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Theorem 2.2. (The prime number theorem for polynomials) Let \( {a}_{n} \) denote the number of monic irreducible polynomials in \( A = \mathbb{F}\left\lbrack T\right\rbrack \) of degree \( n \) . Then,\n\n\[ \n{a}_{n} = \frac{{q}^{n}}{n} + O\left( \frac{{q}^{\frac{n}{2}}}{n}\right)\n\] | Note that if we set \( x = {q}^{n} \) the right-hand side of this equation is \( x/{\log }_{q}\left( x\right) + O\left( {\sqrt{x}/{\log }_{q}\left( x\right) }\right) \) which looks like the conjectured precise form of the classical prime number theorem. This is still not proven. It depends on the truth of the Riemann h... | No |
Proposition 2.4. Let the prime decomposition of \( f \) be given as above. Then,\n\n\[ \Phi \left( f\right) = \left| f\right| \mathop{\prod }\limits_{{P \mid f}}\left( {1 - {\left| P\right| }^{-1}}\right) \]\n\n\[ d\left( f\right) = \left( {{e}_{1} + 1}\right) \left( {{e}_{2} + 1}\right) \ldots \left( {{e}_{t} + 1}\rig... | Proof. The formula for \( \Phi \left( n\right) \) has already been given in Proposition 1.7.\n\nIf \( P \) is a monic irreducible, the only monic divisors of \( {P}^{e} \) are \( 1, P \) , \( {P}^{2},\ldots ,{P}^{e} \) so \( d\left( {P}^{e}\right) = e + 1 \) and the second formula follows.\n\nBy the above paragraph, \(... | Yes |
Proposition 2.5. \( {D}_{d}\left( s\right) = {\zeta }_{A}{\left( s\right) }^{2} = {\left( 1 - qu\right) }^{-2} \) . Consequently, \( D\left( n\right) = \) \( \left( {n + 1}\right) {q}^{n} \) . | Proof.\n\n\[ \n{\zeta }_{A}{\left( s\right) }^{2} = \left( {\mathop{\sum }\limits_{h}\frac{1}{{\left| h\right| }^{s}}}\right) \left( {\mathop{\sum }\limits_{g}\frac{1}{{\left| g\right| }^{s}}}\right) = \n\]\n\n\[ \n\mathop{\sum }\limits_{f}\left( {\mathop{\sum }\limits_{\substack{{h, g} \\ {{hg} = f} }}1}\right) \frac{... | Yes |
Proposition 2.6.\n\n\[ \n{D}_{\lambda }\left( s\right) {D}_{\rho }\left( s\right) = {D}_{\lambda * \rho }\left( s\right) .\n\] | Proof. The calculation is just like that of Proposition 2.5.\n\n\[ \n{D}_{\lambda }\left( s\right) {D}_{\rho }\left( s\right) = \left( {\mathop{\sum }\limits_{h}\frac{\lambda \left( h\right) }{{\left| h\right| }^{s}}}\right) \left( {\mathop{\sum }\limits_{g}\frac{\rho \left( g\right) }{{\left| g\right| }^{s}}}\right) =... | Yes |
Proposition 2.7.\n\n\[ \mathop{\sum }\limits_{\substack{{\deg f = n} \\ {f\text{ monic }} }}\Phi \left( f\right) = {q}^{2n}\left( {1 - {q}^{-1}}\right) . \] | Proof. Let \( A\left( n\right) \) be the left-hand side of the above equation. Then, with the usual transformation \( u = {q}^{-s} \), Equation 6 becomes\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }A\left( n\right) {u}^{n} = \frac{1 - {qu}}{1 - {q}^{2}u}. \]\n\nNow, expand \( {\left( 1 - {q}^{2}u\right) }^{-1} \) in... | Yes |
Proposition 2.8.\n\n\[ \mathop{\sum }\limits_{\substack{{\deg \left( f\right) = n} \\ {f\text{ monic }} }}\sigma \left( f\right) = {q}^{2n} \cdot \frac{1 - {q}^{-n - 1}}{1 - {q}^{-1}}. \] | Proof. Define \( S\left( n\right) \) to be the sum on the left hand side of the above equation. Then, making the substitution \( u = {q}^{-s} \) in Equation 7 we find\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }S\left( n\right) {u}^{n} = {\left( 1 - qu\right) }^{-1}{\left( 1 - {q}^{2}u\right) }^{-1}. \]\n\nExpanding... | Yes |
Proposition 3.1. The d-th power residue symbol has the following properties:\n\n1) \( {\left( \frac{a}{P}\right) }_{d} = {\left( \frac{b}{P}\right) }_{d} \) if \( a \equiv b\left( {\;\operatorname{mod}\;P}\right) \).\n\n2) \( {\left( \frac{ab}{P}\right) }_{d} = {\left( \frac{a}{P}\right) }_{d}{\left( \frac{b}{P}\right)... | Proof. The first assertion follows immediately from the definition. The second follows from the definition and the fact that if two constants are congruent modulo \( P \) then they are equal. The third assertion follows from the definition and Proposition 1.10. Finally, note that the map from \( {\left( A/PA\right) }^{... | Yes |
Proposition 3.2. Let \( \alpha \in \mathbb{F} \) . Then,\n\n\[ \n{\left( \frac{\alpha }{P}\right) }_{d} = {\alpha }^{\frac{q - 1}{d}\deg \left( P\right) }\n\] | Proof. Let \( \delta = \deg \left( P\right) \) . Then,\n\n\[ \n\frac{\left| P\right| - 1}{d} = \frac{{q}^{\delta } - 1}{d} = \left( {1 + q + \cdots + {q}^{\delta - 1}}\right) \frac{q - 1}{d}.\n\]\n\nThe result now follows from the definition and the fact that for all \( \alpha \in \mathbb{F} \) we have \( {\alpha }^{q}... | Yes |
Proposition 3.4. The symbol \( {\left( a/b\right) }_{d} \) has the following properties.\n\n1) If \( {a}_{1} \equiv {a}_{2}\left( {\;\operatorname{mod}\;b}\right) \) then \( {\left( {a}_{1}/b\right) }_{d} = {\left( {a}_{2}/b\right) }_{d} \) .\n\n2) \( {\left( {a}_{1}{a}_{2}/b\right) }_{d} = {\left( {a}_{1}/b\right) }_{... | Proof. Properties 1 -4 follow from the definition and the properties of the symbol \( {\left( a/P\right) }_{d} \) .\n\nTo show property 5, suppose \( {c}^{d} \equiv a\left( {\;\operatorname{mod}\;b}\right) \) . Then, by properties 1 and \( 2,{\left( a/b\right) }_{d} = {\left( {c}^{d}/b\right) }_{d} = {\left( c/b\right)... | Yes |
Theorem 3.5. (The general reciprocity law). Let \( a, b \in A \) be relatively prime, non-zero elements. Then,\n\n\[{\left( \frac{a}{b}\right) }_{d}{\left( \frac{b}{a}\right) }_{d}^{-1} = {\left( -1\right) }^{\frac{q - 1}{d}\deg \left( a\right) \deg \left( b\right) }{\operatorname{sgn}}_{d}{\left( a\right) }^{\deg \lef... | Proof. When \( a \) and \( b \) are monic irreducibles this reduces to Theorem 3.3. In general, the proof proceeds by appealing to Proposition 3.2, Theorem 3.3, the definitions, and the fact that the degree of a product of two polynomials is equal to the sum of their degrees. We omit the details. | No |
Proposition 3.6. With the above assumptions we have\n\n1) If \( \deg \left( m\right) \) is even, \( \left( {q - 1}\right) /d \) is even, or \( p = \operatorname{char}\left( F\right) = 2, m \) is a \( d \) -th power modulo \( P \) iff \( P \equiv {a}_{i}\left( {\;\operatorname{mod}\;m}\right) \) for some \( i = 1,2,\ldo... | Proof. By Theorem 3.5, we have\n\n\[ \n{\left( \frac{m}{P}\right) }_{d} = {\left( -1\right) }^{\frac{q - 1}{d}\deg \left( m\right) \deg \left( P\right) }{\left( \frac{P}{m}\right) }_{d}.\n\]\nIf any of the conditions in Part 1 hold, we have \( {\left( m/P\right) }_{d} = {\left( P/m\right) }_{d} \) and this gives the re... | Yes |
Proposition 4.1. We have\n\n\[ \log {\zeta }_{A}\left( s\right) \approx \log \left( \frac{1}{s - 1}\right) \approx \mathop{\sum }\limits_{P}{\left| P\right| }^{-s}, \]\n\nwhere the sum is over all irreducible monic polynomials \( P \) . | Proof. Since \( {\zeta }_{A}\left( s\right) = {\left( 1 - {q}^{1 - s}\right) }^{-1} \) we see that \( \mathop{\lim }\limits_{{s \rightarrow 1}}\left( {s - 1}\right) {\zeta }_{A}\left( s\right) = 1/\log \left( q\right) \). Thus, \( \log {\zeta }_{A}\left( s\right) - \log {\left( s - 1}\right) }^{-1} \) is bounded as \( ... | Yes |
Proposition 4.2. Let \( \chi \) and \( \psi \) be two Dirichlet characters modulo \( m \) and \( a \) and \( b \) two elements of \( A \) relatively prime to \( m \) . Then\n\n(1) \( \mathop{\sum }\limits_{a}\chi \left( a\right) \overline{\psi \left( a\right) } = \Phi \left( m\right) \delta \left( {\chi ,\psi }\right) ... | The proofs of all these facts are standard. For the corresponding facts over the integers, \( \mathbb{Z} \), the reader can consult, for example, Ireland-Rosen [1], Chapter 16, Section 3. The relations given in the above proposition are called the orthogonality relations. | No |
Proposition 4.3. Let \( \chi \) be a non-trivial Dirichlet character modulo \( m \) . Then, \( L\left( {s,\chi }\right) \) is a polynomial in \( {q}^{-s} \) of degree at most \( \deg \left( m\right) - 1 \) . | Proof. Define\n\n\[ A\left( {n,\chi }\right) = \mathop{\sum }\limits_{{\deg \left( f\right) = n}}\chi \left( f\right) \]\n\n\[ f\text{monic} \]\n\nIt is clear from the definition of \( L\left( {s,\chi }\right) \) that\n\n\[ L\left( {s,\chi }\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }A\left( {n,\chi }\right) {q}... | Yes |
Lemma 4.4. Let \( \chi \) vary over all Dirichlet characters modulo \( m \) . Then, for each prime \( P \) not dividing \( m \), there exist positive integers \( {f}_{P} \) and \( {g}_{P} \) such that \( {f}_{P}{g}_{P} = \Phi \left( m\right) \) and \[ \mathop{\prod }\limits_{\chi }L\left( {s,\chi }\right) = \mathop{\pr... | Proof. For a fixed prime \( P \) not dividing \( m \), the map \( \chi \rightarrow \chi \left( P\right) \) is a homomorphism from the group \( {X}_{m} \rightarrow {\mathbb{C}}^{ * } \) . The image must be a cyclic group of order \( {f}_{P} \), say, generated by \( {\zeta }_{{f}_{p}} \) . If \( {g}_{P} \) is the order o... | Yes |
Lemma 4.5. Suppose \( \chi \) is a complex Dirichlet character modulo \( m \), i.e. \( \bar{\chi } \neq \chi \) . Then, \( L\left( {1,\chi }\right) \neq 0 \) . | Proof. The right-hand side of the equation in the statement of Lemma 4.4 is equal to a Dirichlet series with positive coefficients and constant term 1. Consequently, its value at real numbers \( s \) such that \( s > 1 \) is a real number greater than 1. Suppose \( \chi \) is a complex Dirichlet character and that \( L... | Yes |
Theorem 4.7. Let \( a, m \in A \) be two relatively prime polynomials with \( m \) of positive degree. Consider the set of primes, \( \mathcal{S} = \{ P \in A \mid P \equiv a\left( {\;\operatorname{mod}\;m}\right) \} \) . Then, \( \delta \left( \mathcal{S}\right) = 1/\Phi \left( m\right) \) . In particular, \( \mathcal... | Proof. Using the product formula for \( L\left( {s,\chi }\right) \) and the same technique used in the proof of Proposition 4.1, one finds\n\n\[ \log L\left( {s,\chi }\right) = \mathop{\sum }\limits_{p}\frac{\chi \left( P\right) }{{\left| P\right| }^{s}} + R\left( {s,\chi }\right) ,\]\n\nwhere the function \( R\left( {... | Yes |
Proposition 5.1. Let \( a \in {K}^{ * } \) . Then, \( {\operatorname{ord}}_{P}\left( a\right) = 0 \) for all but finitely many primes \( P \) . Secondly, \( \left( a\right) = 0 \), the zero divisor, if and only if \( a \in {F}^{ * } \), i.e., a is a non-zero constant. Finally, \( \deg {\left( a\right) }_{o} = \deg {\le... | Proof. (Sketch) If \( a \in {F}^{ * } \), it is easy to see from the definitions that \( \left( a\right) = 0 \) . So, suppose \( a \in {K}^{ * } - {F}^{ * } \) . Then, as we have seen, \( K \) is finite over \( F\left( a\right) \) . Let \( R \) be the integral closure of \( F\left\lbrack a\right\rbrack \) in \( K.R \) ... | No |
Lemma 5.2. If \( A \) and \( B \) are linearly equivalent divisors, then \( L\left( A\right) \) and \( L\left( B\right) \) are isomorphic. In particular, \( l\left( A\right) = l\left( B\right) \) . | Proof. Suppose \( A = B + \left( h\right) \) . Then a short calculation shows that \( x \rightarrow {xh} \) is an isomorphism from \( L\left( A\right) \) with \( L\left( B\right) \) . | Yes |
Lemma 5.3. If \( \deg \left( A\right) \leq 0 \) then \( l\left( A\right) = 0 \) unless \( A \sim 0 \) in which case \( l\left( A\right) = 1 \) . | Proof. If \( \deg \left( A\right) < 0 \) and \( x \in L\left( A\right) \), then \( \deg \left( {\left( x\right) + A}\right) \) is both \( < 0 \) and \( \geq 0 \) which is a contradiction. If \( \deg \left( A\right) = 0 \) and \( L\left( A\right) \) is not empty, let \( x \in L\left( A\right) \) . Then \( \left( x\right... | Yes |
Corollary 2. For \( C \in \mathcal{C} \) we have \( l\left( C\right) = g \) . | Proof. Set \( A = 0 \) in the theorem. | Yes |
Corollary 3. For \( C \in \mathcal{C} \) we have \( \deg \left( C\right) = {2g} - 2 \) . | Proof. Set \( A = C \) in the theorem, and use Corollary 2. | No |
Corollary 4. If \( \deg \left( A\right) \geq {2g} - 2 \), then \( l\left( A\right) = \deg \left( A\right) - g + 1 \) except in the \( \operatorname{case}\deg \left( A\right) = {2g} - 2 \) and \( A \in \mathcal{C} \) . | Proof. If \( \deg \left( A\right) \geq {2g} - 2 \), then \( \deg \left( {C - A}\right) \leq 0 \) . Now use Lemma 5.3. | No |
Corollary 5. Suppose that \( {g}^{\prime } \) and \( {C}^{\prime } \) have the same properties as those of \( g \) and \( C \) stated in the theorem. Then, \( g = {g}^{\prime } \) and \( C \sim {C}^{\prime } \) . | Proof. Find a divisor \( A \) whose degree is larger than \( \max \left( {{2g} - 2,2{g}^{\prime } - 2}\right) \) (a large positive multiple of a prime will do). By Corollary \( 4, l\left( A\right) = \deg \left( A\right) - \) \( g + 1 = \deg \left( A\right) - {g}^{\prime } + 1 \) . Thus, \( g = {g}^{\prime } \) . Now se... | Yes |
Lemma 5.5. For any integer \( n \geq 0 \) the number of effective divisors of degree \( n \) is finite. | Proof. (Sketch) Choose an \( x \in K \) such that \( x \) is transcendental over \( \mathbb{F} \) . \( K/\mathbb{F}\left( x\right) \) is finite. The primes of \( \mathbb{F}\left( x\right) \) are in one to one correspondence with the monic irreducible polynomials in \( \mathbb{F}\left\lbrack x\right\rbrack \) with the o... | No |
Lemma 5.6. The number of divisor classes of degree zero, \( {h}_{K} \), is finite. | Proof. Let \( D \) be a divisor of degree 1 . If \( A \) is any divisor of degree 0, then \( \deg \left( {{gD} + A}\right) = g \) and so by Riemann’s inequality, \( l\left( {{gD} + A}\right) \geq g - g + 1 = 1 \) . Let \( f \in L\left( {{gD} + A}\right) \) . Then, \( B = \left( f\right) + {gD} + A \geq 0 \) and so \( A... | Yes |
For any divisor \( A \), the number of effective divisors in \( \bar{A} \) is \( \frac{{q}^{l\left( A\right) } - 1}{q - 1} \) . | We begin by showing that \( \bar{A} \) contains effective divisors if and only if \( l\left( A\right) > 0 \) .\n\nSuppose \( B \in \bar{A} \) and is effective. There is an \( f \in {K}^{ * } \) such that \( \left( f\right) + A = \) \( B \geq 0 \), so \( f \in L\left( A\right) \) and \( l\left( A\right) > 0 \) . The con... | Yes |
Lemma 5.8. Let \( h = {h}_{K} \) . For every integer \( n \), there are \( h \) divisor classes of degree \( n \) . Suppose \( n \geq 0 \) and that \( \left\{ {{\bar{A}}_{1},{\bar{A}}_{2},\ldots ,{\bar{A}}_{h}}\right\} \) are the divisor classes of degree \( n \) . Then the number of effective divisors of degree \( n,{... | Proof. The first assertion follows directly from Lemma 5.6 and the remarks preceding Lemma 5.5. The second follows just as directly from Lemmas 5.6 and 5.7. | No |
Theorem 5.9. Let \( K \) be a global function field in one variable with a finite constant field \( \mathbb{F} \) with \( q \) elements. Suppose that the genus of \( K \) is \( g \) . Then there is a polynomial \( {L}_{K}\left( u\right) \in \mathbb{Z}\left\lbrack u\right\rbrack \) of degree \( {2g} \) such that \[ {\ze... | Proof. It is convenient to work with the variable \( u = {q}^{-s} \) . Then \[ {\zeta }_{K}\left( s\right) \overset{\text{ def }}{ = }{Z}_{K}\left( u\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{b}_{n}{u}^{n} \] We noted earlier that for \( n > {2g} - 2 \) we have \( {b}_{n} = {h}_{K}\frac{{q}^{n - g + 1} - 1}{q ... | Yes |
Proposition 5.11. The number of prime divisors of degree 1 of \( K,{a}_{1} \) , satisfies the inequality \( \left| {{a}_{1} - q - 1}\right| \leq {2g}\sqrt{q} \) . Also, \( {\left( \sqrt{q} - 1\right) }^{2g} \leq {h}_{K} \leq \) \( {\left( \sqrt{q} + 1\right) }^{2g} \) . | Proof. By Theorem 5.9, \( {L}_{K}^{\prime }\left( 0\right) = {a}_{1} - q - 1 \) . From the above factorization of \( {L}_{K}\left( u\right) \) we see \( - {L}_{K}^{\prime }\left( 0\right) = {\pi }_{1} + {\pi }_{2} + \cdots + {\pi }_{2g} \) . The first assertion is immediate from this and Theorem 5.10.\n\nAs for the sec... | Yes |
Theorem 5.12.\n\n\[ \n{a}_{N} = \# \{ P \mid \deg \left( P\right) = N\} = \frac{{q}^{N}}{N} + O\left( \frac{{q}^{\frac{N}{2}}}{N}\right) \n\] | Proof. Using the Euler product decomposition and Theorem 5.9, we see\n\n\[ \n{Z}_{K}\left( u\right) = \frac{\mathop{\prod }\limits_{{i = 1}}^{{2g}}\left( {1 - {\pi }_{i}u}\right) }{\left( {1 - u}\right) \left( {1 - {qu}}\right) } = \mathop{\prod }\limits_{{d = 1}}^{\infty }{\left( 1 - {u}^{d}\right) }^{-{a}_{d}}. \n\]\... | Yes |
Lemma 6.1. Let \( \omega \) be a non-zero meromorphic differential, \( x \in X \), and \( {\omega }_{x} \) the linear functional on \( {M}_{x} \) described above. There is an integer \( N \) such that \( {\omega }_{x} \) vanishes on \( {P}_{x}^{N} \) but not on \( {P}_{x}^{N - 1} \) . This integer is characterized by\n... | Proof. Since we are fixing \( x \) in our considerations we set \( {t}_{x} = t \) and suppose \( \omega \) is expressed in terms of \( t \) as in Equation 1. Assume \( {a}_{-N} \neq 0 \) so that \( {\operatorname{ord}}_{x}\left( \omega \right) = - N \) . From equation (2) it is then clear that \( {\omega }_{x} \) vanis... | Yes |
For every \( f \in M \) we have\n\n\[ \mathop{\sum }\limits_{{x \in X}}{\omega }_{x}\left( f\right) = 0 \] | Proof. Note that \( f \in M \) implies \( f \in {M}_{x} \) for all \( x \in X \), so the terms in the sum make sense. Also, \( f \in {O}_{x} \) for all but finitely many \( x \) (on a compact Riemann surface a meromorphic function has at most finitely many poles). By the corollary to Lemma 6.1, \( {\omega }_{x}\left( f... | Yes |
Lemma 6.4. Let \( D \leq C \) be divisors of \( K \) . Then,\n\n\[ \n{\dim }_{F}{A}_{K}\left( C\right) /{A}_{K}\left( D\right) = \deg C - \deg D.\n\] | Proof. If \( C = D \) the result is clear. Otherwise, \( C \) is obtained from \( D \) by adding finitely many primes, so it suffices to show that\n\n\[ \n{\dim }_{F}{A}_{K}\left( {D + P}\right) /{A}_{K}\left( D\right) = \deg P\n\]\nfor any prime \( P \) .\n\nLet \( \widehat{P} = P{\widehat{O}}_{P} \) . Let \( n = {\op... | Yes |
Lemma 6.5. Let \( D \leq C \) be divisors of \( K \) . Then,\n\n\[ \n{\dim }_{F}\frac{{A}_{K}\left( C\right) + K}{{A}_{K}\left( D\right) + K} = \left( {\deg C - l\left( C\right) }\right) - \left( {\deg D - l\left( D\right) }\right) .\n\] | Proof. Recall that \( {A}_{K}\left( C\right) \cap K = L\left( C\right) \) . Using the first and second laws of isomorphism, the space on the left-hand side of the above equation is seen to be isomorphic to\n\n\[ \n\frac{{A}_{K}\left( C\right) }{{A}_{K}\left( D\right) + L\left( C\right) } \cong \frac{{A}_{K}\left( C\rig... | Yes |
For any divisor \( D \) of \( K \), the space \( {\Omega }_{K}\left( D\right) \) is finite dimensional over \( F \) and\n\n\[ l\left( D\right) = \deg D - g + 1 + {\dim }_{F}{\Omega }_{K}\left( D\right) . \] | Proof. In Lemma 6.5, we are going to fix \( D \) and let \( C \) vary over divisors greater than or equal to \( D \) . By Riemann’s theorem \( l\left( C\right) \geq \deg C - g + 1 \) or what is the same \( \deg C - l\left( C\right) \leq g - 1 \) . So, by Lemma 6.5\n\n\[ \dim \frac{{A}_{K}\left( C\right) + K}{{A}_{K}\le... | Yes |
Corollary 1. Let \( c \) be the constant in Riemann’s theorem. Then if \( D \) is a divisor with \( \deg D \geq c \), we have \( {A}_{K} = {A}_{K}\left( D\right) + K \) . | Proof. We have just shown that \( {\dim }_{F}\left( {{A}_{K}/{A}_{K}\left( D\right) + K}\right) = l\left( D\right) - \deg D + \) \( g - 1 \), which is zero if \( \deg D \geq c \) by Riemann’s theorem. Thus \( {A}_{K} = \) \( {A}_{K}\left( D\right) + K \) in this case. | Yes |
Corollary 2. The genus of \( K, g \), can be characterized as the dimension over \( F \) of the space \( {\Omega }_{K}\left( 0\right) \) . | Proof. The zero divisor, 0 , has degree zero and dimension 1 . From the proposition we derive \( 1 = 0 - g + 1 + {\dim }_{F}{\Omega }_{K}\left( 0\right) \) . This gives the result. | Yes |
Lemma 6.8. Let \( \omega \in {\Omega }_{K} \) be a non-zero differential. Then, there is a unique divisor \( D \) with the property that \( \omega \) vanishes on \( {A}_{K}\left( D\right) \) and if \( {D}^{\prime } \) is any divisor such that \( \omega \) vanishes on \( {A}_{K}\left( {D}^{\prime }\right) \), then \( {D... | Proof. Let \( \mathcal{T} = \left\{ {{D}^{\prime } \mid \omega \left( {{A}_{K}\left( {D}^{\prime }\right) }\right) = 0}\right\} \) . Since \( \omega \) is a differential, \( \mathcal{T} \) is nonempty. By Corollary 1 to Proposition 6.7, we see that \( \deg {D}^{\prime } < c \) for all \( {D}^{\prime } \in \mathcal{T} \... | Yes |
Lemma 6.9. Let \( \omega \in {\Omega }_{K} \) and \( x \in {K}^{ * } \) . Then,\n\n\[ \left( {x\omega }\right) = \left( x\right) + \left( \omega \right) \] | Proof. Suppose \( \omega \in {\Omega }_{K}\left( D\right) \) . If \( \xi \in {A}_{K} \), then \( {x\omega } \) vanishes on \( \xi \) if \( {x\xi } \in \) \( {A}_{K}\left( D\right) \), which is equivalent to \( \xi \in {A}_{K}\left( {\left( x\right) + D}\right) \) . Thus, \( \omega \) vanishes on \( {A}_{K}\left( D\righ... | Yes |
Proposition 6.10. The space of Weil differentials, \( {\Omega }_{K} \), is of dimension one when considered as a vector space over \( K \) . | Proof. Let \( 0 \neq \omega \in {\Omega }_{K} \) and \( x \in L\left( {\left( \omega \right) - D}\right) \) where \( D \) is some divisor. We claim that \( {x\omega } \in {\Omega }_{K}\left( D\right) \) . By the proof of the previous Lemma, we know that \( {x\omega } \) vanishes on \( {A}_{K}\left( {\left( x\right) + \... | Yes |
Corollary 1. Let \( 0 \neq \omega \in {\Omega }_{K} \) and let \( D \) be a divisor. Then there is an F-linear isomorphism between \( L\left( {\left( \omega \right) - D}\right) \) and \( {\Omega }_{K}\left( D\right) \) . | Proof. In the proof of the proposition we showed that \( L\left( {\left( \omega \right) - D}\right) \omega \subseteq \) \( {\Omega }_{K}\left( D\right) \) . So it just remains to show that this inclusion is an equality. Let \( {\omega }^{\prime } \in {\Omega }_{K}\left( D\right) \) . By the proposition, there is an ele... | Yes |
Corollary 2. All the divisors of non-zero differentials fill out a single divisor class. This class is called the canonical class of \( K \) . | Proof. If \( \omega ,{\omega }^{\prime } \in {\Omega }_{K} \) are non-zero, there exists an \( x \in {K}^{ * } \) such that \( {\omega }^{\prime } = {x\omega } \) by the proposition. By Lemma 6.9 we have \( \left( {\omega }^{\prime }\right) = \left( x\right) + \left( \omega \right) \) so that \( \left( {\omega }^{\prim... | Yes |
Proposition 6.11. Let \( \omega \in {\Omega }_{K} \) and \( \xi = \left( {x}_{P}\right) \in {A}_{K} \) . Then, for all but finitely many \( P \) we have \( {\omega }_{P}\left( {x}_{P}\right) = 0 \) and\n\n\[ \omega \left( \xi \right) = \mathop{\sum }\limits_{P}{\omega }_{P}\left( {x}_{P}\right) \] | Proof. Let \( S \) be the finite set of primes where either \( {\operatorname{ord}}_{P}\left( \omega \right) < 0 \) or \( {\operatorname{ord}}_{P}\left( {x}_{P}\right) < 0 \) . If \( P \notin S \) then \( {x}_{P} \in {\widehat{O}}_{P} \) and so \( {\omega }_{P}\left( {x}_{P}\right) = 0 \) by the remark preceding the Pr... | Yes |
Proposition 6.12. Let \( 0 \neq \omega \in {\Omega }_{K} \) . Then, \( N = {\operatorname{ord}}_{P}\left( \omega \right) \) is determined by the following two properties; \( {\omega }_{P} \) vanishes on \( {\widehat{P}}^{-N} \) but does not vanish on \( {\widehat{P}}^{-N - 1} \) . | Proof. We have already seen in the remarks preceding Proposition 6.10 that \( {\omega }_{P} \) vanishes on \( {\widehat{P}}^{-{\operatorname{ord}}_{P}\left( \omega \right) } \) . It remains to show that \( {\omega }_{P} \) doesn’t vanish on \( {\widehat{P}}^{-{\operatorname{ord}}_{P}\left( \omega \right) - 1} \) . We k... | Yes |
Theorem 6.13. Let \( K/F \) be a function field and \( S \subset {\mathcal{S}}_{K} \) a finite set of primes. For each \( P \in S \) let an element \( {a}_{P} \in {\widehat{K}}_{P} \) and a positive integer \( {n}_{P} \) be given. Finally, let’s specify a prime \( Q \notin S \) . Then, there is an element \( a \in K \)... | Proof. Define an adele \( \xi = \left( {x}_{P}\right) \) by the conditions that \( {x}_{P} = {a}_{P} \) for \( P \in S \) and \( {x}_{P} = 0 \) for \( P \notin S \) . Next, define a divisor \( D = {mQ} - \mathop{\sum }\limits_{{P \in S}}{n}_{P}P \) . Choose the integer \( m \) so large that the degree of \( D \) exceed... | Yes |
Proposition 7.1. With the above notations, \( {ef} \leq n = \left\lbrack {L : K}\right\rbrack \), the dimension of \( L \) over \( K \) . | Proof. Let \( \Pi \) be a generator of \( \mathfrak{P} \) and choose \( {\omega }_{1},{\omega }_{2},\ldots ,{\omega }_{m} \) such that their reductions modulo \( \mathfrak{P} \) are linearly independent over \( {O}_{P}/P \) . We will show that the \( {em} \) elements \( {\omega }_{i}{\Pi }^{j} \) with \( 1 \leq i \leq ... | Yes |
Proposition 7.2. Assume \( L/K \) is a finite, separable extension of fields. Then, with the above notations, \( \mathop{\sum }\limits_{{i = 1}}^{g}{e}_{i}{f}_{i} = n = \left\lbrack {L : K}\right\rbrack \) . | Proof. Since \( L/K \) is separable, the trace from \( L \) to \( K, t{r}_{L/K} \), is a nontrivial \( K \) -linear functional on \( L \) . Using this, one can prove that \( R \) is a free module over \( {O}_{P} \) of rank equal to \( n = \left\lbrack {L : K}\right\rbrack \) (see Samuel and Zariski [1]). Thus, \( R/{PR... | Yes |
Lemma 7.3. Let \( L/K \) be a purely inseparable extension of degree \( p \), the characteristic of \( K \) . Assume \( K = {L}^{p} \) (this strange assumption is often correct in function fields). Suppose \( {O}_{P} \subset K \) is a dvr with quotient field \( K \) . Then there is one and only one dvr \( {O}_{\mathfra... | Proof. Let \( R = \left\{ {r \in L \mid {r}^{p} \in {O}_{P}}\right\} \) and \( \mathfrak{P} = \left\{ {r \in L \mid {r}^{p} \in P}\right\} \) . It is easy to see that \( R \) is a ring, \( \mathfrak{P} \) is a prime ideal in \( R \), and \( \mathfrak{P} \cap {O}_{P} = P \) . We will show that \( R \) is a dvr.\n\nLet \... | Yes |
Proposition 7.4. Let \( F \) be a perfect field of positive characteristic \( p \), and \( K \) a function field with constant field \( F \) . Then, \( \left\lbrack {K : {K}^{p}}\right\rbrack = p \) . | Proof. Let \( x \) be an element of \( K \) not in \( F \) . Then \( \left\lbrack {K : F\left( x\right) }\right\rbrack < \infty \) . Consider \( F{\left( x\right) }^{p} = {F}^{p}\left( {x}^{p}\right) = F\left( {x}^{p}\right) \) . It is clear that \( \left\lbrack {F\left( x\right) : F\left( {x}^{p}\right) }\right\rbrack... | Yes |
Proposition 7.5. Let \( K \) be a function field with a perfect constant field \( F \) . Let \( L \) be a finite extension of \( K \), and \( M \), the maximal separable extension of \( K \) in \( L \) . Then, the genus of \( M \) is equal to the genus of \( L \) . Also, for each prime \( \mathfrak{p} \) of \( M \) the... | Proof. The constant field \( E \) of \( M \) is perfect since it is a finite extension of \( F \), which is perfect by assumption.\n\nSince \( L/M \) is purely inseparable, there is a tower of fields\n\n\[ K \subseteq M = {K}_{0} \subset {K}_{1} \subset \cdots \subset {K}_{n - 1} \subset {K}_{n} = L, \]\n\nwhere for ea... | Yes |
Theorem 7.6. Let \( K \) be a function field with perfect constant field \( F \) . Let \( L \) be a finite extension of \( K \) of degree \( n \) . Suppose \( P \) is a prime of \( K \) and \( \left\{ {{\mathfrak{P}}_{1},{\mathfrak{P}}_{2},\ldots ,{\mathfrak{P}}_{g}}\right\} \) the set of primes in \( L \) lying above ... | Proof. Let \( M \) be the maximal separable extension of \( K \) in \( L \) . Let \( {\mathfrak{p}}_{i} \) be the prime of \( M \) lying below \( {\mathfrak{P}}_{i} \) and let \( {e}_{i}^{\prime } \) and \( {f}_{i}^{\prime } \) be the ramification index and relative degree of \( {\mathfrak{p}}_{i} \) over \( P \) . By ... | Yes |
Proposition 7.7. Let \( \mathfrak{A} \in {\mathcal{D}}_{L} \) and \( A \in {\mathcal{D}}_{K} \) . Then\n\n\( {\deg }_{K}{N}_{L/K}\left( \mathfrak{A}\right) = \left\lbrack {E : F}\right\rbrack {\deg }_{L}\mathfrak{A}\; \) and \( \;{\deg }_{L}\left( {{i}_{L/K}\left( A\right) }\right) = \frac{\left\lbrack L : K\right\rbra... | Proof. Both facts follow from the calculation\n\n\[ \left\lbrack {{O}_{\mathfrak{P}}/\mathfrak{P} : F}\right\rbrack = \left\lbrack {{O}_{\mathfrak{P}}/\mathfrak{P} : E}\right\rbrack \left\lbrack {E : F}\right\rbrack = \left\lbrack {{O}_{\mathfrak{P}}/\mathfrak{P} : {O}_{P}/P}\right\rbrack \left\lbrack {{O}_{P}/P : F}\r... | Yes |
(i). If \( a \in {K}^{ * } \), then \( {i}_{L/K}{\left( a\right) }_{K} = {\left( a\right) }_{L} \) . | Proof. To prove the first assertion, one simply computes\n\n\[ \n{i}_{L/K}{\left( a\right) }_{K} = {i}_{L/K}\mathop{\sum }\limits_{P}{\operatorname{ord}}_{P}\left( a\right) P = \mathop{\sum }\limits_{P}{\operatorname{ord}}_{P}\left( a\right) \mathop{\sum }\limits_{{\mathfrak{P} \mid P}}e\left( {\mathfrak{P}/P}\right) \... | Yes |
Proposition 7.9. i) Let \( A \) be a dvr with maximal ideal \( P, K \) its quotient field, \( L \) a finite separable extension, and \( B \) the integral closure of \( A \) in \( L \) . Then, some prime above \( P \) in \( B \) is ramified if and only if \( {\mathfrak{d}}_{B/A} \subset P \) . ii) \( {N}_{L/K}{\mathfrak... | Proof. We have already given the proof of \( i \) ) in the above discussion. For the proof of part \( {ii} \) ) see Serre [2]. | No |
Lemma 7.10. We have \( {\mathfrak{d}}_{B/A}\left( P\right) = {\mathfrak{d}}_{B/A}{\widehat{A}}_{P} \) and \( {\mathfrak{D}}_{B/A}\left( \mathfrak{P}\right) = {\mathfrak{D}}_{B/A}{\widehat{B}}_{\mathfrak{P}} \) . In other words, if \( {\mathfrak{d}}_{B/A} = {P}^{t} \), then \( {\mathfrak{d}}_{B/A}\left( P\right) = {\wid... | For the proof of this result we refer the reader to Serre [2], Chapter 3. | No |
Corollary 1. As in the Lemma, let \( \delta \) be the exact power of \( \mathfrak{P} \) dividing \( {\mathfrak{D}}_{B/A} \) . Then \( \delta \) can be characterized as the largest integer \( m \) such that the trace from \( {\widehat{L}}_{\mathfrak{P}} \) to \( {\widehat{K}}_{P} \) of \( {\widehat{\mathfrak{P}}}^{-m} \... | Proof. From the definition, if \( m \) has the property described in the corollary, the local inverse different is \( {\widehat{\mathfrak{P}}}^{-m} \) and so \( {\mathfrak{D}}_{B/A}\left( \mathfrak{P}\right) = {\widehat{\mathfrak{P}}}^{m} \) . The result is then immediate from the lemma. | Yes |
Corollary 2. With the same notation as Corollary \( 1,\delta \geq e\left( {\mathfrak{P}/P}\right) - 1 \) with equality holding if and only if the characteristic of \( F \) is either zero or does not divide \( e\left( {\mathfrak{P}/P}\right) \) . | Proof. (sketch) Neither \( e\left( {\mathfrak{P}/P}\right) \) nor \( \delta \) changes after passing to the completion (for \( \delta \) this follows from the lemma). So, we can assume \( A \) and \( B \) are complete. Again, nothing essential changes if we replace \( K \) by the maximal unramified extension of \( K \)... | No |
Theorem 7.11. With the above notations and hypotheses, a prime \( \mathfrak{P} \) of \( B \) is ramified over \( A \) if and only if \( \mathfrak{P} \mid {\mathfrak{D}}_{B/A} \) . | Proof. (sketch) The definition of unramifiedness is in two parts; the ramification index must be one, and the residue class field extension must be separable. If one ignores the second condition one can refer to the above Corollary 2 to Lemma 7.10 for a proof of the theorem. We proceed somewhat differently and handle b... | No |
Theorem 7.15. With the hypotheses and notation of Proposition 7.14, we have\n\n\[ \n{\left( {\omega }^{ * }\right) }_{L} = {i}_{L/K}{\left( \omega \right) }_{K} + {D}_{L/K} \n\] | Proof. We are going to use Proposition 6.12 of the last chapter, which shows how to determine the divisor of a differential using properties of its local components.\n\nWe begin by recalling the definition. Let \( \mu \in {\Omega }_{L} \) . For a prime \( \mathfrak{P} \) of \( L \) let \( {i}_{\mathfrak{P}} : {\widehat... | Yes |
Theorem 7.16. (Riemann-Hurwitz) Let \( L/K \) be a finite, separable, geometric extension of function fields. Then,\n\n\[ 2{g}_{L} - 2 = \left\lbrack {L : K}\right\rbrack \left( {2{g}_{K} - 2}\right) + {\deg }_{L}{D}_{L/K}. \] | Proof. Let \( \omega \) be a non-zero differential of \( K \). By the remarks following Corollary 2 to Lemma 6.10, \( {\left( \omega \right) }_{K} \) is in the canonical class of \( K \). By Corollary 3 to Theorem 5.4, every divisor in the canonical class of \( K \) has degree \( 2{g}_{K} - 2 \). Thus, \( {\deg }_{K}{\... | Yes |
Corollary 1. Suppose \( L/K \) is a finite, separable, geometric extension of function fields. Then, \( {g}_{K} \leq {g}_{L} \) . (This need not be true for inseparable extensions!) | Proof. Since the different is an effective divisor (all its coefficients are nonnegative) we see \( 2{g}_{L} - 2 \geq \left\lbrack {L : K}\right\rbrack \left( {2{g}_{K} - 2}\right) \geq 2{g}_{K} - 2 \) . Thus \( {g}_{K} \leq {g}_{L} \) as asserted. | Yes |
Corollary 2. (Luroth’s Theorem) Let \( L = F\left( x\right) \) be a rational function field over \( F \) and \( K \) a subfield properly containing \( F \) . Then, there is a \( u \in K \) such that \( K = F\left( u\right) \) . | Proof. Since \( K \) properly contains \( F \), it is easy to see that \( \left\lbrack {L : K}\right\rbrack < \infty \) . Let \( M \) be the maximal separable extension of \( K \) contained in \( L \) . If the characteristic of \( F \) is zero, then \( M = L \) . If the characteristic of \( F \) is \( p > 0 \) , then \... | Yes |
Corollary 3. Let \( L/K \) be a finite, separable, geometric extension of function fields. Assume \( {g}_{L} = 1 \) . Then, \( {g}_{K} \leq 1 \) with equality holding if and only if \( L/K \) is unramified. | Proof. The inequality \( {g}_{K} \leq 1 \) follows from Corollary 1. From \( {g}_{L} = 1 \) and the theorem we deduce, \( 0 = \left\lbrack {L : K}\right\rbrack \left( {2{g}_{K} - 2}\right) + {\deg }_{L}{D}_{L/K} \) . If \( {g}_{K} = 1 \) the degree of the different is zero and so the different is the zero divisor (reca... | Yes |
Let \( K \) be a function field with a perfect constant field \( F \) . Consider the equation \( {X}^{N} + {Y}^{N} = 1 \) . We assume that \( N \) is not divisible by the characteristic \( p \) of \( F \) . If \( {g}_{K} = 0 \) and \( N \geq 3 \), then there is no nonconstant solution to this equation in \( K \) . If \... | Proof. Suppose that \( \left( {u, v}\right) \in {K}^{2} \) is a non-constant solution. Invoking the ABC theorem we find\n\n\[ \max \left( {{\deg }_{s}{u}^{N},{\deg }_{s}{v}^{N}}\right) \leq 2{g}_{K} - 2 + \mathop{\sum }\limits_{{P \in \operatorname{Supp}\left( {A + B + C}\right) }}{\deg }_{K}P \]\n\nwhere \( A \) is th... | Yes |
Proposition 8.1. Assume \( \left\lbrack {E : F}\right\rbrack < \infty \) . Then, \( \left\lbrack {{KE} : K}\right\rbrack = \left\lbrack {E : F}\right\rbrack \) . Any basis for \( E/F \) is also a basis for \( {KE}/K \) . | Proof. Suppose first that \( E/F \) is a finite, Galois extension. Then, by a standard theorem in Galois theory, \( {KE}/K \) is also Galois and \( \operatorname{Gal}\left( {{KE}/K}\right) \cong \) \( \operatorname{Gal}\left( {E/K \cap E}\right) \) . Since \( F \) is the constant field of \( K, E \cap K = F \) . It fol... | Yes |
Lemma 8.2. (a) Suppose \( L/K \) is a finite extension of fields and that \( K \) contains a field \( F \) which is algebraically closed in \( K \) . If \( \beta \in L \) is algebraic over \( F \), then \( {\operatorname{tr}}_{L/K}\left( \beta \right) \in F \) . (b) Suppose \( L/K \) is a finite extension of fields and... | Proof. This is fairly standard so we merely sketch the proofs.\n\nFor part (a) one considers the minimal polynomial for \( \beta \) over \( K \) and shows that all its roots (in some extension field) are algebraic over \( F \) . Thus the sum of the roots is algebraic over \( F \) and in \( K \), so the sum of the roots... | No |
Proposition 8.3. \( E \) is the exact constant field of \( {KE} \) . | Proof. We have to show that any element of \( {KE} \) which is algebraic over \( E \) is actually in \( E \) .\n\nAssume first that \( \left\lbrack {E : F}\right\rbrack < \infty \), and that \( \left\{ {{\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{n}}\right\} \) is a basis for \( E/F \) . Suppose \( \beta \in {KE} \)... | Yes |
Lemma 8.4. Let \( E/F \) be a finite extension with \( \\left\\{ {{\\alpha }_{1},{\\alpha }_{2},\\cdots ,{\\alpha }_{n}}\\right\\} \) a basis for \( E \) over \( F \) . Let \( P \) be a prime of \( K \) and \( {O}_{P} \) the corresponding valuation ring. Let \( {R}_{P} \) be the integral closure of \( {O}_{P} \) in \( ... | Proof. Since \( F \\subset {O}_{P} \) by definition, and each \( {\\alpha }_{i} \) is algebraic over \( F \), it follows that each \( {\\alpha }_{i} \) is integral over \( {O}_{P} \) .\n\nSuppose \( b \\in {R}_{P} \) . By Proposition 8.1, we can write \( b = \\mathop{\\sum }\\limits_{{i = 1}}^{n}{x}_{i}{\\alpha }_{i} \... | Yes |
Proposition 8.5. Suppose \( E/F \) is a finite extension. Then, \( {KE}/K \) is unramified at all primes. | Proof. Let \( P \) be a prime of \( K,{O}_{P} \) its valuation ring, and \( {R}_{P} \) the integral closure of \( {O}_{P} \) in \( {KE} \) . By Lemma 8.4, any field basis \( \left\{ {{\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{n}}\right\} \) for \( E/F \) is a free basis for \( {R}_{P} \) considered as an \( {O}_{P}... | Yes |
Lemma 8.6. Let \( \\left\\{ {{x}_{1},{x}_{2},\\cdots ,{x}_{m}}\\right\\} \\subset K \) be linearly independent over \( F \) . Then, considered as a subset of \( {KE} \), it remains linearly independent over \( E \) . | Proof. Suppose \( \\mathop{\\sum }\\limits_{{i = 1}}^{m}{\\beta }_{i}{x}_{i} = 0 \) with each \( {\\beta }_{i} \\in E \) . Assuming \( E/F \) is a finite extension, let \( \\left\\{ {{\\alpha }_{1},{\\alpha }_{2},\\cdots ,{\\alpha }_{n}}\\right\\} \) be a basis for \( E/F \) . Then, \( {\\beta }_{i} = \\mathop{\\sum }\... | Yes |
Lemma 8.7. Let \( L/K \) be a finite extension of function fields and \( P \) a prime of \( K \) . Suppose that \( \left\{ {{\mathfrak{P}}_{1},{\mathfrak{P}}_{2},\cdots ,{\mathfrak{P}}_{g}}\right\} \), the primes above \( P \) in \( L \) , are all unramified over \( P \) . Let \( n \in \mathbb{Z} \) be a given integer.... | Proof. Let \( \pi \in K \) be a uniformizing parameter at \( P \) . Then, since each \( {\mathfrak{P}}_{\imath } \) is unramified over \( P \) we have \( 1 = {\operatorname{ord}}_{P}\left( \pi \right) = {\operatorname{ord}}_{{\mathfrak{P}}_{i}}\left( \pi \right) \) for \( 1 \leq i \leq n \) . The inequalities \( {\oper... | Yes |
Proposition 8.9. Let \( E/F \) be a finite extension and \( L = {KE} \) . Then the genus of \( L \), considered as a function field over \( E \), is equal to the genus of \( K \) . (Once more we emphasize that, by hypothesis, \( F \) is perfect). | Proof. Let \( g \) be the genus of \( K \) and \( {g}^{\prime } \) the genus of \( L \) . Choose a divisor \( A \) of \( K \) such that \( {\deg }_{K}\left( A\right) \geq \max \left( {{2g} - 1,2{g}^{\prime } - 1}\right) \), e.g., \( A = {nP} \), where \( P \) is a prime divisor and \( n \) is a sufficiently large posit... | Yes |
Proposition 8.10. Let \( E/F \) be a finite extension and \( L = {KE} \). Let \( \mathfrak{P} \) be a prime of \( L \) and \( P \) the prime lying below it in \( K \). Define \( {E}_{\mathfrak{P}} = {O}_{\mathfrak{P}}/\mathfrak{P} \) and \( {F}_{P} = {O}_{P}/P \). Then, \( {E}_{\mathfrak{P}} \) is the compositum of \( ... | Proof. Let \( \bar{\omega } \in {E}_{\mathfrak{P}} \) and let \( \omega \) be an element of \( {O}_{\mathfrak{P}} \) representing \( \bar{\omega } \). Let’s consider \( \left\{ {{\mathfrak{P}}_{1} = \mathfrak{P},{\mathfrak{P}}_{2},\cdots ,{\mathfrak{P}}_{g}}\right\} \), the primes in \( L \) lying over \( P \). By the ... | Yes |
Proposition 8.11. With the notation of the previous proposition, suppose \( {F}_{P} = F\left\lbrack \theta \right\rbrack \) and that \( h\left( T\right) \in F\left\lbrack T\right\rbrack \) is the irreducible polynomial for \( \theta \) over \( F \) . Let\n\n\[ h\left( T\right) = {h}_{1}\left( T\right) {h}_{2}\left( T\r... | Proof. Lemma 8.4 can be restated to say that \( {R}_{P} \cong {O}_{P}{ \otimes }_{F}E \) . Reducing both sides modulo \( P \) yields, \( {R}_{P}/P{R}_{P} \cong {F}_{P}{ \otimes }_{F}E \) . By hypothesis, \( {F}_{P} = \) \( F\left\lbrack \theta \right\rbrack \cong F\left\lbrack T\right\rbrack /\left( {h\left( T\right) }... | Yes |
Lemma 8.12. The compositum of \( {\mathbb{F}}_{n} \) and \( {\mathbb{F}}_{m} \) is \( {\mathbb{F}}_{\left\lbrack n, m\right\rbrack } \) where \( \left\lbrack {n, m}\right\rbrack \) is the least common multiple of \( n \) and \( m \) . | Proof. Let \( {\mathbb{F}}_{h} \) be the compositum of \( {\mathbb{F}}_{n} \) and \( {\mathbb{F}}_{m} \) inside \( \overline{\mathbb{F}} \) . Since \( {\mathbb{F}}_{n},{\mathbb{F}}_{m} \subseteq \) \( {\mathbb{F}}_{h} \), we have \( n, m \mid h \), which implies \( \left\lbrack {n, m}\right\rbrack \mid h \) . Thus, \( ... | Yes |
Proposition 8.13. Let \( P \) be a prime of \( K \) . Then \( P \) splits into \( \left( {n,{\deg }_{K}P}\right) \) primes in \( {K}_{n} \) . Let \( \mathfrak{P} \) be a prime of \( {K}_{n} \) lying over \( P \) . Then\n\n\[ \n{\deg }_{{K}_{n}}\mathfrak{P} = \frac{{\deg }_{K}P}{\left( n,{\deg }_{K}P\right) }\;\text{ an... | Proof. By definition, the dimension of \( {O}_{P}/P \) over \( \mathbb{F} \) is \( {\deg }_{K}P \) . Thus, by the above lemma, the compositum of \( {O}_{P}/P \) and \( {\mathbb{F}}_{n} \) inside \( {O}_{\mathfrak{P}}/\mathfrak{P} \) has dimension \( \left\lbrack {n,{\deg }_{K}P}\right\rbrack \) over \( \mathbb{F} \) . ... | Yes |
Lemma 8.14. Let \( {\zeta }_{n} \in \mathbb{C} \) be a primitive \( n \) -th root of unity and \( m \) a positive integer. Then\n\n\[ \mathop{\prod }\limits_{{i = 0}}^{{n - 1}}\left( {1 - {\zeta }_{n}^{im}{u}^{m}}\right) = {\left( 1 - {u}^{\left\lbrack n, m\right\rbrack }\right) }^{\left( n, m\right) }.\] | Proof. First consider the case where \( m = 1 \) . The result in this case follows from the identity \( {T}^{n} - 1 = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {T - {\zeta }_{n}^{i}}\right) \) by making the substitution \( T = {u}^{-1} \) and simplifying.\n\nIn the general case, let \( {m}^{\prime } = m/\left( {n, m}\... | Yes |
Theorem 8.15. Let \( u = {q}^{-s} \) and \( {\zeta }_{K}\left( s\right) = {Z}_{K}\left( u\right) \) . Then,\n\n\[ \n{\zeta }_{{K}_{n}}\left( s\right) = {Z}_{{K}_{n}}\left( {u}^{n}\right) = \mathop{\prod }\limits_{{i = 0}}^{{n - 1}}{Z}_{K}\left( {{\zeta }_{n}^{i}u}\right) .\n\] | Proof. Since the constant field of \( {K}_{n} \) is \( {\mathbb{F}}_{n} \) which has \( {q}^{n} \) elements, we have \( {\zeta }_{{K}_{n}}\left( s\right) = {Z}_{{K}_{n}}\left( {u}^{\prime }\right) \), where \( {u}^{\prime } = {\left( {q}^{n}\right) }^{-s} = {q}^{-{ns}} = {u}^{n} \) . Thus, \( {\zeta }_{{K}_{n}}\left( s... | Yes |
Proposition 8.16. Let \( {L}_{K}\left( u\right) = \mathop{\prod }\limits_{{j = 1}}^{{2g}}\left( {1 - {\pi }_{j}u}\right) \) be the factorization of \( {L}_{K}\left( u\right) \) in \( \mathbb{C}\left\lbrack u\right\rbrack \) . Then, \[ {L}_{{K}_{n}}\left( {u}^{\prime }\right) = \mathop{\prod }\limits_{{j = 1}}^{{2g}}\le... | Proof. Using the definitions and Theorem 8.15, we find \[ \frac{{L}_{{K}_{n}}\left( {u}^{n}\right) }{\left( {1 - {u}^{n}}\right) \left( {1 - {q}^{n}{u}^{n}}\right) } = \mathop{\prod }\limits_{{i = 0}}^{{n - 1}}\frac{{L}_{K}\left( {{\zeta }_{n}^{i}u}\right) }{\left( {1 - {\zeta }_{n}^{i}u}\right) \left( {1 - q{\zeta }_{... | Yes |
Proposition 8.17. \( {b}_{m}\left( {K}_{n}\right) = h\left( {K}_{n}\right) \frac{{q}^{{nm} - g + 1} - 1}{{q}^{n} - 1} \), provided \( m > {2g} - 2 \) . | Proof. We proved earlier (see the remarks following Lemma 5.8) that if \( m > {2g} - 2 \), where \( g \) is the genus of \( K \), that\n\n\[ \n{b}_{m}\left( K\right) = h\left( K\right) \frac{{q}^{m - g + 1} - 1}{q - 1}.\n\]\n\nIn this equality, replace \( K \) by \( {K}_{n} \) and \( q \) by \( {q}^{n} \) (since the co... | Yes |
Proposition 8.18. \( {N}_{m}\left( K\right) \) is equal to the number of prime divisors of \( {K}_{m} \) of degree 1. | Proof. It is interesting to note that \( {N}_{1}\left( K\right) = {a}_{1}\left( K\right) \), so the result is certainly true when \( m = 1 \) . To prove the general case, we invoke Proposition 8.16. \[ \frac{\mathop{\prod }\limits_{{j = 1}}^{{2g}}\left( {1 - {\pi }_{j}^{m}{u}^{\prime }}\right) }{\left( {1 - {u}^{\prime... | Yes |
Proposition 8.19. \( {a}_{m}\left( {K}_{n}\right) = {m}^{-1}\mathop{\sum }\limits_{{d \mid m}}\mu \left( d\right) {a}_{1}\left( {K}_{{nm}/d}\right) \) . | Proof. From the definition, \( {N}_{m}\left( {K}_{n}\right) = \mathop{\sum }\limits_{{d \mid m}}d{a}_{d}\left( {K}_{n}\right) \) . Using Möbius inversion, we see that \( m{a}_{m}\left( {K}_{n}\right) = \mathop{\sum }\limits_{{d \mid m}}\mu \left( d\right) {N}_{m/d}\left( {K}_{n}\right) \) . From the relation \( {N}_{m/... | Yes |
Proposition 9.1. The field extension \( E/F \) is Galois and the map \( G \rightarrow \) \( \operatorname{Gal}\left( {E/F}\right) \) obtained by restriction of automorphisms to \( E \) is onto. Let \( N \subseteq \) \( G \) be the kernel of this map. Then the fixed field of \( N \) is \( {KE} \), the maximal constant f... | Proof. If \( \sigma \in G \) and \( \alpha \in E \), then the fact that \( \alpha \) is a root of a polynomial with coefficients in \( F \) shows that \( {\sigma \alpha } \) must be a root of the same polynomial since \( \sigma \) fixes \( F \) . Thus, \( {\sigma \alpha } \) is algebraic over \( F \), which implies \( ... | Yes |
Proposition 9.2. Let \( \left\{ {{\mathfrak{P}}_{1},{\mathfrak{P}}_{2},\cdots ,{\mathfrak{P}}_{g}}\right\} \) be the set of primes of \( L \) lying above \( P \) . The Galois group \( G \) acts transitively on this set. | Proof. For each \( i \) with \( 1 \leq i \leq g \) we need to show there is a \( \sigma \in G \) such that \( \sigma {\mathfrak{P}}_{1} = {\mathfrak{P}}_{i} \). Consider the set \( \left\{ {\sigma {\mathfrak{P}}_{1} \mid \sigma \in G}\right\} \). Suppose some \( {\mathfrak{P}}_{i} \) is not in this set, \( {\mathfrak{P... | Yes |
Proposition 9.3. We continue to use the notation introduced above, except that we now denote the number of primes in \( L \) lying above \( P \) by \( g\left( P\right) \) . We have \( f\left( {{\mathfrak{P}}_{i}/P}\right) = f\left( {{\mathfrak{P}}_{j}/P}\right) \) and \( e\left( {{\mathfrak{P}}_{i}/P}\right) = e\left( ... | Proof. For a given pair \( i \) and \( j \) there is an automorphism \( \sigma \in G \) such that \( \sigma {\mathfrak{P}}_{i} = {\mathfrak{P}}_{j} \) . Map \( {O}_{{\mathfrak{P}}_{i}}/{\mathfrak{P}}_{i} \rightarrow {O}_{{\mathfrak{P}}_{j}}/{\mathfrak{P}}_{j} \) by \( \bar{\omega } \rightarrow \overline{\sigma \omega }... | Yes |
Lemma 9.4. The order of \( Z\left( {\mathfrak{P}/P}\right) \) is \( e\left( {\mathfrak{P}/P}\right) f\left( {\mathfrak{P}/P}\right) \) . | Proof. By Proposition 9.2, the group \( G \) acts transitively on the set of primes of \( L \) lying above \( P \) . The group \( Z\left( {\mathfrak{P}/P}\right) \) is the isotropy group for this action. From this it follows that \( \left\lbrack {G : Z\left( {\mathfrak{P}/P}\right) }\right\rbrack = g\left( P\right) \),... | Yes |
Lemma 9.5. With the above notation, \( \mathfrak{P} \) is the only prime in \( L \) lying above \( \mathfrak{p} \). Moreover, \( e\left( {\mathfrak{p}/P}\right) = f\left( {\mathfrak{p}/P}\right) = 1 \) and \( \left\lbrack {M : K}\right\rbrack = g\left( P\right) \). | Proof. The first assertion follows by applying Proposition 9.2 to the Galois extension \( L/M \) and using the definition of the decomposition group. By Lemma 9.4, \( \# Z\left( {\mathfrak{P}/\mathfrak{p}}\right) = e\left( {\mathfrak{P}/\mathfrak{p}}\right) f\left( {\mathfrak{P}/\mathfrak{p}}\right) \). On the other ha... | Yes |
Proposition 9.7. Suppose \( L/K \) is a Galois extension of function fields and suppose \( \mathfrak{P} \) is a prime of \( L \) lying above a prime \( P \) of \( K \) . Let \( \sigma \in \operatorname{Gal}\left( {L/K}\right) \) . Then, \( Z\left( {\sigma \mathfrak{P}/P}\right) = {\sigma Z}\left( {\mathfrak{P}/P}\right... | Proof. By definition, \( \tau \in Z\left( {\sigma \mathfrak{P}/P}\right) \) if and only if \( {\tau \sigma }\mathfrak{P} = \sigma \mathfrak{P} \) . This is so if and only if \( {\sigma }^{-1}{\tau \sigma }\mathfrak{P} = \mathfrak{P} \), i.e., if and only if \( {\sigma }^{-1}{\tau \sigma } \in Z\left( {\mathfrak{P}/P}\r... | Yes |
Proposition 9.8. Let \( L/K \) be a Galois extension of function fields and \( M \) an arbitrary intermediate field. Let \( \mathfrak{P} \) be a prime of \( L \) and \( \mathfrak{p} \) and \( P \) the primes of \( M \) and \( K \) respectively which lie below \( \mathfrak{P} \) . Set \( H = \operatorname{Gal}\left( {L/... | Proof. Part \( \left( i\right) \) of the proposition follows directly from the definitions.\n\nTo prove part \( \left( {ii}\right) \) we first remark that from the definitions it is easy to prove that \( \rho \) maps \( Z\left( {\mathfrak{P}/P}\right) \) to \( Z\left( {\mathfrak{p}/\mathfrak{p}}\right) \) . The kernel ... | Yes |
Proposition 9.9. Let \( {M}_{1} \) and \( {M}_{2} \) be two Galois extensions of a function field \( K \) and let \( L = {M}_{1}{M}_{2} \) be the compositum. A prime \( P \) of \( K \) splits completely in \( L \) if and only if it splits completely in \( {M}_{1} \) and \( {M}_{2} \) . A prime \( P \) of \( K \) is unr... | Proof. Let \( \mathfrak{P} \) be some prime of \( L \) lying above \( P \) . If \( P \) splits completely in \( L \), then by the previous remarks \( Z\left( {\mathfrak{P}/P}\right) = \left( e\right) \) . Let \( {\mathfrak{p}}_{1} \) and \( {\mathfrak{p}}_{2} \) be the primes of \( {M}_{1} \) and \( {M}_{2} \), respect... | No |
Proposition 9.10. Let \( L/K \) be a Galois extension of global function fields, \( \mathfrak{P} \) a prime of \( L \) and \( P \) the prime of \( K \) lying below it. Suppose \( \mathfrak{P}/P \) is unramified. Then, \( \left( {\mathfrak{P}, L/K}\right) \) is a cyclic generator of \( Z\left( {\mathfrak{P}/P}\right) \)... | Proof. The first assertion is true by the definition of the Frobenius automorphism via the isomorphism \( Z\left( {\mathfrak{P}/P}\right) \cong \operatorname{Gal}\left( {{\mathbb{E}}_{\mathfrak{P}}/{\mathbb{F}}_{P}}\right) \) . To prove the second assertion, recall that for \( \sigma \in G \) we have \( \sigma {O}_{\ma... | Yes |
Proposition 9.11. Let \( L/K \) be a Galois extension of global function fields and \( M \) an arbitrary intermediate field. Let \( \mathfrak{P} \) be a prime of \( L \) and \( \mathfrak{p} \) and \( P \) the primes lying below it in \( M \) and \( K \), respectively. Assume \( \mathfrak{P}/P \) is unramified. Then,\n\... | Proof. Using the characterization\n\n\[ \n\left( {\mathfrak{P}, L/K}\right) \omega \equiv {\omega }^{NP}\;\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \;\forall \omega \in {O}_{\mathfrak{P}},\n\]\n\nwe deduce\n\n\[ \n{\left( \mathfrak{P}, L/K\right) }^{f\left( {\mathfrak{p}/P}\right) }\omega \equiv {\omega }^{N{P... | Yes |
Proposition 9.13. Let \( L/K \) be a Galois extension of global function fields. The Dirichlet density of the set of primes in \( K \) which split in \( L \) is given by \( \delta \left( {\{ L\} }\right) = 1/\left\lbrack {L : K}\right\rbrack \) . If \( {L}_{1} \) and \( {L}_{2} \) are two Galois extensions of \( K \) a... | Proof. We consider the zeta function of \( L \) . We have,\n\n\[ \log {\zeta }_{L}\left( s\right) = \mathop{\sum }\limits_{{\mathfrak{P} \in {\mathcal{S}}_{L}}}\mathop{\sum }\limits_{{k = 1}}^{\infty }{k}^{-1}N{\mathfrak{P}}^{-{ks}} = \mathop{\sum }\limits_{\mathfrak{P}}N{\mathfrak{P}}^{-s} + \mathop{\sum }\limits_{\ma... | Yes |
Lemma 9.14. Let \( \left( {V,\rho }\right) \) be a representation of \( G = \operatorname{Gal}\left( {L/K}\right) \) where \( L/K \) is a Galois extension of global function fields. Let \( P \) be a prime of \( K \) unramified in \( L \) . Then\n\n\[ \log {L}_{P}\left( {s,\chi }\right) = \mathop{\sum }\limits_{{k = 1}}... | Proof. If \( \left\{ {{\alpha }_{1}\left( P\right) ,{\alpha }_{2}\left( P\right) ,\cdots ,{\alpha }_{m}\left( P\right) }\right\} \) are the eigenvalues of \( \left( {\mathfrak{P}, L/K}\right) \), then we showed earlier that \( {L}_{P}{\left( s,\chi \right) }^{-1} = \mathop{\prod }\limits_{{i = 1}}^{m}\left( {1 - {\alph... | Yes |
Proposition 9.15. With the above notation and conventions, \( L\left( {s,\chi }\right) \) converges absolutely in the region \( \Re \left( s\right) > 1 \) and for every \( \delta > 0 \) it converges absolutely and uniformly in the region \( \Re \left( s\right) \geq 1 + \delta \) . Consequently, \( L\left( {s,\chi }\rig... | Proof. An infinite product \( \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 + {a}_{n}}\right) \) converges absolutely if and only if \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\left| {a}_{n}\right| \) converges. Using the local decomposition \( {L}_{P}\left( s\right) = \mathop{\prod }\limits_{{i = 1}}^{m}(1 - \)... | Yes |
Lemma 9.17. Let \( G \) be a finite group and \( C \subset G \) a conjugacy class of \( G \) . Let \( \sigma \in C \) and \( \tau \in G \) . Then\n\n\[ \mathop{\sum }\limits_{\chi }\overline{\chi \left( \sigma \right) }\chi \left( \tau \right) = 0\text{ if }\tau \notin C\text{ and }\frac{\# G}{\# C}\text{ if }\tau \in ... | Proof. This is one of the two standard othogonality relations among characters of finite groups. See Lang [4] or Serre [3]. | No |
Proposition 9.18. The Artin map \( \left( {*, L/K}\right) : {\mathcal{D}}_{K}^{\prime } \rightarrow G \) is onto and the kernel contains the group \( {N}_{L/K}{\mathcal{D}}_{L}^{\prime } \) where \( {\mathcal{D}}_{L}^{\prime } \) is the subgroup of \( {\mathcal{D}}_{L} \) generated by primes of \( L \) unramified over ... | Proof. Let \( {G}^{\prime } \) denote the image of \( \left( {*, L/K}\right) \) and \( M \subset L \) the fixed field of \( {G}^{\prime } \) . If \( P \in {\mathcal{S}}_{K}^{\prime } \) then, by Proposition 9.11, \( {\left. \left( P, L/K\right) \right| }_{M} = \left( {P, M/K}\right) \) . By definition, \( {\left. \left... | Yes |
Proposition 9.19. Let \( L = K\mathbb{E} \) where \( \mathbb{E} \) is an extension of \( \mathbb{F} \) of degree m. Let \( P \) be any prime of \( K \) . Then \( \left( {P, L/K}\right) = {\phi }_{q}^{{\deg }_{K}P} \) . | Proof. Every prime of \( K \) is unramified in \( L \) since \( L \) is a constant field extension. See Proposition 8.5.\n\nSuppose \( \alpha \in \mathbb{E} \) . From the definition, \( \left( {P, L/K}\right) \alpha \equiv {\alpha }^{NP}\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \), where \( \mathfrak{P} \) is ... | Yes |
Proposition 9.20. Maintaining the notation of the previous proposition, the Artin map \( \left( {*, L/K}\right) : {\mathcal{D}}_{K} \rightarrow \operatorname{Gal}\left( {L/K}\right) \cong \operatorname{Gal}\left( {\mathbb{E}/\mathbb{F}}\right) \) is onto and the kernel is the group \( {\mathcal{D}}_{K}^{o}{\mathcal{D}}... | Proof. We already know that the map is onto. To determine the kernel we note that the Artin map is given by \( \left( {D, L/K}\right) = {\phi }_{q}^{{\deg }_{K}D} \) for \( D \in {\mathcal{D}}_{K} \) . This is true for prime divisors by the previous proposition and it follows in general by linearity. From this we see \... | Yes |
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