathop{\sum }\limits_{{i = 1}}^{n}\frac{{x}_{i}}{{a}_{i}} + \mathop{\sum }\limit... | Proof. (Sketch) Let \( F\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) denote the difference between the left-hand side and the right-hand side of the inequality given in the statement of the Lemma. We think of \( F \) as a function on the unit cube and prove that it is non-negative on this domain. That will prove ... | Yes |
Proposition 11.1. Let \( l \) be a prime which divides \( h\left( K\right) \) . If \( l \mid n\frac{{q}^{n} - 1}{q - 1} \) , then \( l \mid h\left( {K}_{n}\right) /h\left( K\right) \) . | Proof. Let \( E \) be the number field obtain by adjoining all the elements \( {\pi }_{\imath } \) to the rational numbers \( \mathbb{Q} \), and let \( \mathcal{L} \) be a prime ideal of \( E \) lying over \( l \) . Then, \( l \mid h\left( K\right) \) implies\n\n\[ \mathop{\prod }\limits_{{i = 1}}^{{2g}}\left( {1 - {\p... | Yes |
Corollary 2. If \( l \mid h\left( K\right) \) and \( n = {l}^{t} \), then \( {l}^{t} \mid h\left( {K}_{n}\right) /h\left( K\right) \) . | Proof. Just use Corollary 1 and induction on \( t \) . | No |
Lemma 11.2. As above, let \( \mathcal{L} \) be a prime of \( E \) lying above \( l \) . If \( {\pi }_{i} \notin \mathcal{L} \) , then \( {\pi }_{i}^{d\left( l\right) } \equiv 1\left( {\;\operatorname{mod}\;\mathcal{L}}\right) \) . | Proof. Let \( {L}_{K}^{ * }\left( u\right) = {u}^{2g}{L}_{K}\left( {1/u}\right) \) . Then, \( {L}_{K}^{ * }\left( u\right) \) is a monic polynomial with integer coefficients whose roots are the numbers \( {\pi }_{i} \) . Suppose \( {\pi }_{i} \notin \mathcal{L} \) and let \( {\bar{\pi }}_{i} \) be the residue of \( {\p... | Yes |
Proposition 11.3. Suppose \( l \) does not divide \( h\left( K\right) \) and that \( n \) is relatively prime to \( d\left( l\right) \) . Then, \( l \) does not divide \( h\left( {K}_{n}\right) \) . | Proof. Suppose \( l \mid h\left( {K}_{n}\right) \) . Then, for some \( k,{\pi }_{k}^{n} - 1 \equiv 0\left( {\;\operatorname{mod}\;\mathcal{L}}\right) \) . By the above lemma, \( {\pi }_{k}^{d\left( l\right) } \equiv 1\left( {\;\operatorname{mod}\;\mathcal{L}}\right) \) . Since \( \left( {n, d\left( l\right) }\right) = ... | Yes |
Proposition 11.4. Let \( n \) be he smallest integer such that \( l \mid h\left( {K}_{n}\right) \) . Then, \( n \mid d\left( l\right) \) . | Proof. There must be an index \( k \) such that \( {\pi }_{k}^{n} \equiv 1\left( {\;\operatorname{mod}\;\mathcal{L}}\right) \) . From the definition of \( n \) it follows that \( n \) is the order of \( {\pi }_{k} \) modulo \( \mathcal{L} \) . By Lemma 11.2, \( {\pi }_{k}^{d\left( l\right) } \equiv 1\left( {\;\operator... | Yes |
Lemma 11.7. Let \( {\bar{L}}_{K}\left( u\right) \) be the reduction of \( {L}_{K}\left( u\right) \) modulo \( p \) . Then \( \lambda \) is equal to the degree of \( {\bar{L}}_{K}\left( u\right) \) . | Proof. \( {L}_{K}\left( u\right) = \mathop{\sum }\limits_{{m = 0}}^{{2g}}{a}_{m}{u}^{m} = \mathop{\prod }\limits_{{i = 1}}^{{2g}}\left( {1 - {\pi }_{i}u}\right) \) . Thus, each \( {a}_{m} \) is the \( m \) -th elementary symmetric function of the \( {\pi }_{i} \) . If \( m > \lambda \), each term of the \( m \) -th ele... | Yes |
Proposition 11.8. The following conditions are equivalent:\n\na) \( K \) is supersingular.\n\nb) All the \( {\pi }_{i} \) are in \( \mathcal{P} \) .\n\nc) The degree of \( {\bar{L}}_{K}\left( u\right) \) is zero.\n\n\[ \text{d)}h\left( {K}_{n}\right) \equiv 1\left( {\;\operatorname{mod}\;\mathcal{P}}\right) \text{for a... | Proof. We have already seen that \( b \) implies \( a \) . To show the reverse, suppose that some \( {\pi }_{i} \notin \mathcal{P} \) . Then, for some \( n > 0 \) we have \( {\pi }_{i}^{n} \equiv 1\left( {\;\operatorname{mod}\;\mathcal{P}}\right) \) . This implies \( p \mid h\left( {K}_{n}\right) \) contrary to \( a \)... | Yes |
Corollary 2. Suppose \( K \) has genus 2. Then, \( K \) is supersingular if and only if the number of primes of degree 1 and the class number, \( h\left( K\right) \), are both congruent to 1 modulo \( p \) . | Both corollaries are consequences of condition \( e \) and the above remark. | No |
Proposition 11.9. Let \( L \) be a finite, unramified, Galois extension of the function field \( K \) . Let \( G = \operatorname{Gal}\left( {L/K}\right) \) . Then, \( {\mathcal{D}}_{L}^{G} = {i}_{L/K}{\mathcal{D}}_{K} \) . Here, \( {\mathcal{D}}_{K} \) and \( {\mathcal{D}}_{L} \) denote the divisor groups of \( K \) an... | Proof. If \( P \) is a prime of \( K \), then, by definition, \( {i}_{L/K}P = \mathop{\sum }\limits_{{\mathfrak{P} \mid P}}e\left( {\mathfrak{P}/P}\right) \mathfrak{P} \) . Since we are assuming that \( L/K \) is unramified, this becomes \( {i}_{L/K}P = \) \( \mathop{\sum }\limits_{{\mathfrak{P} \mid P}}\mathfrak{P} \)... | Yes |
Proposition 11.10. Let \( K \) be a function field with constant field \( F \) . Let \( E \) be a finite, Galois extension of \( F \) and set \( L = {KE} \) . Let \( G = \operatorname{Gal}\left( {E/F}\right) = \) \( \operatorname{Gal}\left( {L/K}\right) \) . Then \( {i}_{L/K} : C{l}_{K} \rightarrow C{l}_{L} \) is one t... | Proof. The proof will use some facts from cohomology of groups.\n\nLet \( {\mathcal{P}}_{L} \) and \( {\mathcal{P}}_{K} \) be the principal divisors of \( L \) and \( K \), respectively. Consider the exact sequence:\n\n\[ \left( 0\right) \rightarrow {E}^{ * } \rightarrow {L}^{ * } \rightarrow {\mathcal{P}}_{L} \rightar... | Yes |
Theorem 11.12. Let \( K/M \) be a function field over an algebraically closed field of constants \( M \) . Set \( J = C{l}_{K}^{o} \) . Then, \( J \) is a divisible group (for all integers \( n \), the map \( x \rightarrow {nx} \) is onto). Denote by \( g \) the genus of \( K \) . If \( l \) is a prime different from t... | In the case where \( M = \mathbb{C} \), the complex numbers, it is a classical theorem (Abel-Jacobi) that \( J \) is isomorphic to the direct sum of \( g \) copies of \( \mathbb{C} \) modulo a lattice \( \Lambda \) of maximal rank, i.e., \( J \cong {\mathbb{C}}^{g}/\Lambda \) . \( \Lambda \) must be a free \( \mathbb{Z... | Yes |
Proposition 11.13. Let \( J\left\lbrack N\right\rbrack \subset J \) be the subgroup of \( J \) consisting of elements whose order divides \( N \) . Then, \( \left\lbrack {\mathbb{F}\left( {J\left\lbrack N\right\rbrack }\right) : \mathbb{F}}\right\rbrack \) is equal to the order of the matrix \( {\rho }_{N}\left( {\phi ... | Proof. Since \( {\rho }_{N} \) is a monomorphism, the order of \( {\rho }_{N}\left( {\phi }_{q}\right) \) is the smallest power of \( {\phi }_{q} \) which is the identity on \( \mathbb{F}\left( {J\left\lbrack N\right\rbrack }\right) \) . By the Galois theory of finite fields, this number is the dimension \( \left\lbrac... | Yes |
Proposition 11.14. For all but finitely many primes \( l \), the dimension \( \left\lbrack {\mathbb{F}\left( {J\left\lbrack l\right\rbrack }\right) : \mathbb{F}}\right\rbrack \) is prime to \( l \) . | Proof. We will need to use one of those basic facts about \( J \) whose proof involves a substantial amount of algebraic geometry. It concerns the characteristic polynomial of the matrix \( {\rho }_{l}\left( {\phi }_{q}\right) \) . We will assume \( l \neq p \), the characteristic of \( \mathbb{F} \) . Let \( {L}_{K}\l... | Yes |
Proposition 11.16. We have the following group isomorphism:\n\n\[ J\left( {\mathbb{F}}_{{l}^{\infty }}\right) \left( l\right) \cong {\bigoplus }_{1}^{{r}_{l}}{\mathbb{Q}}_{l}/{\mathbb{Z}}_{l} \]\n\nwhere \( {r}_{l} \) is the dimension over \( \mathbb{Z}/l\mathbb{Z} \) of \( J\left\lbrack l\right\rbrack \cap J\left( {\m... | Proof. (Sketch) We first show that \( J\left( {\mathbb{F}}_{{l}^{\infty }}\right) \left( l\right) \) is a divisible group. Consider the exact sequence of \( {G}_{\mathbb{F}} \) modules\n\n\[ \left( 0\right) \rightarrow J\left\lbrack l\right\rbrack \rightarrow J\left( l\right) \rightarrow J\left( l\right) \rightarrow \l... | No |
Lemma 11.17. Suppose \( l \) is an odd prime, and that \( A \) is an abelian group isomorphic to \( {\bigoplus }_{1}^{r}{\mathbb{Q}}_{l}/{\mathbb{Z}}_{l} \) . Let \( \phi : A \rightarrow A \) be an endomorphism. Define \( {A}_{0} = \{ x \in A \mid \phi \left( x\right) = x\} \) and \( {A}_{1} = : \left\{ {x \in A \mid {... | Proof. We assume \( l \) is odd and begin by showing there is an endomorphism \( \psi \) of \( A \) such that \( \phi = I + {l\psi } \) where \( I \) is the identity endomorphism.\n\nSince \( A \) is divisible, given \( x \in A \), there is a \( y \in A \) such that \( {ly} = \) \( x \) . Set \( \psi \left( x\right) = ... | Yes |
Corollary 1. In addition to the hypotheses of the lemma, assume that \( {A}_{0} \) is finite. Then,\n\n\[ \n{A}_{0} \cong {\bigoplus }_{i = 1}^{r}{l}^{-{\nu }_{\imath }}{\mathbb{Z}}_{l}/{\mathbb{Z}}_{l}\;\text{ and }\;{A}_{1} \cong {\bigoplus }_{i = 1}^{r}{l}^{-{\nu }_{\imath } - 1}{\mathbb{Z}}_{l}/{\mathbb{Z}}_{l}.\n\... | Proof. Since \( A\left\lbrack l\right\rbrack \subset {A}_{0} \) and \( {A}_{0} \) is finite, it follows that \( {A}_{0} \) is the direct sum of \( r \) cyclic groups each of \( l \) -power order. This gives the first assertion.\n\nRephrasing what has been shown so far, \( {A}_{0} \) has a set of generators \( \left\{ {... | Yes |
Corollary 2. With the notation of Corollary 1, define \( {A}_{n} = \{ x \in \) \( \left. {A \mid {\phi }^{{l}^{n}}\left( x\right) = x}\right\} \) . Then,\n\n\[ \n{A}_{n} \cong {\bigoplus }_{i = 1}^{r}{l}^{-{\nu }_{i} - n}{\mathbb{Z}}_{l}/{\mathbb{Z}}_{l} \n\] | Proof. The proof is by induction on \( n \) . Corollary 1 shows the result is true for \( n = 1 \) . Now assume it is true for \( n - 1 \) . Then apply Corollary 1 with \( {A}_{0} \) replaced by \( {A}_{n - 1} \) and \( \phi \) replaced by \( {\phi }^{{l}^{n - 1}} \) . The result follows. | No |
Theorem 11.18. Let \( K/\mathbb{F} \) be a function field of genus \( g \) over a finite field \( \mathbb{F} \) with \( q \) elements. Let \( J = C{l}^{o}\left( {K\overline{\mathbb{F}}}\right) \) and define \( {r}_{l} \) to be the dimension over \( \mathbb{Z}/l\mathbb{Z} \) of \( J\left\lbrack l\right\rbrack \cap J\lef... | Proof. Define \( {n}_{0} \) by the equation \( {\mathbb{F}}_{{l}^{\infty }} \cap \mathbb{F}\left( {J\left\lbrack l\right\rbrack }\right) = {\mathbb{F}}_{{l}^{{n}_{0}}} \) . It is not hard to see that every element in \( J\left\lbrack l\right\rbrack \cap J\left( {\mathbb{F}}_{{l}^{\infty }}\right) \) is rational over th... | Yes |
Theorem 11.19. Let \( l \) and \( k \) be primes with \( l \neq k \) . Assume \( k \neq p \) . Define \( {n}_{0} \) as follows\n\n\[ \n{n}_{0} = {\operatorname{ord}}_{l}\mathop{\prod }\limits_{{i = 1}}^{{2g}}\left( {{k}^{i} - 1}\right) \n\]\n\nThen, \( J\left( {\mathbb{F}}_{{l}^{\infty }}\right) \cap J\left( k\right) \... | Proof. Let \( P \in J\left( {\mathbb{F}}_{{l}^{\infty }}\right) \) have order \( {k}^{m} \) . Then, \( P \in J\left\lbrack {k}^{m}\right\rbrack \) and so \( P \) is rational over both \( {\mathbb{F}}_{{l}^{\infty }} \) and \( \mathbb{F}\left( {J\left\lbrack {k}^{m}\right\rbrack }\right) \), and is thus rational over th... | Yes |
Proposition 12.2. Let \( k \) be a field and \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) an additive polynomial. If the characteristic of \( k \) is zero then \( f\left( x\right) = {ax} \) for some \( a \in k \) . If the characteristic of \( k \) is \( p > 0 \), then there are elements \( {a}_{i} \in k \) w... | Proof. By definition, if \( f\left( x\right) \) is additive, \( f\left( {x + y}\right) = f\left( x\right) + f\left( y\right) \) . Take the formal partial derivative with respect to \( x \) . Then, \( {\partial }_{x}f\left( {x + y}\right) = {\partial }_{x}f\left( x\right) \) . Setting \( x = 0 \) we see that the formal ... | Yes |
Lemma 12.3. Let \( a \in A, a \neq 0 \) . Let \( M \) be an \( A \) -module and suppose for each \( b \mid a \) that the submodule \( M\left\lbrack b\right\rbrack = \{ m \in M \mid {bm} = 0\} \) has \( {q}^{r\deg \left( b\right) } \) elements. Then\n\n\[ M\left\lbrack a\right\rbrack \cong A/{aA} \oplus A/{aA} \oplus \c... | Proof. Consider the prime decomposition of \( a, a = \alpha {P}_{1}^{{e}_{1}}{P}_{2}^{{e}_{2}}\cdots {P}_{t}^{{e}_{t}} \), where \( \alpha \in {\mathbb{F}}^{ * } \) and the \( {P}_{i} \) run through the monic, irreducible divisors of \( a.M\left\lbrack a\right\rbrack \) is isomorphic to the direct sum of the submodules... | Yes |
Proposition 12.4. Let \( \rho \) be a Drinfeld module of rank \( r \), i.e., for each \( a \in A \) the degree in \( \tau \) of \( {\rho }_{a} \) is \( r\deg \left( a\right) \) . Then, for each \( a \in A, a \neq 0 \) we have\n\n\[ \n{\Lambda }_{\rho }\left\lbrack a\right\rbrack \cong A/{aA} \oplus A/{aA} \oplus \cdots... | Proof. We apply Lemma 12.3 with \( M = {\bar{k}}_{\rho } \) . We have to check that for each \( a \neq 0 \) in \( A \) that \( {\Lambda }_{\rho }\left\lbrack a\right\rbrack \) has \( {q}^{r\deg \left( a\right) } \) elements. From our previous work we see that \( {\rho }_{a}\left( x\right) \) has the form\n\n\[ \n{\rho ... | No |
Proposition 12.5. Define \( {K}_{\rho, a} \) to be the field \( k\left( {{\Lambda }_{\rho }\left\lbrack a\right\rbrack }\right) \) . Then \( {K}_{\rho, a}/k \) is a Galois extension and there is a monomorphism | \[ \operatorname{Gal}\left( {{K}_{\rho, a}/k}\right) \rightarrow G{L}_{r}\left( {A/{aA}}\right) . \] | No |
Let \( {\lambda }_{m} \in {\Lambda }_{m} \) be a generator and suppose \( a \in A \) is relatively prime to \( m \) . Then, \( {C}_{a}\left( {\lambda }_{m}\right) /{\lambda }_{m} \) is a unit in \( {\mathcal{O}}_{m} \) . If \( m \) is divisible by two or more primes, then \( {\lambda }_{m} \) is itself a unit. | Proof. From Equation 3 we see that \( {\lambda }_{m} \) is integral over \( A \) . From the same equation, replacing \( m \) by \( a, d \) by \( \deg \left( a\right) \), and substituting \( x = {\lambda }_{m} \), we see that \( {C}_{a}\left( {\lambda }_{m}\right) /{\lambda }_{m} \in {\mathcal{O}}_{m} \) . We must show ... | Yes |
Let \( P \in A \) be a monic irreducible polynomial and \( e \in \mathbb{Z}, e > 0 \). Then, \( {K}_{{P}^{e}} \) is unramified at every prime ideal \( {QA} \) with \( {QA} \neq {PA} \). The prime \( {PA} \) is totally ramified with ramification index \( \Phi \left( {P}^{e}\right) \). Consequently, \[ \left\lbrack {{K}_... | Proof. Let \( \lambda \) be a primitive generator of \( {\Lambda }_{{P}^{e}} \) and let \( g\left( x\right) \in k\left\lbrack x\right\rbrack \) be the monic irreducible polynomial it satisfies. Then \( g\left( x\right) \) must divide \( {C}_{{P}^{e}}\left( x\right) \). Write \( {C}_{{P}^{e}}\left( x\right) = f\left( x\... | Yes |
Theorem 12.8. \( {K}_{m} \) is the compositum of the fields \( {K}_{{P}_{i}^{{e}_{n}}} \) . The only ideals in \( A \) ramified in \( {\mathcal{O}}_{m} \) are \( {P}_{i}A \) with \( 1 \leq i \leq t \) . We have \( \left\lbrack {{K}_{m} : k}\right\rbrack = \Phi \left( m\right) \) . More precisely,\n\n\[ \operatorname{Ga... | Proof. Define \( {m}_{i} \) to be \( m \) divided by \( {P}_{i}^{{e}_{\imath }} \) . Let \( {\lambda }_{m} \) be a generator of \( {\Lambda }_{m} \) as an \( A \) -module. It is clear that \( {C}_{{m}_{\imath }}\left( {\lambda }_{m}\right) \) is a generator of \( {\Lambda }_{{P}_{\imath }^{{e}_{\imath }}} \) . Set \( {... | Yes |
Theorem 12.10. Let \( m \in A \) have positive degree and let \( P \in A \) be a monic, irreducible polynomial not dividing \( m \) . Then, the Artin automorphism of the prime ideal \( {PA} \) in the extension \( {K}_{m}/k \) is the automorphism \( {\sigma }_{P} \) which takes \( {\lambda }_{m} \) to \( {C}_{P}\left( {... | Proof. Since \( P \) does not divide \( m,{PA} \) is unramified in \( {K}_{m} \) . Let \( \mathfrak{P} \) be any prime ideal in \( {\mathcal{O}}_{m} \) lying above \( {PA} \) . Then, the Artin automorphism is characterized by\n\n\[ \left( {{PA},{K}_{m}/k}\right) \omega \equiv {\omega }^{\left| P\right| }\;\left( {\;\op... | Yes |
Proposition 12.11. Let \( {C}_{m}\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{d}\left\lbrack {m, i}\right\rbrack {x}^{{q}^{i}} \) where each coefficient \( \left\lbrack {m, i}\right\rbrack \in A \) and \( d = \deg \left( m\right) \) . Then the degree of \( \left\lbrack {m, i}\right\rbrack \) as a polynomial in \(... | Proof. If \( i > d \), we set \( \left\lbrack {m, i}\right\rbrack = 0 \) . Notice that \( \left\lbrack {m,0}\right\rbrack = m \) which has degree \( d = {q}^{0}\left( {d - 0}\right) \) . This shows the result is true for \( i = 0 \) . For the rest of the proof we assume \( i > 0 \) .\n\nWe first investigate the special... | Yes |
Proposition 12.13. Let \( {C}_{m}\left( x\right) \in k\left\lbrack x\right\rbrack \subset {k}_{\infty }\left\lbrack x\right\rbrack \) be the Carlitz polynomial corresponding to \( m \in \mathbb{F}\left\lbrack T\right\rbrack \) of degree \( d \) . Let \( {\bar{k}}_{\infty } \) be an algebraic closure of \( {k}_{\infty }... | Proof. Recall that \( {C}_{m}\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{d}\left\lbrack {m, i}\right\rbrack {x}^{{q}^{i}} \) . By Proposition 12.11, \( {\operatorname{ord}}_{\infty }\left\lbrack {m, i}\right\rbrack = - \deg \left\lbrack {m, i}\right\rbrack = - {q}^{i}\left( {d - i}\right) \) . To apply Theorem 1... | Yes |
Lemma 13.2. Suppose \( J \) and \( H \) are non-zero ideals of \( A \) which are relatively prime; i.e., \( J + H = A \) . Then \( M\left\lbrack {JH}\right\rbrack = M\left\lbrack J\right\rbrack \oplus M\left\lbrack H\right\rbrack \) . | Proof. By hypothesis, there exist \( a \in J \) and \( b \in H \) such that \( a + b = 1 \) . For \( m \in M\left\lbrack {JH}\right\rbrack \) we have \( m = {am} + {bm} \) . Since \( {am} \in M\left\lbrack H\right\rbrack \) and \( {bM} \in M\left\lbrack J\right\rbrack \) , we have shown \( M\left\lbrack {JH}\right\rbra... | Yes |
Lemma 13.3. Let \( M \) be a torsion \( A \) -module. Then\n\n\[ M = {\bigoplus }_{P}M\left( P\right) \]\n\nwhere the sum is over all maximal ideals of \( A \) . | Proof. Let \( m \in M \) . Since \( M \) is a torsion module, there is a non-zero \( a \in A \) such that \( {am} = 0 \) . Consider the prime decomposition of the principal ideal (a) and apply Corollary 1 to Lemma 13.2. This shows that \( M = \mathop{\sum }\limits_{P}M\left( P\right) \) .\n\nTo show the sum is direct, ... | Yes |
Lemma 13.4. If \( \left( 0\right) \rightarrow {M}_{1} \rightarrow {M}_{2} \rightarrow {M}_{3} \rightarrow \left( 0\right) \) is an exact sequence of torsion \( A \) -modules and \( P \subset A \) is a maximal ideal, then \( \left( 0\right) \rightarrow {M}_{1}\left( P\right) \rightarrow \) \( {M}_{2}\left( P\right) \rig... | Proof. This follows easily from Lemma 13.3. | No |
Lemma 13.5. Let \( P \) be a maximal ideal of \( A \) and select \( \pi \in P - {P}^{2} \) . Then\n\n\[ M\left\lbrack {P}^{e}\right\rbrack = M\left\lbrack {\pi }^{e}\right\rbrack \left( P\right) . \] | Proof. We have \( \left( {\pi }^{e}\right) = {P}^{e}J \) where \( P \) and \( J \) are relatively prime. By Lemma 13.2,\n\n\[ M\left\lbrack {\pi }^{e}\right\rbrack = M\left\lbrack {P}^{e}\right\rbrack \oplus M\left\lbrack J\right\rbrack \]\n\nTaking the \( P \) -primary component of both sides gives the result, since \... | Yes |
Lemma 13.6. Suppose \( M \) is a divisible \( A \) -module, \( P \subset A \) a maximal ideal, and \( e > 1 \) and integer. The following sequence is exact:\n\n\[ \left( 0\right) \rightarrow M\left\lbrack P\right\rbrack \rightarrow M\left\lbrack {P}^{e}\right\rbrack \rightarrow M\left\lbrack {P}^{e - 1}\right\rbrack \r... | Proof. Choose \( \pi \in P - {P}^{2} \) . Using the divisibility of \( M \), we see that the following sequence is exact:\n\n\[ \left( 0\right) \rightarrow M\left\lbrack \pi \right\rbrack \rightarrow M\left\lbrack {\pi }^{e}\right\rbrack \rightarrow M\left\lbrack {\pi }^{e - 1}\right\rbrack \rightarrow \left( 0\right) ... | Yes |
Proposition 13.7. Let \( \rho \) be an A-Drinfeld module defined over a field \( L \) . Define \( \mu \left( a\right) = - {\deg }_{\tau }{\rho }_{a} \) for all \( a \in A \) . Then there is a positive rational number \( r \) such that \( \mu \left( a\right) = r{\operatorname{ord}}_{\infty }\left( a\right) {d}_{\infty }... | Proof. If we define \( \mu \left( 0\right) = \infty \) we easily check that \( \mu \) gives a map from \( A \) to \( \mathbb{Z} \cup \infty \) such that \( \mu \left( a\right) = \infty \) if and only if \( a = 0,\mu \left( {ab}\right) = \mu \left( a\right) + \mu \left( b\right) \) , and \( \mu \left( {a + b}\right) \ge... | Yes |
For all \( a \in A \) we have \( \deg a = - {\operatorname{ord}}_{\infty }\left( a\right) {d}_{\infty } \) . Thus, the rank of a Drinfeld A-module \( \rho \), can also be defined to be the unique positive, rational number \( r \) such that \( {\deg }_{\tau }\left( {\rho }_{a}\right) = r\deg a \) for all \( a \in A \) . | Let \( {m}_{P} = {\operatorname{ord}}_{P}\left( a\right) \) and \( {aA} = \mathop{\prod }\limits_{P}{P}^{{m}_{P}} \) be the prime decomposition of the principal ideal \( {aA} \) . By the Chinese Remainder Theorem, we have\n\n\[ A/{aA} \cong {\bigoplus }_{P}A/{P}^{{m}_{P}}.\]\n\nFor any maximal ideal \( P \subset A \), ... | Yes |
Proposition 13.9. There is a positive rational number \( h \) with the property that for all \( a \in A \) we have \( \omega \left( a\right) = h{\operatorname{ord}}_{Q}\left( a\right) \deg Q \) . | Proof. The map \( \omega \) takes \( A \) to \( \mathbb{Z} \cup \infty \) and has the following properties: \( \omega \left( a\right) = \infty \) if and only if \( a = 0,\omega \left( {ab}\right) = \omega \left( a\right) + \omega \left( b\right) \), and \( \omega \left( {a + b}\right) \geq \) \( \min \left( {\omega \le... | Yes |
Proposition 13.10. If \( \rho \) and \( {\rho }^{\prime } \) are isogenous Drinfeld modules, then they have the same rank and height. | Proof. Let \( 0 \neq f \in {\operatorname{Hom}}_{L}\left( {\rho ,{\rho }^{\prime }}\right) \) and choose a non-constant element \( a \in A \) . Then, \( f{\rho }_{a} = {\rho }_{a}^{\prime }f \) . Taking the degree with respect to \( \tau \) of both sides shows that \( {\deg }_{\tau }{\rho }_{a} = {\deg }_{\tau }{\rho }... | Yes |
Lemma 13.11. The ring \( L < \tau > \) has a division algorithm on the right. More precisely, if \( f, g \in L < \tau > \) and \( g \neq 0 \), there exist \( s, r \in L < \tau > \) such that \( f = {sg} + r \) with \( r = 0 \) or \( {\deg }_{\tau }r < {\deg }_{\tau }g \) . | Proof. The proof is just about the same as in the case of a commutative polynomial ring. For details see Goss [4], Chapter 1. | No |
Proposition 13.12. Let \( M \) be an algebraically closed field containing \( L \) . Then,\n\n\[ \n{M}_{\rho }\left\lbrack I\right\rbrack = \left\{ {\lambda \in M \mid {\rho }_{I}\left( \lambda \right) = 0}\right\} .\n\] | Proof. Suppose \( \lambda \) is a root of \( {\rho }_{I}\left( x\right) \) . If \( b \in I \), there is an \( {f}_{b} \in L < \tau > \) such that \( {f}_{b}{\rho }_{I} = {\rho }_{b} \) . Thus, \( 0 = {f}_{b}\left( 0\right) = {f}_{b}\left( {{\rho }_{I}\left( \lambda \right) }\right) = {\rho }_{b}\left( \lambda \right) \... | Yes |
Proposition 13.13. Let \( \rho \in {\operatorname{Drin}}_{L}\left( A\right) \) and \( \left( 0\right) \neq I \subset A \) be an ideal. Then there is a uniquely determined Drinfeld module \( I * \rho \in {\operatorname{Drin}}_{L}\left( A\right) \) such that \( {\rho }_{I} \) is an isogeny from \( \rho \) to \( I * \rho ... | Proof. The left ideal \( J \subset L < \tau > \) generated by the set, \( \left\{ {{\rho }_{b} \mid b \in I}\right\} \), is mapped into itself under right multiplication by \( {\rho }_{a} \) for any \( a \in A \) . Thus, for all \( a \in A \), there is a \( {\rho }_{a}^{\prime } \in L < \tau > \) such that \( {\rho }_{... | No |
Proposition 13.14. If \( \rho \in {\operatorname{Drin}}_{L}\left( A\right) \) and \( 0 \neq b \in A \), then \( {\rho }_{\left( b\right) } = {c}^{-1}{\rho }_{b} \) , where \( c \) is the leading coefficient of \( {\rho }_{b} \) . Moreover, \( c{\left\lbrack \left( b\right) * \rho \right\rbrack }_{a} = {\rho }_{a}c \) f... | Proof. We have already proven the first assertion. To prove the second, note that for all \( a \in A \) we have \( {\rho }_{\left( b\right) }{\rho }_{a} = {\left\lbrack \left( b\right) * \rho \right\rbrack }_{a}{\rho }_{\left( b\right) } \) . Thus, \( {c}^{-1}{\rho }_{b}{\rho }_{a} = \) \( {\left\lbrack \left( b\right)... | Yes |
Lemma 13.16. Let \( \rho \in {\operatorname{Drin}}_{L}\left( A\right) \) and let \( I \subset A \) be an ideal prime to the \( A \) -characteristic \( Q \) of \( L \) . Then \( {\rho }_{I}\left( x\right) \) is a separable polynomial. | Proof. Let \( a \in I \) with \( a \) not in the \( A \) -characteristic of \( L \) . We have \( \left( a\right) = {IJ} \) for some ideal \( J \) . By Proposition 13.15, \( {\rho }_{\left( a\right) } = {\left( I * \rho \right) }_{J}{\rho }_{I} \) . Since \( a \) is not in \( Q \) , we know \( {\omega }_{\tau }\left( {\... | Yes |
Proposition 13.17. Let \( \rho \in {\operatorname{Drin}}_{L}\left( A\right) \) be a Drinfeld module of rank \( r \) . Let \( I \subset A \) be a non-zero ideal. Then, \( {\deg }_{\tau }{\rho }_{I} = r\deg I \) . | Proof. To begin with, let us assume that \( I \) is prime to the \( A \) -characteristic of \( L \) . Let \( M \) be an algebraically closed field containing \( L \) and consider \( {M}_{\rho }\left\lbrack I\right\rbrack \) . By Lemma 13.16 and Proposition 13.12, we see that\n\n\[ \n\# {M}_{\rho }\left\lbrack I\right\r... | Yes |
Proposition 13.18. Let \( \rho \in {\operatorname{Drin}}_{L}\left( A\right) \) . Assume that the \( A \) -characteristic of \( L, Q \), is not zero. Let \( J \subset A \) be a non-zero ideal. Then \( {\omega }_{\tau }\left( {\rho }_{J}\right) = \) \( h\deg Q{\operatorname{ord}}_{Q}J \), where \( h \) is the height of \... | Proof. If \( J \) is prime to \( Q \), the \( {\omega }_{\tau }\left( {\rho }_{J}\right) = 0 \) by Proposition 13.16. We also have \( {\operatorname{ord}}_{Q}J = 0 \), so the proposition is proven in this case.\n\nNow, assume \( Q \) divides \( J \) . Then as above we can write \( {aJ} = {bI} \), where \( I \) is an id... | Yes |
Proposition 13.20. Let \( \\Gamma \) be a lattice in \( \\mathbf{C} \) . Then for all \( u, v \\in \\mathbf{C} \) and \( \\alpha \\in \\mathbb{F} \) we have\n\n\[ \n{e}_{\\Gamma }\\left( {u + v}\\right) = {e}_{\\Gamma }\\left( v\\right) + {e}_{\\Gamma }\\left( v\\right) \\;\\text{ and }\\;{e}_{\\Gamma }\\left( {\\alpha... | Proof. For each positive integer \( M \), define \( {\\Gamma }_{M} = \\left\\{ {\\gamma \\in \\Gamma \\left| \\right| \\gamma {\\left. \\right| }_{\\infty } \\leq M}\\right\\} \) . This is readily checked to be a finite \( \\mathbb{F} \) vector space (as we’ve seen, \( {\\left| \\gamma \\right| }_{\\infty } \\rightarro... | Yes |
Lemma 13.21. Let \( V \subset \mathbf{C} \) be a finite, \( \mathbb{F} \) vector space, and set\n\n\[ \n{f}_{V}\left( x\right) = \mathop{\prod }\limits_{{\nu \in V}}\left( {x - \nu }\right) \n\]\n\nThen, \( {P}_{V}\left( x\right) \) is an \( \mathbb{F} \) -linear, additive polynomial in \( x \) . | Proof. We prove this by induction on the dimension of \( V \) . If \( \dim V = 0 \) , then \( V = \left( 0\right) \) and \( {f}_{V}\left( x\right) = x \), so the result is true in this case.\n\nNow assume that the result is true for vector spaces of dimension less than \( n \) and that \( \dim V = n \) . Write \( V = W... | Yes |
Proposition 13.22. The polynomial \( P\left( {x;{\Gamma }^{\prime }/\Gamma }\right) \) is \( \mathbb{F} \) -linear of degree \( \# \left( {{\Gamma }^{\prime }/\Gamma }\right) \) . Its initial term is \( x \) . Moreover, \[ {e}_{{\Gamma }^{\prime }}\left( u\right) = P\left( {{e}_{\Gamma }\left( u\right) ;{\Gamma }^{\pri... | Proof. The first assertion follows from Lemma 13.21, and the second assertion is clear from the definition. To prove the identity, notice that \( P\left( {{e}_{\Gamma }\left( u\right) ;{\Gamma }^{\prime }/\Gamma }\right) \) is zero if and only if \( {e}_{\Gamma }\left( u\right) = {e}_{\Gamma }\left( \mu \right) \) for ... | Yes |
Theorem 13.23. Let \( \Gamma \subset \mathbf{C} \) be a lattice of rank \( r \) . For each \( a \in A, a \neq 0 \) , define \( {\rho }_{a}^{\Gamma } \in \mathbf{C} < \tau > \) by the formula \[ {\rho }_{a}^{\Gamma }\left( x\right) = {aP}\left( {x,{a}^{-1}\Gamma /\Gamma }\right) . \] Then, if we send zero to zero and ma... | Proof. In what follows we regard \( A \) is a subset of \( \mathbf{C} \) via the inclusions \( A \rightarrow k \rightarrow {k}_{\infty } \rightarrow \mathbf{C} \) . Thus the structure map \( \delta : A \rightarrow \mathbf{C} \) is just the inclusion, so the first thing we must show is that \( D\left( {\rho }_{a}^{\Gamm... | No |
Theorem 13.26. The set \( {\operatorname{Drin}}_{A}^{o}\left( {\mathbf{C},1}\right) \) is finite with cardinality equal to the order of the class group of \( A \) . | Proof. By Theorems 13.24 and 13.25, it is equivalent to consider the set of rank 1 \( A \) -lattices in \( \mathbf{C} \) up to isomorphism. Note that two lattice \( \Gamma \) and \( {\Gamma }^{\prime } \) are isomorphic if or only if there is a \( c \in {\mathbf{C}}^{ * } \) such that \( \Gamma = c{\Gamma }^{\prime } \... | Yes |
Proposition 14.1. The following sequences are exact:\n\n(a)\n\n\[ \left( 0\right) \rightarrow {F}^{ * } \rightarrow E\left( S\right) \rightarrow \mathcal{P}\left( S\right) \rightarrow \left( 0\right) ,\] | Proof. The map from \( E\left( S\right) \) to \( \mathcal{P}\left( S\right) \) is given by taking an \( S \) -unit to its divisor. This map is onto by the definition of \( \mathcal{P}\left( S\right) \) . If an \( S \) -unit \( e \) goes to the zero divisor, then \( {\operatorname{ord}}_{P}\left( e\right) = 0 \) for all... | Yes |
Corollary 1. The group \( E\left( S\right) /{F}^{ * } \) is a finitely generated free group of rank at most \( \left| S\right| - 1 \), where \( \left| S\right| \) is the number of elements in \( S \) . | Proof. By the exact sequence \( a \) ) we have \( E\left( S\right) /{F}^{ * } \cong \mathcal{P}\left( S\right) \), which is a subgroup of the free group \( \mathcal{D}{\left( S\right) }^{o} \) on \( \left| S\right| - 1 \) generators. Thus, \( \mathcal{P}\left( S\right) \) is free on at most \( \left| S\right| - 1 \) ge... | Yes |
Corollary 2. \( C{l}_{S} \) is a finite group if \( C{l}^{o} \) is a finite group. Also, \( C{l}_{S} \) is a torsion group if \( C{l}^{o} \) is a torsion group. | Proof. Both statements are immediate consequences of exact sequence \( b \) ). | No |
Proposition 14.2. Let \( K/\mathbb{F} \) be a function field over a finite field \( \mathbb{F} \) . Then, for all finite subsets \( S \subset {\mathcal{S}}_{K} \) we have that \( C{l}_{S} \) is a finite group and \( E\left( S\right) /{\mathbb{F}}^{ * } \) is a free group on \( \left| S\right| - 1 \) generators. | Proof. By Lemma 5.6, \( C{l}^{o} \) is a finite group. By Corollary 2 to Proposition 14.1 we see that \( C{l}_{S} \) is a finite group.\n\nBy exact sequence \( b \) ) of Proposition 14.1 we see that \( \mathcal{D}{\left( S\right) }^{o}/\mathcal{P}\left( S\right) \) is finite. This shows that \( \mathcal{P}\left( S\righ... | Yes |
Theorem 14.4. Let \( K/\mathbb{F} \) be a function field over a finite field \( \mathbb{F} \) with \( q \) elements. Let \( S \subset {\mathcal{S}}_{K} \) be a finite set of primes with \( s \) elements. Then\n\n\[ \n{\zeta }_{S}\left( w\right) = - \frac{{h}_{S}{R}_{S}}{q - 1}{w}^{s - 1} + O\left( {w}^{s}\right) .\n\] | Proof. Referring to Equation 2 we see that everything has already been proved except that in that equation the coefficient of \( {w}^{s - 1} \) is given as\n\n\[ \n- {\left( q - 1\right) }^{-1}{h}_{K}\left( {\mathop{\prod }\limits_{{P \in S}}\deg P}\right) {\left( \ln q\right) }^{s - 1}.\n\]\n\n(4)\n\nOur task is to sh... | Yes |
Theorem 14.4. Let \( K/\mathbb{F} \) be a function field over a finite field \( \mathbb{F} \) with \( q \) elements. Let \( S \subset {\mathcal{S}}_{K} \) be a finite set of primes with \( s \) elements. Then \[ {\zeta }_{S}\left( w\right) = - \frac{{h}_{S}{R}_{S}}{q - 1}{w}^{s - 1} + O\left( {w}^{s}\right) . \] | Proof. Referring to Equation 2 we see that everything has already been proved except that in that equation the coefficient of \( {w}^{s - 1} \) is given as \[ - {\left( q - 1\right) }^{-1}{h}_{K}\left( {\mathop{\prod }\limits_{{P \in S}}\deg P}\right) {\left( \ln q\right) }^{s - 1}. \] (4) Our task is to show that this... | Yes |
Theorem 14.4. Let \( K/\mathbb{F} \) be a function field over a finite field \( \mathbb{F} \) with \( q \) elements. Let \( S \subset {\mathcal{S}}_{K} \) be a finite set of primes with \( s \) elements. Then\n\n\[ \n{\zeta }_{S}\left( w\right) = - \frac{{h}_{S}{R}_{S}}{q - 1}{w}^{s - 1} + O\left( {w}^{s}\right) .\n\] | Proof. Referring to Equation 2 we see that everything has already been proved except that in that equation the coefficient of \( {w}^{s - 1} \) is given as\n\n\[ \n- {\left( q - 1\right) }^{-1}{h}_{K}\left( {\mathop{\prod }\limits_{{P \in S}}\deg P}\right) {\left( \ln q\right) }^{s - 1}.\n\]\n\n(4)\n\nOur task is to sh... | Yes |
Proposition 14.6. Let \( K = k\left( \sqrt{f\left( T\right) }\right) \), where \( f\left( T\right) \in A = \mathbb{F}\left\lbrack T\right\rbrack \) is square-free. Let \( d = \deg f\left( T\right) \) and \( {a}_{d} \) the leading coefficient of \( f\left( T\right) \) . If \( d \) is odd, then \( \infty \) is ramified i... | Proof. Suppose \( d \) is odd. Since \( U \) is a uniformizing parameter at \( \infty \) and \( {a}_{d} \neq 0 \) is the constant term of \( {f}^{ * }\left( U\right) \), we see that \( {f}^{ * }\left( U\right) \) is a unit at infinity. Suppose \( {P}_{\infty } \) is a prime of \( K \) lying above \( \infty \) . Then, s... | Yes |
Proposition 14.7. With the above notation, we have \( {h}_{B} = {h}_{K} \) if \( \infty \) is ramified, \( {h}_{B} = 2{h}_{K} \) if \( \infty \) is inert, and \( {h}_{B}{\log }_{q}{\left| e\right| }_{{P}_{\infty }} = {h}_{K} \) if \( \infty \) splits. In the latter case, e represents a fundamental unit in \( B \), and ... | Proof. In the first two cases the set of primes above \( \infty \) consists of one element \( {P}_{\infty } \) . Thus, \( s = 1 \) and the rank of \( \mathcal{D}{\left( S\right) }^{o} \) is zero. Also, in the first case the degree of \( {P}_{\infty } \) is 1 and in the second case it is 2 . Thus, the first two assertio... | Yes |
Proposition 14.8. The groups \( {\mathcal{O}}_{m}^{ * }/{\mathbb{F}}^{ * } \) and \( {{\mathcal{O}}_{m}^{ + }}^{ * }/{\mathbb{F}}^{ * } \) are free of rank \( \frac{\Phi \left( m\right) }{q - 1} - \) | Proof. With the information already provided, the proof is a straightforward application of Propositions 14.1, 14.2, and Lemma 14.3. | No |
Proposition 14.9. With the above notations, we have\n\n\[ \n{\zeta }_{K}\left( w\right) = {\zeta }_{k}\left( w\right) \mathop{\prod }\limits_{{\chi \neq {\chi }_{o}}}L\left( {w,\chi }\right) .\n\] | Proof. By looking at the product decompositions on both sides we see that it is sufficient to prove the following \ | No |
Corollary 1. Suppose \( K/k \) is abelian and geometric, i.e. that there is no constant field extension. Then\n\n\[ \n{h}_{K} = {h}_{k}\mathop{\prod }\limits_{{\chi \neq {\chi }_{o}}}L\left( {0,\chi }\right) .\n\] | Proof. By Theorem 5.9, we have\n\n\[ \n{\zeta }_{K}\left( w\right) = \frac{{L}_{K}\left( {q}^{-w}\right) }{\left( {1 - {q}^{-w}}\right) \left( {1 - {q}^{1 - w}}\right) },\n\]\n\nwhere \( {L}_{K}\left( u\right) \in \mathbb{Z}\left\lbrack u\right\rbrack \) is such that \( {L}_{K}\left( 1\right) = {h}_{K} \) .\n\nBy the a... | Yes |
For \( \chi \neq {\chi }_{o} \) we have \( L\left( {0,\chi }\right) \neq 0 \) . | Proof. This follows immediately from Corollary 1. | No |
Proposition 14.10. \( {L}_{A}\left( {w,\chi }\right) \) is a polynomial in \( {q}^{-w} \) of degree \( d\left( \chi \right) \) where\n\n\[ d\left( \chi \right) = {2g} - 2 + \deg \mathcal{F}\left( \chi \right) + \mathop{\sum }\limits_{{P \in S\left( \chi \right) }}\deg P. \] | Proof. From the definition of \( {L}_{A}\left( {w,\chi }\right) \) we have\n\n\[ {L}_{A}\left( {w,\chi }\right) = \mathop{\prod }\limits_{{P \in S}}\left( {1 - \chi \left( P\right) N{P}^{-w}}\right) L\left( {w,\chi }\right) . \]\n\nBy a famous result of A. Weil [1], we know that \( L\left( {w,\chi }\right) \) is a poly... | Yes |
Proposition 14.11. We have\n\n\[ \n{\zeta }_{B}\left( w\right) = {\zeta }_{A}\left( w\right) \mathop{\prod }\limits_{{\chi \neq {\chi }_{o}}}{L}_{A}\left( {w,\chi }\right) .\n\] | Proof. This assertion follows immediately from the definitions and the method of proof of Theorem 14.9. The method there uses the semi-local identity given in Equation 6. We simply use that identity for all primes not in \( S \), take the inverse of both sides, and then multiply over all primes not in \( S \) . | No |
Proposition 14.12. Suppose \( \chi \neq {\chi }_{o} \) and let \( m\left( \chi \right) \) denote the order of vanishing of \( {L}_{A}\left( {w,\chi }\right) \) at \( w = 0 \) . Then,\n\n\[ m\left( \chi \right) = \# \{ P \in S \mid \chi \left( {Z\left( P\right) }\right) = 1\} . \] | Proof. From the definition,\n\n\[ {L}_{A}\left( {w,\chi }\right) = \mathop{\prod }\limits_{{P \in S}}\left( {1 - \chi \left( P\right) N{P}^{-w}}\right) L\left( {w,\chi }\right) . \]\n\nSince \( L\left( {0,\chi }\right) \neq 0 \) by Corollary 2 to Proposition 14.9, we see that \( m\left( \chi \right) \) is just the numb... | Yes |
Theorem 14.13. (The Analytic Class Number Formula) Let \( K/k \) be a geometric, abelian extension of global function fields. Let \( S \) be a finite set of primes of \( k, A \) the ring of \( S \) -integers and \( B \) the integral closure of \( A \) in \( K \) . Set \( {R}_{S}^{\left( q\right) } = {R}_{A}^{\left( q\r... | Proof. By Proposition 14.12 we have\n\n\[ \n{L}_{A}\left( {w,\chi }\right) = {c}_{\chi }{w}^{m\left( \chi \right) } + O\left( {w}^{m\left( \chi \right) + 1}\right) , \n\]\n\n(8)\n\nwhere \( {c}_{\chi } \) is a non-zero constant. Combining this with Equation 7 of Proposition 14.11 and the assertion of Theorem 14.4 yield... | Yes |
Proposition 15.1. The group \( \widehat{G} = \operatorname{Hom}\left( {G,{R}^{ * }}\right) \) is isomorphic to \( G \) . | Proof. (Sketch) The proof is very simple in the case that \( G \) is a finite cyclic group. The general case is handled by use of the theorem that a finite abelian group is isomorphic to a direct sum of cyclic groups. See Lang [4] for details. | No |
Lemma 15.2. Let \( G \) be a finite abelian group and \( \sigma \in G,\sigma \neq e \), the identity element of \( G \) . Then there is a \( \chi \in \widehat{G} \) such that \( \chi \left( \sigma \right) \neq 1 \) . | Proof. (Sketch) Suppose \( \chi \left( \sigma \right) = 1 \) for all \( \chi \in \widehat{G} \) . There is a natural homomorphism from \( \widehat{G/\langle \sigma \rangle } \rightarrow \widehat{G} \) which, under our assumptions, would be onto. This contradicts the corollary to Proposition 15.1. | No |
Proposition 15.3. (The Orthogonality Relations) Let \( G \) be a finite abelian group of order \( n \). If \( \sigma ,\tau \in G \), then\n\n(a)\n\n\[ \n\frac{1}{n}\mathop{\sum }\limits_{{\chi \in \widehat{G}}}\chi \left( {\sigma }^{-1}\right) \chi \left( \tau \right) = \delta \left( {\sigma ,\tau }\right) \n\] \n\nwhe... | Proof. To prove the first relation, let \( \gamma \in G \) and set \( T\left( \gamma \right) = \mathop{\sum }\limits_{{\chi \in \widehat{G}}}\chi \left( \gamma \right) \). If \( \gamma = e \), the identity of \( G \), then clearly, \( T\left( e\right) = n \). If \( \gamma \neq e \) there is a \( \psi \in \widehat{G} \)... | Yes |
Proposition 15.5. Let \( G \) be a finite abelian group of order \( n, R \) an integral domain whose units, \( {R}^{ * } \), contains an element of order \( n \) . Assume also that \( n \) is a unit in \( R \) . Let \( V \) be an \( R\left\lbrack G\right\rbrack \) module. Then \( V \) is the direct sum of its isotypic ... | Proof. To begin with, we claim that \( V\left( \chi \right) = \varepsilon \left( \chi \right) V \) . If \( v \in V \), consider \( \varepsilon \left( \chi \right) v \) . By the first part of the above lemma, we see that \( {\sigma \varepsilon }\left( \chi \right) v = \chi \left( \sigma \right) \varepsilon \left( \chi \... | Yes |
\[ \text{1)}g\left( P\right) \in \mathbb{Q}\left( {{\zeta }_{m},{\zeta }_{p}}\right) \text{.} \] | The proof of part 1 is immediate from the definition. | No |
Theorem 15.7. (L. Stickelberger):\n\n\\[ \n\\left( {\\Phi \\left( P\\right) }\\right) = \\mathop{\\prod }\\limits_{\\substack{{t = 1} \\ {\\left( {t, m}\\right) = 1} }}^{{m - 1}}{\\left( {\\sigma }_{t}^{-1}P\\right) }^{t} = \\left( {\\mathop{\\sum }\\limits_{\\substack{{t = 1} \\ {\\left( {t, m}\\right) = 1} }}^{{m - 1... | For the proof of Stickelberger's theorem and some of the many important applications, see Ireland and Rosen [1], Lang [6], and/or Washington [1]. | No |
Proposition 15.9. For all \( \chi \in \widehat{G} \) we have \( \chi \left( {\theta \left( w\right) }\right) = {L}_{S}\left( {w,\bar{\chi }}\right) \) . | Proof. This is an immediate consequence of the definition of \( \theta \left( w\right) \) and Lemma 15.4, part 4. | No |
Proposition 15.10. With the above definitions and notations we have\n\n\[ \n{L}_{S}\left( {w,\chi }\right) = \mathop{\sum }\limits_{{\sigma \in G}}\chi \left( \sigma \right) {\zeta }_{S}\left( {w,\sigma }\right) \n\]\n\n(1)\n\n\[ \n\text{and} \n\]\n\n\[ \n{\zeta }_{S}\left( {w,\sigma }\right) = \frac{1}{n}\mathop{\sum ... | Proof. From the definition of \( {L}_{S}\left( {w,\chi }\right) \), we find (summing over effective divisors or over integral ideals, prime to \( S \) )\n\n\[ \n{L}_{S}\left( {w,\chi }\right) = \mathop{\sum }\limits_{{\left( {D, S}\right) = 1}}\frac{\chi \left( \left( {D, K/k}\right) \right) }{N{D}^{w}} = \mathop{\sum ... | Yes |
Proposition 15.11.\n\n\[ \theta \left( w\right) = {\theta }_{K/k, S}\left( w\right) = \mathop{\sum }\limits_{{\sigma \in G}}{\zeta }_{S}\left( {w,\sigma }\right) {\sigma }^{-1}. \] | Proof. Define \( \widetilde{\theta }\left( w\right) = \mathop{\sum }\limits_{{\sigma \in G}}{\zeta }_{S}\left( {w,\sigma }\right) {\sigma }^{-1} \in R\left\lbrack G\right\rbrack \) . By Equation 1 of the previous proposition, we find that \( \chi \left( {\widetilde{\theta }\left( w\right) }\right) = {L}_{S}\left( {w,\b... | Yes |
Theorem 15.13. Let \( \theta \left( w\right) = \widetilde{\theta }\left( u\right) = {\widetilde{\theta }}_{K/k, S}\left( u\right) \) be the L-function evaluator. Then, \( \left( {1 - {qu}}\right) \widetilde{\theta }\left( u\right) \) is an element of \( \mathbb{Z}\left\lbrack u\right\rbrack \left\lbrack G\right\rbrack ... | The only point which perhaps needs some explanation is the last assertion. Recall that \( u = {q}^{-w} \) . It follows that \( \theta = \theta \left( 0\right) = \widetilde{\theta }\left( 1\right) \) . | Yes |
Proposition 15.16. The element \( \eta \) annihilates \( C{l}_{{K}_{m}}^{o} \) and the element \( {\eta }^{ + } \) annihilates \( C{l}_{{K}_{m}^{ + }}^{o} \) . | We will prove this later as a corollary to the proof of Theorem 15.14. | No |
Proposition 15.17. Let \( {\mathfrak{P}}_{\infty } \) be a prime of \( {K}_{m} \) lying over \( \infty \) in \( k \) . There exists a primitive \( m \) -torsion point \( \lambda \in {\Lambda }_{m} \) such that\n\n\[ \left( \lambda \right) = \left( {\left( {q - 1}\right) {\eta }^{ + } - \eta }\right) {\mathfrak{P}}_{\in... | Proof. Let \( {k}_{\infty } \) be the completion of \( k \) at \( \infty \) and let \( {\bar{k}}_{\infty } \) be its algebraic closure. Let \( {\operatorname{ord}}_{\infty } \) denote the normalized additive valuation of \( {k}_{\infty } \) extended to \( {\bar{k}}_{\infty } \) in the usual way. Finally, let \( \iota :... | Yes |
Proposition 15.19. If a polynomial in \( \phi, f\left( \phi \right) \in \mathbb{Z}\left\lbrack G\right\rbrack \left\lbrack \phi \right\rbrack \), vanishes on \( {V}_{l} \) , then it vanishes on \( J \) . | Proposition 15.19 is a consequence of a far more general result about geometric endomorphisms of abelian varieties. The point is that any such polynomial \( f\left( \phi \right) \) can be thought of as an element of \( {\operatorname{End}}_{\mathbb{F}}\left( J\right) \), regarding \( J \) as an abelian variety over \( ... | No |
Theorem 16.5. Let \( {h}_{p}^{ - } \) be the relative class number of \( \mathbb{Q}\left( {\zeta }_{p}\right) \), where \( p \) is an odd prime number. Then\n\n\[ {h}_{p}^{ - } = \pm \frac{2p}{{2}^{\frac{p - 1}{2}}}\det \left\lbrack {\left\langle \frac{ab}{p}\right\rangle - \left\langle \frac{-{ab}}{p}\right\rangle }\r... | Proof. From Theorem 16.2 and the expression we have derived for the generalized Bernoulli number \( {B}_{1,\chi } \), we find\n\n\[ {h}_{p}^{ - } = \pm \frac{2p}{{2}^{\frac{p - 1}{2}}}\mathop{\prod }\limits_{{\chi \text{ odd }}}\mathop{\sum }\limits_{{a \in {G}_{p}^{ + }}}\chi \left( a\right) \left( {\left\langle \frac... | Yes |
Lemma 16.6. If \( \chi \neq {\chi }_{o} \) is an even character, we have \( {\widetilde{L}}_{A}\left( {1,\chi }\right) = 0 \) . | Proof. If \( \chi \) is even, \( \alpha \in {\mathbb{F}}^{ * } \), and \( a \in A \) is monic, then \( \chi \left( {\alpha a}\right) = \chi \left( a\right) \) . Thus, \[ {\widetilde{L}}_{A}\left( {1,\chi }\right) = \mathop{\sum }\limits_{\substack{{a\text{ monic }} \\ {\deg a < {M}_{\chi }} }}\chi \left( a\right) = {\l... | Yes |
Proposition 16.7. We have\n\n(a)\n\n\[ \n{\zeta }_{{\mathcal{O}}_{m}}\left( w\right) = {\zeta }_{A}\left( w\right) \mathop{\prod }\limits_{{\chi \neq {\chi }_{o}}}{L}_{A}\left( {w,\chi }\right) .\n\]\n\n\( \left( b\right) \)\n\n\[ \n{\zeta }_{{\mathcal{O}}_{m}^{ + }}\left( w\right) = \mathop{\prod }\limits_{\substack{{... | Proof. Both formulas are special cases of Proposition 14.11. To justify the second formula, note that even characters are the characters of \( {\left( A/m\right) }^{ * }/{\mathbb{F}}^{ * } \cong \) \( {G}_{m}/J \cong {G}_{m}^{ + } = \operatorname{Gal}\left( {{K}_{m}^{ + }/k}\right) . \)\n\nIt is possible to give a proo... | Yes |
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