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Proposition 6.20. Let \( \mathcal{F} \) be the Borel subsets of \( \mathbb{C} \) and suppose \( E : \mathcal{F} \rightarrow \mathcal{B}\left( \mathcal{H}\right) \) is a spectral measure with compact support. Set \( A = \int {zdE} \) and suppose \( {\lambda }_{0} \in \mathbb{C} \) . We have \[ E\left( \left\{ {\lambda }... | Proof. Let \( D \) be a closed disk in \( \mathbb{C} \) containing both \( {\lambda }_{0} \) and the support of \( E \) . Clearly \( D \) is a carrier of \( E \) . If we set \( f = z{\chi }_{D} \) , \[ A = \int {fdE} = \pi \left( f\right) \] By Proposition 6.19, \[ \ker \left( {A - {\lambda }_{0}I}\right) = \ker \pi \l... | Yes |
Theorem 6.23 (Spectral Theorem, Spectral Measure Version). Let A be a normal operator in \( \mathcal{B}\left( \mathcal{H}\right) \), where \( \mathcal{H} \) is a separable Hilbert space. Let \( \mathcal{F} \) denote the Borel subsets of \( \sigma \left( A\right) \) . There is a unique spectral measure \( F : \mathcal{F... | Proof. We address the existence part of the statement first. By Theorem 6.2 we know that there is a \( \sigma \) -finite measure space \( \left( {X, v}\right) \), a unitary operator \( W : {L}^{2}\left( {X, v}\right) \rightarrow \) \( \mathcal{H} \), and a \( \varphi \in {L}^{\infty }\left( {X, v}\right) \) so that \( ... | No |
The simplest example of an algebraic function field is the rational function field; \( F/K \) is called rational if \( F = K\left( x\right) \) for some \( x \in F \) which is transcendental over \( K \) . Each element \( 0 \neq z \in K\left( x\right) \) has a unique representation | \[ z = a \cdot \mathop{\prod }\limits_{i}{p}_{i}{\left( x\right) }^{{n}_{i}} \] in which \( 0 \neq a \in K \), the polynomials \( {p}_{i}\left( x\right) \in K\left\lbrack x\right\rbrack \) are monic, pairwise distinct and irreducible and \( {n}_{i} \in \mathbb{Z} \). | Yes |
Proposition 1.1.5. Let \( \mathcal{O} \) be a valuation ring of the function field \( F/K \) . Then the following hold:\n\n(a) \( \mathcal{O} \) is a local ring; i.e., \( \mathcal{O} \) has a unique maximal ideal \( P = \mathcal{O} \smallsetminus {\mathcal{O}}^{ \times } \), where \( {\mathcal{O}}^{ \times } = \{ z \in... | Proof. (a) We claim that \( P \mathrel{\text{:=}} \mathcal{O} \smallsetminus {\mathcal{O}}^{ \times } \) is an ideal of \( \mathcal{O} \) (from this it follows at once that \( P \) is the unique maximal ideal since a proper ideal of \( \mathcal{O} \) cannot contain a unit).\n\n(1) Let \( x \in P, z \in \mathcal{O} \) .... | Yes |
Lemma 1.1.7. Let \( \mathcal{O} \) be a valuation ring of the algebraic function field \( F/K \) , let \( P \) be its maximal ideal and \( 0 \neq x \in P \) . Let \( {x}_{1},\ldots ,{x}_{n} \in P \) be such that \( {x}_{1} = x \) and \( {x}_{i} \in {x}_{i + 1}P \) for \( i = 1,\ldots, n - 1 \) . Then we have\n\n\[ n \l... | Proof. From Remark 1.1.2 and Proposition 1.1.5(c) follows that \( F/K\left( x\right) \) is a finite extension, so it is sufficient to prove that \( {x}_{1},\ldots ,{x}_{n} \) are linearly independent over \( K\left( x\right) \) . Suppose there is a non-trivial linear combination \( \mathop{\sum }\limits_{{i = 1}}^{n}{\... | Yes |
Lemma 1.1.11 (Strict Triangle Inequality). Let \( v \) be a discrete valuation of \( F/K \) and let \( x, y \in F \) with \( v\left( x\right) \neq v\left( y\right) \) . Then \( v\left( {x + y}\right) = \min \{ v\left( x\right), v\left( y\right) \} \) . | Proof. Observe that \( v\left( {ay}\right) = v\left( y\right) \) for \( 0 \neq a \in K \) (by (2) and (5)), in particular \( v\left( {-y}\right) = v\left( y\right) \) . Since \( v\left( x\right) \neq v\left( y\right) \) we can assume \( v\left( x\right) < v\left( y\right) \) . Suppose that \( v\left( {x + y}\right) \ne... | Yes |
Theorem 1.1.13. Let \( F/K \) be a function field.\n\n(a) For a place \( P \in {\mathbb{P}}_{F} \), the function \( {v}_{P} \) defined above is a discrete valuation of \( F/K \). Moreover we have\n\n\[ \n{\mathcal{O}}_{P} = \left\{ {z \in F \mid {v}_{P}\left( z\right) \geq 0}\right\} \n\]\n\n\[ \n{\mathcal{O}}_{P}^{ \t... | Proof. (a) Obviously \( {v}_{P} \) has the properties (1),(2),(4) and (5) of Definition 1.1.9. In order to prove the Triangle Inequality (3) consider \( x, y \in F \) with \( {v}_{P}\left( x\right) = n,{v}_{P}\left( y\right) = m \) . We can assume that \( n \leq m < \infty \), thus \( x = {t}^{n}{u}_{1} \) and \( y = {... | Yes |
Proposition 1.1.15. If \( P \) is a place of \( F/K \) and \( 0 \neq x \in P \) then\n\n\[ \deg P \leq \left\lbrack {F : K\left( x\right) }\right\rbrack < \infty . \] | Proof. First we observe that \( \left\lbrack {F : K\left( x\right) }\right\rbrack < \infty \) by Remark 1.1.2. Thus it suffices to show that any elements \( {z}_{1},\ldots ,{z}_{n} \in {\mathcal{O}}_{P} \), whose residue classes \( {z}_{1}\left( P\right) ,\ldots ,{z}_{n}\left( P\right) \in {F}_{P} \) are linearly indep... | Yes |
Corollary 1.1.16. The field \( \widetilde{K} \) of constants of \( F/K \) is a finite field extension of \( K \) . | Proof. We use the fact that \( {\mathbb{P}}_{F} \neq \varnothing \) (which will be proved only in Corollary 1.1.20). Choose some \( P \in {\mathbb{P}}_{F} \) . Since \( \widetilde{K} \) is embedded into \( {F}_{P} \) via the residue class map \( {\mathcal{O}}_{P} \rightarrow {F}_{P} \), it follows that \( \left\lbrack ... | No |
Corollary 1.1.20. Let \( F/K \) be a function field, \( z \in F \) transcendental over \( K \) . Then \( z \) has at least one zero and one pole. In particular \( {\mathbb{P}}_{F} \neq \varnothing \) . | Proof. Consider the ring \( R = K\left\lbrack z\right\rbrack \) and the ideal \( I = {zK}\left\lbrack z\right\rbrack \) . Theorem 1.1.19 ensures that there is a place \( P \in {\mathbb{P}}_{F} \) with \( z \in P \), hence \( P \) is a zero of \( z \) . The same argument proves that \( {z}^{-1} \) has a zero \( Q \in {\... | Yes |
Proposition 1.2.1. Let \( F = K\left( x\right) \) be the rational function field.\n\n(a) Let \( P = {P}_{p\left( x\right) } \in {\mathbb{P}}_{K\left( x\right) } \) be the place defined by (1.8), where \( p\left( x\right) \in K\left\lbrack x\right\rbrack \) is an irreducible polynomial. Then \( p\left( x\right) \) is a ... | Proof. We prove only some essentials of this proposition; the remaining parts of the proof are straightforward.\n\n(a) Let \( P = {P}_{p\left( x\right) }, p\left( x\right) \in K\left\lbrack x\right\rbrack \) irreducible. The ideal \( {P}_{p\left( x\right) } \subseteq {\mathcal{O}}_{p\left( x\right) } \) is obviously ge... | No |
Corollary 1.2.3. The places of \( K\left( x\right) /K \) of degree one are in \( 1 - 1 - \) correspondence with \( K \cup \{ \infty \} \) . | The corollary is obvious by Proposition 1.2.1 and Theorem 1.2.2. In terms of algebraic geometry (cf. Appendix B) \( K \cup \{ \infty \} \) is usually interpreted as the projective line \( {\mathbf{P}}^{1}\left( K\right) \) over \( K \), hence the places of \( K\left( x\right) /K \) of degree one correspond in a one-to-... | No |
Corollary 1.3.2. Every function field has infinitely many places. | Proof of Corollary 1.3.2. Suppose there are only finitely many places, say \( {P}_{1},\ldots ,{P}_{n} \) . By Theorem 1.3.1 we find a non-zero element \( x \in F \) with \( {v}_{{P}_{i}}\left( x\right) > \) 0 for \( i = 1,\ldots, n \) . Then \( x \) is transcendental over \( K \) since it has zeros. But \( x \) has no ... | Yes |
Corollary 1.3.4. In a function field \( F/K \) every element \( 0 \neq x \in F \) has only finitely many zeros and poles. | Proof. If \( x \) is constant, \( x \) has neither zeros nor poles. If \( x \) is transcendental over \( K \), the number of zeros is \( \leq \left\lbrack {F : K\left( x\right) }\right\rbrack \) by Proposition 1.3.3. The same argument shows that \( {x}^{-1} \) has only a finite number of zeros. | Yes |
Lemma 1.4.6. Let \( A \in \operatorname{Div}\left( F\right) \) . Then we have:\n\n(a) \( \mathcal{L}\left( A\right) \) is a vector space over \( K \) .\n\n(b) If \( {A}^{\prime } \) is a divisor equivalent to \( A \), then \( \mathcal{L}\left( A\right) \simeq \mathcal{L}\left( {A}^{\prime }\right) \) (isomorphic as vec... | Proof. (a) Let \( x, y \in \mathcal{L}\left( A\right) \) and \( a \in K \) . Then for all \( P \in {\mathbb{P}}_{F},{v}_{P}\left( {x + y}\right) \geq \) \( \min \left\{ {{v}_{P}\left( x\right) ,{v}_{P}\left( y\right) }\right\} \geq - {v}_{P}\left( A\right) \) and \( {v}_{P}\left( {ax}\right) = {v}_{P}\left( a\right) + ... | Yes |
Lemma 1.4.7. (a) \( \mathcal{L}\left( 0\right) = K \) .\n\n(b) If \( A < 0 \) then \( \mathcal{L}\left( A\right) = \{ 0\} \) . | Proof. (a) We have \( \left( x\right) = 0 \) for \( 0 \neq x \in K \), therefore \( K \subseteq \mathcal{L}\left( 0\right) \) . Conversely, if \( 0 \neq x \in \mathcal{L}\left( 0\right) \) then \( \left( x\right) \geq 0 \) . This means that \( x \) has no pole, so \( x \in K \) by Corollary 1.1.20.\n\n(b) Assume there ... | Yes |
Lemma 1.4.8. Let \( A, B \) be divisors of \( F/K \) with \( A \leq B \) . Then we have \( \mathcal{L}\left( A\right) \subseteq \mathcal{L}\left( B\right) \) and\n\n\[ \dim \left( {\mathcal{L}\left( B\right) /\mathcal{L}\left( A\right) }\right) \leq \deg B - \deg A. \] | Proof. \( \mathcal{L}\left( A\right) \subseteq \mathcal{L}\left( B\right) \) is trivial. In order to prove the other assertion we can assume that \( B = A + P \) for some \( P \in {\mathbb{P}}_{F} \) ; the general case follows then by induction. Choose an element \( t \in F \) with \( {v}_{P}\left( t\right) = {v}_{P}\l... | Yes |
Proposition 1.4.9. For each divisor \( A \in \operatorname{Div}\left( F\right) \) the space \( \mathcal{L}\left( A\right) \) is a finite-dimensional vector space over \( K \) . More precisely: if \( A = {A}_{ + } - {A}_{ - } \) with positive divisors \( {A}_{ + } \) and \( {A}_{ - } \), then\n\n\[ \dim \mathcal{L}\left... | Proof. Since \( \mathcal{L}\left( A\right) \subseteq \mathcal{L}\left( {A}_{ + }\right) \), it is sufficient to show that\n\n\[ \dim \mathcal{L}\left( {A}_{ + }\right) \leq \deg {A}_{ + } + 1 \]\n\nWe have \( 0 \leq {A}_{ + } \), so Lemma 1.4.8 yields \( \dim \left( {\mathcal{L}\left( {A}_{ + }\right) /\mathcal{L}\left... | Yes |
Theorem 1.4.11. All principal divisors have degree zero. More precisely: let \( x \in F \smallsetminus K \) and \( {\left( x\right) }_{0} \) resp. \( {\left( x\right) }_{\infty } \) denote the zero resp. pole divisor of \( x \) . Then | \[ \deg {\left( x\right) }_{0} = \deg {\left( x\right) }_{\infty } = \left\lbrack {F : K\left( x\right) }\right\rbrack . \] | No |
Corollary 1.4.12. (a) Let \( A,{A}^{\prime } \) be divisors with \( A \sim {A}^{\prime } \) . Then we have \( \ell \left( A\right) = \ell \left( {A}^{\prime }\right) \) and \( \deg A = \deg {A}^{\prime } \) . | Proof of Corollary 1.4.12. (a) follows immediately from Lemma 1.4.6 and Theorem 1.4.11. | Yes |
Proposition 1.4.14. There is a constant \( \gamma \in \mathbb{Z} \) such that for all divisors \( A \in \operatorname{Div}\left( F\right) \) the following holds:\n\n\[ \deg A - \ell \left( A\right) \leq \gamma \] | Proof. To begin with, observe that\n\n\[ {A}_{1} \leq {A}_{2} \Rightarrow \deg {A}_{1} - \ell \left( {A}_{1}\right) \leq \deg {A}_{2} - \ell \left( {A}_{2}\right) \]\n\n(1.22)\n\nby Lemma 1.4.8. We fix an element \( x \in F \smallsetminus K \) and consider the specific divisor \( B \mathrel{\text{:=}} {\left( x\right) ... | Yes |
Corollary 1.4.16. The genus of \( F/K \) is a non-negative integer. | Proof. In the definition of \( g \), put \( A = 0 \) . Then \( \deg \left( 0\right) - \ell \left( 0\right) + 1 = 0 \), hence \( g \geq 0 \) . | Yes |
Theorem 1.4.17 (Riemann's Theorem). Let \( F/K \) be a function field of genus \( g \) . Then we have:\n\n(a) For all divisors \( A \in \operatorname{Div}\left( F\right) \) ,\n\n\[ \ell \left( A\right) \geq \deg A + 1 - g.\]\n\n(b) There is an integer \( c \), depending only on the function field \( F/K \), such that\n... | Proof. (a) This is just the definition of the genus.\n\n(b) Choose a divisor \( {A}_{0} \) with \( g = \deg {A}_{0} - \ell \left( {A}_{0}\right) + 1 \) and set \( c \mathrel{\text{:=}} \deg {A}_{0} + g \) . If \( \deg A \geq c \) then\n\n\[ \ell \left( {A - {A}_{0}}\right) \geq \deg \left( {A - {A}_{0}}\right) + 1 - g ... | Yes |
We want to show that the rational function field \( K\left( x\right) /K \) has genus \( g = 0 \). | In order to prove this, let \( {P}_{\infty } \) denote the pole divisor of \( x \) (notation as in Proposition 1.2.1). Consider for \( r \geq 0 \) the vector space \( \mathcal{L}\left( {r{P}_{\infty }}\right) \) . Obviously the elements \( 1, x,\ldots ,{x}^{r} \) are in \( \mathcal{L}\left( {r{P}_{\infty }}\right) \), ... | Yes |
Corollary 1.5.5. \( \;g = \dim \left( {{\mathcal{A}}_{F}/\left( {{\mathcal{A}}_{F}\left( 0\right) + F}\right) }\right) \) . | Proof of Corollary 1.5.5. \( \;i\left( 0\right) = \ell \left( 0\right) - \deg \left( 0\right) + g - 1 = 1 - 0 + g - 1 = g \) . | Yes |
Lemma 1.5.7. For \( A \in \operatorname{Div}\left( F\right) \) we have \( \dim {\Omega }_{F}\left( A\right) = i\left( A\right) \) . | Proof. \( {\Omega }_{F}\left( A\right) \) is in a natural way isomorphic to the space of linear forms on \( {\mathcal{A}}_{F}/\left( {{\mathcal{A}}_{F}\left( A\right) + F}\right) \) . Since \( {\mathcal{A}}_{F}/\left( {{\mathcal{A}}_{F}\left( A\right) + F}\right) \) is finite-dimensional of dimension \( i\left( A\right... | Yes |
Proposition 1.5.9. \( {\Omega }_{F} \) is a one-dimensional vector space over \( F \). | Proof. Choose \( 0 \neq {\omega }_{1} \in {\Omega }_{F} \) (we already know that \( {\Omega }_{F} \neq 0 \) ). It has to be shown that for every \( {\omega }_{2} \in {\Omega }_{F} \) there is some \( z \in F \) with \( {\omega }_{2} = z{\omega }_{1} \) . We can assume that \( {\omega }_{2} \neq 0 \) . Choose \( {A}_{1}... | Yes |
Lemma 1.5.10. Let \( 0 \neq \omega \in {\Omega }_{F} \) . Then there is a uniquely determined divisor \( W \in M\left( \omega \right) \) such that \( A \leq W \) for all \( A \in M\left( \omega \right) \) . | Proof. By Riemann’s Theorem there exists a constant \( c \), depending only on the function field \( F/K \), with the property \( i\left( A\right) = 0 \) for all \( A \in \operatorname{Div}\left( F\right) \) of degree \( \geq c \) . Since \( \dim \left( {{\mathcal{A}}_{F}/\left( {{\mathcal{A}}_{F}\left( A\right) + F}\r... | Yes |
Proposition 1.5.13. (a) For \( 0 \neq x \in F \) and \( 0 \neq \omega \in {\Omega }_{F} \) we have \( \left( {x\omega }\right) = \) \( \left( x\right) + \left( \omega \right) \) . | Proof of Proposition 1.5.13. If \( \omega \) vanishes on \( {\mathcal{A}}_{F}\left( A\right) + F \) then \( {x\omega } \) vanishes on \( {\mathcal{A}}_{F}\left( {A + \left( x\right) }\right) + F \), consequently\n\n\[ \left( \omega \right) + \left( x\right) \leq \left( {x\omega }\right) . \]\n\nLikewise \( \left( {x\om... | Yes |
Theorem 1.5.14 (Duality Theorem). Let \( A \) be an arbitrary divisor and \( W = \left( \omega \right) \) be a canonical divisor of \( F/K \) . Then the mapping \[ \mu : \left\{ \begin{matrix} \mathcal{L}\left( {W - A}\right) & \rightarrow & {\Omega }_{F}\left( A\right) , \\ x & \mapsto & {x\omega } \end{matrix}\right.... | Proof. For \( x \in \mathcal{L}\left( {W - A}\right) \) we have \[ \left( {x\omega }\right) = \left( x\right) + \left( \omega \right) \geq - \left( {W - A}\right) + W = A, \] hence \( {x\omega } \in {\Omega }_{F}\left( A\right) \) by Remark 1.5.12. Therefore \( \mu \) maps \( \mathcal{L}\left( {W - A}\right) \) into \(... | Yes |
Theorem 1.5.15 (Riemann-Roch Theorem). Let \( W \) be a canonical divisor of \( F/K \) . Then for each divisor \( A \in \operatorname{Div}\left( F\right) \) , \[ \ell \left( A\right) = \deg A + 1 - g + \ell \left( {W - A}\right) . \] | Proof. This is an immediate consequence of Theorem 1.5.14 and the definition of \( i\left( A\right) \) . | No |
Corollary 1.5.16. For a canonical divisor \( W \) we have\n\n\[ \deg W = {2g} - 2\;\text{ and }\;\ell \left( W\right) = g. \] | Proof. For \( A = 0 \), the Riemann-Roch Theorem and Lemma 1.4.7 give\n\n\[ 1 = \ell \left( 0\right) = \deg 0 + 1 - g + \ell \left( {W - 0}\right) . \]\n\nThus \( \ell \left( W\right) = g \) . Setting \( A = W \) we obtain\n\n\[ g = \ell \left( W\right) = \deg W + 1 - g + \ell \left( {W - W}\right) = \deg W + 2 - g. \]... | Yes |
Theorem 1.5.17. If \( A \) is a divisor of \( F/K \) of degree \( \deg A \geq {2g} - 1 \) then\n\n\[ \ell \left( A\right) = \deg A + 1 - g. \] | Proof. We have \( \ell \left( A\right) = \deg A + 1 - g + \ell \left( {W - A}\right) \), where \( W \) is a canonical divisor. Since \( \deg A \geq {2g} - 1 \) and \( \deg W = {2g} - 2 \), we conclude that \( \deg \left( {W - A}\right) < 0 \) . It follows from Corollary 1.4.12 that \( \ell \left( {W - A}\right) = 0 \). | Yes |
Proposition 1.6.1. Suppose that \( {g}_{0} \in \mathbb{Z} \) and \( {W}_{0} \in \operatorname{Div}\left( F\right) \) satisfy\n\n\[ \ell \left( A\right) = \deg A + 1 - {g}_{0} + \ell \left( {{W}_{0} - A}\right) \]\n\n(1.33)\n\nfor all \( A \in \operatorname{Div}\left( F\right) \) . Then \( {g}_{0} = g \), and \( {W}_{0}... | Proof. Setting \( A = 0 \) resp. \( A = {W}_{0} \) in (1.33) we obtain \( \ell \left( {W}_{0}\right) = {g}_{0} \) and \( \deg {W}_{0} = 2{g}_{0} - 2 \) (cf. the proof of Corollary 1.5.16). Let \( W \) be a canonical divisor of \( F/K \) . We choose a divisor \( A \) with \( \deg A > \max \left\{ {{2g} - 2,2{g}_{0} - 2}... | Yes |
Proposition 1.6.2. A divisor \( B \) is canonical if and only if \( \deg B = {2g} - 2 \) and \( \ell \left( B\right) \geq g \) . | Proof. Suppose that \( \deg B = {2g} - 2 \) and \( \ell \left( B\right) \geq g \) . Choose a canonical divisor \( W \) . Then\n\n\[ g \leq \ell \left( B\right) = \deg B + 1 - g + \ell \left( {W - B}\right) = g - 1 + \ell \left( {W - B}\right) ,\]\n\ntherefore \( \ell \left( {W - B}\right) \geq 1 \) . Since \( \deg \lef... | Yes |
Proposition 1.6.3. For a function field \( F/K \) the following conditions are equivalent:\n\n(1) \( F/K \) is rational; i.e., \( F = K\\left( x\\right) \) for some \( x \) which is transcendental over the field \( K \) .\n\n(2) \( F/K \) has genus 0, and there is some divisor \( A \\in \\operatorname{Div}\\left( F\\ri... | Proof. \( \\left( 1\\right) \\Rightarrow \\left( 2\\right) \) : See Example 1.4.18.\n\n\( \\left( 2\\right) \\Rightarrow \\left( 1\\right) \) : Let \( g = 0 \) and \( \\deg A = 1 \) . As \( \\deg A \\geq {2g} - 1 \) we have that \( \\ell \\left( A\\right) = \\deg A + 1 - g = 2 \) by Theorem 1.5.17. Thus \( A \\sim {A}^... | Yes |
Theorem 1.6.5 (Strong Approximation Theorem). Let \( S \subsetneqq {\mathbb{P}}_{F} \) be a proper subset of \( {\mathbb{P}}_{F} \) and \( {P}_{1},\ldots ,{P}_{r} \in S \) . Suppose there are given elements \( {x}_{1},\ldots ,{x}_{r} \in F \) and integers \( {n}_{1},\ldots ,{n}_{r} \in \mathbb{Z} \) . Then there exists... | Proof. Consider the adele \( \alpha = {\left( {\alpha }_{P}\right) }_{P \in {\mathbb{P}}_{F}} \) with\n\n\[ \n{\alpha }_{P} \mathrel{\text{:=}} \left\{ \begin{array}{ll} {x}_{i} & \text{ for }P = {P}_{i}, i = 1,\ldots, r, \\ 0 & \text{ otherwise } \end{array}\right. \n\]\n\nChoose a place \( Q \in {\mathbb{P}}_{F} \sma... | Yes |
Proposition 1.6.6. Let \( P \in {\mathbb{P}}_{F} \) . Then for each \( n \geq {2g} \) there exists an element \( x \in F \) with pole divisor \( {\left( x\right) }_{\infty } = {nP} \) . | Proof. By Theorem 1.5.17 we know that \( \ell \left( {\left( {n - 1}\right) P}\right) = \left( {n - 1}\right) \deg P + 1 - g \) and \( \ell \left( {nP}\right) = n \cdot \deg P + 1 - g \), hence \( \mathcal{L}\left( {\left( {n - 1}\right) P}\right) \subsetneqq \mathcal{L}\left( {nP}\right) \) . Every element \( x \in \m... | Yes |
Theorem 1.6.8 (Weierstrass Gap Theorem). Suppose that \( F/K \) has genus \( g > 0 \) and \( P \) is a place of degree one. Then there are exactly \( g \) gap numbers \( {i}_{1} < \ldots < {i}_{g} \) of \( P \) . We have\n\n\[ \n{i}_{1} = 1\;\text{ and }\;{i}_{g} \leq {2g} - 1.\n\] | Proof. Each gap number of \( P \) is \( \leq {2g} - 1 \) by Proposition 1.6.6, and 0 is a pole number. We have the following obvious characterization of gap numbers:\n\n\[ \ni\text{is a gap number of}P \Leftrightarrow \mathcal{L}\left( {\left( {i - 1}\right) P}\right) = \mathcal{L}\left( {iP}\right) \text{.}\]\n\nConsi... | Yes |
Proposition 1.6.12. Suppose that \( T \subseteq {\mathbb{P}}_{F} \) is a set of places of degree one such that \( \left| T\right| \geq g \) . Then there exists a non-special divisor \( B \geq 0 \) with \( \deg B = \) \( g \) and \( \operatorname{supp}B \subseteq T \) . | Proof. The crucial step of the proof is the following claim:\n\nClaim. Given \( g \) distinct places \( {P}_{1},\ldots ,{P}_{g} \in T \) and a divisor \( A \geq 0 \) with \( \ell \left( A\right) = 1 \) and \( \deg A \leq g - 1 \), there is an index \( j \in \{ 1,\ldots, g\} \) such that \( \ell \left( {A + {P}_{j}}\rig... | Yes |
Theorem 1.6.13 (Clifford’s Theorem). For all divisors \( A \) with \( 0 \leq \) \( \deg A \leq {2g} - 2 \) holds\n\n\[ \ell \left( A\right) \leq 1 + \frac{1}{2} \cdot \deg A \] | The main step in the proof of Clifford's Theorem is the following result.\n\nLemma 1.6.14. Suppose t | No |
Lemma 1.6.14. Suppose that \( A \) and \( B \) are divisors such that \( \ell \left( A\right) > 0 \) and \( \ell \left( B\right) > 0 \) . Then\n\n\[ \ell \left( A\right) + \ell \left( B\right) \leq 1 + \ell \left( {A + B}\right) . \] | Proof of Lemma 1.6.14. Since \( \ell \left( A\right) > 0 \) and \( \ell \left( B\right) > 0 \) we can find \( {A}_{0},{B}_{0} \geq 0 \) with \( A \sim {A}_{0} \) and \( B \sim {B}_{0} \) (cf. Remark 1.4.5). The set\n\n\[ X \mathrel{\text{:=}} \left\{ {D \in \operatorname{Div}\left( F\right) \mid D \leq {A}_{0}\text{ an... | Yes |
Proposition 1.7.2. Let \( \omega \in {\Omega }_{F} \) and \( \alpha = \left( {\alpha }_{P}\right) \in {\mathcal{A}}_{F} \) . Then \( {\omega }_{P}\left( {\alpha }_{P}\right) \neq 0 \) for at most finitely many places \( P \), and\n\n\[ \omega \left( \alpha \right) = \mathop{\sum }\limits_{{P \in {\mathbb{P}}_{F}}}{\ome... | Proof. We can assume that \( \omega \neq 0 \) and we set \( W \mathrel{\text{:=}} \left( \omega \right) \), the divisor of \( \omega \) (see Definition 1.5.11). There is a finite set \( S \subseteq {\mathbb{P}}_{F} \) such that\n\n\[ {v}_{P}\left( W\right) = 0\;\text{ and }\;{v}_{P}\left( {\alpha }_{P}\right) \geq 0\;\... | Yes |
Proposition 1.7.3. (a) Let \( \omega \neq 0 \) be a Weil differential of \( F/K \) and \( P \in {\mathbb{P}}_{F} \) . Then\n\n\[ \n{v}_{P}\left( \omega \right) = \max \left\{ {r \in \mathbb{Z} \mid {\omega }_{P}\left( x\right) = 0\text{ for all }x \in F\text{ with }{v}_{P}\left( x\right) \geq - r}\right\} .\n\]\n\nIn p... | Proof. (a) Recall that, by definition, \( {v}_{P}\left( \omega \right) = {v}_{P}\left( W\right) \) where \( W = \left( \omega \right) \) denotes the divisor of \( \omega \) . Let \( s \mathrel{\text{:=}} {v}_{P}\left( \omega \right) \) . For \( x \in F \) with \( {v}_{P}\left( x\right) \geq - s \) we have \( {\iota }_{... | Yes |
Proposition 2.1.8 (Singleton Bound). For an \( \left\lbrack {n, k, d}\right\rbrack \) code \( C \) holds\n\n\[ k + d \leq n + 1 \] | Proof. Consider the linear subspace \( E \subseteq {\mathbb{F}}_{q}^{n} \) given by\n\n\[ E \mathrel{\text{:=}} \left\{ {\left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {\mathbb{F}}_{q}^{n} \mid {a}_{i} = 0\;\text{ for all }\;i \geq d}\right\} . \]\n\nEvery \( a \in E \) has weight \( \leq d - 1 \), hence \( E \cap C = 0 \... | Yes |
Theorem 2.2.2. \( {C}_{\mathcal{L}}\left( {D, G}\right) \) is an \( \left\lbrack {n, k, d}\right\rbrack \) code with parameters\n\n\[ k = \ell \left( G\right) - \ell \left( {G - D}\right) \;\text{ and }\;d \geq n - \deg G. \] | Proof. The evaluation map (2.4) is a surjective linear map from \( \mathcal{L}\left( G\right) \) to \( {C}_{\mathcal{L}}\left( {D, G}\right) \) with kernel\n\n\[ \operatorname{Ker}\left( {\operatorname{ev}}_{D}\right) = \left\{ {x \in \mathcal{L}\left( G\right) \mid {v}_{{P}_{i}}\left( x\right) > 0\text{ for }i = 1,\ld... | Yes |
Corollary 2.2.3. Suppose that the degree of \( G \) is strictly less than \( n \) . Then the evaluation map \( {\operatorname{ev}}_{D} : \mathcal{L}\left( G\right) \rightarrow {C}_{\mathcal{L}}\left( {D, G}\right) \) is injective, and we have:\n\n(a) \( {C}_{\mathcal{L}}\left( {D, G}\right) \) is an \( \left\lbrack {n,... | Proof. By assumption we have \( \deg \left( {G - D}\right) = \deg G - n < 0 \), so \( \mathcal{L}\left( {G - D}\right) = 0 \) . Since \( \mathcal{L}\left( {G - D}\right) \) is the kernel of the evaluation map, this is an injective mapping. The remaining assertions are trivial consequences of Theorem 2.2.2 and the Riema... | Yes |
Theorem 2.2.8. The codes \( {C}_{\mathcal{L}}\left( {D, G}\right) \) and \( {C}_{\Omega }\left( {D, G}\right) \) are dual to each other; i.e., \[ {C}_{\Omega }\left( {D, G}\right) = {C}_{\mathcal{L}}{\left( D, G\right) }^{ \bot }.\] | Proof. First we note the following fact: Consider a place \( P \in {\mathbb{P}}_{F} \) of degree one, a Weil differential \( \omega \) with \( {v}_{P}\left( \omega \right) \geq - 1 \) and an element \( x \in F \) with \( {v}_{P}\left( x\right) \geq 0 \) . Then \[ {\omega }_{P}\left( x\right) = x\left( P\right) \cdot {\... | Yes |
Lemma 2.2.9. There exists a Weil differential \( \eta \) such that\n\n\[ \n{v}_{{P}_{i}}\left( \eta \right) = - 1\;\text{ and }\;{\eta }_{{P}_{i}}\left( 1\right) = 1\;\text{ for }i = 1,\ldots, n.\n\] | Proof. Choose an arbitrary Weil differential \( {\omega }_{0} \neq 0 \) . By the Weak Approximation Theorem there is an element \( z \in F \) with \( {v}_{{P}_{i}}\left( z\right) = - {v}_{{P}_{i}}\left( {\omega }_{0}\right) - 1 \) for \( i = 1,\ldots, n \) . Setting \( \omega \mathrel{\text{:=}} z{\omega }_{0} \) we ob... | Yes |
Proposition 2.2.10. Let \( \eta \) be a Weil differential such that \( {v}_{{P}_{i}}\left( \eta \right) = - 1 \) and \( {\eta }_{{P}_{i}}\left( 1\right) = 1 \) for \( i = 1,\ldots, n \) . Then\n\n\[ \n{C}_{\mathcal{L}}{\left( D, G\right) }^{ \bot } = {C}_{\Omega }\left( {D, G}\right) = {C}_{\mathcal{L}}\left( {D, H}\ri... | Proof. The equality \( {C}_{\mathcal{L}}{\left( D, G\right) }^{ \bot } = {C}_{\Omega }\left( {D, G}\right) \) was already shown in Theorem 2.2.8. Observe that \( \operatorname{supp}\left( {D - G + \left( \eta \right) }\right) \cap \operatorname{supp}D = \varnothing \) since \( {v}_{{P}_{i}}\left( \eta \right) = - 1 \) ... | Yes |
Corollary 2.2.11. Suppose there is a Weil differential \( \eta \) such that\n\n\[ \n{2G} - D \leq \left( \eta \right) \;\text{ and }\;{\eta }_{{P}_{i}}\left( 1\right) = 1\;\text{ for }\;i = 1,\ldots, n.\n\]\n\nThen the code \( {C}_{\mathcal{L}}\left( {D, G}\right) \) is self-orthogonal; i.e., \( {C}_{\mathcal{L}}\left(... | Proof. The assumption \( {2G} - D \leq \left( \eta \right) \) is equivalent to \( G \leq D - G + \left( \eta \right) \) . Hence Proposition 2.2.10 implies\n\n\[ \n{C}_{\mathcal{L}}{\left( D, G\right) }^{ \bot } = {C}_{\mathcal{L}}\left( {D, D - G + \left( \eta \right) }\right) \supseteq {C}_{\mathcal{L}}\left( {D, G}\r... | Yes |
Proposition 2.2.14. (a) Suppose \( {G}_{1} \) and \( {G}_{2} \) are divisors with \( {G}_{1} \sim {G}_{2} \) and \( \operatorname{supp}{G}_{1} \cap \operatorname{supp}D = \operatorname{supp}{G}_{2} \cap \operatorname{supp}D = \varnothing \) . Then the codes \( {C}_{\mathcal{L}}\left( {D,{G}_{1}}\right) \) and \( {C}_{\... | Proof. (a) By assumption we have that \( {G}_{2} = {G}_{1} - \left( z\right) \) with \( {v}_{{P}_{i}}\left( z\right) = 0 \) for \( i = 1,\ldots, n \) . Hence \( a \mathrel{\text{:=}} \left( {z\left( {P}_{1}\right) ,\ldots, z\left( {P}_{n}\right) }\right) \) is in \( {\left( {\mathbb{F}}_{q}^{ \times }\right) }^{n} \), ... | Yes |
Proposition 2.3.3. Let \( C = {C}_{\mathcal{L}}\left( {D, G}\right) \) be a rational AG code over \( {\mathbb{F}}_{q} \) with parameters \( n, k \) and \( d \) .\n\n(a) If \( n \leq q \) then there exist pairwise distinct elements \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in {\mathbb{F}}_{q} \) and \( {v}_{1},\ldots ,{v}... | Proof. (a) Let \( D = {P}_{1} + \ldots + {P}_{n} \) . As \( n \leq q \), there is a place \( P \) of degree one which is not in the support of \( D \) . Choose a place \( Q \neq P \) of degree one (e.g., \( Q = {P}_{1} \) ). By Riemann-Roch, \( \ell \left( {Q - P}\right) = 1 \), hence \( Q - P \) is a principal divisor... | Yes |
Proposition 2.3.5. Every generalized Reed-Solomon code \( {\operatorname{GRS}}_{k}\left( {\alpha, v}\right) \) can be represented as a rational \( \mathrm{{AG}} \) code. | Proof. Let \( \alpha = \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) with \( {\alpha }_{i} \in {\mathbb{F}}_{q} \) and \( v = \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) with \( {v}_{i} \in {\mathbb{F}}_{q}^{ \times } \) . Consider the rational function field \( F = {\mathbb{F}}_{q}\left( z\right) \) . Denote b... | Yes |
Proposition 2.3.9. Let \( n \mid \left( {{q}^{m} - 1}\right) \) and let \( \beta \in {\mathbb{F}}_{{q}^{m}} \) be a primitive \( n \) -th root of unity. Let \( F = {\mathbb{F}}_{{q}^{m}}\left( z\right) \) be the rational function field over \( {\mathbb{F}}_{{q}^{m}} \) and \( {P}_{0} \) (resp. \( {P}_{\infty } \) ) be ... | Proof. (a) We consider the code \( {C}_{\mathcal{L}}\left( {{D}_{\beta }, a{P}_{0} + b{P}_{\infty }}\right) \) where \( 0 \leq a + b \leq n - 2 \) . The elements \( {z}^{-a} \cdot {z}^{j} \) with \( 0 \leq j \leq a + b \) constitute a basis of \( \mathcal{L}\left( {a{P}_{0} + b{P}_{\infty }}\right) \) . Hence the matri... | Yes |
Proposition 2.3.11. In addition to the notation of Definition 2.3.10, let \( {P}_{i} \) denote the zero of \( z - {\alpha }_{i} \) (for \( {\alpha }_{i} \in L \) ), \( {P}_{\infty } \) the pole of \( z \) and \( {D}_{L} \mathrel{\text{:=}} {P}_{1} + \ldots + {P}_{n} \) . Let \( {G}_{0} \) be the zero divisor of \( g\le... | Proof. It is sufficient to prove (2.22). For \( 0 \leq j \leq t - 1 \), the element \( {z}^{j}g{\left( z\right) }^{-1} \) is in \( \mathcal{L}\left( {{G}_{0} - {P}_{\infty }}\right) \) because\n\n\[ \left( {{z}^{j}g{\left( z\right) }^{-1}}\right) = j\left( {{P}_{0} - {P}_{\infty }}\right) - \left( {{G}_{0} - t{P}_{\inf... | Yes |
Corollary 2.3.12. (a) (BCH Bound.) The minimum distance of a BCH code with designed distance \( \delta \) is at least \( \delta \) . | Proof. (a) Using notation as in Proposition 2.3.9 we represent the BCH code in the form \( {\left. C = {C}_{\mathcal{L}}\left( {D}_{\beta }, r{P}_{0} + s{P}_{\infty }\right) \right| }_{{\mathbb{F}}_{q}} \) . The minimum distance of the code \( {C}_{\mathcal{L}}\left( {{D}_{\beta }, r{P}_{0} + s{P}_{\infty }}\right) \) ... | Yes |
Lemma 3.1.2. Let \( {F}^{\prime }/{K}^{\prime } \) be an algebraic extension of \( F/K \) . Then the following hold:\n\n(a) \( {K}^{\prime }/K \) is algebraic and \( F \cap {K}^{\prime } = K \) .\n\n(b) \( {F}^{\prime }/{K}^{\prime } \) is a finite extension of \( F/K \) if and only if \( \left\lbrack {{K}^{\prime } : ... | Proof. (a) and (c) are trivial. As to (b), we assume first that \( {F}^{\prime }/{K}^{\prime } \) is a finite extension of \( F/K \) . Then \( {F}^{\prime } \) can be considered as an algebraic function field over \( K \) whose full constant field is \( {K}^{\prime } \) . By Corollary 1.1.16 we conclude that \( \left\l... | Yes |
Proposition 3.1.4. Let \( {F}^{\prime }/{K}^{\prime } \) be an algebraic extension of \( F/K \) . Suppose that \( P \) (resp. \( {P}^{\prime } \) ) is a place of \( F/K \) (resp. \( {F}^{\prime }/{K}^{\prime } \) ), and let \( {\mathcal{O}}_{P} \subseteq F \) (resp. \( \left. {{\mathcal{O}}_{{P}^{\prime }} \subseteq {F... | Proof. (1) \( \Rightarrow \) (2): Suppose that \( {P}^{\prime } \mid P \) but \( {\mathcal{O}}_{P} \nsubseteq {\mathcal{O}}_{{P}^{\prime }} \) . Then there is some \( u \in F \) with \( {v}_{P}\left( u\right) \geq 0 \) and \( {v}_{{P}^{\prime }}\left( u\right) < 0 \) . As \( P \subseteq {P}^{\prime } \) we conclude \( ... | Yes |
Proposition 3.1.6. Let \( {F}^{\prime }/{K}^{\prime } \) be an algebraic extension of \( F/K \) and let \( {P}^{\prime } \) be a place of \( {F}^{\prime }/{K}^{\prime } \) lying over \( P \in {\mathbb{P}}_{F} \). Then\n\n(a) \( f\left( {{P}^{\prime } \mid P}\right) < \infty \Leftrightarrow \left\lbrack {{F}^{\prime } :... | Proof. (a) Consider the natural embeddings \( K \subseteq {F}_{P} \subseteq {F}_{{P}^{\prime }}^{\prime } \) and \( K \subseteq {K}^{\prime } \subseteq {F}_{{P}^{\prime }}^{\prime } \), where \( \left\lbrack {{F}_{P} : K}\right\rbrack < \infty \) and \( \left\lbrack {{F}_{{P}^{\prime }}^{\prime } : {K}^{\prime }}\right... | Yes |
Proposition 3.1.7. Let \( {F}^{\prime }/{K}^{\prime } \) be an algebraic extension of \( F/K \) .\n\n(a) For each place \( {P}^{\prime } \in {\mathbb{P}}_{{F}^{\prime }} \) there is exactly one place \( P \in {\mathbb{P}}_{F} \) such that \( {P}^{\prime } \mid P \) , namely \( P = {P}^{\prime } \cap F \) .\n\n(b) Conve... | Proof. (a) The main step of the proof is the following.\n\nClaim. There is some \( z \in F, z \neq 0 \), with \( {v}_{{P}^{\prime }}\left( z\right) \neq 0 \).\n\n(3.3)\n\nAssume this is false. Choose \( t \in {F}^{\prime } \) with \( {v}_{{P}^{\prime }}\left( t\right) > 0 \) . Since \( {F}^{\prime }/F \) is algebraic, ... | Yes |
Proposition 3.1.9. Let \( {F}^{\prime }/{K}^{\prime } \) be an algebraic extension of the function field \( F/K \) . For \( 0 \neq x \in F \) let \( {\left( x\right) }_{0}^{F},{\left( x\right) }_{\infty }^{F},{\left( x\right) }^{F} \) resp. \( {\left( x\right) }_{0}^{{F}^{\prime }},{\left( x\right) }_{\infty }^{{F}^{\p... | Proof. From the definition of the principal divisor of \( x \) follows that\n\n\[{\left( x\right) }^{{F}^{\prime }} = \mathop{\sum }\limits_{{{P}^{\prime } \in {\mathbb{P}}_{{F}^{\prime }}}}{v}_{{P}^{\prime }}\left( x\right) \cdot {P}^{\prime } = \mathop{\sum }\limits_{{P \in {\mathbb{P}}_{F}}}\mathop{\sum }\limits_{{{... | Yes |
Lemma 3.1.10. Let \( {K}^{\prime }/K \) be a finite field extension and let \( x \) be transcendental over \( K \) . Then\n\n\[ \left\lbrack {{K}^{\prime }\left( x\right) : K\left( x\right) }\right\rbrack = \left\lbrack {{K}^{\prime } : K}\right\rbrack . \]\n | Proof. We can assume that \( {K}^{\prime } = K\left( \alpha \right) \) for some element \( \alpha \in {K}^{\prime } \) . Clearly \( \left\lbrack {{K}^{\prime }\left( x\right) : K\left( x\right) }\right\rbrack \leq \left\lbrack {{K}^{\prime } : K}\right\rbrack \) since \( {K}^{\prime }\left( x\right) = K\left( x\right) ... | Yes |
Theorem 3.1.11 (Fundamental Equality). Let \( {F}^{\prime }/{K}^{\prime } \) be a finite extension of \( F/K \), let \( P \) be a place of \( F/K \) and let \( {P}_{1},\ldots ,{P}_{m} \) be all the places of \( {F}^{\prime }/{K}^{\prime } \) lying over \( P \) . Let \( {e}_{i} \mathrel{\text{:=}} e\left( {{P}_{i} \mid ... | Proof. Choose \( x \in F \) such that \( P \) is the only zero of \( x \) in \( F/K \), and let \( {v}_{P}\left( x\right) = : r > 0 \) . Then the places \( {P}_{1},\ldots ,{P}_{m} \in {\mathbb{P}}_{{F}^{\prime }} \) are exactly the zeros of \( x \) in \( {F}^{\prime }/{K}^{\prime } \) by (3.4). Now we evaluate the degr... | Yes |
Corollary 3.1.14. Let \( {F}^{\prime }/{K}^{\prime } \) be a finite extension of \( F/K \) . Then for each divisor \( A \in \operatorname{Div}\left( F\right) \), \[ \deg {\operatorname{Con}}_{{F}^{\prime }/F}\left( A\right) = \frac{\left\lbrack {F}^{\prime } : F\right\rbrack }{\left\lbrack {K}^{\prime } : K\right\rbrac... | Proof. It is sufficient to consider a prime divisor \( A = P \in {\mathbb{P}}_{F} \). We have \[ \deg {\operatorname{Con}}_{{F}^{\prime }/F}\left( P\right) = \deg \left( {\mathop{\sum }\limits_{{{P}^{\prime } \mid P}}e\left( {{P}^{\prime } \mid P}\right) \cdot {P}^{\prime }}\right) \] \[ = \mathop{\sum }\limits_{{{P}^{... | Yes |
Proposition 3.1.15. Consider a function field \( F/K \) and a polynomial\n\n\[ \varphi \left( T\right) = {a}_{n}{T}^{n} + {a}_{n - 1}{T}^{n - 1} + \ldots + {a}_{1}T + {a}_{0} \]\n\nwith coefficients \( {a}_{i} \in F \) . Assume that there exists a place \( P \in {\mathbb{P}}_{F} \) such that one of the following condit... | Proof. We consider an extension field \( {F}^{\prime } = F\left( y\right) \) with \( \varphi \left( y\right) = 0 \) . The degree of \( {F}^{\prime }/F \) is \( \left\lbrack {{F}^{\prime } : F}\right\rbrack \leq \deg \varphi \left( T\right) = n \), with equality if and only if \( \varphi \left( T\right) \) is irreducibl... | Yes |
Lemma 3.2.3. (b) Every holomorphy ring \( {\mathcal{O}}_{S} \) is a subring of \( F/K \) . | Proof. (b) Since \( {\mathcal{O}}_{S} \) is a ring with \( K \subseteq {\mathcal{O}}_{S} \subseteq F \) we have only to show that it is not a field. Choose a place \( {P}_{1} \in S \) . As \( S \neq {\mathbb{P}}_{F} \), the Strong Approximation Theorem yields an element \( 0 \neq x \in F \) such that\n\n\[{v}_{{P}_{1}}... | Yes |
Proposition 3.2.5. Let \( {\mathcal{O}}_{S} \) be a holomorphy ring of \( F/K \) . Then\n\n(a) \( F \) is the quotient field of \( {\mathcal{O}}_{S} \) .\n\n(b) \( {\mathcal{O}}_{S} \) is integrally closed. | Proof. (a) Let \( x \in F, x \neq 0 \) . Choose a place \( {P}_{0} \in S \) . By the Strong Approximation Theorem there is an element \( z \in F \) such that\n\n\[ \n{v}_{{P}_{0}}\left( z\right) = \max \left\{ {0,{v}_{{P}_{0}}\left( {x}^{-1}\right) }\right. \text{and}{v}_{P}\left( z\right) \geq \max \left\{ {0,{v}_{P}\... | Yes |
Theorem 3.2.6. Let \( R \) be a subring of \( F/K \) and\n\n\[ S\left( R\right) \mathrel{\text{:=}} \left\{ {P \in {\mathbb{P}}_{F} \mid R \subseteq {\mathcal{O}}_{P}}\right\} .\n\]\n\nThen the following hold:\n\n(a) \( \varnothing \neq S\left( R\right) \subsetneqq {\mathbb{P}}_{F} \) .\n\n(b) The integral closure of \... | Proof. (a) Since \( R \) is not a field we can find a proper ideal \( I \subsetneqq R \), and by Theorem 1.1.19 there exists a place \( P \in {\mathbb{P}}_{F} \) such that \( I \subseteq P \) and \( R \subseteq {\mathcal{O}}_{P} \) . Therefore \( S\left( R\right) \neq \varnothing \) . On the other hand, consider an ele... | Yes |
Proposition 3.2.9. Let \( {\mathcal{O}}_{S} \) be a holomorphy ring of \( F/K \) . Then there is a 1-1-correspondence between \( S \) and the set of maximal ideals of \( {\mathcal{O}}_{S} \), given by\n\n\[ \nP \mapsto {M}_{P} \mathrel{\text{:=}} P \cap {\mathcal{O}}_{S}\;\left( {\text{ for }P \in S}\right) .\n\]\n\nMo... | Proof. Consider for \( P \in S \) the ring homomorphism\n\n\[ \n\phi : \left\{ \begin{array}{ll} {\mathcal{O}}_{S} & \rightarrow {F}_{P}, \\ x & \mapsto x + P. \end{array}\right.\n\]\n\nWe claim that \( \phi \) is surjective. In fact, let \( z + P \in {F}_{P} \) with \( z \in {\mathcal{O}}_{P} \) . By the Strong Approx... | Yes |
Proposition 3.2.10. If \( S \subseteq {\mathbb{P}}_{F} \) is a non-empty finite set of places of \( F/K \) , then \( {\mathcal{O}}_{S} \) is a principal ideal domain. | Proof. Let \( S = \left\{ {{P}_{1},\ldots ,{P}_{s}}\right\} \) and let \( \{ 0\} \neq I \subseteq {\mathcal{O}}_{S} \) be an ideal of \( {\mathcal{O}}_{S} \) . For \( i = 1,\ldots, s \) choose \( {x}_{i} \in I \) such that\n\n\[ \n{v}_{{P}_{i}}\left( {x}_{i}\right) = : {n}_{i} \leq {v}_{{P}_{i}}\left( u\right) \;\text{... | Yes |
Proposition 3.3.1. Let \( R \) be an integrally closed subring of \( F/K \) such that \( F \) is the quotient field of \( R \) (i.e., \( R \) is a holomorphy ring of \( F/K \) ). For \( z \in {F}^{\prime } \) let \( \varphi \left( T\right) \in F\left\lbrack T\right\rbrack \) denote its minimal polynomial over \( F \) .... | Proof. By definition, \( \varphi \left( T\right) \) is the unique irreducible monic polynomial with coefficients in \( F \) such that \( \varphi \left( z\right) = 0 \) . If \( \varphi \left( T\right) \in R\left\lbrack T\right\rbrack \) then \( z \) is clearly integral over \( R \) .\n\nThe converse is not so evident. I... | Yes |
Corollary 3.3.2. Notation as in Proposition 3.3.1. Let \( {\operatorname{Tr}}_{{F}^{\prime }/F} : {F}^{\prime } \rightarrow F \) denote the trace map from \( {F}^{\prime } \) to \( F \) and let \( x \in {F}^{\prime } \) be integral over \( R \) . Then \( {\operatorname{Tr}}_{{F}^{\prime }/F}\left( x\right) \in R \) . | This corollary follows easily from well-known properties of the trace mapping. Let us briefly recall some of these properties which will be of use in the sequel. We consider a finite field extension \( M/L \) of degree \( n \) . If \( M/L \) is not separable, the trace map \( {\operatorname{Tr}}_{M/L} : M \rightarrow L... | Yes |
Proposition 3.3.3. Let \( M/L \) be a finite separable field extension, and consider a basis \( \left\{ {{z}_{1},\ldots ,{z}_{n}}\right\} \) of \( M/L \) . Then there are uniquely determined elements \( {z}_{1}^{ * },\ldots ,{z}_{n}^{ * } \in M \), such that\n\n\[ \n{\operatorname{Tr}}_{M/L}\left( {{z}_{i}{z}_{j}^{ * }... | Proof. We consider the dual space \( {M}^{ \land } \) of \( M \) over \( L \) ; i.e., \( {M}^{ \land } \) is the space of all \( L \) -linear maps \( \lambda : M \rightarrow L \) . It is well-known from linear algebra that \( {M}^{ \land } \) is an \( n \) -dimensional vector space over \( L \) . For \( z \in M \) and ... | Yes |
Theorem 3.3.4. Let \( R \) be an integrally closed subring of \( F/K \) with quotient field \( F \), and \( {F}^{\prime }/F \) be a finite separable extension of degree \( n \) . Let \( {R}^{\prime } = {\operatorname{ic}}_{{F}^{\prime }}\left( R\right) \) denote the integral closure of \( R \) in \( {F}^{\prime } \) . ... | Proof. (a) It must be shown that for every \( x \in {F}^{\prime } \) there is some element \( 0 \neq a \in R \) such that \( {ax} \) satisfies an integral equation over \( R \) . Since \( {F}^{\prime }/F \) is algebraic and \( F \) is the quotient field of \( R \), there are elements \( {a}_{i},{b}_{i} \in R \) with \(... | Yes |
Corollary 3.3.5. Let \( {F}^{\prime }/F \) be a finite separable extension of the function field \( F/K \) and let \( P \in {\mathbb{P}}_{F} \) be a place of \( F/K \) . Then the integral closure \( {\mathcal{O}}_{P}^{\prime } \) of \( {\mathcal{O}}_{P} \) in \( {F}^{\prime } \) is\n\n\[{\mathcal{O}}_{P}^{\prime } = \m... | Proof. This is clear by Theorem 3.2.6(b), Remark 3.2.7 and Theorem 3.3.4 (observe that \( {\mathcal{O}}_{P} \) is a principal ideal domain). | No |
Theorem 3.3.6. Let \( F/K \) be a function field, \( {F}^{\prime }/F \) be a finite separable extension field. Then each basis \( \left\{ {{z}_{1},\ldots .{z}_{n}}\right\} \) of \( {F}^{\prime }/F \) is an integral basis for almost all (i.e., all but finitely many) places \( P \in {\mathbb{P}}_{F} \) . | Proof. We consider the dual basis \( \left\{ {{z}_{1}^{ * },\ldots ,{z}_{n}^{ * }}\right\} \) of \( \left\{ {{z}_{1},\ldots ,{z}_{n}}\right\} \) . The minimal polynomials of \( {z}_{1},\ldots ,{z}_{n},{z}_{1}^{ * },\ldots ,{z}_{n}^{ * } \) over \( F \) involve only finitely many coefficients. Let \( S \subseteq {\mathb... | Yes |
Theorem 3.3.7 (Kummer). Suppose that \( {F}^{\prime } = F\left( y\right) \) where \( y \) is integral over \( {\mathcal{O}}_{P} \), and consider the minimal polynomial \( \varphi \left( T\right) \in {\mathcal{O}}_{P}\left\lbrack T\right\rbrack \) of \( y \) over \( F \) . Let\n\n\[ \bar{\varphi }\left( T\right) = \math... | Proof. We set \( {\bar{F}}_{i} \mathrel{\text{:=}} \bar{F}\left\lbrack T\right\rbrack /\left( {{\gamma }_{i}\left( T\right) }\right) \) . Since \( {\gamma }_{i}\left( T\right) \) is irreducible, \( {\bar{F}}_{i} \) is an extension field of \( \bar{F} \) of degree\n\n\[ \left\lbrack {{\bar{F}}_{i} : \bar{F}}\right\rbrac... | Yes |
Corollary 3.3.8. Let \( \varphi \left( T\right) = {T}^{n} + {f}_{n - 1}\left( x\right) {T}^{n - 1} + \cdots + {f}_{0}\left( x\right) \in K\left( x\right) \left\lbrack T\right\rbrack \) be an irreducible polynomial over the rational function field \( K\left( x\right) \) . We consider the function field \( K\left( {x, y}... | Proof. We set \( F \mathrel{\text{:=}} K\left( x\right) \) and \( {F}^{\prime } \mathrel{\text{:=}} K\left( {x, y}\right) \) . The assumption \( {f}_{j}\left( \alpha \right) \neq \infty \) implies that \( y \) is integral over the valuation ring of \( {P}_{\alpha } \), and the polynomial \( {\varphi }_{\alpha }\left( T... | Yes |
Proposition 3.4.2. With notation as in Definition 3.4.1 the following hold: (a) \( {\mathcal{C}}_{P} \) is an \( {\mathcal{O}}_{P}^{\prime } \) -module and \( {\mathcal{O}}_{P}^{\prime } \subseteq {\mathcal{C}}_{P} \) . | Proof. (a) The assertion that \( {\mathcal{C}}_{P} \) is an \( {\mathcal{O}}_{P}^{\prime } \) -module is trivial. Since the trace of an element \( y \in {\mathcal{O}}_{P}^{\prime } \) is in \( {\mathcal{O}}_{P} \), by Corollary 3.3.2, we have \( {\mathcal{O}}_{P}^{\prime } \subseteq {\mathcal{C}}_{P} \) . | Yes |
Lemma 3.4.9. For each \( {C}^{\prime } \in \operatorname{Div}\left( {F}^{\prime }\right) \) we have \( {\mathcal{A}}_{{F}^{\prime }} = {\mathcal{A}}_{{F}^{\prime }/F} + {\mathcal{A}}_{{F}^{\prime }}\left( {C}^{\prime }\right) \) . | Proof. Let \( \alpha = {\left( {\alpha }_{{P}^{\prime }}\right) }_{{P}^{\prime } \in {\mathbb{P}}_{{F}^{\prime }}} \) be an adele of \( {F}^{\prime } \) . For all \( P \in {\mathbb{P}}_{F} \) there exists by the Approximation Theorem an element \( {x}_{P} \in {F}^{\prime } \) with\n\n\[ \n{v}_{{P}^{\prime }}\left( {{\a... | Yes |
Lemma 3.4.10. Let \( M/L \) be a finite separable field extension, \( V \) a vector space over \( M \) and \( \mu : V \rightarrow L \) be an \( L \) -linear map. Then there is a unique \( M \) -linear map \( {\mu }^{\prime } : V \rightarrow M \) such that \( {\operatorname{Tr}}_{M/L} \circ {\mu }^{\prime } = \mu \) . | Proof. As in the proof of Proposition 3.3.3 we consider the space of linear forms \( {M}^{ \land } = \{ \lambda : M \rightarrow L \mid \lambda \) is \( L \) -linear \( \} \) as a vector space over \( M \) by setting \( \left( {z \cdot \lambda }\right) \left( w\right) = \lambda \left( {z \cdot w}\right) \) for \( \lambd... | Yes |
Proposition 3.4.11. (a) If \( \omega ,{\omega }_{1} \) and \( {\omega }_{2} \) are Weil differentials of \( F/K \) and \( x \in F \), then\n\n\[ \n{\operatorname{Cotr}}_{{F}^{\prime }/F}\left( {{\omega }_{1} + {\omega }_{2}}\right) = {\operatorname{Cotr}}_{{F}^{\prime }/F}\left( {\omega }_{1}\right) + {\operatorname{Co... | Proof. Keeping in mind the uniqueness assertion in Theorem 3.4.6, it is sufficient to show that\n\n\[ \n{\operatorname{Tr}}_{{K}^{\prime }/K}\left( {\left( {{\operatorname{Cotr}}_{{F}^{\prime }/F}\left( {\omega }_{1}\right) + {\operatorname{Cotr}}_{{F}^{\prime }/F}\left( {\omega }_{2}\right) }\right) \left( \alpha \rig... | Yes |
Corollary 3.4.12 (Transitivity of the Different). If \( {F}^{\prime \prime } \supseteq {F}^{\prime } \supseteq F \) are finite separable extensions, the following hold:\n\n(a) \( \operatorname{Diff}\left( {{F}^{\prime \prime }/F}\right) = {\operatorname{Con}}_{{F}^{\prime \prime }/{F}^{\prime }}\left( {\operatorname{Di... | Proof. (b) is merely a reformulation of (a), so we only prove (a). Choose a Weil differential \( \omega \neq 0 \) of \( F/K \) . Then the divisor of \( {\operatorname{Cotr}}_{{F}^{\prime \prime }/F}\left( \omega \right) \) is\n\n\[ \left( {{\operatorname{Cotr}}_{{F}^{\prime \prime }/F}\left( \omega \right) }\right) = {... | Yes |
Theorem 3.4.13 (Hurwitz Genus Formula). Let \( F/K \) be an algebraic function field of genus \( g \) and let \( {F}^{\prime }/F \) be a finite separable extension. Let \( {K}^{\prime } \) denote the constant field of \( {F}^{\prime } \) and \( {g}^{\prime } \) the genus of \( {F}^{\prime }/{K}^{\prime } \). Then we ha... | Proof. Choose a Weil differential \( \omega \neq 0 \) of \( F/K \). It follows from Theorem 3.4.6 that\n\n\[ \left( {{\operatorname{Cotr}}_{{F}^{\prime }/F}\left( \omega \right) }\right) = {\operatorname{Con}}_{{F}^{\prime }/F}\left( \left( \omega \right) \right) + \operatorname{Diff}\left( {{F}^{\prime }/F}\right) . \... | Yes |
Theorem 3.5.1 (Dedekind's Different Theorem). With notation as above we have for all \( {P}^{\prime } \mid P \)\n\n(a) \( d\left( {{P}^{\prime } \mid P}\right) \geq e\left( {{P}^{\prime } \mid P}\right) - 1 \) .\n\n(b) \( d\left( {{P}^{\prime } \mid P}\right) = e\left( {{P}^{\prime } \mid P}\right) - 1 \) if and only i... | We shall first prove part (a) of Dedekind's Theorem; the proof will require an understanding of the action of automorphisms on the places of a function field. More precisely we need:\n\nLemma 3.5.2. Let \( {F} | No |
Lemma 3.5.2. Let \( {F}^{ * }/F \) be an algebraic extension of function fields, \( P \in {\mathbb{P}}_{F} \) and \( {P}^{ * } \in {\mathbb{P}}_{{F}^{ * }} \) with \( {P}^{ * } \mid P \) . Consider an automorphism \( \sigma \) of \( {F}^{ * }/F \) . Then \( \sigma \left( {P}^{ * }\right) \mathrel{\text{:=}} \left\{ {\s... | Proof of the Lemma. Clearly \( \sigma \left( {\mathcal{O}}_{{P}^{ * }}\right) \) is a valuation ring of \( {F}^{ * } \) and \( \sigma \left( {P}^{ * }\right) \) is its maximal ideal; therefore \( \sigma \left( {P}^{ * }\right) \) is a place of \( {F}^{ * } \), and the corresponding valuation ring is \( {\mathcal{O}}_{\... | Yes |
Corollary 3.5.5. Let \( {F}^{\prime }/F \) be a finite separable extension of algebraic function fields.\n\n(a) If \( P \in {\mathbb{P}}_{F} \) and \( {P}^{\prime } \in {\mathbb{P}}_{{F}^{\prime }} \) such that \( {P}^{\prime } \mid P \), then \( {P}^{\prime } \mid P \) is ramified if and only if \( {P}^{\prime } \leq ... | If \( {P}^{\prime } \mid P \) is ramified, then\n\n\[ d\left( {{P}^{\prime } \mid P}\right) = e\left( {{P}^{\prime } \mid P}\right) - 1 \Leftrightarrow {P}^{\prime } \mid P\text{ is tamely ramified,}\]\n\n\[ d\left( {{P}^{\prime } \mid P}\right) \geq e\left( {{P}^{\prime } \mid P}\right) \Leftrightarrow {P}^{\prime } \... | No |
Corollary 3.5.6. Suppose that \( {F}^{\prime }/F \) is a finite separable extension of algebraic function fields having the same constant field \( K \) . Let \( g \) (resp. \( {g}^{\prime } \) ) denote the genus of \( F/K \) (resp. \( {F}^{\prime }/K \) ). Then\n\n\[ 2{g}^{\prime } - 2 \geq \left\lbrack {{F}^{\prime } ... | Proof. Trivial by Theorems 3.4.13 and 3.5.1. | No |
Corollary 3.5.8. Let \( F/K\\left( x\\right) \) be a finite separable extension of the rational function field of degree \( \\left\\lbrack {F : K\\left( x\\right) }\\right\\rbrack > 1 \) such that \( K \) is the constant field of \( F \) . Then \( F/K\\left( x\\right) \) is ramified. | Proof. The Hurwitz Genus Formula yields\n\n\[ \n{2g} - 2 = - 2\\left\\lbrack {F : K\\left( x\\right) }\\right\\rbrack + \\deg \\operatorname{Diff}\\left( {F/K\\left( x\\right) }\\right) ,\n\]\n\nwhere \( g \) is the genus of \( F/K \) . Therefore\n\n\[ \n\\deg \\operatorname{Diff}\\left( {F/K\\left( x\\right) }\\right)... | Yes |
Proposition 3.5.9 (Lüroth's Theorem). Every subfield of a rational function field is rational; i.e., if \( K \subsetneqq {F}_{0} \subseteq K\left( x\right) \) then \( {F}_{0} = K\left( y\right) \) for some \( y \in {F}_{0} \) . | Proof. Suppose first that \( K\left( x\right) /{F}_{0} \) is separable. Let \( {g}_{0} \) denote the genus of \( {F}_{0}/K \) . Then\n\n\[ - 2 = \left\lbrack {K\left( x\right) : {F}_{0}}\right\rbrack \cdot \left( {2{g}_{0} - 2}\right) + \deg \operatorname{Diff}\left( {K\left( x\right) /{F}_{0}}\right) ,\]\n\nwhich impl... | Yes |
Theorem 3.5.10. Suppose \( {F}^{\prime } = F\left( y\right) \) is a finite separable extension of a function field \( F \) of degree \( \left\lbrack {{F}^{\prime } : F}\right\rbrack = n \) . Let \( P \in {\mathbb{P}}_{F} \) be such that the minimal polynomial \( \varphi \left( T\right) \) of \( y \) over \( F \) has co... | Proof. The dual basis of \( \left\{ {1, y,\ldots ,{y}^{n - 1}}\right\} \) is closely related to the different exponents \( d\left( {{P}_{i} \mid P}\right) \) by Proposition 3.4.2, therefore our first aim is to determine this dual basis. Since \( \varphi \left( y\right) = 0 \), the polynomial \( \varphi \left( T\right) ... | Yes |
Corollary 3.5.11. Let \( {F}^{\prime } = F\left( y\right) \) be a finite separable extension of function fields of degree \( \left\lbrack {{F}^{\prime } : F}\right\rbrack = n \), and let \( \varphi \left( T\right) \in F\left\lbrack T\right\rbrack \) be the minimal polynomial of \( y \) over \( F \) . Suppose \( P \in {... | Proof. We have by Theorem 3.5.10\n\n\[ 0 \leq d\left( {{P}^{\prime } \mid P}\right) \leq {v}_{{P}^{\prime }}\left( {{\varphi }^{\prime }\left( y\right) }\right) \leq 0 \]\n\nfor all \( {P}^{\prime } \mid P \), hence \( {v}_{{P}^{\prime }}\left( {{\varphi }^{\prime }\left( y\right) }\right) = d\left( {{P}^{\prime } \mid... | Yes |
Proposition 3.5.12. Let \( {F}^{\prime }/F \) be a finite separable extension of function fields, \( P \in {\mathbb{P}}_{F} \) and \( {P}^{\prime } \in {\mathbb{P}}_{{F}^{\prime }} \) with \( {P}^{\prime } \mid P \) . Suppose that \( {P}^{\prime } \mid P \) is totally ramified; i.e., \( e\left( {{P}^{\prime } \mid P}\r... | Proof. First we claim that \( 1, t,\ldots ,{t}^{n - 1} \) are linearly independent over \( F \) . Assume the contrary, so that\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}{r}_{i}{t}^{i} = 0\;\text{ with }\;{r}_{i} \in F,\text{ not all }{r}_{i} = 0. \]\n\nFor \( {r}_{i} \neq 0 \) we have\n\n\[ {v}_{{P}^{\prime }}\lef... | Yes |
Lemma 3.6.2. Suppose \( \alpha \in \Phi \) is algebraic over \( K \) . Then \( \left\lbrack {K\left( \alpha \right) : K}\right\rbrack = \) \( \left\lbrack {F\left( \alpha \right) : F}\right\rbrack \) . | Proof of the Lemma. The inequality \( \left\lbrack {F\left( \alpha \right) : F}\right\rbrack \leq \left\lbrack {K\left( \alpha \right) : K}\right\rbrack \) being trivial, we only have to prove that the minimal polynomial \( \varphi \left( T\right) \in K\left\lbrack T\right\rbrack \) of \( \alpha \) over \( K \) remains... | Yes |
Theorem 3.6.3. In an algebraic constant field extension \( {F}^{\prime } = F{K}^{\prime } \) of \( F/K \) the following hold:\n\n(a) \( {F}^{\prime }/F \) is unramified (i.e., \( e\left( {{P}^{\prime } \mid P}\right) = 1 \) for all \( P \in {\mathbb{P}}_{F} \) and all \( {P}^{\prime } \in {\mathbb{P}}_{{F}^{\prime }} \... | Proof. The proof is organized as follows: first we discuss (a) and (b) in the case of a finite constant field extension, then we prove (h), (a), (c), (b), (d), (e), (f) and (g) in the general case. To begin with, we assume that\n\n\[ {K}^{\prime } = K\left( \alpha \right) \text{is a finite extension of}K\text{.}\]\n\n(... | Yes |
Corollary 3.6.4. Let \( {F}^{\prime }/{K}^{\prime } \) be an algebraic extension of \( F/K \) (not necessarily a constant field extension). Then we have for each divisor \( A \in \operatorname{Div}\left( F\right) \), \[ \deg {\operatorname{Con}}_{{F}^{\prime }/F}\left( A\right) = \left\lbrack {{F}^{\prime } : F{K}^{\pr... | Proof. By Lemma 3.1.2 we know that \( \left\lbrack {{F}^{\prime } : F{K}^{\prime }}\right\rbrack < \infty \), and \( F{K}^{\prime }/{K}^{\prime } \) is a constant field extension of \( F/K \). Since \[ {\operatorname{Con}}_{{F}^{\prime }/F}\left( A\right) = {\operatorname{Con}}_{{F}^{\prime }/F{K}^{\prime }}\left( {{\o... | Yes |
Proposition 3.6.6. Let \( F/K \) be a function field with constant field \( K \) . Suppose that \( {F}^{\prime }/F \) is a finite extension field, with constant field \( {K}^{\prime } \) . Let \( \bar{K} \subseteq \Phi \) denote the algebraic closure of \( K \) . Then\n\n\[ \left\lbrack {{F}^{\prime } : F}\right\rbrack... | Proof. Since \( F \subseteq F{K}^{\prime } \subseteq {F}^{\prime } \), we have\n\n\[ \left\lbrack {{F}^{\prime } : F}\right\rbrack = \left\lbrack {{F}^{\prime } : F{K}^{\prime }}\right\rbrack \cdot \left\lbrack {F{K}^{\prime } : F}\right\rbrack . \]\n\n(3.88)\n\nThe extension \( {K}^{\prime }/K \) is separable and of f... | Yes |
Proposition 3.6.1(c) shows that for each \( x \in F \smallsetminus K \) , \[ \left\lbrack {F{K}^{\prime } : {K}^{\prime }\left( x\right) }\right\rbrack = \left\lbrack {F\bar{K} : \bar{K}\left( x\right) }\right\rbrack \text{ and }\left\lbrack {{F}^{\prime } : {K}^{\prime }\left( x\right) }\right\rbrack = \left\lbrack {{... | This implies \[ \left\lbrack {{F}^{\prime } : F{K}^{\prime }}\right\rbrack = \left\lbrack {{F}^{\prime }\bar{K} : F\bar{K}}\right\rbrack \] (3.90) Substituting (3.89) and (3.90) into (3.88) yields (3.87). | No |
Corollary 3.6.7. Let \( F/K \) be a function field and let \( {F}^{\prime }/F \) be a finite extension such that \( K \) is the full constant field of \( F \) and of \( {F}^{\prime } \) . Let \( L/K \) be an algebraic extension. Then \( L \) is the full constant field of \( {FL} \) and \( {F}^{\prime }L \), and we have... | Proof. It was already shown in Proposition 3.6.1 that \( L \) is the full constant field of \( {FL} \) and of \( {F}^{\prime }L \) . Let \( \bar{K} \supseteq L \) be the algebraic closure of \( K \), then we obtain from Proposition 3.6.6 \[ \left\lbrack {{F}^{\prime } : F}\right\rbrack = \left\lbrack {{F}^{\prime }\bar... | Yes |
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