Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Theorem 7.2. Let \( E \) be a separable extension of \( k \) . Then \( E \) is solvable by radicals if and only if \( E/k \) is solvable. | Proof. Assume that \( E/k \) is solvable, and let \( K \) be a finite solvable Galois extension of \( k \) containing \( E \) . Let \( m \) be the product of all primes unequal to the characteristic dividing the degree \( \left\lbrack {K : k}\right\rbrack \), and let \( F = k\left( \zeta \right) \) where \( \zeta \) is... | Yes |
Theorem 8.1. Let \( k \) be a field, \( m \) an integer \( > 0 \) prime to the characteristic of \( k \) (if not 0 ). We assume that \( k \) contains \( {\mathbf{\mu }}_{m} \) . Let \( B \) be a subgroup of \( {k}^{ * } \) containing \( {k}^{*m} \) and let \( {K}_{B} = k\left( {B}^{1/m}\right) \) . Then \( {K}_{B} \) i... | Proof. Let \( \sigma \in G \) . Suppose \( \langle \sigma, a\rangle = 1 \) for all \( a \in B \) . Then for every generator \( \alpha \) of \( {K}_{B} \) such that \( {\alpha }^{m} = a \in B \) we have \( {\sigma \alpha } = \alpha \) . Hence \( \sigma \) induces the identity on \( {K}_{B} \) and the kernel on the left ... | Yes |
Theorem 8.2. Notation being as in Theorem 8.1, the map \( B \mapsto {K}_{B} \) gives a bijection of the set of subgroups of \( {k}^{ * } \) containing \( {k}^{*m} \) and the abelian extensions of \( k \) of exponent \( m \) . | Proof. Let \( {B}_{1},{B}_{2} \) be subgroups of \( {k}^{ * } \) containing \( {k}^{*m} \) . If \( {B}_{1} \subset {B}_{2} \) then \( k\left( {B}_{1}^{1/m}\right) \subset k\left( {B}_{2}^{1/m}\right) \) . Conversely, assume that \( k\left( {B}_{1}^{1/m}\right) \subset k\left( {B}_{2}^{1/m}\right) \) . We wish to prove ... | Yes |
Theorem 8.3. Let \( k \) be a field of characteristic \( p \) . The map \( B \mapsto k\left( {{\wp }^{-1}B}\right) \) is a bijection between subgroups of \( k \) containing \( \wp k \) and abelian extensions of \( k \) of exponent \( p \) . Let \( K = {K}_{B} = k\left( {{\wp }^{-1}B}\right) \), and let \( G \) be its G... | Proof. The proof is entirely similar to the proof of Theorems 8.1 and 8.2. It can be obtained by replacing multiplication by addition, and using the \ | No |
Theorem 9.1. Let \( k \) be a field and \( n \) an integer \( \geqq 2 \) . Let \( a \in k, a \neq 0 \) . Assume that for all prime numbers \( p \) such that \( p \mid n \) we have \( a \notin {k}^{p} \), and if \( 4 \mid n \) then \( a \notin - 4{k}^{4} \) . Then \( {X}^{n} - a \) is irreducible in \( k\left\lbrack X\r... | Proof. Our first assumption means that \( a \) is not a \( p \) -th power in \( k \) . We shall reduce our theorem to the case when \( n \) is a prime power, by induction.\n\nWrite \( n = {p}^{r}m \) with \( p \) prime to \( m \), and \( p \) odd. Let\n\n\[ \n{X}^{m} - a = \mathop{\prod }\limits_{{v = 1}}^{m}\left( {X ... | Yes |
Corollary 9.2. Let \( k \) be a field and assume that \( a \in k, a \neq 0 \), and that \( a \) is not a p-th power for some prime p. If \( p \) is equal to the characteristic, or if \( p \) is odd, then for every integer \( r \geqq 1 \) the polynomial \( {X}^{{p}^{r}} - a \) is irreducible over \( k \) . | Proof. The assertion is logically weaker than the assertion of the theorem. | No |
Corollary 9.3. Let \( k \) be a field and assume that the algebraic closure \( {k}^{a} \) of \( k \) is of finite degree \( > 1 \) over \( k \) . Then \( {k}^{\mathrm{a}} = k\left( i\right) \) where \( {i}^{2} = - 1 \), and \( k \) has characteristic 0 . | Proof. We note that \( {k}^{\mathrm{a}} \) is normal over \( k \) . If \( {k}^{\mathrm{a}} \) is not separable over \( k \), so char \( k = p > 0 \), then \( {k}^{\mathrm{a}} \) is purely inseparable over some subfield of degree \( > \) 1 (by Chapter V,§6), and hence there is a subfield \( E \) containing \( k \), and ... | Yes |
Let \( k \) be a field. Let \( n \) be an odd positive integer prime to the characteristic, and assume that \( \left\lbrack {k\left( {\mathbf{\mu }}_{n}\right) : k}\right\rbrack = \varphi \left( n\right) \) . Let \( a \in k \), and suppose that for each prime \( p \mid n \) the element \( a \) is not a p-th power in \(... | This is a special case of the general theory of \( \$ {11} \), and Exercise 39, taking into account the representation of \( {G}_{K/k} \) in the group of matrices. One need only use the fact that the order of \( {G}_{K/k} \) is \( {n\varphi }\left( n\right) \), according to that exercise, and so \( \# \left( {G}_{K/k}\... | No |
Theorem 10.1. Let \( K/k \) be a finite Galois extension with Galois group \( G \) . Then for the operation of \( G \) on \( {K}^{ * } \) we have \( {H}^{1}\left( {G,{K}^{ * }}\right) = 1 \), and for the operation of \( G \) on the additive group of \( K \) we have \( {H}^{1}\left( {G, K}\right) = 0 \) . In other words... | Proof. Let \( {\left\{ {\alpha }_{\sigma }\right\} }_{\sigma \in G} \) be a 1-cocycle of \( G \) in \( {K}^{ * } \) . The multiplicative cocycle relation reads\n\n\[ \n{\alpha }_{\sigma }{\alpha }_{\tau }^{\sigma } = {\alpha }_{\sigma \tau }\n\]\n\nBy the linear independence of characters, there exists \( \theta \in K ... | Yes |
Lemma 10.2. (Sah). Let \( G \) be a group and let \( E \) be a \( G \) -module. Let \( \tau \) be in the center of \( G \) . Then \( {H}^{1}\left( {G, E}\right) \) is annihilated by the map \( x \mapsto {\tau x} - x \) on \( E \) . In particular, if this map is an automorphism of \( E \), then \( {H}^{1}\left( {G, E}\r... | Proof. Let \( f \) be a 1-cocycle of \( G \) in \( E \) . Then\n\n\[ f\left( \sigma \right) = f\left( {{\tau \sigma }{\tau }^{-1}}\right) = f\left( \tau \right) + \tau \left( {f\left( {\sigma {\tau }^{-1}}\right) }\right) \]\n\n\[ = f\left( \tau \right) + \tau \left\lbrack {f\left( \sigma \right) + {\sigma f}\left( {\t... | Yes |
Theorem 11.1. Let \( M \mid N \) . Let \( \varphi \) be the homomorphism\n\n\[ \varphi : \Gamma \rightarrow \operatorname{Hom}\left( {{H}_{\Gamma }\left( {M, N}\right) ,{A}_{M}}\right) \]\n\nand let \( {\Gamma }_{\varphi } \) be its kernel. Let \( {e}_{M}\left( \Gamma \right) = \) g.c.d. \( \left( {e\left( {{\Gamma }^{... | Proof. Let \( x \in \Gamma \) and suppose \( {\varphi }_{x} = 0 \) . Let \( {My} = x \) . For \( \sigma \in G \) let\n\n\[ {y}_{\sigma } = {\sigma y} - y \]\n\nThen \( \left\{ {y}_{\sigma }\right\} \) is a 1-cocycle of \( G \) in \( {A}_{M} \), and by the hypothesis that \( {\varphi }_{x} = 0 \), this cocycle depends o... | Yes |
Assume that \( M \) is prime to \( 2\left( {{\Gamma }^{\prime } : \Gamma }\right) \) and is not divisible by any primes of the special set \( S \) . Then we have an injection\n\n\[ \varphi : \Gamma /{M\Gamma } \rightarrow \operatorname{Hom}\left( {{H}_{\Gamma }\left( {M, N}\right) ,{A}_{M}}\right) \]\n\nIf in addition ... | Under the hypotheses of the corollary, we have \( c\left( M\right) = 1 \) and \( {c}_{M}\left( \Gamma \right) = 1 \) in the theorem. | No |
Theorem 12.1. (Artin). Let \( {\lambda }_{1},\ldots ,{\lambda }_{n} : A \rightarrow K \) be additive homomorphisms of an additive group into a field. If these homomorphisms are algebraically dependent over \( K \), then there exists an additive polynomial\n\n\[ f\left( {{X}_{1},\ldots ,{X}_{n}}\right) \neq 0 \]\n\nin \... | Proof. Let \( f\left( X\right) = f\left( {{X}_{1},\ldots ,{X}_{n}}\right) \in K\left\lbrack X\right\rbrack \) be a reduced polynomial of lowest possible degree such that \( f \neq 0 \) but for all \( x \in A, f\left( {\Lambda \left( x\right) }\right) = 0 \), where \( \Lambda \left( x\right) \) is the vector \( \left( {... | Yes |
Theorem 12.2. Let \( K \) be an infinite field, and let \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) be the distinct elements of a finite group of automorphisms of \( K \) . Then \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) are algebraically independent over \( K \) . | Proof. (Artin). In characteristic 0, Theorem 12.1 and the linear independence of characters show that our assertion is true. Let the characteristic be \( p > 0 \), and assume that \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) are algebraically dependent.\n\nThere exists an additive polynomial \( f\left( {{X}_{1},\ldots ,{X... | Yes |
Theorem 13.1. Let \( K/k \) be a finite Galois extension of degree \( n \) . Let \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) be the elements of the Galois group \( G \) . Then there exists an element \( w \in K \) such that \( {\sigma }_{1}w,\ldots ,{\sigma }_{n}w \) form a basis of \( K \) over \( k \) . | Proof. We prove this here only when \( k \) is infinite. The case when \( k \) is finite can be proved later by methods of linear algebra, as an exercise.\n\nFor each \( \sigma \in G \), let \( {X}_{\sigma } \) be a variable, and let \( {t}_{\sigma ,\tau } = {X}_{{\sigma }^{-1}\tau } \) . Let \( {X}_{i} = {X}_{{\sigma ... | No |
Theorem 14.1. The homomorphism \( G \rightarrow \lim G/H \) is an isomorphism. | Proof. First the kernel is trivial, because if \( \sigma \) is in the kernel, then \( \sigma \) restricted to every finite subextension of \( K \) is trivial, and so is trivial on \( K \) . Recall that an element of the inverse limit is a family \( \left\{ {\sigma }_{H}\right\} \) with \( {\sigma }_{H} \in G/H \), sati... | Yes |
Theorem 15.2. For each prime \( l \) there exists a unique Galois extension \( K \) of \( \mathbf{Q} \), with Galois group \( G \), and an injective homomorphism\n\n\[ \rho : G \rightarrow G{L}_{2}\left( {\mathbf{F}}_{l}\right) \]\n\nhaving the following property. For all but a finite number of primes \( p \), if \( {a... | The above theorem was conjectured by Serre in 1968 [Se 68]. A proof of the existence as in the first statement was given by Deligne [De 68]. The second statement, describing how big the Galois group actually is in the group of matrices \( G{L}_{2}\left( {\mathbf{F}}_{l}\right) \) is due to Serre and Swinnerton-Dyer [Se... | Yes |
Proposition 1.1. Let \( A \) be an entire ring and \( K \) its quotient field. Let \( \alpha \) be algebraic over \( K \) . Then there exists an element \( c \neq 0 \) in \( A \) such that \( {c\alpha } \) is integral over \( A \) . | Proof. There exists an equation\n\n\[ \n{a}_{n}{\alpha }^{n} + {a}_{n - 1}{\alpha }^{n - 1} + \cdots + {a}_{0} = 0 \n\]\n\nwith \( {a}_{i} \in A \) and \( {a}_{n} \neq 0 \) . Multiply it by \( {a}_{n}^{n - 1} \) . Then\n\n\[ \n{\left( {a}_{n}\alpha \right) }^{n} + \cdots + {a}_{0}{a}_{n}^{n - 1} = 0 \n\]\n\nis an integ... | Yes |
Proposition 1.2. If \( B \) is integral over \( A \) and finitely generated as an \( A \) -algebra, then \( B \) is finitely generated as an \( A \) -module. | Proof. We may prove this by induction on the number of ring generators, and thus we may assume that \( B = A\left\lbrack \alpha \right\rbrack \) for some element \( \alpha \) integral over \( A \), by considering a tower\n\n\[ A \subset A\left\lbrack {\alpha }_{1}\right\rbrack \subset A\left\lbrack {{\alpha }_{1},{\alp... | Yes |
Proposition 1.3. Integral ring extensions form a distinguished class. | Proof. Let \( A \subset B \subset C \) be a tower of rings. If \( C \) is integral over \( A \), then it is clear that \( B \) is integral over \( A \) and \( C \) is integral over \( B \) . Conversely, assume that each step in the tower is integral. Let \( \alpha \in C \) . Then \( \alpha \) satisfies an integral equa... | Yes |
Proposition 1.4. Let \( A \) be a subring of \( C \) . Then the elements of \( C \) which are integral over \( A \) form a subring of \( C \) . | Proof. Let \( \alpha ,\beta \in C \) be integral over \( A \) . Let \( M = A\left\lbrack \alpha \right\rbrack \) and \( N = A\left\lbrack \beta \right\rbrack \) . Then \( {MN} \) contains 1, and is therefore faithful as an \( A \) -module. Furthermore, \( {\alpha M} \subset M \) and \( {\beta N} \subset N \) . Hence \(... | No |
Proposition 1.5. Let \( A \subset B \) be an extension ring, and let \( B \) be integral over \( A \) . Let \( \sigma \) be a homomorphism of \( B \) . Then \( \sigma \left( B\right) \) is integral over \( \sigma \left( A\right) \) . | Proof. Let \( \alpha \in B \), and let\n\n\[ \n{\alpha }^{n} + {a}_{n - 1}{\alpha }^{n - 1} + \cdots + {a}_{0} = 0 \n\]\n\nbe an integral equation for \( \alpha \) over \( A \) . Applying \( \sigma \) yields\n\n\[ \n\sigma {\left( \alpha \right) }^{n} + \sigma \left( {a}_{n - 1}\right) \sigma {\left( \alpha \right) }^{... | Yes |
Corollary 1.6. Let \( A \) be an entire ring, \( k \) its quotient field, and \( E \) a finite extension of \( k \) . Let \( \alpha \in E \) be integral over \( A \) . Then the norm and trace of \( \alpha \) (from \( E \) to \( k \) ) are integral over \( A \), and so are the coefficients of the irreducible polynomial ... | Proof. For each embedding \( \sigma \) of \( E \) over \( k,{\sigma \alpha } \) is integral over \( A \) . Since the norm is the product of \( {\sigma \alpha } \) over all such \( \sigma \) (raised to a power of the characteristic), it follows that the norm is integral over \( A \) . Similarly for the trace, and simila... | Yes |
Proposition 1.7. Let \( A \) be entire and factorial. Then \( A \) is integrally closed. | Proof. Suppose that there exists a quotient \( a/b \) with \( a, b \in A \) which is integral over \( A \), and a prime element \( p \) in \( A \) which divides \( b \) but not \( a \) . We have, for some integer \( n \geqq 1 \), and \( {a}_{i} \in A \), \[ {\left( a/b\right) }^{n} + {a}_{n - 1}{\left( a/b\right) }^{n ... | Yes |
Proposition 1.8. Let \( f : A \rightarrow B \) be integral, and let \( S \) be a multiplicative subset of \( A \) . Then \( {S}^{-1}f : {S}^{-1}A \rightarrow {S}^{-1}B \) is integral. | Proof. If \( \alpha \in B \) is integral over \( f\left( A\right) \), then writing \( {\alpha \beta } \) instead of \( f\left( a\right) \beta \) for \( a \in A \) and \( \beta \in B \) we have\n\n\[ \n{\alpha }^{n} + {a}_{n - 1}{\alpha }^{n - 1} + \cdots + {a}_{0} = 0 \n\]\n\nwith \( {a}_{i} \in A \) . Taking the canon... | Yes |
Proposition 1.9. Let \( A \) be entire and integrally closed. Let \( S \) be a multiplicative subset of \( A,0 \notin S \) . Then \( {S}^{-1}A \) is integrally closed. | Proof. Let \( \alpha \) be an element of the quotient field, integral over \( {S}^{-1}A \) . We have an equation\n\n\[ \n{\alpha }^{n} + \frac{{a}_{n - 1}}{{s}_{n - 1}}{\alpha }^{n - 1} + \cdots + \frac{{a}_{0}}{{s}_{0}} = 0 \n\]\n\n\( {a}_{i} \in A \) and \( {s}_{i} \in S \) . Let \( s \) be the product \( {s}_{n - 1}... | Yes |
Proposition 1.10. Let \( A \) be a subring of \( B \), let \( \mathfrak{p} \) be a prime ideal of \( A \), and assume \( B \) integral over \( A \) . Then \( \mathfrak{p}B \neq B \) and there exists a prime ideal \( \mathfrak{P} \) of B lying above \( \mathfrak{p} \) . | Proof. We know that \( {B}_{\mathfrak{p}} \) is integral over \( {A}_{\mathfrak{p}} \) and that \( {A}_{\mathfrak{p}} \) is a local ring with maximal ideal \( {m}_{\mathfrak{p}} = {S}^{-1}\mathfrak{p} \), where \( S = A - \mathfrak{p} \) . Since we obviously have\n\n\[ \mathfrak{p}{B}_{\mathfrak{p}} = \mathfrak{p}{A}_{... | Yes |
Proposition 1.11. Let \( A \) be a subring of \( B \), and assume that \( B \) is integral over \( A \) . Let \( \mathfrak{P} \) be a prime ideal of \( B \) lying over a prime ideal \( \mathfrak{p} \) of \( A \) . Then \( \mathfrak{P} \) is maximal if and only if \( \mathfrak{p} \) is maximal. | Proof. Assume \( \mathfrak{p} \) maximal in \( A \) . Then \( A/\mathfrak{p} \) is a field, and \( B/\mathfrak{P} \) is an entire ring, integral over \( A/\mathfrak{p} \) . If \( \alpha \in B/\mathfrak{P} \), then \( \alpha \) is algebraic over \( A/\mathfrak{p} \), and we know that \( A/\mathfrak{p}\left\lbrack \alpha... | Yes |
Proposition 2.1. Let \( A \) be an entire ring, integrally closed in its quotient field \( K \) . Let \( L \) be a finite Galois extension of \( K \) with group \( G \) . Let \( \mathfrak{p} \) be a maximal ideal of \( A \), and let \( \mathfrak{P} \) , \( \mathfrak{Q} \) be prime ideals of the integral closure \( B \)... | Proof. Suppose that \( \mathfrak{Q} \neq \sigma \mathfrak{P} \) for any \( \sigma \in G \) . Then \( \tau \mathfrak{Q} \neq \sigma \mathfrak{P} \) for any pair of elements \( \sigma ,\tau \in G \) . There exists an element \( x \in B \) such that \[ x \equiv 0\;\left( {{\;\operatorname{mod}\;\sigma }\mathfrak{P}}\right... | Yes |
Corollary 2.2 Let \( A \) be integrally closed in its quotient field \( K \). Let \( E \) be a finite separable extension of \( K \), and \( B \) the integral closure of \( A \) in \( E \). Let \( \mathfrak{p} \) be a maximal ideal of \( A \). Then there exists only a finite number of prime ideals of B lying above \( \... | Proof. Let \( L \) be the smallest Galois extension of \( K \) containing \( E \). If \( {\mathfrak{Q}}_{1} \), \( {\mathfrak{Q}}_{2} \) are two distinct prime ideals of \( B \) lying above \( \mathfrak{p} \), and \( {\mathfrak{P}}_{1},{\mathfrak{P}}_{2} \) are two prime ideals of the integral closure of \( A \) in \( ... | Yes |
Proposition 2.3. The field \( {L}^{\text{dec }} \) is the smallest subfield \( E \) of \( L \) containing \( K \) such that \( \mathfrak{P} \) is the only prime of \( B \) lying above \( \mathfrak{P} \cap E \) (which is prime in \( B \cap E) \) . | Proof. Let \( E \) be as above, and let \( H \) be the Galois group of \( L \) over \( E \) . Let \( \mathfrak{q} = \mathfrak{P} \cap E \) . By Proposition 2.1, all primes of \( B \) lying above \( \mathfrak{q} \) are conjugate by elements of \( H \) . Since there is only one prime, namely \( \mathfrak{P} \), it means ... | Yes |
Proposition 2.4. Notation being as above, we have \( A/\mathfrak{p} = {B}^{\mathrm{{dec}}}/\mathfrak{Q} \) (under the canonical injection \( A/\mathfrak{p} \rightarrow {B}^{\mathrm{{dec}}}/\mathfrak{Q} \) ). | Proof. If \( \sigma \) is an element of \( G \), not in \( {G}_{\mathfrak{P}} \), then \( \sigma \mathfrak{P} \neq \mathfrak{P} \) and \( {\sigma }^{-1}\mathfrak{P} \neq \mathfrak{P} \) . Let\n\n\[ \n{\mathfrak{Q}}_{\sigma } = {\sigma }^{-1}\mathfrak{P} \cap {B}^{\mathrm{{dec}}} \n\]\n\nThen \( {\mathfrak{Q}}_{\sigma }... | Yes |
Corollary 2.6. Let \( A \) be integrally closed in its quotient field \( K \) . Let \( L \) be a finite Galois extension of \( K \), and \( B \) the integral closure of \( A \) in \( L \) . Let \( \mathfrak{p} \) be a maximal ideal of \( A \) . Let \( \varphi : A \rightarrow A/\mathfrak{p} \) be the canonical homomorph... | Proof. The kernels of \( {\psi }_{1},{\psi }_{2} \) are prime ideals of \( B \) which are conjugate by Proposition 2.1. Hence there exists an element \( \tau \) of the Galois group \( G \) such that \( {\psi }_{1},{\psi }_{2} \circ \tau \) have the same kernel. Without loss of generality, we may therefore assume that \... | Yes |
Corollary 2.7. Let the assumptions be as in Corollary 2.6 and assume that \( \mathfrak{P} \) is the only prime of \( B \) lying above \( \mathfrak{p} \) . Let \( f\left( X\right) \) be a polynomial in \( A\left\lbrack X\right\rbrack \) with leading coefficient 1. Assume that \( f \) is irreducible in \( K\left\lbrack X... | Proof. By Corollary 2.6, we know that any two roots of \( \bar{f} \) are conjugate under some isomorphism of \( \bar{B} \) over \( \bar{A} \), and hence that \( \bar{f} \) cannot split into relative prime polynomials. Therefore, \( \bar{f} \) is a power of an irreducible polynomial. | Yes |
Proposition 2.8. Let \( A \) be an entire ring, integrally closed in its quotient field \( K \) . Let \( L \) be a finite Galois extension of \( K \) . Let \( L = K\left( \alpha \right) \), where \( \alpha \) is integral over \( A \), and let\n\n\[ f\left( X\right) = {X}^{n} + {a}_{n - 1}{X}^{n - 1} + \cdots + {a}_{0} ... | Proof. Let\n\n\[ f\left( X\right) = \prod \left( {X - {x}_{i}}\right) \]\n\nbe the factorization of \( f \) in \( L \) . We know that all \( {x}_{i} \in B \) . If \( \sigma \in {G}_{\mathfrak{P}} \), then we denote by \( \bar{\sigma } \) the homomorphic image of \( \sigma \) in the group \( {\bar{G}}_{\mathfrak{P}} \),... | Yes |
Theorem 2.9. Let \( A \) be an entire ring, integrally closed in its quotient field \( K \) . Let \( f\left( X\right) \in A\left\lbrack X\right\rbrack \) have leading coefficient 1 and be irreducible over \( K \) (or \( A \), it’s the same thing). Let \( \mathfrak{p} \) be a maximal ideal of \( A \) and let \( \bar{f} ... | Proof. Let \( \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) be the roots of \( f \) in \( B \) and let \( \left( {{\bar{\alpha }}_{1},\ldots ,{\bar{\alpha }}_{n}}\right) \) be their reductions mod \( \mathfrak{P} \) . Since\n\n\[ \nf\left( X\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {X - {\alpha }_{i... | Yes |
Proposition 3.1. Let \( A \) be a subring of \( B \) and assume that \( B \) is integral over A. Let \( \varphi : A \rightarrow L \) be a homomorphism into a field \( L \) which is algebraically closed. Then \( \varphi \) has an extension to a homomorphism of \( B \) into \( L \) . | Proof. Let \( \mathfrak{p} \) be the kernel of \( \varphi \) and let \( S \) be the complement of \( \mathfrak{p} \) in \( A \) . Then we have a commutative diagram\n\n\n\nand \( \varphi \) can be factored through th... | No |
Theorem 3.2. Let \( A \) be a subring of a field \( K \) and let \( x \in K, x \neq 0 \) . Let \( \varphi : A \rightarrow L \) be a homomorphism of \( A \) into an algebraically closed field \( L \) . Then \( \varphi \) has an extension to a homomorphism of \( A\left\lbrack x\right\rbrack \) or \( A\left\lbrack {x}^{-1... | Proof. We may first extend \( \varphi \) to a homomorphism of the local ring \( {A}_{\mathfrak{p}} \) , where \( \mathfrak{p} \) is the kernel of \( \varphi \) . Thus without loss of generality, we may assume that \( A \) is a local ring with maximal ideal \( m \) . Suppose that\n\n\[ \n{mA}\left\lbrack {x}^{-1}\right\... | Yes |
Corollary 3.3. Let \( A \) be a subring of a field \( K \) and let \( L \) be an algebraically closed field. Let \( \varphi : A \rightarrow L \) be a homomorphism. Let \( B \) be a maximal subring of \( K \) to which \( \varphi \) has an extension homomorphism into \( L \) . Then \( B \) is a local ring and if \( x \in... | Proof. Let \( S \) be the set of pairs \( \left( {C,\psi }\right) \) where \( C \) is a subring of \( K \) and \( \psi : C \rightarrow L \) is a homomorphism extending \( \varphi \) . Then \( S \) is not empty (containing \( \left( {A,\varphi }\right) \rbrack \), and is partially ordered by ascending inclusion and rest... | Yes |
Proposition 3.4. Let \( \varphi : K \rightarrow \{ L,\infty \} \) be an L-valued place of \( K \) . Given a finite number of non-zero elements \( {x}_{1},\ldots ,{x}_{n} \in K \) there exists an index \( j \) such that \( \varphi \) is finite on \( {x}_{i}/{x}_{j} \) for \( i = 1,\ldots, n \) . | Proof. Let \( B \) be the valuation ring of the place. Define \( {x}_{i} \leqq {x}_{j} \) to mean that \( {x}_{i}/{x}_{j} \in B \) . Then the relation \( \leqq \) is transitive, that is if \( {x}_{i} \leqq {x}_{j} \) and \( {x}_{j} \leqq {x}_{r} \) then \( {x}_{i} \leqq {x}_{r} \) . Furthermore, by the property of a va... | Yes |
Proposition 3.4. Let \( \varphi : K \rightarrow \{ L,\infty \} \) be an L-valued place of \( K \) . Given a finite number of non-zero elements \( {x}_{1},\ldots ,{x}_{n} \in K \) there exists an index \( j \) such that \( \varphi \) is finite on \( {x}_{i}/{x}_{j} \) for \( i = 1,\ldots, n \) . | Proof. Let \( B \) be the valuation ring of the place. Define \( {x}_{i} \leqq {x}_{j} \) to mean that \( {x}_{i}/{x}_{j} \in B \) . Then the relation \( \leqq \) is transitive, that is if \( {x}_{i} \leqq {x}_{j} \) and \( {x}_{j} \leqq {x}_{r} \) then \( {x}_{i} \leqq {x}_{r} \) . Furthermore, by the property of a va... | Yes |
Proposition 3.4. Let \( \varphi : K \rightarrow \{ L,\infty \} \) be an L-valued place of \( K \) . Given a finite number of non-zero elements \( {x}_{1},\ldots ,{x}_{n} \in K \) there exists an index \( j \) such that \( \varphi \) is finite on \( {x}_{i}/{x}_{j} \) for \( i = 1,\ldots, n \) . | Proof. Let \( B \) be the valuation ring of the place. Define \( {x}_{i} \leqq {x}_{j} \) to mean that \( {x}_{i}/{x}_{j} \in B \) . Then the relation \( \leqq \) is transitive, that is if \( {x}_{i} \leqq {x}_{j} \) and \( {x}_{j} \leqq {x}_{r} \) then \( {x}_{i} \leqq {x}_{r} \) . Furthermore, by the property of a va... | Yes |
Proposition 3.4. Let \( \varphi : K \rightarrow \{ L,\infty \} \) be an L-valued place of \( K \) . Given a finite number of non-zero elements \( {x}_{1},\ldots ,{x}_{n} \in K \) there exists an index \( j \) such that \( \varphi \) is finite on \( {x}_{i}/{x}_{j} \) for \( i = 1,\ldots, n \) . | Proof. Let \( B \) be the valuation ring of the place. Define \( {x}_{i} \leqq {x}_{j} \) to mean that \( {x}_{i}/{x}_{j} \in B \) . Then the relation \( \leqq \) is transitive, that is if \( {x}_{i} \leqq {x}_{j} \) and \( {x}_{j} \leqq {x}_{r} \) then \( {x}_{i} \leqq {x}_{r} \) . Furthermore, by the property of a va... | Yes |
Proposition 3.5. Let \( \mathfrak{o} \) be a local ring contained in a field \( L \) . An element \( x \) of \( L \) is integral over \( \mathfrak{o} \) if and only if \( x \) lies in every valuation ring \( \mathfrak{O} \) of \( L \) lying above \( \mathfrak{o} \) . | Proof. Assume that \( x \) is not integral over \( \mathfrak{o} \) . Let \( \mathfrak{m} \) be the maximal ideal of \( \mathfrak{o} \) . Then the ideal \( \left( {m,1/x}\right) \) of \( \mathfrak{o}\left\lbrack {1/x}\right\rbrack \) cannot be the entire ring, otherwise we can write\n\n\[ - 1 = {a}_{n}{\left( 1/x\right)... | Yes |
Proposition 3.6. Let \( A \) be a ring contained in a field \( L \) . An element \( x \) of \( L \) is integral over \( A \) if and only if \( x \) lies in every valuation ring \( \mathfrak{O} \) of \( L \) containing A. In terms of places, \( x \) is integral over \( A \) if and only if every place of \( L \) finite o... | Proof. Assume that every place finite on \( A \) is finite on \( x \) . We may assume \( x \neq 0 \) . If \( 1/x \) is a unit in \( A\left\lbrack {1/x}\right\rbrack \) then we can write\n\n\[ x = {c}_{0} + {c}_{1}\left( {1/x}\right) + \cdots + {c}_{n - 1}{\left( 1/x\right) }^{n - 1} \]\n\nwith \( {c}_{i} \in A \) and s... | Yes |
Theorem 3.7. General Integrality Criterion. Let \( A \) be an entire ring. Let \( {z}_{1},\ldots ,{z}_{m} \) be elements of some extension field of its quotient field \( K \) . Assume that each \( {z}_{s}\left( {s = 1,\ldots, m}\right) \) satisfies a polynomial relation\n\n\[ \n{z}_{s}^{{d}_{s}} + {g}_{s}\left( {{z}_{1... | Proof. We apply Proposition 3.6. Suppose some \( {z}_{s} \) is not integral over \( A \) . There exists a place \( \varphi \) of \( K \), finite on \( A \), such that \( \varphi \left( {z}_{s}\right) = \infty \) for some \( s \) . By Proposition 3.4 we can pick an index \( s \) such that \( \varphi \left( {{z}_{j}/{z}_... | Yes |
Theorem 2.1. Let \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack = k\left\lbrack x\right\rbrack \) be a finitely generated entire ring over a field \( k \), and assume that \( k\left( x\right) \) has transcendence degree \( r \) . Then there exist elements \( {y}_{1},\ldots ,{y}_{r} \) in \( k\left\lbrack x\rig... | Proof. If \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) are already algebraically independent over \( k \), we are done. If not, there is a non-trivial relation \[ \sum {a}_{\left( j\right) }{x}_{1}^{{j}_{1}}\cdots {x}_{n}^{{j}_{n}} = 0 \] with each coefficient \( {a}_{\left( j\right) } \in k \) and \( {a}_{\left( j\ri... | Yes |
Theorem 2.2. With the above notation, \( {k}_{u}\left\lbrack x\right\rbrack \) is integral over \( {k}_{u}\left\lbrack {{y}_{1},\ldots ,{y}_{r}}\right\rbrack \) . | Proof. Suppose some \( {x}_{i} \) is not integral over \( {k}_{u}\left\lbrack {{y}_{1},\ldots ,{y}_{r}}\right\rbrack \) . Then there exists a place \( \varphi \) of \( {k}_{u}\left( y\right) \) finite on \( {k}_{u}\left\lbrack {{y}_{1},\ldots ,{y}_{r}}\right\rbrack \) but taking the value \( \infty \) on some \( {x}_{i... | Yes |
Corollary 2.3. Let \( k \) be a field, and let \( k\left( x\right) \) be a finitely generated extension of transcendence degree \( r \) . There exists a polynomial \( P\left( u\right) = \) \( P\left( {u}_{ij}\right) \in k\left\lbrack u\right\rbrack \) such that if \( \left( c\right) = \left( {c}_{ij}\right) \) is a fam... | Proof. By Theorem 2.2, each \( {x}_{i} \) is integral over \( {k}_{u}\left\lbrack {{y}_{1},\ldots ,{y}_{r}}\right\rbrack \) . The coefficients of an integral equation are rational functions in \( {k}_{u} \) . We let \( P\left( u\right) \) be a common denominator for these rational functions. If \( P\left( c\right) \neq... | Yes |
Proposition 3.1. Let \( K \) be a field containing another field \( k \), and let \( L \supset E \) be two other extensions of \( k \) . Then \( K \) and \( L \) are linearly disjoint over \( k \) if and only if \( K \) and \( E \) are linearly disjoint over \( k \) and \( {KE}, L \) are linearly disjoint over \( E \) ... | Proof. Assume first that \( K, E \) are linearly disjoint over \( k \), and \( {KE}, L \) are linearly disjoint over \( E \) . Let \( \{ \kappa \} \) be a basis of \( K \) as vector space over \( k \) (we use the elements of this basis as their own indexing set), and let \( \{ \alpha \} \) be a basis of \( E \) over \(... | No |
Proposition 3.2. If \( K \) and \( L \) are linearly disjoint over \( k \), then they are free over \( k \) . | Proof. Let \( {x}_{1},\ldots ,{x}_{n} \) be elements of \( K \) algebraically independent over \( k \) . Suppose they become algebraically dependent over \( L \) . We get a relation\n\n\[ \sum {y}_{a}{M}_{\alpha }\left( x\right) = 0 \]\n\nbetween monomials \( {M}_{\alpha }\left( x\right) \) with coefficients \( {y}_{\a... | Yes |
Proposition 3.3. Let \( L \) be an extension of \( k \), and let \( \left( u\right) = \left( {{u}_{1},\ldots ,{u}_{r}}\right) \) be a set of quantities algebraically independent over \( L \) . Then the field \( k\left( u\right) \) is linearly disjoint from \( L \) over \( k \) . | Proof. According to the criteria for linear disjointness, it suffices to prove that the elements of a basis for the ring \( k\left\lbrack u\right\rbrack \) that are linearly independent over \( k \) remain so over \( L \) . In fact the monomials \( M\left( u\right) \) give a basis of \( k\left\lbrack u\right\rbrack \) ... | Yes |
Corollary 4.3. Let \( E \) be a separable extension of \( k \), and \( K \) a separable extension of \( E \) . Then \( K \) is a separable extension of \( k \) . | Proof. Apply Proposition 3.1 and the definition of separability. | No |
Corollary 4.5. Let \( K \) be a separable extension of \( k \), and free from an extension \( L \) of \( k \) . Then \( {KL} \) is a separable extension of \( L \) . | Proof. An element of \( {KL} \) has an expression in terms of a finite number of elements of \( K \) and \( L \) . Hence any finitely generated subfield of \( {KL} \) containing \( L \) is contained in a composite field \( {FL} \), where \( F \) is a subfield of \( K \) finitely generated over \( k \) . By Corollary 4.... | Yes |
Corollary 4.5. Let \( K \) be a separable extension of \( k \), and free from an extension \( L \) of \( k \) . Then \( {KL} \) is a separable extension of \( L \) . | Proof. An element of \( {KL} \) has an expression in terms of a finite number of elements of \( K \) and \( L \) . Hence any finitely generated subfield of \( {KL} \) containing \( L \) is contained in a composite field \( {FL} \), where \( F \) is a subfield of \( K \) finitely generated over \( k \) . By Corollary 4.... | Yes |
Corollary 4.6. Let \( K \) and \( L \) be two separable extensions of \( k \), free from each other over \( k \) . Then \( {KL} \) is separable over \( k \) . | Proof. Use Corollaries 4.5 and 4.3. | No |
Corollary 4.7. Let \( K, L \) be two extensions of \( k \), linearly disjoint over \( k \) . Then \( K \) is separable over \( k \) if and only if \( {KL} \) is separable over \( L \) . | Proof. If \( K \) is not separable over \( k \), it is not linearly disjoint from \( {k}^{1/p} \) over \( k \), and hence a fortiori it is not linearly disjoint from \( L{k}^{1/p} \) over \( k \) . By Proposition 4.1, this implies that \( {KL} \) is not linearly disjoint from \( L{k}^{1/p} \) over \( L \), and hence th... | Yes |
Proposition 4.9. Let \( K \) be a finitely generated extension of a field \( k \). If \( {K}^{{p}^{m}}k = K \) for some \( m \), then \( K \) is separably algebraic over \( k \). Conversely, if \( K \) is separably algebraic over \( k \), then \( {K}^{{p}^{m}}k = K \) for all \( m \). | Proof. If \( K/k \) is separably algebraic, then the conclusion follows from the elementary theory of finite algebraic extensions. Conversely, if \( K/k \) is finite algebraic but not separable, then the maximal separable extension of \( k \) in \( K \) cannot be all of \( K \), and hence \( {K}^{p}k \) cannot be equal... | Yes |
Proposition 4.9. Let \( K \) be a finitely generated extension of a field \( k \) . If \( {K}^{{p}^{m}}k = K \) for some \( m \), then \( K \) is separably algebraic over \( k \) . Conversely, if \( K \) is separably algebraic over \( k \), then \( {K}^{{p}^{m}}k = K \) for all \( m \) . | Proof. If \( K/k \) is separably algebraic, then the conclusion follows from the elementary theory of finite algebraic extensions. Conversely, if \( K/k \) is finite algebraic but not separable, then the maximal separable extension of \( k \) in \( K \) cannot be all of \( K \), and hence \( {K}^{p}k \) cannot be equal... | Yes |
Lemma 4.10. Let \( k \) be algebraically closed in extension \( K \) . Let \( x \) be some element of an extension of \( K \), but algebraic over \( k \) . Then \( k\left( x\right) \) and \( K \) are linearly disjoint over \( k \), and \( \left\lbrack {k\left( x\right) : k}\right\rbrack = \left\lbrack {K\left( x\right)... | Proof. Let \( f\left( X\right) \) be the irreducible polynomial for \( x \) over \( k \) . Then \( f \) remains irreducible over \( K \) ; otherwise, its factors would have coefficients algebraic over \( k \), hence in \( k \) . Powers of \( x \) form a basis of \( k\left( x\right) \) over \( k \), hence the same power... | Yes |
Proposition 4.11.\n\n(a) Let \( K \) be a regular extension of \( k \), and let \( E \) be a subfield of \( K \) containing \( k \) . Then \( E \) is regular over \( k \) .\n\n(b) Let \( E \) be a regular extension of \( k \), and \( K \) a regular extension of \( E \) . Then \( K \) is a regular extension of \( k \) .... | Proof. Each assertion is immediate from the definition conditions REG 1 and REG 2. | No |
Theorem 4.12. Let \( K \) be a regular extension of \( k \), let \( L \) be an arbitrary extension of \( k \), both contained in some larger field, and assume that \( K, L \) are free over \( k \) . Then \( K, L \) are linearly disjoint over \( k \) . | Proof (Artin). Without loss of generality, we may assume that \( K \) is finitely generated over \( k \) . Let \( {x}_{1},\ldots ,{x}_{n} \) be elements of \( K \) linearly independent over \( k \) . Suppose we have a relation of linear dependence\n\n\[ \n{x}_{1}{y}_{1} + \cdots + {x}_{n}{y}_{n} = 0 \n\]\n\nwith \( {y}... | Yes |
Theorem 4.13. Let \( K \) be a regular extension of \( k \), free from an extension \( L \) of \( k \) over \( k \) . Then \( {KL} \) is a regular extension of \( L \) . | Proof. From the hypothesis, we deduce that \( K \) is free from the algebraic closure \( {L}^{\mathrm{a}} \) of \( L \) over \( k \) . By Theorem 4.12, \( K \) is linearly disjoint from \( {L}^{\mathrm{a}} \) over \( k \) . By Proposition 3.1, \( {KL} \) is linearly disjoint from \( {L}^{\mathrm{a}} \) over \( L \), an... | Yes |
Corollary 4.14. Let \( K, L \) be regular extensions of \( k \), free from each other over \( k \) . Then \( {KL} \) is a regular extension of \( k \) . | Proof. Use Corollary 4.13 and Proposition 4.11(b). | No |
Let \( K = k\left( x\right) \) be a finitely generated regular extension, free from an extension \( L \) of \( k \), and both contained in some larger field. Then the natural \( k \) -algebra homomorphism \[ L{ \otimes }_{k}k\left\lbrack x\right\rbrack \rightarrow L\left\lbrack x\right\rbrack \] is an isomorphism. | Proof. By Theorem 4.12 the homomorphism is injective, and it is obviously surjective, whence the corollary follows. | No |
Corollary 4.16. Let \( k\left( x\right) \) be a finitely generated regular extension, and let \( \mathfrak{p} \) be the prime ideal in \( k\left\lbrack X\right\rbrack \) vanishing on \( \left( x\right) \), that is, consisting of all polynomials \( f\left( X\right) \in k\left\lbrack X\right\rbrack \) such that \( f\left... | Proof. Consider the exact sequence\n\n\[ 0 \rightarrow \mathfrak{p} \rightarrow k\left\lbrack X\right\rbrack \rightarrow k\left\lbrack x\right\rbrack \rightarrow 0. \]\n\nSince we are dealing with vector spaces over a field, the sequence remains exact when tensored with any \( k \) -space, so we get an exact sequence\n... | Yes |
Theorem 5.1. Let \( D \) be a derivation of a field \( K \). Let\n\n\[ \left( x\right) = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \]\n\nbe a finite family of elements in an extension of \( K \). Let \( \left\{ {{f}_{\alpha }\left( X\right) }\right\} \) be a set of generators for the ideal determined by \( \left( x\right... | Proof. The necessity has been shown above. Conversely, if \( g\left( x\right), h\left( x\right) \) are in \( K\left\lbrack x\right\rbrack \), and \( h\left( x\right) \neq 0 \), one verifies immediately that the mapping \( {D}^{ * } \) defined by the formulas\n\n\[ {D}^{ * }g\left( x\right) = {g}^{D}\left( x\right) + \s... | Yes |
Proposition 5.2. A finitely generated extension \( K\\left( x\\right) \) over \( K \) is separable algebraic if and only if every derivation \( D \) of \( K\\left( x\\right) \) which is trivial on \( K \) is trivial on \( K\\left( x\\right) \) . | Proof. If \( K\\left( x\\right) \) is separable algebraic over \( K \), this is Case 1. Conversely, if it is not, we can make a tower of extensions between \( K \) and \( K\\left( x\\right) \), such that each step is covered by one of the three above cases. At least one step will be covered by Case 2 or 3. Taking the u... | No |
Proposition 5.3. Given \( K \) and elements \( \left( x\right) = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) in some extension field, assume that there exist \( n \) polynomials \( {f}_{i} \in K\left\lbrack X\right\rbrack \) such that:\n\n(i) \( {f}_{i}\left( x\right) = 0 \), and\n\n(ii) \( \det \left( {\partial {f}_{i}... | Proof. Let \( D \) be a derivation on \( K\left( x\right) \), trivial on \( K \) . Having \( {f}_{i}\left( x\right) = 0 \) we must have \( D{f}_{i}\left( x\right) = 0 \), whence the \( D{x}_{i} \) satisfy \( n \) linear equations such that the coefficient matrix has non-zero determinant. Hence \( D{x}_{i} = 0 \), so \(... | Yes |
Proposition 5.4. Let \( K = k\left( x\right) \) be a finitely generated extension of \( k \) . An element \( z \) of \( K \) is in \( {K}^{p}k \) if and only if every derivation \( D \) of \( K \) over \( k \) is such that \( {Dz} = 0 \) . | Proof. If \( z \) is in \( {K}^{p}k \), then it is obvious that every derivation \( D \) of \( K \) over \( k \) vanishes on \( z \) . Conversely, if \( z \notin {K}^{p}k \), then \( z \) is purely inseparable over \( {K}^{p}k \), and by Case 3 of the extension theorem, we can find a derivation \( D \) trivial on \( {K... | Yes |
Proposition 5.5. Assume that \( K \) is a separably generated and finitely generated extension of \( k \) of transcendence degree \( r \) . Then the vector space \( \mathfrak{D} \) (over \( K \) ) of derivations of \( K \) over \( k \) has dimension \( r \) . Elements \( {t}_{1},\ldots ,{t}_{r} \) of \( K \) from a sep... | Proof. If \( {t}_{1},\ldots ,{t}_{r} \) is a separating transcendence base for \( K \) over \( k \), then we can find derivations \( {D}_{1},\ldots ,{D}_{r} \) of \( K \) over \( k \) such that \( {D}_{i}{t}_{j} = {\delta }_{ij} \), by Cases 1 and 2 of the extension theorem. Given \( D \in \mathfrak{D} \), let \( {w}_{... | Yes |
Corollary 5.6. Let \( K \) be a finitely generated and separably generated extension of \( k \) . Let \( z \) be an element of \( K \) transcendental over \( k \) . Then \( K \) is separable over \( k\left( z\right) \) if and only if there exists a derivation \( D \) of \( K \) over \( k \) such that \( {Dz} \neq 0 \) ... | Proof. If \( K \) is separable over \( k\left( z\right) \), then \( z \) can be completed to a separating base of \( K \) over \( k \) and we can apply the proposition. If \( {Dz} \neq 0 \), then \( {dz} \neq 0 \), and we can complete \( {dz} \) to a basis of \( \mathfrak{F} \) over \( K \) . Again from the proposition... | No |
Theorem 5.7. (Zariski-Matsusaka). Let \( K \) be a finitely generated separable extension of a field \( k \). Let \( y, z \in K \) and \( z \notin {K}^{p}k \) if the characteristic is \( p > 0 \). Let \( u \) be transcendental over \( K \), and put \( {k}_{u} = k\left( u\right) ,{K}_{u} = K\left( u\right) \). (a) For a... | Proof. We shall use throughout the fact that a subfield of a finitely generated extension is also finitely generated (see Exercise 4). If \( w \) is an element of \( K \), and if there exists a derivation \( D \) of \( K \) over \( k \) such that \( {Dw} \neq 0 \), then \( K \) is separable over \( k\left( w\right) \),... | Yes |
Theorem 5.8. Let \( K = k\left( {{x}_{1},\ldots ,{x}_{n}}\right) = k\left( x\right) \) be a finitely generated regular extension of a field \( k \) . Let \( {u}_{1},\ldots ,{u}_{n} \) be algebraically independent over \( k\left( x\right) \) . Let\n\n\[ \n{u}_{n + 1} = {u}_{1}{x}_{1} + \cdots + {u}_{n}{x}_{n} \n\]\n\nan... | Proof. By the separability of \( k\left( x\right) \) over \( k \), some \( {x}_{i} \) does not lie in \( {K}^{p}k \) , say \( {x}_{n} \notin {K}^{p}k \) . Then we take\n\n\[ \ny = {u}_{1}{x}_{1} + \cdots + {u}_{n - 1}{x}_{n - 1}\;\text{ and }\;z = {x}_{n}, \n\]\n\nso that \( {u}_{n + 1} = y + {u}_{n}z \), and we apply ... | Yes |
Theorem 1.1. Let \( k \) be a field, and let \( k\left\lbrack x\right\rbrack = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be a finitely generated ring over \( k \) . Let \( \varphi : k \rightarrow L \) be an embedding of \( k \) into an algebraically closed field \( L \) . Then there exists an extension of... | Proof. Let \( \mathfrak{M} \) be a maximal ideal of \( k\left\lbrack x\right\rbrack \) . Let \( \sigma \) be the canonical homomorphism \( \sigma : k\left\lbrack x\right\rbrack \rightarrow k\left\lbrack x\right\rbrack /\mathfrak{M} \) . Then \( {\sigma k}\left\lbrack {\sigma {x}_{1},\ldots ,\sigma {x}_{n}}\right\rbrack... | Yes |
Corollary 1.2. Let \( k \) be a field and \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) a finitely generated extension ring of \( k \) . If \( k\left\lbrack x\right\rbrack \) is a field, then \( k\left\lbrack x\right\rbrack \) is algebraic over \( k \) . | Proof. All homomorphisms of a field are isomorphisms (onto the image), and there exists a homomorphism of \( k\left\lbrack x\right\rbrack \) over \( k \) into the algebraic closure of \( k \) . | No |
Corollary 1.3. Let \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be a finitely generated entire ring over a field \( k \), and let \( {y}_{1},\ldots ,{y}_{m} \) be non-zero elements of this ring. Then there exists a homomorphism\n\n\[ \psi : k\left\lbrack x\right\rbrack \rightarrow {k}^{\mathrm{a}} \]\n\no... | Proof. Consider the ring \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n},{y}_{1}^{-1},\ldots ,{y}_{m}^{-1}}\right\rbrack \) and apply the theorem to this ring. | No |
Theorem 1.4. Let \( \mathfrak{a} \) be an ideal in \( k\left\lbrack X\right\rbrack = k\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . Then either \( \mathfrak{a} = k\left\lbrack X\right\rbrack \) or \( \mathfrak{a} \) has a zero in \( {k}^{\mathrm{a}} \) . | Proof. Suppose \( \mathfrak{a} \neq k\left\lbrack X\right\rbrack \) . Then \( \mathfrak{a} \) is contained in some maximal ideal \( \mathfrak{m} \), and \( k\left\lbrack X\right\rbrack /\mathfrak{m} \) is a field, which is a finitely generated extension of \( k \), because it is generated by the images of \( {X}_{1},\l... | Yes |
Theorem 1.5. (Hilbert’s Nullstellensatz). Let \( \mathfrak{a} \) be an ideal in \( k\left\lbrack X\right\rbrack \) . Let \( f \) be a polynomial in \( k\left\lbrack X\right\rbrack \) such that \( f\left( c\right) = 0 \) for every zero \( \left( c\right) = \left( {{c}_{1},\ldots ,{c}_{n}}\right) \) of \( \mathfrak{a} \)... | Proof. We may assume that \( f \neq 0 \) . We use the Rabinowitsch trick of introducing a new variable \( Y \), and of considering the ideal \( {\mathfrak{a}}^{\prime } \) generated by a and \( 1 - {Yf} \) in \( k\left\lbrack {X, Y}\right\rbrack \) . By Theorem 1.4, and the current assumption, the ideal \( {\mathfrak{a... | Yes |
Theorem 2.1. The finite union and the finite intersection of algebraic sets are algebraic sets. If \( A, B \) are the algebraic sets of zeros of ideals \( \mathfrak{a},\mathfrak{b} \), respectively, then \( A \cup B \) is the set of zeros of \( \mathfrak{a} \cap \mathfrak{b} \) and \( A \cap B \) is the set of zeros of... | Proof. We first consider \( A \cup B \) . Let \( \left( x\right) \in A \cup B \) . Then \( \left( x\right) \) is a zero of \( \mathfrak{a} \cap \mathfrak{b} \) . Conversely, let \( \left( x\right) \) be a zero of \( \mathfrak{a} \cap \mathfrak{b} \), and suppose \( \left( x\right) \notin A \) . There exists a polynomia... | Yes |
Theorem 2.2. Let \( A \) be an algebraic set.\n\n(i) Then \( A \) can be expressed as a finite union of irreducible algebraic sets \( A = {V}_{1} \cup \ldots \cup {V}_{r} \)\n\n(ii) If there is no inclusion relation among the \( {V}_{i} \), i.e. if \( {V}_{i} ⊄ {V}_{j} \) for \( i \neq j \), then the representation is ... | Proof. We first show existence. Suppose the set of algebraic sets which cannot be represented as a finite union of irreducible ones is not empty. Let \( V \) be a minimal element in its. Then \( V \) cannot be irreducible, and we can write \( V = A \cup B \) where \( A, B \) are algebraic sets, but \( A \neq V \) and \... | Yes |
Theorem 2.4. Let \( R \) be a factorial ring, and let \( {W}_{1},\ldots ,{W}_{m} \) be \( m \) independent variables over its quotient field \( k \) . Let \( k\left( {{w}_{1},\ldots ,{w}_{m}}\right) \) be an extension of transcendence degree \( m - 1 \) . Then the ideal in \( R\left\lbrack W\right\rbrack \) vanishing o... | Proof. By hypothesis there is some polynomial \( P\left( W\right) \in R\left\lbrack W\right\rbrack \) of degree \( \geqq 1 \) vanishing on \( \left( w\right) \), and after taking an irreducible factor we may assume that this polynomial is irreducible, and so is a prime element in the factorial ring \( R\left\lbrack W\r... | Yes |
Proposition 2.5. An ideal \( \mathfrak{a} \) is homogeneous if and only if \( \mathfrak{a} \) has a set of generators over \( k\left\lbrack X\right\rbrack \) consisting of forms. | Proof. Suppose \( \mathfrak{a} \) is homogeneous and that \( {f}_{1},\ldots ,{f}_{r} \) are generators. By hypothesis, for each integer \( d \geqq 0 \) the homogeneous components \( {f}_{i}^{\left( d\right) } \) also lie in \( \mathfrak{a} \), and the set of such \( {f}_{i}^{\left( d\right) } \) (for all \( i, d \) ) f... | Yes |
Proposition 2.7. Let \( \mathcal{L} \) be a homogeneous algebraic space. Then each irreducible component \( V \) of \( \mathcal{L} \) is also homogeneous. | Proof. Let \( V = {V}_{1},\ldots ,{V}_{r} \) be the irreducible components of \( \mathcal{Z} \), without inclusion relation. By Remark 3.3 we know that \( {V}_{1} ⊄ {V}_{2} \cup \ldots \cup {V}_{r} \), so there is a point \( \left( x\right) \in {V}_{1} \) such that \( \left( x\right) \notin {V}_{1} \) for \( i = 2,\ldo... | No |
Theorem 3.1. Let \( \left( W\right) = \left( {{W}_{1},\ldots ,{W}_{m}}\right) \) and \( \left( X\right) = \left( {{X}_{1},\ldots ,{X}_{n}}\right) \) be independent families of variables. Let \( \mathfrak{p} \) be a prime ideal in \( k\left\lbrack {W, X}\right\rbrack \) (resp. \( \mathbf{Z}\left\lbrack {W, X}\right\rbra... | Proof. Let \( V \) have generic point \( \left( {w, x}\right) \) . We have to prove that every zero \( \left( {w}^{\prime }\right) \) of \( {\mathfrak{p}}_{1} \) in a field is the projection of some zero \( \left( {{w}^{\prime },{x}^{\prime }}\right) \) of \( \mathfrak{p} \) such that not all the coordinates of \( \lef... | Yes |
Theorem 3.2. (Fundamental theorem of elimination theory.) Given degrees \( {d}_{1},\ldots ,{d}_{r} \), the set of all forms \( \left( {{f}_{1},\ldots ,{f}_{r}}\right) \) in \( n \) variables having a non-trivial common zero is an algebraic subspace of \( {\mathbf{A}}^{m} \) over \( \mathbf{Z} \) . | Proof. Let \( \left( W\right) = \left( {W}_{i,\left( \nu \right) }\right) \) be a family of variables independent of \( \left( X\right) \) . Let \( \left( F\right) = \left( {{F}_{1},\ldots ,{F}_{r}}\right) \) be the family of polynomials in \( \mathbf{Z}\left\lbrack {W, X}\right\rbrack \) given by\n\n(2)\n\n\[ \n{F}_{i... | Yes |
Corollary 3.3. Let \( \left( f\right) \) be a family of \( n \) forms in \( n \) variables, and assume that \( {\left( w\right) }_{f} \) is a generic point of \( {\mathbf{A}}^{m} \), i.e. that the coefficients of these forms are algebraically independent. Then \( \left( f\right) \) does not have a non-trivial zero. | Proof. There exists a specialization of \( \left( f\right) \) which has only the trivial zero, namely \( {f}_{1}^{\prime } = {X}_{1}^{{d}_{1}},\ldots ,{f}_{n}^{\prime } = {X}_{n}^{{d}_{n}} \) . | No |
Assume \( r = n \), so we deal with \( n \) forms in \( n \) variables. Then \( {\mathfrak{p}}_{1} \) is principal, generated by a single polynomial, so \( {\mathbf{Q}}_{1} \) is what one calls a hypersurface. If \( \left( w\right) \) is a generic point of \( {\mathbf{Q}}_{1} \) over a field \( k \), then the transcend... | We prove the second statement first, and use the same notation as in the proof of Theorem 3.4. Let \( {u}_{j} = {x}_{j}/{x}_{n} \) . Then \( {u}_{n} = 1 \) and \( \left( y\right) ,\left( {{u}_{1},\ldots ,{u}_{n - 1}}\right) \) are algebraically independent. By (3), we have \( {z}_{i} = - {F}_{i}^{ * }\left( {y, u}\righ... | Yes |
Lemma 3.6. There is a positive integer \( s \) with the following properties. Fix an index \( i \) with \( 1 \leqq i \leqq n - 1 \) . For each pair of \( n \) -tuples of integers \( \geqq 0 \)\n\n\[ \left( \alpha \right) = \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \;\text{ and }\;\left( \beta \right) = \left(... | To see this, we use the fact from Theorem 3.4 that for some \( s \), \n\n\[ {X}_{n}^{s}R\left( W\right) = {Q}_{1}{F}_{1} + \cdots + {Q}_{n}{F}_{n}\text{ with }{Q}_{j} \in \mathbf{Z}\left\lbrack {W, X}\right\rbrack . \]\n\nDifferentiating with respect to \( {W}_{i,\left( \beta \right) } \) we get\n\n\[ {X}_{n}^{s}\frac{... | Yes |
Lemma 3.9. Let \( G, H \) be generic independent forms with \( \deg \left( {GH}\right) = {d}_{1} \). Then \( R\left( {{GH},{F}_{2},\ldots ,{F}_{n}}\right) \) is divisible by \( R\left( {G,{F}_{2},\ldots ,{F}_{n}}\right) R\left( {H,{F}_{2},\ldots ,{F}_{n}}\right) \). | Proof. By Theorem 3.5, there is an expression \[ {X}_{n}^{s}R\left( {{F}_{1},\ldots ,{F}_{n}}\right) = {Q}_{1}{F}_{1} + \cdots + {Q}_{n}{F}_{n}\text{with}{Q}_{i} \in \mathbf{Z}\left\lbrack {W, X}\right\rbrack \text{.} \] Let \( {W}_{G},{W}_{H},{W}_{{F}_{2}},\ldots ,{W}_{{F}_{n}} \) be the coefficients of \( G, H,{F}_{2... | Yes |
Theorem 3.10. Let \( {f}_{1} = {gh} \) be a product of forms such that \( \deg \left( {gh}\right) = {d}_{1} \) . Let \( {f}_{2},\ldots ,{f}_{n} \) be arbitrary forms of degrees \( {d}_{2},\ldots ,{d}_{n} \) . Then\n\n\[ \n\operatorname{Res}\left( {{gh},{f}_{2},\ldots ,{f}_{n}}\right) = \operatorname{Res}\left( {g,{f}_{... | Proof. From the fact that the degrees have to add in a product of polynomials, together with Theorem 3.8(a) and (b), we now see in Lemma 3.9 that we must have the precise equality in what was only a divisibility before we knew the precise degree of \( R \) in each set of variables. | No |
Theorem 3.11. Let \( {F}_{1},\ldots ,{F}_{n} \) be the generic forms in \( n \) variables, and let \( {\bar{F}}_{1},\ldots ,{\bar{F}}_{n} \) be the forms obtained by substituting \( {X}_{n} = 0 \), so that \( {\bar{F}}_{1},\ldots ,{\bar{F}}_{n - 1} \) are the generic forms in \( n - 1 \) variables. Let \( n \geqq 2 \) ... | Proof. By Theorem 3.10 it suffices to prove the assertion when \( {d}_{n} = 1 \) . By Theorem 3.4, for each \( i = 1,\ldots, n - 1 \) we have an expression (*)\[ {X}_{i}^{s}\operatorname{Res}\left( {{F}_{1},\ldots ,{F}_{n - 1},{X}_{n}}\right) = {Q}_{1}{F}_{1} + \cdots + {Q}_{n - 1}{F}_{n - 1} + {Q}_{n}{X}_{n} \] with \... | Yes |
Lemma 3.12. Let \( A \) be a commutative ring. Let \( {f}_{1},\ldots ,{f}_{n},{g}_{1},\ldots ,{g}_{n} \) be homogeneous polynomials in \( A\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . Assume that\n\n\[ \left( {{g}_{1},\ldots ,{g}_{n}}\right) \subset \left( {{f}_{1},\ldots ,{f}_{n}}\right) \]\n\n as ideals i... | Proof. Express each \( {g}_{i} = \sum {h}_{ij}{f}_{j} \) with \( {h}_{ij} \) homogeneous in \( A\left\lbrack X\right\rbrack \) . By specialization, we may then assume that \( {g}_{i} = \sum {H}_{ij}{F}_{j} \) where \( {H}_{ij} \) and \( {F}_{j} \) have algebraically independent coefficients over \( \mathbf{Z} \) . By T... | Yes |
Corollary 3.14. Let \( C = \left( {c}_{ij}\right) \) be a square matrix with coefficients in \( A \) . Let \( {f}_{i}\left( X\right) = {F}_{i}\left( {CX}\right) \) (where \( {CX} \) is multiplication of matrices, viewing \( X \) as a column vector). Then | \[ \operatorname{Res}\left( {{f}_{1},\ldots ,{f}_{n}}\right) = \det {\left( C\right) }^{{d}_{1}\ldots {d}_{n}}\operatorname{Res}\left( {{F}_{1},\ldots ,{F}_{n}}\right) . \] Proof. This is the case when \( d = 1 \) and \( {g}_{i} \) is a linear form for each \( i \) . | Yes |
Theorem 3.15. Let \( {f}_{1},\ldots ,{f}_{n} \) be homogeneous in \( A\left\lbrack X\right\rbrack \), and suppose \( {d}_{n} \geqq {d}_{i} \) for all \( i \) . Let \( {h}_{i} \) be homogeneous of degree \( {d}_{n} - {d}_{i} \) in \( A\left\lbrack X\right\rbrack \) . Then\n\n\[ \n\operatorname{Res}\left( {{f}_{1},\ldots... | Proof. We may assume \( {f}_{i} = {F}_{i} \) are the generic forms, \( {H}_{i} \) are forms generic independent from \( {F}_{1},\ldots ,{F}_{n} \), and \( A = \mathbf{Z}\left\lbrack {{W}_{F},{W}_{H}}\right\rbrack \), where \( \left( {W}_{F}\right) \) and \( \left( {W}_{H}\right) \) are the coefficients of the respectiv... | Yes |
Theorem 3.16. Let \( \pi \) be a permutation of \( \{ 1,\ldots, n\} \), and let \( \varepsilon \left( \pi \right) \) be its sign. Then\n\n\[ \operatorname{Res}\left( {{F}_{\pi \left( 1\right) },\ldots ,{F}_{\pi \left( n\right) }}\right) = \varepsilon {\left( \pi \right) }^{{d}_{1}\ldots {d}_{n}}\operatorname{Res}\left(... | Proof. Again using Lemma 3.12 with the ideals \( \left( {{F}_{1},\ldots ,{F}_{n}}\right) \) and \( \left( {{F}_{\pi \left( 1\right) },\ldots ,{F}_{\pi \left( n\right) }}\right) \), which are equal, we conclude the desired equality up to a factor \( \pm 1 \), in \( \mathbf{Z}\left\lbrack {W}_{F}\right\rbrack \) . We det... | Yes |
Theorem 3.17. Let \( {L}_{1},\ldots ,{L}_{n - 1}, F \) be generic forms in \( n \) variables, such that \( {L}_{1},\ldots ,{L}_{n - 1} \) are of degree 1, and \( F \) has degree \( d = {d}_{n} \) . Let\n\n\[ \n{\Delta }_{j}\left( {j = 1,\ldots, n}\right) \n\]\n\nbe \( {\left( -1\right) }^{n - j} \) times the \( j \) -t... | Proof. We first claim that for all \( j = 1,\ldots, n \) we have the congruence\n\n(*)\n\n\[ \n{X}_{n}{\Delta }_{j} - {X}_{j}{\Delta }_{n} \equiv 0{\;\operatorname{mod}\;\left( {{L}_{1},\ldots ,{L}_{n - 1}}\right) }\mathbf{Z}\left\lbrack {W, X}\right\rbrack , \n\]\n\nwhere as usual, \( \left( W\right) \) are the coeffi... | Yes |
Theorem 4.1. Given degrees \( {d}_{1},\ldots ,{d}_{r} \geqq 1 \), and positive integers \( m, n \) . Let \( \left( W\right) = \left( {W}_{i,\left( \nu \right) }\right) \) be the variables as in \( §3 \) ,(2) viewed as algebraically independent elements over the integers \( \mathbf{Z} \) . There exists an effectively de... | A finite family \( \left\{ {R}_{\rho }\right\} \) having the property stated in Theorem 4.1 will be called a resultant system for the given degrees. According to van der Waerden (Moderne Algebra, first and second edition, §80), the following technique of proof using resultants goes back to Kronecker elimination, and to... | Yes |
Proposition 4.2. Let \( {f}_{a},{g}_{b} \) be homogeneous polynomials with coefficients in a field \( K \) . Then \( R\left( {a, b}\right) = 0 \) if and only if the system of equations \[ {f}_{a}\left( X\right) = 0,\;{g}_{b}\left( X\right) = 0 \] has a non-trivial zero in some extension of \( K \) (which can be taken t... | If \( {a}_{0} = 0 \) then a zero of \( {g}_{b} \) is also a zero of \( {f}_{a} \) ; and if \( {b}_{0} = 0 \) then a zero of \( {f}_{a} \) is also a zero of \( {g}_{b} \) . If \( {a}_{0}{b}_{0} \neq 0 \) then from the expression of the resultant as a product of the difference of roots \( \left( {{\alpha }_{i} - {\beta }... | No |
Proposition 4.3. The system \( \left\{ {{Q}_{\mu }\left( W\right) }\right\} \) just constructed satisfies the property of Theorem 4.1, i.e. it is a resultant system for \( r \) forms of the same degree \( d \) . | Proof. Suppose that there is a non-trivial solution of a special system \( {F}_{j}\left( {W, X}\right) = 0 \) with \( \left( w\right) \) in some field \( k \) . Then \( \left( {w, T, U}\right) \) is a common non-trivial zero of \( f, g \), so \( \operatorname{Res}\left( {f, g}\right) = 0 \) and therefore \( {Q}_{\mu }\... | Yes |
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