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Corollary 1.2.9 (Hensel’s Lemma). Let \( R \) be complete at an ideal 1 and let \( f\left( X\right) \in R\left\lbrack X\right\rbrack \) . Suppose, for some \( u \in R \), that \( f\left( u\right) \equiv 0{\;\operatorname{mod}\;I} \) and that \( {f}^{\prime }\left( u\right) \) is invertible modulo I. Then there exists a... | Proof. By (1.2.6) every element of \( R \) congruent to \( {f}^{\prime }\left( u\right) {\;\operatorname{mod}\;I} \) is invertible. Put \( {u}_{1} = u \) and apply (1.2.7) to obtain an element \( {u}_{2} \equiv {u}_{1}{\;\operatorname{mod}\;I} \) with \( f\left( {u}_{2}\right) \equiv 0 \) \( {\;\operatorname{mod}\;{I}^... | Yes |
Theorem 1.2.10. Let \( \mathcal{O} \) be a discrete valuation ring with maximal ideal \( P \) and local parameter \( t \) . Then \( {\widehat{\mathcal{O}}}_{P} \) is also a discrete valuation ring with local parameter \( t \), and the natural projection \( {\pi }_{1} : {\widehat{\mathcal{O}}}_{P} \rightarrow \mathcal{O... | Proof. We have \( \widehat{P} = \ker {\pi }_{1} \) by (1.2.4), so \( {\pi }_{1} \) induces an isomorphism \( {\widehat{\mathcal{O}}}_{P}/\widehat{P} \simeq \) \( {\mathcal{O}}_{P}/P \) . Now (1.2.6) implies that every element of \( {\widehat{\mathcal{O}}}_{P} \smallsetminus \widehat{P} \) is a unit.\n\nSince \( P = {\m... | Yes |
Lemma 1.2.12. Suppose that the \( k \) -algebra \( \mathcal{O} \) is a complete discrete \( k \) -valuation ring with residue class map \( \eta : \mathcal{O} \rightarrow F \) . Assume further that \( F \) is a finite extension of \( k \) . Let \( {F}^{\text{sep }}/k \) be the maximal separable subextension of \( k \) .... | Proof. Let \( {F}^{\text{sep }} = k\left( u\right) \), where \( u \) is a root of the separable irreducible polynomial \( f\left( X\right) \in k\left\lbrack X\right\rbrack \) and \( \deg \left( f\right) = n \) . Since \( f\left( X\right) \) is separable, we have \( {f}^{\prime }\left( u\right) \neq 0 \), so Hensel’s Le... | Yes |
Lemma 1.2.13. Let \( F \) be a field. Then \( F\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) is a complete discrete valuation ring with local parameter \( X \) and residue field \( F \) . | Proof. Define \( v\left( {\mathop{\sum }\limits_{i}{a}_{i}{X}^{i}}\right) = n \) if \( {a}_{i} = 0 \) for \( i < n \) and \( {a}_{n} \neq 0 \) . It is trivial to verify that \( v \) is a discrete valuation, so \( F\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) is a valuation ring with maximal ideal \( M \) c... | Yes |
Theorem 1.2.14. Suppose that the \( k \) -algebra \( \mathcal{O} \) is a complete discrete \( k \) -valuation ring with residue class map \( \eta : \mathcal{O} \rightarrow F \) . Assume further that \( F \) is a finite separable extension of \( k \) . Given any local parameter \( t \), there is a unique isometric isomo... | Proof. Let \( \eta : \mathcal{O} \rightarrow F \) be the residue class map, and let \( \mu : F \rightarrow \mathcal{O} \) be the unique splitting given by (1.2.12). Define \( \widehat{\mu } : F\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \rightarrow \mathcal{O} \) via\n\n\[ \mu \left( {\mathop{\sum }\limits_{... | Yes |
Corollary 1.2.15. For every power series\n\n\[ s = \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{t}^{n} \in k\left\lbrack \left\lbrack t\right\rbrack \right\rbrack \]\n\nwith \( {a}_{1} \neq 0 \), there is a unique automorphism \( {\phi }_{s} \) of \( k\left\lbrack \left\lbrack t\right\rbrack \right\rbrack \) that i... | Proof. This is immediate from (1.2.14) because \( s \) is a local parameter. | No |
Theorem 1.3.2. Let \( K \) be a \( k \) -algebra over a commutative ring \( k \) , \( M \) an \( K \) -module, and \( \delta : K \rightarrow M \) a \( k \) -derivation. Then there exists a unique homomorphism \( \phi \) : \( {\Omega }_{K/k} \rightarrow M \) with \( \delta = \phi \circ {d}_{K/k}. \) | Proof. Let \( {\phi }^{\prime }\left( {x, y}\right) = {x\delta }\left( y\right) \) . Then \( {\phi }^{\prime } \) is \( k \) -bilinear so it factors uniquely through \( K{ \otimes }_{k}K \) by the universal property of tensor products. From the product rule, \( {\phi }^{\prime }\left( D\right) = 0 \) and the rest is ob... | No |
Lemma 1.3.3. Suppose \( \phi : K \rightarrow {K}^{\prime } \) is a \( k \) -algebra map. Then there exists a unique map \( {d\phi } \) making the diagram\n\n\n\ncommute. Moreover, given another \( k \) -algebra map \( ... | Proof. The composition \( {d}_{{K}^{\prime }} \circ \phi : K \rightarrow {\Omega }_{{K}^{\prime }} \) is certainly a derivation, so there is a unique map \( {d\phi } : {\Omega }_{K} \rightarrow {\ddot{\Omega }}_{{K}^{\prime }} \) making the above diagram commute. It is easy to check that\n\n![4861ec54-7ace-4a2a-805d-bd... | Yes |
Theorem 1.3.4. Suppose that \( k \subseteq K \subseteq {K}_{1} \) are fields, \( {K}_{1}/K \) is finite and separable, \( A \) is complete graded \( k \) -algebra, and \( D : K \rightarrow A \) is a \( k \) -algebra homomorphism. Given any extension \( {D}_{1}^{\left( 0\right) } \) of \( {D}^{\left( 0\right) } \) to \(... | Proof. By (A.0.17) we have \( {K}_{1} = K\left( u\right) \) for some element \( u \) with separable minimum polynomial \( f\left( X\right) \in K\left\lbrack X\right\rbrack \) . Put \( v \mathrel{\text{:=}} {D}_{1}^{\left( 0\right) }\left( u\right) \in {A}_{0} \) . Then \( v \) is a root of \( {f}_{1} \mathrel{\text{:=}... | Yes |
Corollary 1.3.5. Let \( {K}_{1}/K \) be a finite extension of fields. Then \( {K}_{1}/K \) is separable if and only if every derivation of \( K \) into a \( {K}_{1} \) -module \( M \) extends uniquely to \( {K}_{1} \) . | Proof. One implication is immediate from the theorem. Conversely, if \( {K}_{1}/K \) is inseparable, there is a subfield \( K \subseteq E \subseteq {K}_{1} \) where \( {K}_{1}/E \) is purely inseparable of degree \( p = \operatorname{char}\left( K\right) \) (see (A.0.9)). Thus, we have \( {K}_{1} \simeq E\left\lbrack X... | Yes |
Corollary 1.3.6. Suppose that \( k \subseteq K \) are fields and \( x \) is a separating variable for \( K/k \) . Then \( {\dim }_{K}{\Omega }_{K/k} = 1 \) and \( {d}_{K/k}\left( x\right) \neq 0 \) . | Proof. If \( K = k\left( x\right) \), the formal derivative is a nonzero derivation, so \( {dx} \neq 0 \) . From the sum, product, and quotient rules, every derivation on \( k\left( x\right) \) is determined by its value at \( x \), so the universal property of \( {dx} \) implies that \( {dx} \) is a \( k\left( x\right... | Yes |
Lemma 1.3.9. Suppose that \( R \) is an integral domain and \( D : R \rightarrow A \) is a homomorphism for some complete graded algebra A. If \( {A}^{\left( 0\right) } \) is a field and \( {D}^{\left( 0\right) } \) is an embedding, then \( D \) extends uniquely to the field of fractions of \( R \) . | Proof. Since \( {A}^{\left( 0\right) } \) is a field, every element of \( A \) with a nonzero component in degree zero is invertible by (1.2.6). Since we are assuming that \( {D}^{\left( 0\right) }\left( r\right) \neq 0 \) for all nonzero \( r \in R, D \) extends uniquely to the field of fractions. | No |
Lemma 1.3.13. If \( \operatorname{char}\left( K\right) = 0 \) or if \( n < \operatorname{char}\left( K\right) \) and \( x \in K \) is a separating variable, then\n\n\[ \n{D}_{x}^{\left( n\right) } = \frac{1}{n!}{D}_{x}^{n} \n\]\n\nwhere \( {D}_{x}^{n} \) is the \( n \) -fold iterated first Hasse derivative. | Proof. Let \( \phi \) be the automorphism of \( K\left\lbrack \left\lbrack t\right\rbrack \right\rbrack \) described above, and put\n\n\[ \ns \mathrel{\text{:=}} \phi \left( t\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }{D}_{x}^{\left( i\right) }\left( y\right) {t}^{i} \n\]\n\nThen for any \( f \in K \) we have\n... | Yes |
Lemma 1.3.14. Suppose that \( k \subseteq K \) are fields and \( x, y \in K \) are separating variables for \( K/k \) . Then there are functions \( {d}_{1},\ldots ,{d}_{n - 1} \in K \) that are polynomial expressions in \( {D}_{x}^{\left( i\right) }\left( y\right) \) for \( 1 \leq i \leq n \), such that for any \( f \i... | Proof. Let \( \phi \) be the automorphism of \( K\left\lbrack \left\lbrack t\right\rbrack \right\rbrack \) described above, and put\n\n\[ \ns \mathrel{\text{:=}} \phi \left( t\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }{D}_{x}^{\left( i\right) }\left( y\right) {t}^{i} \]\n\nThen for any \( f \in K \) we have\n\n... | Yes |
Corollary 1.3.15. With the notation of the lemma, suppose that \( \operatorname{char}\left( K\right) = p \) and that \( f \in {K}^{p} \) . Then\n\n\[ \n{D}_{x}^{\left( p\right) }\left( f\right) = {\left( {D}_{x}^{\left( 1\right) }\left( y\right) \right) }^{p}{D}_{y}^{\left( p\right) }\left( f\right) .\n\] | Proof. By (1.3.13) we have \( {D}_{y}^{\left( i\right) }\left( f\right) = 0 \) for \( 0 < i < p \) . | No |
Lemma 1.4.1. Suppose \( y : V \rightarrow V \) is \( k \) -linear and \( W, U \) are core subspaces. Then \( {\operatorname{tr}}_{W}\left( y\right) = {\operatorname{tr}}_{U}\left( y\right) \) | Proof. We may assume, without loss of generality, that \( V = W + U \) is finite-dimensional. Then \( {\operatorname{tr}}_{V}\left( y\right) = {\operatorname{tr}}_{V/W}\left( y\right) + {\operatorname{tr}}_{W}\left( y\right) = {\operatorname{tr}}_{W}\left( y\right) \) and similarly, \( {\operatorname{tr}}_{V}\left( y\r... | No |
Lemma 1.4.3. If \( y \) and \( x \) are any two \( k \) -linear maps on \( V \) and \( {yx} \) is finitepotent, then \( {xy} \) is also finite potent and \( {\operatorname{tr}}_{V}\left( {yx}\right) = {\operatorname{tr}}_{V}\left( {xy}\right) \) . | Proof. If \( W = {\left( yx\right) }^{n}\left( V\right) \) is finite-dimensional, then\n\n\[ U \mathrel{\text{:=}} {\left( xy\right) }^{n + 1}\left( V\right) = x \circ {\left( yx\right) }^{n} \circ y\left( V\right) \subseteq x\left( W\right) ,\]\n\nso \( U \) is also finite-dimensional. Moreover, by choosing \( n \) la... | No |
Lemma 1.4.3. If \( y \) and \( x \) are any two \( k \) -linear maps on \( V \) and \( {yx} \) is finitepotent, then \( {xy} \) is also finite potent and \( {\operatorname{tr}}_{V}\left( {yx}\right) = {\operatorname{tr}}_{V}\left( {xy}\right) \) . | Proof. If \( W = {\left( yx\right) }^{n}\left( V\right) \) is finite-dimensional, then\n\n\[ U \mathrel{\text{:=}} {\left( xy\right) }^{n + 1}\left( V\right) = x \circ {\left( yx\right) }^{n} \circ y\left( V\right) \subseteq x\left( W\right) ,\]\n\nso \( U \) is also finite-dimensional. Moreover, by choosing \( n \) la... | Yes |
Lemma 1.4.4. If \( E \) is a finitepotent subspace of \( {\operatorname{End}}_{k}\left( V\right) \), then \( \operatorname{tr} : E \rightarrow k \) is \( k \) - linear. | Proof. Take \( y, x \in E \) and for any nonnegative integer \( n \), put\n\n\[ \n{V}_{n} \mathrel{\text{:=}} \mathop{\sum }\limits_{w}w\left( V\right)\n\]\n\nwhere the sum is taken over all words \( w \) of length \( n \) in \( x \) and \( y \) . If \( {w}_{0} \) is any initial segment of \( w \), then \( w\left( V\ri... | No |
Lemma 1.4.7. \( {E}_{V}\left( {W,{W}^{\prime }}\right) \) is a \( k \) -subspace of \( {\operatorname{End}}_{k}\left( V\right) \) . If \( y \in {E}_{V}\left( {W,{W}^{\prime }}\right) ,{W}^{\prime } \preccurlyeq U \) , and \( x \in {E}_{V}\left( {U,{U}^{\prime }}\right) \), then \( {xy} \in {E}_{V}\left( {W,{U}^{\prime ... | Proof. Let \( y, x \in {E}_{V}\left( {W,{W}^{\prime }}\right) \) and \( \alpha \in k \) . Then \( \left( {{\alpha y} + x}\right) \left( W\right) \subseteq y\left( W\right) + x\left( W\right) \preccurlyeq {W}^{\prime } \) by (1.4.6). Moreover, if \( y \in {E}_{V}\left( {W,{W}^{\prime }}\right) ,{W}^{\prime } \preccurlye... | Yes |
Lemma 1.4.9. If \( W \subseteq V \) is nearly invariant under commuting maps \( y, x \), and \( \pi : V \rightarrow W \) is any projection, then \( \left\lbrack {{\pi y},{\pi x}}\right\rbrack \) nearly stabilizes the chain \( V \supseteq W \supseteq 0 \) , and \( {\operatorname{tr}}_{V}\left\lbrack {{\pi y},{\pi x}}\ri... | \[ {\operatorname{tr}}_{V}\left\lbrack {{\pi y},{\pi x}}\right\rbrack = {\operatorname{tr}}_{{W}_{0}}\left\lbrack {{\pi y},{\pi x}}\right\rbrack \] | No |
Lemma 1.4.10. If \( V \) is a \( K \) -submodule of \( {V}^{\prime } \), then \( \langle y, x{\rangle }_{{V}^{\prime }, W} = \langle y, x{\rangle }_{V, W} \) for all \( y, x \in K \) . If \( {W}^{\prime } \subseteq V \) and \( {W}^{\prime } \sim W \), then \( {W}^{\prime } \) is a near \( K \) -submodule and \( \langle... | Proof. Since core subspaces for all finitepotent maps under consideration lie in \( W \), enlarging \( V \) has no effect, and the first statement is immediate. The second easily reduces to the case that \( {W}^{\prime } \subseteq W \), since \( W \) and \( {W}^{\prime } \) both have finite index in \( W + {W}^{\prime ... | Yes |
Theorem 1.4.12. Let \( K \) be a \( k \) -algebra, \( V \) a \( K \) -module, and \( W \subseteq V \) a near submodule. Then there is a \( k \) -linear function \( {\operatorname{Res}}_{W}^{V} : {\Omega }_{K/k} \rightarrow k \) that vanishes on exact differential forms such that\n\n\[ \n{\operatorname{Res}}_{W}^{V}\lef... | Proof. Let \( y, x, w \in K \) and decompose each of them using \( {E}_{V}\left( {W, W}\right) = {E}_{1} + {E}_{2} \) as above. Then we have the following identities:\n\n\[ \n\left\lbrack {{y}_{1},{x}_{1}{w}_{1}}\right\rbrack = {y}_{1}{x}_{1}{w}_{1} - {x}_{1}{w}_{1}{y}_{1}\n\]\n\n\[ \n\left\lbrack {{y}_{1}{x}_{1},{w}_{... | Yes |
Theorem 1.4.14. Let \( K \) be a \( k \) -algebra, \( V \) a \( K \) -module, and \( W \subseteq V \) a near submodule. Suppose that \( K \subseteq {K}^{\prime } \), where \( {K}^{\prime } \) is a commutative \( k \) -algebra that has a \( K \) -basis \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) . Put\n\n\[ \n{V}^{\... | Proof. If we put\n\n\[ \n\widetilde{W} \mathrel{\text{:=}} \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{j}{x}_{j} \otimes W = \mathop{\sum }\limits_{{j = 1}}^{n}{x}_{j} \otimes {a}_{j}W\n\]\n\nfor any \( {a}_{j} \in K\left( {1 \leq j \leq n}\right) \), we have \( \widetilde{W} \preccurlyeq W \) by (1.4.6). From this it foll... | Yes |
Lemma 1.4.16. Suppose that \( K \) contains a finite extension \( {k}^{\prime } \) of \( k \), and that the near \( K \) -submodule \( W \) of \( V \) is \( {k}^{\prime } \) -invariant. Then\n\n\[ \langle y, x{\rangle }_{V, W} = {\operatorname{tr}}_{{k}^{\prime }/k}\left( {\langle y, x{\rangle }_{V, W}^{\prime }}\right... | Proof. Since \( V \) is a \( K \) -module, it is a \( {k}^{\prime } \) -vector space, and we are assuming that \( W \) is \( {k}^{\prime } \) -invariant. Since the residue form is independent of the choice of projection map \( \pi \), we can compute \( \langle y, x{\rangle }_{W} \) using a \( {k}^{\prime } \) -linear p... | Yes |
Lemma 2.1.1. Let \( P \) be a prime divisor of \( K \) and suppose that \( x \in K \) vanishes at \( P \) . Then \( {v}_{P} \) divides the \( x \) -adic valuation \( {v}_{x} \) of \( k\left( x\right) \) . In particular, \( {F}_{P} \) is a finite extension of \( k \) of degree \( f\left( {{v}_{p} \mid {v}_{\left( x\righ... | Proof. Since \( {v}_{P}\left( x\right) > 0 \), it follows immediately from (1.1.14) that \( {v}_{P} \mid {v}_{x} \) . Since the residue field of \( {v}_{x} \) is just \( k \), the result follows. | Yes |
Lemma 2.1.2. If \( {P}_{i} \) is a prime divisor of \( K \) of degree \( {f}_{i} \) and \( x \in {K}^{ \times } \) with \( {v}_{{P}_{i}}\left( x\right) = {e}_{i} \) for \( 1 \leq i \leq s \), then\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{s}{e}_{i}{f}_{i} \leq \left| {K : k\left( x\right) }\right| \]\n\nIn particular, \( ... | Proof. If \( x \in k \), both sides of the inequality are zero. For \( x \notin k \), this is a straightforward application of (1.1.22), viewing \( K \) as a finite extension of \( k\left( x\right) \) . Namely, put \( {v}_{i} \mathrel{\text{:=}} {v}_{{P}_{i}} \) and let \( {v}_{\left( x\right) } \) be the valuation of ... | Yes |
Lemma 2.1.4. Any nonconstant element of \( K \) has at least one zero and one pole. Hence, any two elements of \( K \) with the same divisor differ by a constant multiple. | Proof. Since any nonconstant function in \( K \) is transcendental over \( k \) ,(1.1.7) yields prime divisors \( P, Q \) with \( {v}_{P}\left( x\right) > 0 \) and \( {v}_{Q}\left( {x}^{-1}\right) > 0 \), so \( x \) has a zero at \( P \) and a pole at \( Q \) . Since \( \left\lbrack {xy}\right\rbrack = \left\lbrack x\r... | Yes |
Lemma 2.1.5. Let \( D \) be a divisor on \( K \) . Then\n\n1. \( L\left( D\right) \) is a \( k \) -linear subspace of \( K \) .\n\n2. If \( {D}_{1} \sim {D}_{2} \), then \( L\left( {D}_{1}\right) \cong L\left( {D}_{2}\right) \) .\n\n3. \( L\left( D\right) \neq 0 \) iff there is a nonnegative divisor \( {D}^{\prime } \s... | Proof. 1. This follows from the ultrametric inequality (1.1.1).\n\n2. Suppose \( {D}_{2} = {D}_{1} + \left\lbrack x\right\rbrack \), then multiplication by \( x \) is an isomorphism from \( L\left( {D}_{1}\right) \) to \( L\left( {D}_{2}\right) \) .\n\n3. \( \left\lbrack x\right\rbrack \geq - D \) iff \( \left\lbrack x... | Yes |
Lemma 2.1.6. Suppose \( {D}_{1} \leq {D}_{2} \) are divisors on \( K \) . Then \( {A}_{K}\left( {D}_{1}\right) \subseteq {A}_{K}\left( {D}_{2}\right) \), and\n\n\[ \dim \left( {{A}_{K}\left( {D}_{2}\right) /{A}_{K}\left( {D}_{1}\right) }\right) = \deg \left( {D}_{2}\right) - \deg \left( {D}_{1}\right) . \] | Proof. It is immediate from the definitions that \( {A}_{K}\left( {D}_{1}\right) \subseteq {A}_{K}\left( {D}_{2}\right) \) . By induction on \( \deg \left( {D}_{2}\right) - \deg \left( {D}_{1}\right) \), we may assume that \( {D}_{2} = {D}_{1} + P \) for some prime divisor \( P \), and prove that \( \dim \left( {{A}_{K... | Yes |
Lemma 2.1.7. Given any two divisors \( {D}_{1},{D}_{2} \) we have\n\n1. \( {A}_{K}\left( {{D}_{1} \cap {D}_{2}}\right) = {A}_{K}\left( {D}_{1}\right) \cap {A}_{K}\left( {D}_{2}\right) \) ,\n\n2. \( {A}_{K}\left( {{D}_{1} \cup {D}_{2}}\right) = {A}_{K}\left( {D}_{1}\right) + {A}_{K}\left( {D}_{2}\right) \) . | Proof. 1. This is immediate from the definitions: \( \alpha \in {A}_{K}\left( {{D}_{1} \cap {D}_{2}}\right) \) iff \( - v\left( \alpha \right) \leq v\left( {{D}_{1} \cap {D}_{2}}\right) = \min \{ v\left( {D}_{1}\right), v\left( {D}_{2}\right) \} \) for all \( v \) iff \( \alpha \in {A}_{K}\left( {D}_{1}\right) \cap \) ... | Yes |
Lemma 2.1.8. Suppose \( {D}_{1} \leq {D}_{2} \) are divisors on \( K \) . Then there is a natural short exact sequence\n\n\[ 0 \rightarrow L\left( {D}_{2}\right) /L\left( {D}_{1}\right) \rightarrow {A}_{K}\left( {D}_{2}\right) /{A}_{K}\left( {D}_{1}\right) \rightarrow \overline{{A}_{K}\left( {D}_{2}\right) }/\overline{... | Proof. This is an exercise in using the isomorphism theorems \( {}^{2} \) . Let\n\n\[ \phi : {A}_{K}\left( {D}_{2}\right) \rightarrow \overline{{A}_{K}\left( {D}_{2}\right) } \]\n\nbe the natural map, with kernel \( L\left( {D}_{2}\right) \) . Then \( {\phi }^{-1}\left( \overline{{A}_{K}\left( {D}_{1}\right) }\right) =... | No |
Corollary 2.1.10. \( L\left( D\right) \) is finite dimensional, for any divisor \( D \) . If \( {D}_{1} \leq {D}_{2} \) are divisors, then\n\n\[ \dim L\left( {D}_{2}\right) /L\left( {D}_{1}\right) \leq \deg {D}_{2} - \deg {D}_{1} \] | Proof. Inequality (2.1.11) is immediate from (2.1.6) and (2.1.9). Setting \( {D}_{2} = D \) and \( {D}_{1} = 0 \), we have \( \dim L\left( D\right) - \dim L\left( 0\right) \leq \deg L\left( D\right) \) . But \( L\left( 0\right) = k \) by (2.1.4), whence \( L\left( D\right) \) has finite dimension at most equal to \( \d... | Yes |
Theorem 2.1.14. Let \( x \in K \) . Then \( \deg \left\lbrack x\right\rbrack = 0 \), and there is an integer \( B \) depending on \( x \) such that \( \delta \left( {\left\lbrack {x}^{m}\right\rbrack }_{\infty }\right) \leq B \) for all \( m \geq 0 \) . Furthermore, if \( {\left\lbrack x\right\rbrack }_{0} = \mathop{\s... | Proof. Let \( \left\{ {{u}_{1},\ldots ,{u}_{n}}\right\} \) be a basis for \( K/k\left( x\right) \), and let \( D \) be a nonnegative divisor such that \( \left\lbrack {u}_{j}\right\rbrack \geq - D \) for all \( j \) . Thus \( {u}_{j} \in L\left( D\right) \) for all \( j \) . For any positive integer \( m \) , the funct... | Yes |
Corollary 2.1.16.\n\n1. If \( D \sim {D}^{\prime } \), then \( \deg D = \deg {D}^{\prime } \) and \( \delta \left( D\right) = \delta \left( {D}^{\prime }\right) \) .\n\n2. \( L\left( D\right) = 0 \) for all divisors \( D \) with \( \deg D < 0 \) . | Proof.\n\n1. If \( {D}^{\prime } = D + \left\lbrack x\right\rbrack \) for some principal divisor \( \left\lbrack x\right\rbrack \), then \( \deg {D}^{\prime } = \deg D + \deg \left\lbrack x\right\rbrack = \) \( \deg D \), and \( \dim L\left( D\right) = \dim L\left( {D}^{\prime }\right) \) by (2.1.5).\n\n2. If \( 0 \neq... | Yes |
Corollary 2.1.17. Let \( K/k \) be a function field and let \( {K}^{\prime } \) be a finite extension of \( K \) . Suppose that \( P \) is a prime divisor of \( K \), and let \( {Q}_{1},\ldots ,{Q}_{r} \) be the set of all distinct primes of \( {K}^{\prime } \) dividing \( P \) . Then\n\n\[\n\mathop{\sum }\limits_{{i =... | Proof. Choose \( 0 \neq x \in P \), let \( \left\{ {P = {P}_{1},{P}_{2},\ldots ,{P}_{s}}\right\} \) be the set of all prime divisors of \( \left( x\right) \) in \( K \) and let \( {e}_{i} \mathrel{\text{:=}} e\left( {{P}_{i} \mid \left( x\right) }\right) \) . The \( {P}_{i} \) are the zeros of \( x \) in \( {\mathbb{P}... | Yes |
Corollary 2.1.18. Let \( P \) be a prime of \( K \) . Then the integral closure of \( {\mathcal{O}}_{P} \) in \( {K}^{\prime } \) is a finitely generated \( {\mathcal{O}}_{P} \) -module. | Proof. This is immediate from (1.1.8), (2.1.17) and (1.1.22). | No |
Theorem 2.2.1. If \( K \) has genus \( g \), then\n\n\[ \n\dim {A}_{K}/\left( {{A}_{K}\left( D\right) + K}\right) = g - \delta \left( D\right) \n\]\n\nfor any divisor \( D \) of \( K \) . | Proof. Let \( D \) be a nonspecial divisor and let \( \alpha \) be an adele. There is certainly a divisor \( {D}_{1} \geq D \) such that \( \alpha \in {A}_{K}\left( {D}_{1}\right) \) . From (2.1.13) we see that \( {D}_{1} \) is also nonspecial and that \( {A}_{K}\left( D\right) + K = {A}_{K}\left( {D}_{1}\right) + K \)... | Yes |
Lemma 2.2.3. Let \( w \) be a nonzero Weil differential. Then there is a unique divisor \( D \) of maximum degree such that \( w \in {W}_{K}\left( D\right) \) . Moreover, for any divisor \( E \) we have \( w \in {W}_{K}\left( E\right) \) iff \( E \leq D. \) | Proof. This is an easy consequence of (2.1.7): If \( w \) vanishes on \( {A}_{K}\left( {D}_{1}\right) \) and \( {A}_{K}\left( {D}_{2}\right) \), then it vanishes on \( {A}_{K}\left( {D}_{1}\right) + {A}_{K}\left( {D}_{2}\right) = {A}_{K}\left( {{D}_{1} \cup {D}_{2}}\right) \) . | No |
Lemma 2.2.5. \( L\left( D\right) {W}_{K}\left( C\right) \subseteq {W}_{K}\left( {C - D}\right) \) for any divisors \( C, D \) . Moreover, we have\n\n(2.2.6)\n\n\[ \left\lbrack {xw}\right\rbrack = \left\lbrack x\right\rbrack + \left\lbrack w\right\rbrack \] \n\nfor any \( x \in K \) and \( w \in {W}_{K} \) . | Proof. It is immediate from the definitions that \( {A}_{K}\left( C\right) {A}_{K}\left( D\right) \subseteq {A}_{K}\left( {C + D}\right) \) for any divisors \( C, D \) . Thus, for \( x \in {L}_{K}\left( D\right) \) we have \( x{A}_{K}\left( C\right) \subseteq {A}_{K}\left( {C + D}\right) \) . This implies that \( x{W}_... | Yes |
Theorem 2.2.8. Let \( K \) be a function field. Then \( {\dim }_{K}\left( {W}_{K}\right) = 1 \) . Any two canonical divisors are linearly equivalent. | Proof. The second statement is immediate from the first and (2.2.6). Choose any two nonzero Weil differentials \( {w}_{1},{w}_{2} \) . For \( i = 1,2 \) suppose that \( {w}_{i} \in {W}_{K}\left( {D}_{i}\right) \) . Then the map \( x \mapsto x{w}_{i} \) defines for any divisor \( D \) an embedding \( {\phi }_{i, D} : L\... | Yes |
Theorem 2.2.9 (Riemann-Roch). Let \( K \) be a function field of genus \( g \) and let \( C \) be a canonical divisor on \( K \) . Then for any divisor \( D,{L}_{K}\left( {C - D}\right) \simeq {W}_{K}\left( D\right) \), and we have\n\n\[ \dim {L}_{K}\left( D\right) = \deg D + 1 - g + \dim {L}_{K}\left( {C - D}\right) .... | Proof. A restatement of the formula is \( \dim L\left( {C - D}\right) = g - \delta \left( D\right) \), so by (2.2.2) the formula follows from the \( k \) -isomorphism \( L\left( {C - D}\right) \simeq {W}_{K}\left( D\right) \) .\n\nBy (2.2.8) we can take \( C = \left\lbrack w\right\rbrack \) for any nonzero Weil differe... | Yes |
Corollary 2.2.10. Let \( K \) be a function field of genus \( g \) and let \( C \) be a divisor on \( K \) . Then \( C \) is a canonical divisor if and only if \( \dim L\left( C\right) = g \) and \( \deg C = {2g} - 2 \) . In particular, all divisors of degee at least \( {2g} - 1 \) are nonspecial. | Proof. Suppose \( C \) is canonical and put \( D = 0 \) in (2.2.9). This yields \( \dim C = g \) . Now put \( D = C \) and obtain \( \deg C = {2g} - 2 \) . Conversely, assume \( \dim L\left( C\right) = g \) and \( \deg C = {2g} - 2 \) . Then \( \delta \left( C\right) = g - 1 \), so \( \dim \overline{{A}_{K}}/\overline{... | Yes |
Corollary 2.2.11. The following conditions are equivalent for a function field \( K \) :\n\n1. K has genus 0 and has a prime divisor \( P \) of degree one.\n\n2. \( K \) has an element \( x \) with \( \deg {\\left\\lbrack x\\right\\rbrack }_{\\infty } = 1 \) .\n\n3. \( K = k\\left( x\\right) \) for some \( x \\in K \) ... | Proof. \( 1 \\Rightarrow 2 : \) By (2.2.10) canonical divisors have degree -2, so the Riemann-Roch theorem gives \( \\dim L\\left( P\\right) = 2 \) . Let \( x \) be a nonconstant function in \( L\\left( P\\right) \) .\n\nThen \( x \) has exactly one pole of order at most 1 at \( P \) . But \( x \) must have a pole, so ... | Yes |
Corollary 2.2.12. If \( g\left( K\right) > 0 \), then \( \psi \) is injective. | Proof. The condition \( P - {P}^{\prime } = \left\lbrack x\right\rbrack \) implies that \( {\left\lbrack x\right\rbrack }_{\infty } = {P}^{\prime } \), so \( {g}_{K} = 0 \) by (2.2.11). | No |
Theorem 2.2.13 (Strong Approximation Theorem). Suppose that\n\n\[ S \mathrel{\text{:=}} \left\{ {{P}_{\infty },{P}_{1},\ldots ,{P}_{n}}\right\} \subseteq {\mathbb{P}}_{K} \]\n\n\( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \subseteq K \), and \( \left\{ {{m}_{1},\ldots ,{m}_{n}}\right\} \subseteq \mathbb{Z} \) . Put \( ... | Proof. Consider the divisor \( D \mathrel{\text{:=}} N{P}_{\infty } - \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{m}_{i} + 1}\right) {P}_{i} \) where \( N \gg 0 \), and the adele\n\n\[ {\alpha }_{P} \mathrel{\text{:=}} \left\{ \begin{array}{ll} {x}_{i} & \text{ if }P = {P}_{i},1 \leq i \leq n \\ 0 & \text{ otherwise. }... | Yes |
Lemma 2.4.2. Let \( k \subseteq K \) be fields such that \( {k}^{\prime }{ \otimes }_{k}K \) is a field for every finite extension \( {k}^{\prime } \) of \( k \) . Then \( k \) is algebraically closed in \( K \) . | Proof. If \( k \) is not algebraically closed in \( K \), there is a finite simple extension \( {k}^{\prime } = \) \( k\left( u\right) \) for some \( u \in K \smallsetminus k \) . Since \( {k}^{\prime } \) is a direct summand of \( K \) as a \( k \) -vector space, \( {k}^{\prime }{ \otimes }_{k}K \) contains the finite... | Yes |
Corollary 2.4.4. Let \( K/k \) be a geometric function field and let \( {k}^{\prime } \) be a finite extension of \( K \) . Then \( {k}^{\prime }{ \otimes }_{k}K \) is a geometric function field over \( {k}^{\prime } . | Proof. For any finite extension \( {k}^{\prime \prime } \) of \( {k}^{\prime } \) we have\n\n\[ \n{k}^{\prime \prime }{ \otimes }_{k}^{\prime }\left( {{k}^{\prime }{ \otimes }_{k}K}\right) = \left( {{k}^{\prime \prime }{ \otimes }_{k}^{\prime }{k}^{\prime }}\right) { \otimes }_{k}K = {k}^{\prime \prime }{ \otimes }_{k}... | Yes |
Corollary 2.4.4. Let \( K/k \) be a geometric function field and let \( {k}^{\prime } \) be a finite extension of \( K \) . Then \( {k}^{\prime }{ \otimes }_{k}K \) is a geometric function field over \( {k}^{\prime } \) . | Proof. For any finite extension \( {k}^{\prime \prime } \) of \( {k}^{\prime } \) we have\n\n\[ \n{k}^{\prime \prime }{ \otimes }_{k}^{\prime }\left( {{k}^{\prime }{ \otimes }_{k}K}\right) = \left( {{k}^{\prime \prime }{ \otimes }_{k}^{\prime }{k}^{\prime }}\right) { \otimes }_{k}K = {k}^{\prime \prime }{ \otimes }_{k}... | Yes |
Lemma 2.4.5. Let \( K \) be a function field over a perfect ground field \( k \). Then \( K \) is geometric. | Proof. Let \( {k}^{\prime } \) be a finite extension of \( k \). Then \( {k}^{\prime }/k \) is separable, so \( {k}^{\prime } = k\left( u\right) \) for some \( u \in {k}^{\prime } \) by (A.0.17). Moreover, \( u \) satisfies an irreducible separable polynomial \( f\left( X\right) \in k\left\lbrack X\right\rbrack \) of d... | Yes |
Theorem 2.4.9. Let \( {K}^{\prime }/K \) be a finite separable extension of function fields with \( {P}^{\prime } \in {\mathbb{P}}_{{K}^{\prime }} \), and put \( P \mathrel{\text{:=}} {P}^{\prime } \cap K \) . Then for almost all \( {P}^{\prime }, e\left( {{P}^{\prime } \mid P}\right) = 1 \) and \( {F}_{{P}^{\prime }}/... | Proof. By (A.0.17), \( {K}^{\prime } = K\left( u\right) \) for some element \( u \in {K}^{\prime } \) . Let \( f\left( X\right) \mathrel{\text{:=}} {X}^{n} + \) \( {a}_{1}{X}^{n - 1} + \cdots + {a}_{n} \) be the minimum polynomial of \( u \) over \( K \) . Then \( f\left( X\right) \) has distinct roots \( u = {u}_{1},{... | Yes |
Lemma 2.4.11. Let \( K/k \) be a geometric function field of characteristic \( p > 0 \) . Then \( k{K}^{p} \) is the unique subfield of \( K \) containing \( k \) for which \( K/{K}_{0} \) is purely inseparable of degree \( p \) . If \( {K}_{0}/{k}_{0} \) is any geometric subfield of \( K \) of finite index, then the n... | Proof. If \( K/{K}_{0} \) is purely inseparable of degree \( p \), then \( {K}^{p} \subseteq {K}_{0} \), and since \( \left| {K : k{K}^{p}}\right| = p \) by (2.4.6), the first assertion follows. For any geometric \( {K}_{0} \subseteq K \) , the natural map is either zero or an embedding, because \( {\dim }_{{K}_{0}}{\O... | No |
Lemma 2.4.12. Let \( {K}^{\prime }/{k}^{\prime } \) be a weakly separable finite extension of \( K/k \) . Then \( {K}^{\prime }/{k}^{\prime }K \) is separable. | Proof. We may assume that \( \operatorname{char}\left( K\right) = : p > 0 \) . Since \( {k}^{\prime } \) is the full field of constants of \( {K}^{\prime } \), it is also the full field of constants of \( {k}^{\prime }K \), so replacing \( K \) by \( {k}^{\prime }K \), we may as well assume that \( {k}^{\prime } = k \)... | No |
Lemma 2.5.1. If \( u, v \in {\widehat{\mathcal{O}}}_{P} \), then \( {\operatorname{Res}}_{P}\left( {udv}\right) = 0 \) . In particular, if \( \alpha \) is an adele and \( v \in K \), then \( {\operatorname{Res}}_{P}\left( {{\alpha }_{P}{dv}}\right) = 0 \) for almost all \( P \in {\mathbb{P}}_{K} \) . | Proof. The first statement is immediate from (1.4.9) because \( {\widehat{\mathcal{O}}}_{P} \) is invariant under \( u \) and \( v \) . The second follows because \( \alpha \) and \( v \) have only finitely many poles. | Yes |
Corollary 2.5.4 (Residue Theorem). Let \( K/k \) be a function field and let \( \omega \in {\Omega }_{K} \). Then\n\n\[ \mathop{\sum }\limits_{{P \in {\mathbf{P}}_{K}}}{\operatorname{Res}}_{P}\left( \omega \right) = 0 \] | Proof. If we take \( S = {\mathbb{P}}_{K} \) in the theorem, we get \( {\mathcal{O}}_{S} = k \sim 0 \), and the result follows by (1.4.10). | No |
Let \( K \) be a geometric function field, let \( P \in {\mathbb{P}}_{K} \), and let \( x \in K \) be a separating variable. Then \( {\operatorname{Res}}_{P}\left( {ydx}\right) \neq 0 \) for some \( y \in {\widehat{R}}_{P} \) . If \( {F}^{\text{sep }} \) is the maximal separable subfield of \( {F}_{P} \), then \( {F}^{... | Proof. Since \( {\nu }_{P}\left( {dx}\right) \) is finite, there must be an adele \( \alpha \) with \( {\operatorname{Res}}_{P}\left( {{\alpha }_{P}{dx}}\right) \neq 0 \) , proving the first statement. For the second, we note that any finite extension of \( k \) contained in \( {\widehat{K}}_{P} \) lies in \( {\widehat... | Yes |
Corollary 2.5.10. Let \( K \) be a geometric function field of genus \( g \), and let \( {\Omega }_{K}\left( 0\right) \) denote the space of regular differential forms on \( K \) . Then \( {\dim }_{k}{\Omega }_{K}\left( 0\right) = g \) . | Proof. This is immediate from (2.5.7) and (2.2.2). | No |
Theorem 2.5.13. Suppose that \( K \) is geometric and \( P \in {P}_{K}^{\text{sep }} \) with local parameter t. If the power series expansion of \( x \in {\mathcal{O}}_{P} \) at \( P \) is\n\n\[ x = \mathop{\sum }\limits_{{m = 0}}^{\infty }{a}_{m}{t}^{m} \]\n\nwith \( {a}_{m} \in {F}_{P} \), then the power series expan... | Proof. This follows from (2.5.13) by observing that the constant term in the power series expansion of \( {D}_{t}^{\left( n\right) }\left( x\right) \) at \( t \) is \( {a}_{n} \), the coefficient of \( {t}^{n} \) in the expansion of \( x \) . | No |
Corollary 2.5.14 (Taylor’s Theorem). Suppose that \( P \in {P}_{K}^{\text{sep }} \) with local parameter \( t \) and \( x \in {\mathcal{O}}_{P} \) . Then\n\n\[ x = \mathop{\sum }\limits_{{n = 0}}^{\infty }{D}_{t}^{\left( n\right) }\left( x\right) \left( P\right) {t}^{n} \] | Proof. This follows from (2.5.13) by observing that the constant term in the power series expansion of \( {D}_{t}^{\left( n\right) }\left( x\right) \) at \( t \) is \( {a}_{n} \), the coefficient of \( {t}^{n} \) in the expansion of \( x \) . | Yes |
Lemma 2.5.15. Let \( K/k \) be a geometric function field and let \( P \in {\mathbb{P}}_{K}^{\text{sep }} \) with local parameter \( t \) and residue field \( F \) . Let \( u, v \in K \) . Then\n\n\[ \n{\operatorname{Res}}_{P}\left( {udv}\right) = {\operatorname{tr}}_{F/k}\left( {a}_{-1}\right)\n\]\n\nwhere \( {a}_{-1}... | Proof. Put \( x \mathrel{\text{:=}} u\left( {{dv}/{dt}}\right) \), so that \( {udv} = {xdt} \), and use (1.2.14) to write\n\n\[ \nx = \mathop{\sum }\limits_{{i = - n}}^{\infty }{a}_{i}{t}^{i}\n\]\n\nwith \( {a}_{i} \in F \) . We need to show that \( {\operatorname{Res}}_{P}\left( {xdt}\right) = {\operatorname{tr}}_{F/k... | Yes |
Lemma 3.0.1. Let \( K/k \) be a geometric function field and let \( {K}^{\prime }/{k}^{\prime } \) be a finite extension of \( K/k \) . Then \( {K}^{\prime }/{k}^{\prime } \) is geometric. | Proof. Since \( K/k \) is geometric, \( {k}^{\prime }{ \otimes }_{k}K \) is a field isomorphic to the subfield \( {k}^{\prime }K \subseteq \) \( {K}^{\prime } \) . By (2.4.4) \( {k}^{\prime }K/{k}^{\prime } \) is geometric, so we may as well assume that \( {k}^{\prime } = k \) .\n\nNow let \( k \subseteq {k}_{0} \subse... | Yes |
Lemma 3.1.2. Let \( {K}^{\prime }/{k}^{\prime } \) be a finite extension of \( K/k \), let \( {D}^{\prime } \in \operatorname{Div}\left( {K}^{\prime }\right) \), and let \( D \in \operatorname{Div}\left( K\right) \). Then\n\n\[ \deg {N}_{{K}^{\prime }/K}\left( {D}^{\prime }\right) = \left| {{k}^{\prime } : k}\right| \d... | Proof. Let \( P \) be a prime divisor of \( K \), and let \( \left\{ {{Q}_{1},\ldots ,{Q}_{r}}\right\} \) be the set of all distinct prime divisors of \( P \) in \( {K}^{\prime } \). Put \( {e}_{i} \mathrel{\text{:=}} e\left( {{Q}_{i} \mid P}\right) \) and \( {f}_{i} \mathrel{\text{:=}} f\left( {{Q}_{i} \mid P}\right) ... | Yes |
Theorem 3.1.3. Let \( {K}^{\prime } \) be a finite extension of \( K \), and let \( {\left\lbrack x\right\rbrack }_{K} \) (resp. \( {\left\lbrack x\right\rbrack }_{{K}^{\prime }} \) ) denote the principal divisor of \( x \) in \( K \) (resp. \( {K}^{\prime } \) .) Then\n\n\[ \n{N}_{{K}^{\prime }/K}\left( {\left\lbrack ... | Proof. Suppose that \( x = {yz} \) . Dropping subscripts, we have \( N\left( x\right) = N\left( y\right) N\left( z\right) \) and \( N\left( \left\lbrack x\right\rbrack \right) = N\left( \left\lbrack y\right\rbrack \right) + N\left( \left\lbrack z\right\rbrack \right) \) . Therefore, if we can prove the formula for \( y... | No |
Theorem 3.1.5 (Trace Formula). Let \( K/k \) be a geometric function field, let \( {K}^{\prime }/k \) be a finite separable extension, let \( P \in {\mathbb{P}}_{K} \), and let \( \omega \in {\Omega }_{{K}^{\prime }} \). Then\n\n\[ \n{\operatorname{Res}}_{P}\left( {{\operatorname{tr}}_{{K}^{\prime }/K}\left( \omega \ri... | Proof. Let \( x \in K \) be a separating variable as above, and put \( \omega = {ydx} \) for some \( y \in {K}^{\prime } \). If we take \( S \) in Tate’s theorem (2.5.2) to be the set of prime divisors \( Q \) of \( P \) in \( {K}^{\prime } \), we get\n\n\[ \n{\operatorname{Res}}_{{\mathcal{O}}_{S}}^{{K}^{\prime }} = \... | Yes |
Lemma 3.2.2. Let \( P \in {\mathbb{P}}_{K} \) and let \( {k}^{\prime }/k \) be a finite extension. If either \( {F}_{P}/k \) or \( {k}^{\prime }/k \) is purely inseparable, then there is a unique divisor of \( P \) in \( {K}^{\prime } \mathrel{\text{:=}} {k}^{\prime }{ \otimes }_{k}K \) . | Proof. Case 1): \( {k}^{\prime }/k \) is purely inseparable. Then for some power \( q = {p}^{n} \) the map \( x \mapsto {x}^{q} \) is an isomorphism of \( {K}^{\prime } \) onto a subfield of \( K \) . For \( x \in {K}^{\prime } \) define\n\n\[ \n{v}^{\prime }\left( x\right) \mathrel{\text{:=}} {v}_{p}\left( {x}^{q}\rig... | Yes |
Theorem 3.2.3. Suppose that \( K/k \) is a geometric function field and that \( {k}^{\prime }/k \) is finite and separable. Put \( {K}^{\prime } \mathrel{\text{:=}} {k}^{\prime }{ \otimes }_{k}K \) and let \( Q \in {\mathbb{P}}_{{K}^{\prime }} \) with \( P \mathrel{\text{:=}} Q \cap K \) . Then \( {k}^{\prime }{ \otime... | Proof. We have \( {k}^{\prime } = k\left( \alpha \right) \) for some \( \alpha \in {k}^{\prime } \) by (A.0.17). Let \( f\left( X\right) \) be the minimal polynomial of \( \alpha \) over \( k \) . Then \( f\left( X\right) \) is irreducible over \( K \) by (3.2.1). Let \( R \) be the integral closure of \( {\mathcal{O}}... | Yes |
Lemma 3.2.4. Suppose that \( P \in {\mathbb{P}}_{K} \) is nonsingular with respect to \( {k}^{\prime } \) . Put \( {K}^{\prime } \mathrel{\text{:=}} {k}^{\prime }{ \otimes }_{k}K \), and let \( Q \in {\mathbb{P}}_{{K}^{\prime }} \) with \( Q \mid P \) . Then \( {F}_{Q} = {k}^{\prime }{F}_{P} \) . In particular, if \( P... | Proof. Let \( R \) be the integral closure of \( {\mathcal{O}}_{P} \) in \( {K}^{\prime } \) . Then (1.1.22) yields \( {\mathcal{O}}_{Q} = R + Q \) , from which it follows immediately that \( {F}_{Q} = {k}^{\prime }{F}_{P} \) . In particular, \( {F}_{Q} = {k}^{\prime } \) when \( {k}^{\prime } \supseteq {F}_{P} \ | Yes |
Theorem 3.2.6. Let \( K/k \) be a geometric function field and let \( P \) be a nonsingular prime divisor of \( K \). Let \( {k}^{\prime }/k \) be a finite extension that is a splitting field for \( P \), and put \( {K}^{\prime } \mathrel{\text{:=}} {k}^{\prime }{ \otimes }_{k}K \). Then:\n\n1. There are exactly \( \le... | Proof. 1) Let \( {P}^{\prime } \) be a divisor of \( P \) in \( {\mathbb{P}}_{{K}^{\prime }} \). By hypothesis and (3.2.4) we have \( {k}^{\prime } = {F}_{{P}^{\prime }} = {k}^{\prime }{F}_{P} \) and therefore \( {k}^{\prime } \supseteq {F}_{P} \). Put \( {k}_{0}^{\prime } \mathrel{\text{:=}} {F}_{P}^{\text{sep }} \) a... | Yes |
Corollary 3.2.7. Let \( D \) be a divisor on \( \bar{K} \). Then \( D \) is defined over some finite extension \( {k}^{\prime } \) of \( k \). | Proof. We may assume without loss of generality that \( D \) is a point \( Q \). Let \( P \mathrel{\text{:=}} \) \( Q \cap K \), let \( {k}^{\prime } \) be a splitting field for \( P \), and apply (3.2.6). | No |
Lemma 3.3.3. Suppose that \( K/k \subseteq {K}^{\prime }/{k}^{\prime } \subseteq {K}^{\prime \prime }/{k}^{\prime \prime } \) are function fields with \( {K}^{\prime \prime } \) weakly separable over \( K \) . Then\n\n\[ \n{\mathcal{D}}_{{K}^{\prime \prime }/K} = {\mathcal{D}}_{{K}^{\prime \prime }/{K}^{\prime }} + {N}... | Proof. By (3.3.2) we have\n\n\[ \n{\left\lbrack \omega \right\rbrack }_{{K}^{\prime }} = {N}_{{K}^{\prime }/K}^{ * }\left( {\left\lbrack \omega \right\rbrack }_{K}\right) + {\mathcal{D}}_{{K}^{\prime }/K} \n\]\n\n\[ \n{\left\lbrack \omega \right\rbrack }_{{K}^{\prime \prime }} = {N}_{{K}^{\prime \prime }/{K}^{\prime }}... | Yes |
Theorem 3.3.7. Let \( {K}^{\prime } \) be a finite weakly separable extension of \( K \) and suppose that \( Q \) is a separable prime of \( {K}^{\prime } \) dividing the separable prime \( P \) of \( K \) with \( f\left( {Q \mid P}\right) = 1 \) . If \( s \) is a local parameter at \( Q \) such that \( {v}_{{Q}^{\prim... | Proof. Put \( {K}_{1} \mathrel{\text{:=}} K\left( s\right) \) and \( {Q}_{1} \mathrel{\text{:=}} Q \cap {K}_{1} \) . Then \( e\left( {Q \mid {Q}_{1}}\right) = 1 \) and \( Q \) is the unique prime divisor of \( {Q}_{1} \) in \( {\mathbb{P}}_{K} \) because \( s \) is a local unit at every other prime divisor of \( P \) .... | Yes |
Lemma 3.3.10. Suppose that \( {K}^{\prime } = K\left( y\right) \), where \( y \) is separable over \( K \) and integral over \( {\mathcal{O}}_{P} \) for some \( P \in {\mathbb{P}}_{K} \) . Then \( {\mathcal{O}}_{P}{\left\lbrack y\right\rbrack }^{ * }{\delta }_{K}\left( y\right) = {\mathcal{O}}_{P}\left\lbrack y\right\r... | Proof. Because \( y \) is integral, \( {\mathcal{O}}_{P}\left\lbrack y\right\rbrack \subseteq {\mathcal{O}}_{P}{\left\lbrack y\right\rbrack }^{ * } \) and \( {\mathcal{O}}_{P}{\left\lbrack y\right\rbrack }^{ * } \) is an \( {\mathcal{O}}_{P}\left\lbrack y\right\rbrack \) -module. We therefore get \( {\delta }_{K}{\left... | Yes |
Corollary 3.3.12. Suppose \( P \in {\mathbb{P}}_{K}^{\text{sep }},{K}^{\prime } = K\left( y\right) \) for some separable element \( y \in \) \( {K}^{\prime } \) that is integral over \( {\mathcal{O}}_{P} \), and \( Q \in {\mathbb{P}}_{{K}^{\prime }}^{\text{sep }} \) divides \( P \) with \( f\left( {Q \mid P}\right) = 1... | Proof. Put \( M \mathrel{\text{:=}} Q \cap {\mathcal{O}}_{P}\left\lbrack y\right\rbrack \) . From (3.3.10) we have \( {R}_{P}^{ * }{\delta }_{K}\left( y\right) = {C}_{P}\left( y\right) \) and the inequality is immediate from (3.3.11). Moreover, equality holds precisely when \( {C}_{P}\left( y\right) \nsubseteq M \) .\n... | Yes |
Lemma 3.4.1. Let \( K/k \) be a geometric function field with \( P \in {\mathbb{P}}_{K} \) and let \( {k}^{\prime }/k \) be a finite extension. Let \( R \) be the integral closure of \( {\mathcal{O}}_{P} \) in \( {K}^{\prime } \mathrel{\text{:=}} \ddot{{k}^{\prime }}{ \otimes }_{k}K \) and let \( \widetilde{R} \mathrel... | Proof. (1) Since both \( R \) and \( \widetilde{R} \) are free \( {\mathcal{O}}_{P} \) -modules of rank \( \left| {{k}^{\prime } : k}\right| \) (see (2.1.18) and (1.1.9)), the \( {\mathcal{O}}_{P} \) -module \( R/\widetilde{R} \) has finite length by (1.1.12) and is therefore finite \( k \) -dimensional. This implies t... | Yes |
Lemma 3.4.3. Let \( {k}^{\prime } \) be a finite extension of \( k \) with \( {K}^{\prime } \mathrel{\text{:=}} {k}^{\prime }{ \otimes }_{k}K \) . Let \( Q \in {\mathbb{P}}_{{K}^{\prime }} \) and put \( P \mathrel{\text{:=}} Q \cap K \) . Suppose that the subfields \( {k}^{\prime } \) and \( {\widehat{K}}_{P} \) of \( ... | Proof. We first apply (1.2.11) to conclude that \( {\widehat{{K}^{\prime }}}_{Q} = {k}^{\prime }{\widehat{K}}_{P} \) . Thus, our hypothesis says that there is an isomorphism\n\n(*)\n\n\[ \n{\widehat{{K}^{\prime }}}_{Q} \simeq {k}^{\prime }{ \otimes }_{{k}_{Q}^{\prime }}{\widehat{K}}_{P} \n\]\n\nThis puts us in a positi... | Yes |
Corollary 3.4.6. Let \( K/k \) be a geometric function field. Then \( K \) has at most finitely many singular prime divisors. Moreover, there exists a purely inseparable finite extension \( {k}^{\prime }/k \) such that \( {k}^{\prime }{ \otimes }_{k}K \) is nonsingular. In particular, every prime divisor of \( K \) has... | Proof. Let \( P \in {\mathbb{P}}_{K} \) and suppose that \( {K}^{\prime } \) is a scalar extension of \( K \) . If \( d\left( {Q \mid P}\right) < 0 \) for some \( Q \in {\mathbb{P}}_{{K}^{\prime }} \), then \( Q \) divides \( {\mathcal{D}}_{{K}^{\prime }/K} \) . Moreover, \( d\left( {{Q}^{\prime } \mid P}\right) < 0 \)... | Yes |
Theorem 3.4.7. Let \( K/k \) be a nonsingular geometric function field and suppose that \( {K}^{\prime }/{k}^{\prime } \) is a finite weakly separable extension. Then \( {\mathcal{D}}_{{K}^{\prime }/K} \geq 0 \) . If \( Q \in {\mathbb{P}}_{{K}^{\prime }} \) is singular, then \( d\left( {Q \mid Q \cap K}\right) \neq 0 \... | Proof. If \( {K}^{\prime } \) is a scalar extension of \( K \), then \( {K}^{\prime } \) is nonsingular by definition, and hence \( {\mathcal{D}}_{{K}^{\prime }/K} = 0 \) . Thus, if we can prove the theorem when \( k = {k}^{\prime } \), it will follow in general by (3.3.4), so we may as well assume that \( {k}^{\prime ... | Yes |
Theorem 3.5.1. Let \( {Q}_{1} \) and \( {Q}_{2} \) be prime divisors of \( {K}^{\prime } \) with \( P \mathrel{\text{:=}} {Q}_{1} \cap K = {Q}_{2} \cap K \) . Then there exists \( \sigma \in \operatorname{Gal}\left( {{K}^{\prime }/K}\right) \) with \( {Q}_{1}^{\sigma } = {Q}_{2} \) . | Proof. By the weak approximation theorem there exists an element \( x \in {K}^{\prime } \) such that \( {v}_{{Q}_{1}}\left( x\right) = 1 \) while \( {v}_{Q}\left( x\right) = 0 \) for every other prime \( Q \) dividing \( P \) . Then for every prime divisor \( Q \) of \( P \) and every \( \sigma \in \operatorname{Gal}\l... | Yes |
Corollary 3.5.3. Assume (3.5.2), then \( \left| {G : {G}_{0}}\right| \) is the number of distinct prime divisors of \( P \) in \( {K}^{\prime } \) . For any prime divisor \( {Q}^{\prime } \) of \( P \) in \( {K}^{\prime } \) we have \( e\left( {{Q}^{\prime } \mid P}\right) = e \) and \( f\left( {{Q}^{\prime } \mid P}\r... | Proof. Let \( \left\{ {{Q}_{1},\ldots ,{Q}_{r}}\right\} \) be the set of all distinct prime divisors of \( P \) in \( {K}^{\prime } \) . As a consequence of (3.5.1), there is a bijection between cosets \( {G}_{0}\sigma \) of \( {G}_{0} \) in \( G \) and prime divisors \( {Q}^{\sigma } \) of \( P \), whence \( \left| {G... | Yes |
Theorem 3.5.4. Assume (3.5.2), then \( {F}_{Q}/{F}_{P} \) is Galois, and the natural map \( {G}_{0}/{G}_{1} \rightarrow \operatorname{Gal}\left( {{F}_{Q}/{F}_{P}}\right) \) is an isomorphism. In particular, \( \left| {G}_{1}\right| = e \) . | Proof. By (3.5.3) we have \( {F}_{{Q}_{0}} = {F}_{P} \), so it suffices to prove this result in the special case that \( G = {G}_{0} \) and \( K = {K}_{0} \) . Now, by (3.5.1) \( Q \) is the unique prime divisor of \( P \) in \( {K}^{\prime } \), so \( {\mathcal{O}}_{Q} \) is the integral closure of \( {\mathcal{O}}_{P... | Yes |
Corollary 3.5.5. Assume (3.5.2), and suppose \( K \subseteq E \subseteq {K}^{\prime } \) is an intermediate field. Put \( {Q}_{E} \mathrel{\text{:=}} Q \cap E \) . Then \( e\left( {{Q}_{E} \mid P}\right) = 1 \) if and only if \( E \subseteq {K}_{1} \) . In particular, \( {K}^{\prime }/{K}_{1} \) is totally ramified at ... | Proof. From (A.0.16) we have \( \operatorname{Gal}\left( {{K}^{\prime }/E}\right) = {G}_{E} \subseteq G \), where \( {G}_{E} \) is the subgroup of \( G \) fixing \( E \) elementwise. By definition, \( {G}_{E} \cap {G}_{1} \) is the inertia group of \( Q \) over \( E \) . By (1.1.25), \( e\left( {{Q}_{E} \mid P}\right) ... | Yes |
Lemma 3.5.6. Assume (3.5.2), let \( \sigma \in {G}_{1} \) and let \( t \) be a local parameter at \( Q \) . Then for every integer \( j \geq 1 \) we have \( \sigma \in {\dot{G}}_{j} \) iff \( \sigma \left( t\right) \equiv t{\;\operatorname{mod}\;{Q}^{j}} \) . | Proof. Suppose \( \sigma \left( t\right) \equiv t{\;\operatorname{mod}\;{Q}^{j}} \) for some \( \sigma \in {G}_{1} \) and some \( j \geq 1 \) . We only need to show that \( \sigma \in {G}_{j} \) since the converse is trivial.\n\nApplying (1.1.24) to the extension \( {K}^{\prime }/{K}_{1} \) which is totally ramified by... | Yes |
Theorem 3.5.7. Assume (3.5.2). The map \( \sigma \mapsto {u}_{\sigma } \mathrel{\text{:=}} \sigma \left( t\right) /t \) defines a homomorphism \( {G}_{1} \rightarrow {F}_{Q}^{ * } \) whose kernel is \( {G}_{2} \) . In particular, \( {G}_{1}/{G}_{2} \) is cyclic. | Proof. For any \( {\sigma }^{\prime } \in {G}_{1} \) we have\n\n\[ \n{u}_{{\sigma }^{\prime }\sigma }t = {\sigma }^{\prime }\left( {\sigma \left( t\right) }\right) = {\sigma }^{\prime }\left( {u}_{\sigma }\right) {\sigma }^{\prime }\left( t\right) = {\sigma }^{\prime }\left( {u}_{\sigma }\right) {u}_{{\sigma }^{\prime ... | Yes |
Theorem 3.5.8. Assume (3.5.2). For \( \sigma \in {G}_{i} \) and \( i \geq 2 \), the map \( \sigma \mapsto {x}_{\sigma } \mathrel{\text{:=}} \) \( \left( {\sigma \left( t\right) - t}\right) {t}^{-i} \) defines a homomorphism of \( {G}_{i} \) into the additive subgroup of \( {F}_{O} \) whose kernel is \( {G}_{i + 1} \) .... | Proof. For any \( {\sigma }^{\prime } \in {G}_{i} \) we have\n\n\[ t + {x}_{{\sigma }^{\prime }\sigma }{t}^{i} = {\sigma }^{\prime }\left( t\right) + {\sigma }^{\prime }\left( {x}_{\sigma }\right) {\sigma }^{\prime }{\left( t\right) }^{i} = t + {x}_{{\sigma }^{\prime }}{t}^{i} + \left( {{x}_{\sigma } + y{t}^{i}}\right)... | Yes |
Theorem 3.5.9. Assume (3.5.2), and let \( d\left( {Q \mid P}\right) \) be the different exponent. Then\n\n\[ d\left( {Q \mid P}\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\left( {\left| {G}_{i}\right| - 1}\right) \]\n\nIn particular, \( {G}_{2} = 1 \) if and only if \( Q \mid P \) is tamely ramified. | Proof. Note that the sum is finite, since \( {G}_{i} = 1 \) for almost all \( i \) . By (3.5.5) we have \( e\left( {{Q}_{1} \mid P}\right) = 1 \) and therefore \( d\left( {{Q}_{1} \mid P}\right) = 0 \) by (3.3.6). Now (3.3.3) implies that \( d\left( {Q \mid P}\right) = d\left( {Q \mid {Q}_{1}}\right) \).\n\nLet \( t \)... | Yes |
Theorem 3.6.2. Suppose that \( K = k\left( {x, y}\right) ,\operatorname{char}\left( K\right) \neq 2 \), and\n\n\[ \n{y}^{2} = f\left( x\right) = \mathop{\prod }\limits_{{i = 1}}^{r}{p}_{i}\left( x\right) \n\] \n\nwhere the \( {p}_{i} \) are distinct irreducible polynomials and \( r \geq 1 \) . Then \( k \) is the full ... | Proof. From (3.3.5) we obtain\n\n\[ \n2{g}_{K} - 2 = - 4 + \deg {\mathcal{D}}_{K/k\left( x\right) } \n\] \n\nand the result follows from (3.6.2). | No |
Corollary 3.6.3. Assume the hypotheses of (3.6.2). Then\n\n\[ \n{g}_{K} = \left\{ \begin{array}{ll} \frac{1}{2}\left( {\deg f - 2}\right) & \text{ if }\deg f\text{ is even,} \\ \frac{1}{2}\left( {\deg f - 1}\right) & \text{ if }\deg f\text{ is odd. } \end{array}\right. \n\] | Proof. From (3.3.5) we obtain\n\n\[ \n2{g}_{K} - 2 = - 4 + \deg {\mathcal{D}}_{K/k\left( x\right) }\n\]\n\nand the result follows from (3.6.2). | No |
Lemma 3.6.5. Suppose that \( \operatorname{char}\left( k\right) = 2 \) and \( K/k\left( x\right) \) is separable of degree 2. If \( k{\left( x\right) }^{\left( 2\right) } \) denotes the additive subgroup of \( k\left( x\right) \) consisting of all rational functions of the form \( a{\left( x\right) }^{2} + a\left( x\ri... | We first consider the (uninteresting) case that \( K \) is a scalar extension of \( k\left( x\right) \) . Then there is a constant \( \alpha \in K \smallsetminus k\left( x\right) \) . Since \( {T}_{K/k\left( x\right) }\left( \alpha \right) \) is a nonzero element of \( k \) , we can divide to get a scalar of trace 1 . ... | No |
Lemma 3.6.7. Suppose that \( k \) is perfect of characteristic \( 2, a\left( x\right), p\left( x\right) \in k\left\lbrack x\right\rbrack, p\left( x\right) \) is irreducible, and\n\n\[ f\left( x\right) \mathrel{\text{:=}} \frac{a\left( x\right) }{p{\left( x\right) }^{2e}} \]\n\nfor some positive integer e. Then there ex... | Proof. Separating terms of even and odd degree and using that fact that \( {k}^{2} = k \) , we can write \( a\left( x\right) = {a}_{0}{\left( x\right) }^{2} + x{a}_{1}{\left( x\right) }^{2} \) and \( p\left( x\right) = {p}_{0}{\left( x\right) }^{2} + x{p}_{1}{\left( x\right) }^{2} \) where \( {a}_{i} \) and \( {p}_{i} ... | Yes |
Theorem 3.6.10. Let \( k \) be perfect of characteristic 2, and let \( K \mathrel{\text{:=}} k\left( {x, y}\right) \), where \( {y}^{2} + y = f\left( x\right) \) and the pole divisor of \( f\left( x\right) \) in \( k\left( x\right) \) is given by\n\n\[ {\left\lbrack f\left( x\right) \right\rbrack }_{\infty } = \mathop{... | Proof. The zero divisor of \( \phi \left( x\right) \) in \( K \) is\n\n\[ {\left\lbrack \phi \left( x\right) \right\rbrack }_{0} = \mathop{\sum }\limits_{{{P}_{i} \neq \infty }}2{d}_{i}{P}_{i} = \mathcal{D} - 2{d}_{\infty }{P}_{\infty } \]\n\nwhere we set \( {d}_{\infty } = 0 \) if \( f\left( x\right) \) has no pole at... | Yes |
Theorem 3.6.10. Let \( k \) be perfect of characteristic 2, and let \( K \mathrel{\text{:=}} k\left( {x, y}\right) \), where \( {y}^{2} + y = f\left( x\right) \) and the pole divisor of \( f\left( x\right) \) in \( k\left( x\right) \) is given by\n\n\[{\left\lbrack f\left( x\right) \right\rbrack }_{\infty } = \mathop{\... | Proof. The zero divisor of \( \phi \left( x\right) \) in \( K \) is\n\n\[{\left\lbrack \phi \left( x\right) \right\rbrack }_{0} = \mathop{\sum }\limits_{{{P}_{i} \neq \infty }}2{d}_{i}{P}_{i} = \mathcal{D} - 2{d}_{\infty }{P}_{\infty }\]\n\nwhere we set \( {d}_{\infty } = 0 \) if \( f\left( x\right) \) has no pole at i... | Yes |
Theorem 4.1.2. If the finitely generated \( k \) -algebra \( K \) is a field, then \( \left| {K : k}\right| \) is finite. | Proof. Let \( K = k\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) and put \( {k}_{1} \mathrel{\text{:=}} k\left( {x}_{1}\right) \subseteq K \) . Then \( K = {k}_{1}\left\lbrack {{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) , and \( \left| {K : {k}_{1}}\right| \) is finite by induction on \( n \) . If \( {x}... | Yes |
Corollary 4.1.3. Suppose that \( M \) is a maximal ideal of the polynomial ring \( A = k\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . Then there exist \( {a}_{1},\ldots ,{a}_{n} \in k \) such that \( M = \left( {{X}_{1} - {a}_{1},{X}_{2} - }\right. \) \( \left. {{a}_{2},\ldots ,{X}_{n} - {a}_{n}}\right) \) .... | Proof. \( A/M \) satisfies the hypotheses of (4.1.2). Since \( k \) is algebraically closed, we conclude that \( A/M = k \) . Let \( {a}_{i} \) be the image of \( {X}_{i} \) in \( A/M \) . Then the natural map \( A \rightarrow A/M = k \) is just evaluation at the point \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) . | Yes |
Corollary 4.1.4 (Hilbert Nullstellensatz). If \( J \) is an ideal of \( A \mathrel{\text{:=}} k\left\lbrack {{X}_{0},{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) , then \( \mathbf{I}\left( {\mathbf{V}\left( J\right) }\right) = \sqrt{J} \) . | Proof. If \( {f}^{r} \in J \), it is clear that \( f \) vanishes at every zero of \( J \) . The converse argument is a well-known but clever trick. We adjoin an additional indeterminate \( Y \) to \( A \) . Then the ideal \( J + \left( {1 - {fY}}\right) \subseteq A\left\lbrack Y\right\rbrack \) has no zeros at all; hen... | Yes |
Corollary 4.1.6. The mappings \( I \mapsto \mathbf{V}\left( I\right) \) and \( V \mapsto \mathbf{I}\left( V\right) \) define an inclusion-reversing bijection between proper graded radical ideals \( I \subseteq k\left\lbrack {{X}_{0},\ldots ,{X}_{n}}\right\rbrack \) and closed subsets \( V \subseteq {\mathbb{P}}^{n} \) ... | Proof. Because the two mappings are inclusion-reversing, we have \( \mathbf{V}\left( {\mathbf{I}\left( {\mathbf{V}\left( J\right) }\right) }\right) = \) \( \mathbf{V}\left( J\right) \) and \( \mathbf{I}\left( {\mathbf{V}\left( {\mathbf{I}\left( S\right) }\right) }\right) = \mathbf{I}\left( S\right) \) for any ideal \( ... | Yes |
Lemma 4.1.8. Let \( V \subseteq {\mathbb{P}}^{n} \) be a projective variety. Relabeling the \( {X}_{i} \) if necessary so that \( V \nsubseteq \mathbf{V}\left( {X}_{0}\right), k\left( V\right) \) is generated as a field over \( k \) by the restrictions of the functions \( {X}_{i}/{X}_{0} \) to \( V \) for \( 1 \leq i \... | Proof. Abusing notation, we continue to denote by \( {X}_{i}/{X}_{0} \) the restriction of \( {X}_{i}/{X}_{0} \) to \( V \) . Put \( {K}_{0} \mathrel{\text{:=}} k\left( {{X}_{1}/{X}_{0},\ldots ,{X}_{n}/{X}_{0}}\right) \) . Then clearly \( {K}_{0} \subseteq k\left( V\right) \) . However, if \( p\left( X\right) \) is hom... | Yes |
Lemma 4.1.10. If \( V \subseteq {\mathbb{P}}^{2} \) is a projective curve, then \( \mathbf{I}\left( V\right) = \left( f\right) \) for some irreducible homogeneous polynomial \( f \) . | Proof. Choose \( f \in \mathbf{I}\left( V\right) \) homogeneous of minimal degree \( d \) . Since it is easy to see that the product of inhomogeneous polynomials is inhomogeneous, the minimal-ity of \( d \) makes \( f \) irreducible. Put \( x \mathrel{\text{:=}} {X}_{1}/{X}_{0} \) and \( y \mathrel{\text{:=}} {X}_{2}/{... | Yes |
Theorem 4.2.2. Let \( K/k \) be a function field and let \( \left( {{\phi }_{0},\ldots ,{\phi }_{n}}\right) \in K \) with \( {\phi }_{0} \neq 0 \) and \( {\phi }_{i}/{\phi }_{0} \) nonconstant for some \( i \) . For any point \( P \) of \( K \), let \( {t}_{P} \) be a local parameter at \( P \) and put \( {e}_{P} \math... | \[ \phi \left( P\right) \mathrel{\text{:=}} \left( {{t}_{P}^{{e}_{P}}{\phi }_{0}\left( P\right) : {t}_{P}^{{e}_{P}}{\phi }_{1}\left( P\right) : \cdots : {t}_{P}^{{e}_{P}}{\phi }_{n}\left( P\right) }\right) \in {\mathbb{P}}^{n} \] is well-defined, is independent of the choice of \( {t}_{P} \), and \[ V \mathrel{\text{:=... | Yes |
Lemma 4.2.3. Let \( \phi : {\mathbb{P}}_{K} \rightarrow V \subseteq {\mathbb{P}}^{n} \) be an effective birational map. Then there exists a separating variable \( x \in \langle \phi \rangle \), and for each such \( x \) there exists \( y \in \langle \phi \rangle \) such that \( \left( {1, x, y}\right) \) is birational. | Proof. Since \( K = k\left( {\langle \phi \rangle }\right) \), we have \( \langle \phi \rangle \nsubseteq {K}^{p} \) ; hence there is an element \( x \in \langle \phi \rangle \) such that \( K/k\left( x\right) \) is separable by (2.4.6). In particular, there are only finitely many intermediate fields, all of which cont... | No |
Lemma 4.2.4. Let \( \phi : {\mathbb{P}}_{K} \rightarrow V \) be a projective map. If \( K \supseteq {K}^{\prime } \supseteq {\phi }^{ * }\left( {k\left( V\right) }\right) \) for some subfield \( {K}^{\prime } \) of \( K \), there is a uniquely determined projective map \( {\phi }^{\prime } : {\mathbb{P}}_{{K}^{\prime }... | Proof. If \( Q \in {\mathbb{P}}_{K} \) and \( P \mathrel{\text{:=}} Q \cap {K}^{\prime } \), then \( f\left( Q\right) = f\left( P\right) \) for any function \( f \in {K}^{\prime } \) by definition. Since all functions \( {\phi }_{i}/{\phi }_{0} \) lie in \( {K}^{\prime } \), we conclude that \( \phi \) is constant on a... | Yes |
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