Q
stringlengths 4
3.96k
| A
stringlengths 1
3k
| Result
stringclasses 4
values |
|---|---|---|
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 a primitive \( m \) -th root of unity. Then \( F/k \) is abelian. We lift \( K \) over \( F \) . Then \( {KF} \) is solvable over \( F \) . There is a tower of subfields between \( F \) and \( {KF} \) such that each step is cyclic of prime order, because every solvable group admits a tower of sub-groups of the same type, and we can use Theorem 1.10. By Theorems 6.2 and 6.4, we conclude that \( {KF} \) is solvable by radicals over \( F \), and hence is solvable by radicals over \( k \) . This proves that \( E/k \) is solvable by radicals.\n\nConversely, assume that \( E/k \) is solvable by radicals. For any embedding \( \sigma \) of \( E \) in \( {E}^{\mathrm{a}} \) over \( k \), the extension \( {\sigma E}/k \) is also solvable by radicals. Hence the smallest Galois extension \( K \) of \( E \) containing \( k \), which is a composite of \( E \) and its conjugates is solvable by radicals. Let \( m \) be the product of all primes unequal to the characteristic dividing the degree \( \left\lbrack {K : k}\right\rbrack \) and again let \( F = k\left( \zeta \right) \) where \( \zeta \) is a primitive \( m \) -th root of unity. It will suffice to prove that \( {KF} \) is solvable over \( F \), because it follows then that \( {KF} \) is solvable over \( k \) and hence \( G\left( {K/k}\right) \) is solvable because it is a homomorphic image of \( G\left( {{KF}/k}\right) \) . But \( {KF}/F \) can be decomposed into a tower of extensions, such that each step is of prime degree and of the type described in Theorem 6.2 or Theorem 6.4, and the corresponding root of unity is in the field \( F \) . Hence \( {KF}/F \) is solvable, and our theorem is proved.
|
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 \( {K}_{B} \) is Galois, and abelian of exponent \( m \) . Let \( G \) be its Galois group. We have a bilinear map\n\n\[ G \times B \rightarrow {\mathbf{\mu }}_{m}\;\text{ given by }\;\left( {\sigma, a}\right) \mapsto \langle \sigma, a\rangle . \]\n\nIf \( \sigma \in G \) and \( a \in B \), and \( {\alpha }^{m} = a \) then \( \langle \sigma, a\rangle = {\sigma \alpha }/\alpha \) . The kernel on the left is 1 and the kernel on the right is \( {k}^{*m} \) . The extension \( {K}_{B}/k \) is finite if and only if \( \left( {B : {k}^{*m}}\right) \) is finite. If that is the case, then\n\n\[ B/{k}^{*m} \approx {G}^{ \land }, \]\n\nand in particular we have the equality\n\n\[ \left\lbrack {{K}_{B} : k}\right\rbrack = \left( {B : {k}^{*m}}\right) . \]
|
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 is 1 . Let \( a \in B \) and suppose \( \langle \sigma, a\rangle = 1 \) for all \( \sigma \in G \) . Consider the subfield \( k\left( {a}^{1/m}\right) \) of \( {K}_{B} \) . If \( {a}^{1/m} \) is not in \( k \), there exists an automorphism of \( k\left( {a}^{1/m}\right) \) over \( k \) which is not the identity. Extend this automorphism to \( {K}_{B} \), and call this extension \( \sigma \) . Then clearly \( \langle \sigma, a\rangle \neq 1 \) . This proves our contention.\n\nBy the duality theorem of Chapter I,§9 we see that \( G \) is finite if and only if \( B/{k}^{*m} \) is finite, and in that case we have the isomorphism as stated, so that in particular the order of \( G \) is equal to \( \left( {B : {k}^{*m}}\right) \), thereby proving the theorem.
|
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 \( {B}_{1} \subset {B}_{2} \) . Let \( b \in {B}_{1} \) . Then \( k\left( {b}^{1/m}\right) \subset k\left( {B}_{2}^{1/m}\right) \) and \( k\left( {b}^{1/m}\right) \) is contained in a finitely generated subextension of \( k\left( {B}_{2}^{1/m}\right) \) . Thus we may assume without loss of generality that \( {B}_{2}/{k}^{*m} \) is finitely generated, hence finite. Let \( {B}_{3} \) be the subgroup of \( {k}^{ * } \) generated by \( {B}_{2} \) and \( b \) . Then \( k\left( {B}_{2}^{1/m}\right) = k\left( {B}_{3}^{1/m}\right) \) and from what we saw above, the degree of this field over \( k \) is precisely\n\n\[ \n\left( {{B}_{2} : {k}^{*m}}\right) \text{ or }\left( {{B}_{3} : {k}^{*m}}\right) .\n\]\n\nThus these two indices are equal, and \( {B}_{2} = {B}_{3} \) . This proves that \( {B}_{1} \subset {B}_{2} \) .\n\nWe now have obtained an injection of our set of groups \( B \) into the set of abelian extensions of \( k \) of exponent \( m \) . Assume finally that \( K \) is an abelian extension of \( k \) of exponent \( m \) . Any finite subextension is a composite of cyclic extensions of exponent \( m \) because any finite abelian group is a product of cyclic groups, and we can apply Corollary 1.16. By Theorem 6.2, every cyclic extension can be obtained by adjoining an \( m \) -th root. Hence \( K \) can be obtained by adjoining a family of \( m \) -th roots, say \( m \) -th roots of elements \( {\left\{ {b}_{j}\right\} }_{j \in J} \) with \( {b}_{j} \in {k}^{ * } \) . Let \( B \) be the subgroup of \( {k}^{ * } \) generated by all \( {b}_{j} \) and \( {k}^{*m} \) . If \( {b}^{\prime } = b{a}^{m} \) with \( a, b \in k \) then obviously\n\n\[ \nk\left( {b}^{\prime 1/m}\right) = k\left( {b}^{1/m}\right) \n\]\n\nHence \( k\left( {B}^{1/m}\right) = K \), as desired.
|
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 Galois group. If \( \sigma \in G \) and \( a \in B \), and \( \wp \alpha = a \), let \( \langle \sigma, a\rangle = {\sigma \alpha } - \alpha \) . Then we have a bilinear map\n\n\[ \nG \times B \rightarrow \mathbf{Z}/p\mathbf{Z}\;\text{ given by }\;\left( {\sigma, a}\right) \rightarrow \langle \sigma, a\rangle .\n\]\n\nThe kernel on the left is 1 and the kernel on the right is \( \wp k \) . The extension \( {K}_{B}/k \) is finite if and only if \( \left( {B : \wp k}\right) \) is finite and if that is the case, then\n\n\[ \n\left\lbrack {{K}_{B} : k}\right\rbrack = \left( {B : \wp k}\right) .\n\]
|
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 \( k\left\lbrack X\right\rbrack \) .
|
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\[ \n{X}^{m} - a = \mathop{\prod }\limits_{{v = 1}}^{m}\left( {X - {\alpha }_{v}}\right)\n\]\n\nbe the factorization of \( {X}^{m} - a \) into linear factors, and say \( \alpha = {\alpha }_{1} \) . Substituting \( {X}^{{p}^{r}} \) for \( X \) we get\n\n\[ \n{X}^{n} - a = {X}^{{p}^{r}m} - a = \mathop{\prod }\limits_{{v = 1}}^{m}\left( {{X}^{{p}^{r}} - {\alpha }_{v}}\right) .\n\]\n\nWe may assume inductively that \( {X}^{m} - a \) is irreducible in \( k\left\lbrack X\right\rbrack \) . We contend that \( \alpha \) is not a \( p \) -th power in \( k\left( \alpha \right) \) . Otherwise, \( \alpha = {\beta }^{p},\beta \in k\left( \alpha \right) \) . Let \( N \) be the norm from \( k\left( \alpha \right) \) to \( k \) . Then\n\n\[ \n- a = {\left( -1\right) }^{m}N\left( \alpha \right) = {\left( -1\right) }^{m}N\left( {\beta }^{p}\right) = {\left( -1\right) }^{m}N{\left( \beta \right) }^{p}.\n\]\n\nIf \( m \) is odd, \( a \) is a \( p \) -th power, which is impossible. Similarly, if \( m \) is even and \( p \) is odd, we also get a contradiction. This proves our contention, because \( m \) is prime to \( p \) . If we know our theorem for prime powers, then we conclude that \( {X}^{{p}^{r}} - \alpha \) is irreducible over \( k\left( \alpha \right) \) . If \( A \) is a root of \( {X}^{{p}^{r}} - \alpha \) then \( k \subset k\left( \alpha \right) \subset k\left( A\right) \) gives a tower, of which the bottom step has degree \( m \) and the top step has degree \( {p}^{r} \) . It follows that \( A \) has degree \( n \) over \( k \) and hence that \( {X}^{n} - a \) is irreducible.
|
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 an element \( a \in E \) such that \( {X}^{p} - a \) is irreducible over \( E \) . By Corollary 9.2, \( {k}^{\mathrm{a}} \) cannot be of finite degree over \( E \) . (The reader may restrict his or her attention to characteristic 0 if Chapter V, §6 was omitted.)\n\nWe may therefore assume that \( {k}^{\mathrm{a}} \) is Galois over \( k \) . Let \( {k}_{1} = k\left( i\right) \) . Then \( {k}^{\mathrm{a}} \) is also Galois over \( {k}_{1} \) . Let \( G \) be the Galois group of \( {k}^{\mathrm{a}}/{k}_{1} \) . Suppose that there is a prime number \( p \) dividing the order of \( G \), and let \( H \) be a subgroup of order \( p \) . Let \( F \) be its fixed field. Then \( \left\lbrack {{k}^{a} : F}\right\rbrack = p \) . If \( p \) is the characteristic, then Exercise 29 at the end of the chapter will give the contradiction. We may assume that \( p \) is not the characteristic. The \( p \) -th roots of unity \( \neq 1 \) are the roots of a polynomial of degree \( \leqq p - 1 \) (namely \( {X}^{p - 1} + \cdots + 1 \) ), and hence must lie in \( F \) . By Theorem 6.2, it follows that \( {k}^{\mathrm{a}} \) is the splitting field of some polynomial \( {X}^{p} - a \) with \( a \in F \) . The polynomial \( {X}^{{p}^{2}} - a \) is necessarily reducible. By the theorem, we must have \( p = 2 \) and \( a = - 4{b}^{4} \) with \( b \in F \) . This implies\n\n\[ \n{k}^{\mathrm{a}} = F\left( {a}^{1/2}\right) = F\left( i\right) .\n\]\n\nBut we assumed \( i \in {k}_{1} \), contradiction.\n\nThus we have proved \( {k}^{\mathrm{a}} = k\left( i\right) \) . It remains to prove that char \( k = 0 \), and for this I use an argument shown to me by Keith Conrad. We first show that a sum of squares in \( k \) is a square. It suffices to prove this for a sum of two squares, and in this case we write an element \( x + {iy} \in k\left( i\right) = {k}^{\mathrm{a}} \) as a square.\n\n\[ \nx + {iy} = {\left( u + iv\right) }^{2},\;x, y, u, v \in k, \n\]\n\nand then \( {x}^{2} + {y}^{2} = {\left( {u}^{2} + {v}^{2}\right) }^{2} \) . Then to prove \( k \) has characteristic 0, we merely observe that if the characteristic is \( > 0 \), then -1 is a finite sum \( 1 + \ldots + 1 \) , whence a square by what we have just shown, but \( {k}^{\mathrm{a}} = k\left( i\right) \), so this concludes the proof.
|
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 \( k \) . Let \( K \) be the splitting field of \( {X}^{n} - a \) over \( k \) . Then the above homomorphism \( \sigma \mapsto M\left( \sigma \right) \) is an isomorphism of \( {G}_{K/k} \) with \( G\left( n\right) \) . The commutator group is \( \operatorname{Gal}\left( {K/k\left( {\mathbf{\mu }}_{n}\right) }\right) \), so \( k\left( {\mathbf{\mu }}_{n}\right) \) is the maximal abelian subextension of \( K \) .
|
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 \( \# \left( {G}_{K/k}\right) = \# G\left( n\right) \), so \( {G}_{K/k} = G\left( n\right) \) . However, we shall given an independent proof as an example of techniques of Galois theory. We prove the theorem by induction.\n\nSuppose first \( n = p \) is prime. Since \( \left\lbrack {k\left( {\mathbf{\mu }}_{p}\right) : k}\right\rbrack = p - 1 \) is prime to \( p \), it follows that if \( \alpha \) is a root of \( {X}^{p} - a \), then \( k\left( \alpha \right) \cap k\left( {\mathbf{\mu }}_{p}\right) = k \) because \( \left\lbrack {k\left( \alpha \right) : k}\right\rbrack = p \) . Hence \( \left\lbrack {K : k}\right\rbrack = p\left( {p - 1}\right) \), so \( {G}_{K/k} = G\left( p\right) \) .\n\nA direct computation of a commutator of elements in \( G\left( n\right) \) for arbitrary \( n \) shows that the commutator subgroup is contained in the group of matrices\n\n\[ \left( \begin{array}{ll} 1 & 0 \\ b & 1 \end{array}\right), b \in \mathbf{Z}/n\mathbf{Z} \]\n\nand so must be that subgroup because its factor group is isomorphic to \( {\left( \mathbf{Z}/n\mathbf{Z}\right) }^{ * } \) under the projection
|
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, the first cohomology group is trivial in both cases.
|
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 \( \theta \in K \) such that the element\n\n\[ \n\beta = \mathop{\sum }\limits_{{\tau \in G}}{\alpha }_{\tau }\tau \left( \theta \right)\n\]\n\nis \( \neq 0 \) . Then\n\n\[ \n{\sigma \beta } = \mathop{\sum }\limits_{{\tau \in G}}{\alpha }_{\tau }^{\sigma }{\sigma \tau }\left( \theta \right) = \mathop{\sum }\limits_{{\tau \in G}}{\alpha }_{\sigma \tau }{\alpha }_{\sigma }^{-1}{\sigma \tau }\left( \theta \right)\n\]\n\n\[ \n= {\alpha }_{\sigma }^{-1}\mathop{\sum }\limits_{{\tau \in G}}{\alpha }_{\sigma \tau }{\sigma \tau }\left( \theta \right) = {\alpha }_{\sigma }^{-1}\beta\n\]\n\nWe get \( {\alpha }_{\sigma } = \beta /{\sigma \beta } \), and using \( {\beta }^{-1} \) instead of \( \beta \) gives what we want.\n\nFor the additive part of the theorem, we find an element \( \theta \in K \) such that the trace \( \operatorname{Tr}\left( \theta \right) \) is not equal to 0 . Given a 1-cocycle \( \left\{ {\alpha }_{\sigma }\right\} \) in the additive group of \( K \) , we let\n\n\[ \n\beta = \frac{1}{\operatorname{Tr}\left( \theta \right) }\mathop{\sum }\limits_{{\tau \in G}}{\alpha }_{\tau }\tau \left( \theta \right)\n\]\n\nIt follows at once that \( {\alpha }_{\sigma } = \beta - {\sigma \beta } \), as desired.
|
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 \( {H}^{1}\left( {G, E}\right) = 0 \) .
|
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\[ = f\left( \tau \right) + \tau \left\lbrack {f\left( \sigma \right) + {\sigma f}\left( {\tau }^{-1}\right) }\right\rbrack . \]\n\nTherefore\n\n\[ {\tau f}\left( \sigma \right) - f\left( \sigma \right) = - {\sigma \tau f}\left( {\tau }^{-1}\right) - f\left( \tau \right) \]\n\nBut \( f\left( 1\right) = f\left( 1\right) + f\left( 1\right) \) implies \( f\left( 1\right) = 0 \), and\n\n\[ 0 = f\left( 1\right) = f\left( {\tau {\tau }^{-1}}\right) = f\left( \tau \right) + {\tau f}\left( {\tau }^{-1}\right) . \]\n\nThis shows that \( \left( {\tau - 1}\right) f\left( \sigma \right) = \left( {\sigma - 1}\right) f\left( \tau \right) \), so \( \left( {\tau - 1}\right) f \) is a coboundary. This proves the lemma.
|
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. \( \left( {e\left( {{\Gamma }^{\prime }/\Gamma }\right), M}\right) \) . Under the nondegeneracy assumption, we have\n\n\[ c\left( M\right) {e}_{M}\left( \Gamma \right) {\Gamma }_{\varphi } \subset {M\Gamma } \]
|
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 only on the class of \( \sigma \) modulo the subgroup of \( G \) leaving the elements of \( {A}_{N} \) fixed. In other words, we may view \( \left\{ {y}_{\sigma }\right\} \) as a cocycle of \( G\left( N\right) \) in \( {A}_{M} \) . Let \( c = c\left( N\right) \) . By Lemma 10.2, it follows that \( \left\{ {c{y}_{\sigma }}\right\} \) splits as a cocycle of \( G\left( N\right) \) in \( {A}_{M} \) . In other words, there exists \( {t}_{0} \in {A}_{M} \) such that\n\n\[ c{y}_{\sigma } = \sigma {t}_{0} - {t}_{0} \]\n\nand this equation in fact holds for \( \sigma \in G \) . Let \( t \) be such that \( {ct} = {t}_{0} \) . Then\n\n\[ {c\sigma y} - {cy} = {\sigma ct} - {cy}, \]\n\nwhence \( c\left( {y - t}\right) \) is fixed by all \( \sigma \in G \), and therefore lies in \( \frac{1}{N}\Gamma \) . Therefore\n\n\[ e\left( {{\Gamma }^{\prime }/\Gamma }\right) c\left( {y - t}\right) \in \Gamma \]\n\nWe multiply both sides by \( M \) and observe that \( {cM}\left( {y - t}\right) = {cMy} = {cx} \) . This shows that\n\n\[ c\left( N\right) e\left( {{\Gamma }^{\prime }/\Gamma }\right) {\Gamma }_{\varphi } \subset {M\Gamma } \]\n\nSince \( \Gamma /{M\Gamma } \) has exponent \( M \), we may replace \( e\left( {{\Gamma }^{\prime }/\Gamma }\right) \) by the greatest common divisor as stated in the theorem, and we can replace \( c\left( N\right) \) by \( c\left( M\right) \) to conclude the proof.
|
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 \( \Gamma \) is free with basis \( \left\{ {{a}_{1},\ldots ,{a}_{r}}\right\} \), and we let \( {\varphi }_{i} = {\varphi }_{{a}_{i}} \), then the map\n\n\[ {H}_{\Gamma }\left( {M, N}\right) \rightarrow {A}_{M}^{\left( r\right) }\;\text{ given by }\;\tau \rightarrow \left( {{\varphi }_{1}\left( \tau \right) ,\ldots ,{\varphi }_{r}\left( \tau \right) }\right) \]\n\nis injective. If \( {A}_{M} \) is cyclic of order \( M \), this map is an isomorphism.
|
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 \( K\left\lbrack X\right\rbrack \) such that\n\n\[ f\left( {{\lambda }_{1}\left( x\right) ,\ldots ,{\lambda }_{n}\left( x\right) }\right) = 0 \]\n\nfor all \( x \in A \) .
|
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 \( \left( {{\lambda }_{1}\left( x\right) ,\ldots ,{\lambda }_{n}\left( x\right) }\right) \) . We shall prove that \( f \) is additive.\n\nLet \( g\left( {X, Y}\right) = f\left( {X + Y}\right) - f\left( X\right) - f\left( Y\right) \) . Then\n\n\[ g\left( {\Lambda \left( x\right) ,\Lambda \left( y\right) }\right) = f\left( {\Lambda \left( {x + y}\right) }\right) - f\left( {\Lambda \left( x\right) }\right) - f\left( {\Lambda \left( y\right) }\right) = 0 \]\n\nfor all \( x, y \in A \) . We shall prove that \( g \) induces the zero function on \( {K}^{\left( n\right) } \times {K}^{\left( n\right) } \) . Assume otherwise. We have two cases.\n\nCase 1. We have \( g\left( {\xi ,\Lambda \left( y\right) }\right) = 0 \) for all \( \xi \in {K}^{\left( n\right) } \) and all \( y \in A \) . By hypothesis, there exists \( {\xi }^{\prime } \in {K}^{\left( n\right) } \) such that \( g\left( {{\xi }^{\prime }, Y}\right) \) is not identically 0 . Let \( P\left( Y\right) = g\left( {{\xi }^{\prime }, Y}\right) \) . Since the degree of \( g \) in \( \left( Y\right) \) is strictly smaller than the degree of \( f \), we have a contradiction.\n\nCase 2. There exist \( {\xi }^{\prime } \in {K}^{\left( n\right) } \) and \( {y}^{\prime } \in A \) such that \( g\left( {{\xi }^{\prime },\Lambda \left( {y}^{\prime }\right) }\right) \neq 0 \) . Let \( P\left( X\right) = g\left( {X,\Lambda \left( {y}^{\prime }\right) }\right) \) . Then \( P \) is not the zero polynomial, but \( P\left( {\Lambda \left( x\right) }\right) = 0 \) for all \( x \in A \), again a contradiction.\n\nWe conclude that \( g \) induces the zero function on \( {K}^{\left( n\right) } \times {K}^{\left( n\right) } \), which proves what we wanted, namely that \( f \) is additive.
|
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 \( f\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) in \( K\left\lbrack X\right\rbrack \) which is reduced, \( f \neq 0 \), and such that\n\n\[ f\left( {{\sigma }_{1}\left( x\right) ,\ldots ,{\sigma }_{n}\left( x\right) }\right) = 0 \]\n\nfor all \( x \in K \) . By what we saw above, we can write this relation in the form\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}\mathop{\sum }\limits_{{r = 1}}^{m}{a}_{ir}{\sigma }_{i}{\left( x\right) }^{{p}^{r}} = 0 \]\n\nfor all \( x \in K \), and with not all coefficients \( {a}_{ir} \) equal to 0 . Therefore by the linear independence of characters, the automorphisms\n\n\[ \left\{ {\sigma }_{i}^{{p}^{r}}\right\} \;\text{ with }\;i = 1,\ldots, n\;\text{ and }\;r = 1,\ldots, m \]\n\ncannot be all distinct. Hence we have\n\n\[ {\sigma }_{i}^{{p}^{r}} = {\sigma }_{j}^{{p}^{s}} \]\n\nwith either \( i \neq j \) or \( r \neq s \) . Say \( r \leqq s \) . For all \( x \in K \) we have\n\n\[ {\sigma }_{i}{\left( x\right) }^{{p}^{r}} = {\sigma }_{j}{\left( x\right) }^{{p}^{s}} \]\n\nExtracting \( p \) -th roots in characteristic \( p \) is unique. Hence\n\n\[ {\sigma }_{i}\left( x\right) = {\sigma }_{j}{\left( x\right) }^{{ps} - r} = {\sigma }_{j}\left( {x}^{{ps} - r}\right) \]\n\nfor all \( x \in K \) . Let \( \sigma = {\sigma }_{j}^{-1}{\sigma }_{i} \) . Then\n\n\[ \sigma \left( x\right) = {x}^{{p}^{s - r}} \]\n\nfor all \( x \in K \) . Taking \( {\sigma }^{n} = \) id shows that\n\n\[ x = {x}^{{p}^{n\left( {s - r}\right) }} \]\n\nfor all \( x \in K \) . Since \( K \) is infinite, this can hold only if \( s = r \) . But in that case, \( {\sigma }_{i} = {\sigma }_{j} \), contradicting the fact that we started with distinct automorphisms.
|
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 \( {X}_{i} = {X}_{{\sigma }_{i}} \) . Let\n\n\[ f\left( {{X}_{1},\ldots ,{X}_{n}}\right) = \det \left( {t}_{{\sigma }_{i},{\sigma }_{j}}\right) . \]\n\nThen \( f \) is not identically 0, as one sees by substituting 1 for \( {X}_{\mathrm{{id}}} \) and 0 for \( {X}_{\sigma } \) if \( \sigma \neq \mathrm{{id}} \) . Since \( k \) is infinite, \( f \) is reduced. Hence the determinant will not be 0 for all \( x \in K \) if we substitute \( {\sigma }_{i}\left( x\right) \) for \( {X}_{i} \) in \( f \) . Hence there exists \( w \in K \) such that\n\n\[ \det \left( {{\sigma }_{i}^{-1}{\sigma }_{j}\left( w\right) }\right) \neq 0. \]\n\nSuppose \( {a}_{1},\ldots ,{a}_{n} \in k \) are such that\n\n\[ {a}_{1}{\sigma }_{1}\left( w\right) + \cdots + {a}_{n}{\sigma }_{n}\left( w\right) = 0. \]\n\nApply \( {\sigma }_{i}^{-1} \) to this relation for each \( i = 1,\ldots, n \) . Since \( {a}_{j} \in k \) we get a system of linear equations, regarding the \( {a}_{j} \) as unknowns. Since the determinant of the coefficients is \( \neq 0 \), it follows that\n\n\[ {a}_{j} = 0\;\text{ for }\;j = 1,\ldots, n \]\n\nand hence that \( w \) is the desired element.
|
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 \( {\sigma }_{H} \in G/H \), satisfying a certain compatibility condition. This compatibility condition means that we may define an element \( \sigma \) of \( G \) as follows. Let \( \alpha \in K \) . Then \( \alpha \) is contained in some finite Galois extension \( F \subset K \) . Let \( H = \operatorname{Gal}\left( {K/F}\right) \) . Let \( {\sigma \alpha } = {\sigma }_{H}\alpha \) . The compatibility condition means that \( {\sigma }_{H}\alpha \) is independent of the choice of \( F \) . Then it is immediately verified that \( \sigma \) is an automorphism of \( K \) over \( k \), which maps to each \( {\sigma }_{H} \) in the canonical map of \( G \) into \( G/H \) . Hence the map \( G \rightarrow \underline{\lim }G/H \) is surjective, thereby proving the theorem.
|
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 \( {a}_{p} \) is the coefficient of \( {q}^{p} \) in \( \Delta \left( q\right) \), then we have\n\n\[ \operatorname{tr}\rho \left( {\mathrm{{Fr}}}_{p}\right) \equiv {a}_{p}{\;\operatorname{mod}\;l}\;\text{ and }\;\det \rho \left( {\mathrm{{Fr}}}_{p}\right) \equiv {p}^{11}{\;\operatorname{mod}\;l}. \]\n\nFurthermore, for all primes \( l \neq 2,3,5,7,{23},{691} \), the image \( \rho \left( G\right) \) in \( G{L}_{2}\left( {\mathbf{F}}_{l}\right) \) consists of those matrices \( M \in G{L}_{2}\left( {\mathbf{F}}_{l}\right) \) such that \( \det M \) is an eleventh power in \( {\mathbf{F}}_{l}^{ * } \) .
|
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 72],[SwD 73].
|
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 integral equation for \( {a}_{n}\alpha \) over \( A \) . This proves the proposition.
|
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\[ A \subset A\left\lbrack {\alpha }_{1}\right\rbrack \subset A\left\lbrack {{\alpha }_{1},{\alpha }_{2}}\right\rbrack \subset \cdots \subset A\left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\rbrack = B.\]\n\nBut we have already seen that our assertion is true in that case, this being part of the definition of integrality.
|
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 equation\n\n\[ \n{\alpha }^{n} + {b}_{n - 1}{\alpha }^{n - 1} + \cdots + {b}_{0} = 0 \n\]\n\nwith \( {b}_{i} \in B \) . Let \( {B}_{1} = A\left\lbrack {{b}_{0},\ldots ,{b}_{n - 1}}\right\rbrack \) . Then \( {B}_{1} \) is a finitely generated \( A \) - module by Proposition 1.2, and is obviously faithful. Then \( {B}_{1}\left\lbrack \alpha \right\rbrack \) is finite over \( {B}_{1} \), hence over \( A \), and hence \( \alpha \) is integral over \( A \) . Hence \( C \) is integral over \( A \) . Finally let \( B, C \) be extension rings of \( A \) and assume \( B \) integral over \( A \) . Assume that \( B, C \) are subrings of some ring. Then \( C\left\lbrack B\right\rbrack \) is generated by elements of \( B \) over \( C \), and each element of \( B \) is integral over \( C \) . That \( C\left\lbrack B\right\rbrack \) is integral over \( C \) will follow immediately from our next proposition.
|
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 \( {MN} \) is mapped into itself by multiplication with \( \alpha \pm \beta \) and \( {\alpha \beta } \) . Furthermore \( {MN} \) is finitely generated over \( A \) (if \( \left\{ {w}_{i}\right\} \) are generators of \( M \) and \( \left\{ {v}_{j}\right\} \) are generators of \( N \) then \( \left\{ {{w}_{i}{v}_{j}}\right\} \) are generators of \( {MN}) \) . This proves our proposition.
|
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\[ \n\sigma {\left( \alpha \right) }^{n} + \sigma \left( {a}_{n - 1}\right) \sigma {\left( \alpha \right) }^{n - 1} + \cdots + \sigma \left( {a}_{0}\right) = 0, \n\]\n\nthereby proving our assertion.
|
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 satisfied by \( \alpha \) over \( k \) .
|
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 similarly for the coefficients of \( \operatorname{Irr}\left( {\alpha, k, X}\right) \), which are elementary symmetric functions of the roots.
|
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 \( {a}_{i} \in A \), \[ {\left( a/b\right) }^{n} + {a}_{n - 1}{\left( a/b\right) }^{n - 1} + \cdots + {a}_{0} = 0 \] whence \[ {a}^{n} + {a}_{n - 1}b{a}^{n - 1} + \cdots + {a}_{0}{b}^{n} = 0. \] Since \( p \) divides \( b \), it must divide \( {a}^{n} \), and hence must divide \( a \), contradiction.
|
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 canonical image in \( {S}^{-1}A \) and \( {S}^{-1}B \) respectively, we see that this relation proves the integrality of \( \alpha /1 \) over \( {S}^{-1}A \), the coefficients being now \( {a}_{i}/1 \) .
|
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 \( {s}_{n - 1}\cdots {s}_{0} \) . Then it is clear that \( {s\alpha } \) is integral over \( A \), whence in \( A \) . Hence \( \alpha \) lies in \( {S}^{-1}A \), and \( {S}^{-1}A \) is integrally closed.
|
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\[ \mathfrak{p}{B}_{\mathfrak{p}} = \mathfrak{p}{A}_{\mathfrak{p}}{B}_{\mathfrak{p}} = {\mathfrak{m}}_{\mathfrak{p}}{B}_{\mathfrak{p}} \]\n\nit will suffice to prove our first assertion when \( A \) is a local ring. (Note that the existence of a prime ideal \( \mathfrak{p} \) implies that \( 1 \neq 0 \), and \( \mathfrak{p}B = B \) if and only if \( 1 \in \mathfrak{p}B \) .) In that case, if \( \mathfrak{p}B = B \), then 1 has an expression as a finite linear combination of elements of \( B \) with coefficients in \( \mathfrak{p} \) ,\n\n\[ 1 = {a}_{1}{b}_{1} + \cdots + {a}_{n}{b}_{n} \]\n\nwith \( {a}_{i} \in \mathfrak{p} \) and \( {b}_{i} \in B \) . We shall now use notation as if \( {A}_{\mathfrak{p}} \subset {B}_{\mathfrak{p}} \) . We leave it to the reader as an exercise to verify that our arguments are valid when we deal only with a canonical homomorphism \( {A}_{\mathfrak{p}} \rightarrow {B}_{\mathfrak{p}} \) . Let \( {B}_{0} = A\left\lbrack {{b}_{1},\ldots ,{b}_{n}}\right\rbrack \) . Then \( \mathfrak{p}{B}_{0} = {B}_{0} \) and \( {B}_{0} \) is a finite \( A \) -module by Proposition 1.2. Hence \( {B}_{0} = 0 \) by Nakayama's lemma, contradiction. (See Lemma 4.1 of Chapter X.)\n\nTo prove our second assertion, note the following commutative diagram:\n\n\n\nWe have just proved \( {m}_{\mathfrak{p}}{B}_{\mathfrak{p}} \neq {B}_{\mathfrak{p}} \) . Hence \( {m}_{\mathfrak{p}}{B}_{\mathfrak{p}} \) is contained in a maximal ideal \( \mathfrak{M} \) of \( {B}_{\mathfrak{p}} \) . Taking inverse images, we see that the inverse image of \( \mathfrak{M} \) in \( {A}_{\mathfrak{p}} \) is an ideal containing \( {\mathfrak{m}}_{\mathfrak{p}} \) (in the case of an inclusion \( {A}_{\mathfrak{p}} \subset {B}_{\mathfrak{p}} \) the inverse image is \( \mathfrak{M} \cap {A}_{\mathfrak{p}} \) ). Since \( {\mathfrak{m}}_{\mathfrak{p}} \) is maximal, we have \( \mathfrak{M} \cap {A}_{\mathfrak{p}} = {\mathfrak{m}}_{\mathfrak{p}} \) . Let \( \mathfrak{P} \) be the inverse image of \( \mathfrak{M} \) in \( B \) (in the case of inclusion, \( \mathfrak{P} = \mathfrak{M} \cap B \) ). Then \( \mathfrak{P} \) is a prime ideal of \( B \) . The inverse image of \( {m}_{\mathfrak{p}} \) in \( A \) is simply \( \mathfrak{p} \) . Taking the inverse image of \( \mathfrak{M} \) going around both ways in the diagram, we find that\n\n\[ \mathfrak{P} \cap A = \mathfrak{p} \]\n\nas was to be shown.
|
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 \( A/\mathfrak{p}\left\lbrack \alpha \right\rbrack \) is a field. Hence every non-zero element of \( B/\mathfrak{P} \) is invertible in \( B/\mathfrak{P} \), which is therefore a field. Conversely, assume that \( \mathfrak{P} \) is maximal in \( B \) . Then \( B/\mathfrak{P} \) is a field, which is integral over the entire ring \( A/\mathfrak{p} \) . If \( A/\mathfrak{p} \) is not a field, it has a non-zero maximal ideal \( m \) . By Proposition 1.10, there exists a prime ideal \( \mathfrak{M} \) of \( B/\mathfrak{P} \) lying above \( m,\mathfrak{M} \neq 0 \), contradiction.
|
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 \( B \) of \( A \) in \( L \) lying above \( \mathfrak{p} \) . Then there exists \( \sigma \in G \) such that \( \sigma \mathfrak{P} = \mathfrak{Q} \) .
|
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 \[ x \equiv 0\;\left( {{\;\operatorname{mod}\;\sigma }\mathfrak{P}}\right) ,\;\text{ all }\sigma \in G \] \[ x \equiv 1\;\left( {{\;\operatorname{mod}\;\sigma }\mathfrak{Q}}\right) ,\;\text{ all }\sigma \in G \] (use the Chinese remainder theorem). The norm \[ {N}_{K}^{L}\left( x\right) = \mathop{\prod }\limits_{{\sigma \in G}}{\sigma x} \] lies in \( B \cap K = A \) (because \( A \) is integrally closed), and lies in \( \mathfrak{P} \cap A = \mathfrak{p} \) . But \( x \notin \sigma \mathfrak{Q} \) for all \( \sigma \in G \), so that \( {\sigma x} \notin \mathfrak{Q} \) for all \( \sigma \in G \) . This contradicts the fact that the norm of \( x \) lies in \( \mathfrak{p} = \mathfrak{Q} \cap A \) .
|
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 \( \mathfrak{p \).
|
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 \( L \) lying above \( {\mathfrak{Q}}_{1} \) and \( {\mathfrak{Q}}_{2} \) respectively, then \( {\mathfrak{P}}_{1} \neq {\mathfrak{P}}_{2} \). This argument reduces our assertion to the case that \( E \) is Galois over \( K \), and it then becomes an immediate consequence of the proposition.
|
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 that \( H \) leaves \( \mathfrak{P} \) invariant. Hence \( G \subset {G}_{\mathfrak{P}} \) and \( E \supset {L}^{\text{dec }} \) . We have already observed that \( {L}^{\text{dec }} \) has the required property.
|
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 \( {\mathfrak{Q}}_{\sigma } \neq \mathfrak{Q} \) . Let \( x \) be an element of \( {B}^{\text{dec }} \) . There exists an element \( y \) of \( {B}^{\text{dec }} \) such that\n\n\[ \ny \equiv x\;\left( {\;\operatorname{mod}\;\Omega }\right) \n\]\n\n\[ \ny \equiv 1\;\left( {\;\operatorname{mod}\;{\mathfrak{Q}}_{\sigma }}\right) \n\]\n\nfor each \( \sigma \) in \( G \), but not in \( {G}_{\mathfrak{P}} \) . Hence in particular,\n\n\[ \ny \equiv x\;\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \n\]\n\n\[ \ny \equiv 1\;\left( {{\;\operatorname{mod}\;{\sigma }^{-1}}\mathfrak{P}}\right) \n\]\n\nfor each \( \sigma \) not in \( {G}_{\mathfrak{P}} \) . This second congruence yields\n\n\[ \n{\sigma y} \equiv 1\;\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \n\]\n\nfor all \( \sigma \notin {G}_{\mathfrak{P}} \) . The norm of \( y \) from \( {L}^{\text{dec }} \) to \( K \) is a product of \( y \) and other factors \( {\sigma y} \) with \( \sigma \notin {G}_{\mathfrak{P}} \) . Thus we obtain\n\n\[ \n{N}_{K}^{{L}^{\mathrm{{dec}}}}\left( y\right) \equiv x\;\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) . \n\]\n\nBut the norm lies in \( K \), and even in \( A \), since it is a product of elements integral over \( A \) . This last congruence holds mod \( \mathfrak{Q} \), since both \( x \) and the norm lie in \( {B}^{\text{dec }} \) . This is precisely the meaning of the assertion in our proposition.
|
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 homomorphism, and let \( {\psi }_{1},{\psi }_{2} \) be two homomorphisms of \( B \) extending \( \varphi \) in a given algebraic closure of \( A/\mathfrak{p} \) . Then there exists an automorphism \( \sigma \) of \( L \) over \( K \) such that\n\n\[{\psi }_{1} = {\psi }_{2} \circ \sigma\]
|
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 \( {\psi }_{1},{\psi }_{2} \) have the same kernel \( \mathfrak{P} \) . Hence there exists an automorphism \( \omega \) of \( {\psi }_{1}\left( B\right) \) onto \( {\psi }_{2}\left( B\right) \) such that \( \omega \circ {\psi }_{1} = {\psi }_{2} \) . There exists an element \( \sigma \) of \( {G}_{\mathfrak{P}} \) such that \( \omega \circ {\psi }_{1} = {\psi }_{1} \circ \sigma \), by the preceding proposition. This proves what we wanted.
|
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 \( K\left\lbrack X\right\rbrack \), and has a root \( \alpha \) in B. Then the reduced polynomial \( \bar{f} \) is a power of an irreducible polynomial in \( \bar{A}\left\lbrack X\right\rbrack \) .
|
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\[ f\left( X\right) = {X}^{n} + {a}_{n - 1}{X}^{n - 1} + \cdots + {a}_{0} \]\n\nbe the irreducible polynomial of \( \alpha \) over \( k \), with \( {a}_{i} \in A \) . Let \( \mathfrak{p} \) be a maximal ideal in \( A \), let \( \mathfrak{P} \) be a prime ideal of the integral closure \( B \) of \( A \) in \( L \) , \( \mathfrak{P} \) lying above \( \mathfrak{p} \) . Let \( \bar{f}\left( X\right) \) be the reduced polynomial with coefficients in \( A/\mathfrak{p} \) . Let \( {G}_{\mathfrak{P}} \) be the decomposition group. If \( \bar{f} \) has no multiple roots, then the map \( \sigma \mapsto \bar{\sigma } \) has trivial kernel, and is an isomorphism of \( {G}_{\mathfrak{P}} \) on the Galois group of \( \bar{f} \) over \( A/\mathfrak{p} \) .
|
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 \( {\bar{G}}_{\mathfrak{P}} \), as before. We have\n\n\[ \bar{f}\left( X\right) = \prod \left( {X - {\bar{x}}_{i}}\right) \]\n\nSuppose that \( \bar{\sigma }{\bar{x}}_{i} = {\bar{x}}_{i} \) for all \( i \) . Since \( \left( \overline{\sigma {x}_{i}}\right) = \bar{\sigma }{\bar{x}}_{i} \), and since \( \bar{f} \) has no multiple roots, it follows that \( \sigma \) is also the identity. Hence our map is injective, the inertia group is trivial. The field \( \bar{A}\left\lbrack {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n}}\right\rbrack \) is a subfield of \( \bar{B} \) and any automorphism of \( \bar{B} \) over \( \bar{A} \) which restricts to the identity on this subfield must be the identity, because the map \( {G}_{\mathfrak{P}} \rightarrow {\bar{G}}_{\mathfrak{P}} \) is onto the Galois group of \( \bar{B} \) over \( \bar{A} \) . Hence \( \bar{B} \) is purely inseparable over \( \bar{A}\left\lbrack {{\bar{x}}_{1},\ldots ,{\bar{x}}_{n}}\right\rbrack \) and therefore \( {G}_{\mathfrak{P}} \) is isomorphic to the Galois group of \( \bar{f} \) over \( \bar{A} \) .
|
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 \( \bar{f} = f{\;\operatorname{mod}\;\mathfrak{p}} \) . Suppose that \( \bar{f} \) has no multiple roots in an algebraic closure of \( A/\mathfrak{p} \) . Let \( L \) be a splitting field for \( f \) over \( K \), and let \( B \) be the integral closure of \( A \) in L. Let \( \mathfrak{P} \) be any prime of \( B \) above \( \mathfrak{p} \) and let a bar denote reduction mod \( \mathfrak{p} \) . Then the map\n\n\[ \n{G}_{\mathfrak{P}} \rightarrow {\bar{G}}_{\mathfrak{P}} \n\]\n\nis an isomorphism of \( {G}_{\mathfrak{P}} \) with the Galois group of \( \bar{f} \) over \( \bar{A} \) .
|
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\[ \nf\left( X\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {X - {\alpha }_{i}}\right) \n\]\n\nit follows that\n\n\[ \n\bar{f}\left( X\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {X - {\bar{\alpha }}_{i}}\right) \n\]\n\nAny element of \( G \) is determined by its effect as a permutation of the roots, and for \( \sigma \in {G}_{\mathfrak{P}} \), we have\n\n\[ \n\bar{\sigma }{\bar{\alpha }}_{i} = \overline{\sigma {\alpha }_{i}}. \n\]\n\nHence if \( \bar{\sigma } = \mathrm{{id}} \) then \( \sigma = \mathrm{{id}} \), so the map \( {G}_{\mathfrak{P}} \rightarrow {\bar{G}}_{\mathfrak{P}} \) is injective. It is surjective by Proposition 2.5, so the theorem is proved.
|
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 the canonical homomorphism of \( A \) into \( {S}^{-1}A \) . Furthermore, \( {S}^{-1}B \) is integral over \( {S}^{-1}A \) . This reduces the question to the case when we deal with a local ring, which has just been discussed above.
|
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 \( A\left\lbrack {x}^{-1}\right\rbrack \) into \( L \) .
|
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\[ \n{mA}\left\lbrack {x}^{-1}\right\rbrack = A\left\lbrack {x}^{-1}\right\rbrack .\n\]\n\nThen we can write\n\n\[ \n1 = {a}_{0} + {a}_{1}{x}^{-1} + \cdots + {a}_{n}{x}^{-n}\n\]\n\nwith \( {a}_{i} \in \mathfrak{m} \) . Multiplying by \( {x}^{n} \) we obtain\n\n\[ \n\left( {1 - {a}_{0}}\right) {x}^{n} + {b}_{n - 1}{x}^{n - 1} + \cdots + {b}_{0} = 0\n\]\n\nwith suitable elements \( {b}_{i} \in A \) . Since \( {a}_{0} \in \mathfrak{m} \), it follows that \( 1 - {a}_{0} \notin \mathfrak{m} \) and hence \( 1 - {a}_{0} \) is a unit in \( A \) because \( A \) is assumed to be a local ring. Dividing by \( 1 - {a}_{0} \) we see that \( x \) is integral over \( A \), and hence that our homomorphism has an extension to \( A\left\lbrack x\right\rbrack \) by Proposition 3.1.\n\nIf on the other hand we have\n\n\[ \n{mA}\left\lbrack {x}^{-1}\right\rbrack \neq A\left\lbrack {x}^{-1}\right\rbrack\n\]\n\nthen \( \mathfrak{m}A\left\lbrack {x}^{-1}\right\rbrack \) is contained in some maximal ideal \( \mathfrak{P} \) of \( A\left\lbrack {x}^{-1}\right\rbrack \) and \( \mathfrak{P} \cap A \) contains \( m \) . Since \( m \) is maximal, we must have \( \mathfrak{P} \cap A = m \) . Since \( \varphi \) and the canonical map \( A \rightarrow A/\mathfrak{m} \) have the same kernel, namely \( \mathfrak{m} \), we can find an embedding \( \psi \) of \( A/m \) into \( L \) such that the composite map\n\n\[ \nA \rightarrow A/\mathfrak{m}\xrightarrow[]{\psi }L\n\]\n\nis equal to \( \varphi \) . We note that \( A/\mathfrak{m} \) is canonically embedded in \( B/\mathfrak{P} \) where \( B = A\left\lbrack {x}^{-1}\right\rbrack \), and extend \( \psi \) to a homomorphism of \( B/\mathfrak{P} \) into \( L \), which we can do whether the image of \( {x}^{-1} \) in \( B/\mathfrak{P} \) is transcendental or algebraic over \( A/\mathfrak{m} \) . The composite \( B \rightarrow B/\mathfrak{P} \rightarrow L \) gives us what we want.
|
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 \( x \in K, x \neq 0 \), then \( x \in B \) or \( {x}^{-1} \in B \) .
|
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 restriction. In other words, \( \left( {C,\psi }\right) \leqq \left( {{C}^{\prime },{\psi }^{\prime }}\right) \) if \( C \subset {C}^{\prime } \) and the restriction of \( {\psi }^{\prime } \) to \( C \) is equal to \( \psi \) . It is clear that \( S \) is inductively ordered, and by Zorn’s lemma there exists a maximal element, say \( \left( {B,{\psi }_{0}}\right) \) . Then first \( B \) is a local ring, otherwise \( {\psi }_{0} \) extends to the local ring arising from the kernel, and second, \( B \) has the desired property according to Theorem 3.2.
|
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 valuation ring, we always have \( {x}_{i} \leqq {x}_{j} \) or \( {x}_{j} \leqq {x}_{i} \) for all pairs of indices \( i, j \) . Hence we may order our elements, and we select the index \( j \) such that \( {x}_{i} \leqq {x}_{j} \) for all \( i \) . This index \( j \) satisfies the requirement of the proposition.
|
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 valuation ring, we always have \( {x}_{i} \leqq {x}_{j} \) or \( {x}_{j} \leqq {x}_{i} \) for all pairs of indices \( i, j \) . Hence we may order our elements, and we select the index \( j \) such that \( {x}_{i} \leqq {x}_{j} \) for all \( i \) . This index \( j \) satisfies the requirement of the proposition.
|
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 valuation ring, we always have \( {x}_{i} \leqq {x}_{j} \) or \( {x}_{j} \leqq {x}_{i} \) for all pairs of indices \( i, j \) . Hence we may order our elements, and we select the index \( j \) such that \( {x}_{i} \leqq {x}_{j} \) for all \( i \) . This index \( j \) satisfies the requirement of the proposition.
|
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 valuation ring, we always have \( {x}_{i} \leqq {x}_{j} \) or \( {x}_{j} \leqq {x}_{i} \) for all pairs of indices \( i, j \) . Hence we may order our elements, and we select the index \( j \) such that \( {x}_{i} \leqq {x}_{j} \) for all \( i \) . This index \( j \) satisfies the requirement of the proposition.
|
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\[ - 1 = {a}_{n}{\left( 1/x\right) }^{n} + \cdots + {a}_{1}\left( {1/x}\right) + y \]\n\nwith \( y \in \mathfrak{m} \) and \( {a}_{i} \in \mathfrak{o} \) . From this we get\n\n\[ \left( {1 + y}\right) {x}^{n} + \cdots + {a}_{n} = 0. \]\n\nBut \( 1 + y \) is not in \( m \), hence is a unit of \( \mathfrak{o} \) . We divide the equation by \( 1 + y \) to conclude that \( x \) is integral over \( \mathfrak{o} \), contrary to our hypothesis. Thus \( \left( {\mathfrak{m},1/x}\right) \) is not the entire ring, and is contained in a maximal ideal \( \mathfrak{P} \), whose intersection with 0 contains \( m \) and hence must be equal to \( m \) . Extending the canonical homomorphism \( \mathfrak{o}\left\lbrack {1/x}\right\rbrack \rightarrow \mathfrak{o}\left\lbrack {1/x}\right\rbrack /\mathfrak{P} \) to a homomorphism of a valuation ring \( \mathfrak{O} \) of \( L \) , we see that the image of \( 1/x \) is 0 and hence that \( x \) cannot be in this valuation ring.\n\nConversely, assume that \( x \) is integral over \( \mathfrak{o} \), and let\n\n\[ {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} = 0 \]\n\nbe an integral equation for \( x \) with coefficients in \( \mathfrak{o} \) . Let \( \mathfrak{O} \) be any valuation ring of \( L \) lying above \( \mathfrak{o} \) . Suppose \( x \notin \mathfrak{O} \) . Let \( \varphi \) be the place given by the canonical homomorphism of \( \mathfrak{O} \) modulo its maximal ideal. Then \( \varphi \left( x\right) = \infty \) so \( \varphi \left( {1/x}\right) = 0 \) . Divide the above equation by \( {x}^{n} \), and apply \( \varphi \) . Then each term except the first maps to 0 under \( \varphi \), so we get \( \varphi \left( 1\right) = 0 \), a contradiction which proves the proposition.
|
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 on \( A \) is finite on \( x \) .
|
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 some \( n \) . Multiplying by \( {x}^{n - 1} \) we conclude that \( x \) is integral over A. If \( 1/x \) is not a unit in \( A\left\lbrack {1/x}\right\rbrack \), then \( 1/x \) generates a proper principal ideal. By Zorn’s lemma this ideal is contained in a maximal ideal \( \mathfrak{M} \) . The homomorphism \( A\left\lbrack {1/x}\right\rbrack \rightarrow A\left\lbrack {1/x}\right\rbrack /\mathfrak{M} \) can be extended to a place which is a finite on \( A \) but maps \( 1/x \) on 0, so \( x \) on \( \infty \), which contradicts the possibility that \( 1/x \) is not a unit in \( A\left\lbrack {1/x}\right\rbrack \) and proves that \( x \) is integral over \( A \) . The converse implication is proved just as in the second part of Proposition 3.5.
|
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\[ \n{z}_{s}^{{d}_{s}} + {g}_{s}\left( {{z}_{1},\ldots ,{z}_{m}}\right) = 0 \n\]\n\nwhere \( {g}_{s}\left( {{Z}_{1},\ldots ,{Z}_{m}}\right) \in A\left\lbrack {{Z}_{1},\ldots ,{Z}_{m}}\right\rbrack \) is a polynomial of total degree \( < {d}_{s} \) , and that any pure power of \( {Z}_{s} \) occuring with non-zero coefficient in \( {g}_{s} \) occurs with a power strictly less than \( {d}_{s} \) . Then \( {z}_{1},\ldots ,{z}_{m} \) are integral over \( A \) .
|
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 \( \varphi \left( {{z}_{j}/{z}_{s}}\right) \neq \infty \) for all \( j \) . We divide the polynomial relation of the hypothesis in the lemma by \( {z}_{s}^{{d}_{s}} \) and apply the place. By the hypothesis on \( {g}_{s} \), it follows that \( \varphi \left( {{g}_{s}\left( z\right) /{z}_{s}^{{d}_{s}}}\right) = 0 \), whence we get \( 1 = 0 \), a contradiction which proves the theorem.
|
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 \( k\left\lbrack x\right\rbrack \) such that \( k\left\lbrack x\right\rbrack \) is integral over \[ k\left\lbrack y\right\rbrack = k\left\lbrack {{y}_{1},\ldots ,{y}_{r}}\right\rbrack . \]
|
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 \( {a}_{\left( j\right) } \neq 0 \) . The sum is taken over a finite number of distinct \( n \) -tuples of integers \( \left( {{j}_{1},\ldots ,{j}_{n}}\right) ,{j}_{v} \geqq 0 \) . Let \( {m}_{2},\ldots ,{m}_{n} \) be positive integers, and put \[ {y}_{2} = {x}_{2} - {x}_{1}^{{m}_{2}},\ldots ,{y}_{n} = {x}_{n} - {x}_{1}^{{m}_{n}}. \] Substitute \( {x}_{i} = {y}_{i} + {x}_{1}^{{m}_{i}}\left( {i = 2,\ldots, n}\right) \) in the above equation. Using vector notation, we put \( \left( m\right) = \left( {1,{m}_{2},\ldots ,{m}_{n}}\right) \) and use the dot product \( \left( j\right) \cdot \left( m\right) \) to denote \( {j}_{1} + {m}_{2}{j}_{2} + \cdots + {m}_{n}{j}_{n} \) . If we expand the relation after making the above substitution, we get \[ \sum {c}_{\left( j\right) }{x}_{1}^{\left( j\right) \cdot \left( m\right) } + f\left( {{x}_{1},{y}_{2},\ldots ,{y}_{n}}\right) = 0 \] where \( f \) is a polynomial in which no pure power of \( {x}_{1} \) appears. We now select \( d \) to be a large integer [say greater than any component of a vector \( \left( j\right) \) such that \( \left. {{c}_{\left( j\right) } \neq 0}\right\rbrack \) and take \[ \left( m\right) = \left( {1, d,{d}^{2},\ldots ,{d}^{n}}\right) . \] Then all \( \left( j\right) \cdot \left( m\right) \) are distinct for those \( \left( j\right) \) such that \( {c}_{\left( j\right) } \neq 0 \) . In this way we obtain an integral equation for \( {x}_{1} \) over \( k\left\lbrack {{y}_{2},\ldots ,{y}_{n}}\right\rbrack \) . Since each \( {x}_{i}\left( {i > 1}\right) \) is integral over \( k\left\lbrack {{x}_{1},{y}_{2},\ldots ,{y}_{n}}\right\rbrack \), it follows that \( k\left\lbrack x\right\rbrack \) is integral over \( k\left\lbrack {{y}_{2},\ldots ,{y}_{n}}\right\rbrack \) . We can now proceed inductively, using the transitivity of integral extensions to shrink the number of \( y \) ’s until we reach an algebraically independent set of \( y \) ’s.
|
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 \( {x}_{i} \) . Using Proposition 3.4 of Chapter VII, and renumbering the indices if necessary, say \( \varphi \left( {{x}_{j}/{x}_{n}}\right) \) is finite for all \( i \) . Let \( {z}_{j}^{\prime } = \varphi \left( {{x}_{j}/{x}_{n}}\right) \) for \( j = 1,\ldots, n \) . Then dividing the equations \( {y}_{i} = \sum {u}_{ij}{x}_{j} \) by \( {x}_{n} \) (for \( i = 1,\ldots, r \) ) and applying the place, we get\n\n\[ 0 = {u}_{11}{z}_{1}^{\prime } + {u}_{12}{z}_{2}^{\prime } + \cdots + {u}_{1n} \]\n\n\[ 0 = {u}_{r1}{z}_{1}^{\prime } + {u}_{r2}{z}_{2}^{\prime } + \cdots + {u}_{rn} \]\n\nThe transcendence degree of \( k\left( {z}^{\prime }\right) \) over \( k \) cannot be \( r \), for otherwise, the place \( \varphi \) would be an isomorphism of \( k\left( x\right) \) on its image. [Indeed, if, say, \( {z}_{1}^{\prime },\ldots ,{z}_{r}^{\prime } \) are algebraically independent and \( {z}_{i} = {x}_{i}/{x}_{n} \), then \( {z}_{1},\ldots ,{z}_{r} \) are also algebraically independent, and so form a transcendence base for \( k\left( x\right) \) over \( k \) . Then the place is an isomorphism from \( k\left( {{z}_{1},\ldots ,{z}_{r}}\right) \) to \( k\left( {{z}_{1}^{\prime },\ldots ,{z}_{r}^{\prime }}\right) \), and hence is an isomorphism from \( k\left( x\right) \) to its image.] We then conclude that\n\n\[ {u}_{1n},\ldots ,{u}_{rn} \in k\left( {{u}_{ij},{z}^{\prime }}\right) \;\text{ with }\;i = 1,\ldots, r;\;j = 1,\ldots, n - 1. \]\n\nHence the transcendence degree of \( k\left( u\right) \) over \( k \) would be \( \leqq {rn} - 1 \), which is a contradiction, proving the theorem.
|
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 family of elements \( {c}_{ij} \in k \) satisfying \( P\left( c\right) \neq 0 \), and we let \( {y}_{i}^{\prime } = \sum {c}_{ij}{x}_{j} \), then \( k\left\lbrack x\right\rbrack \) is integral over \( k\left\lbrack {{y}_{1}^{\prime },\ldots ,{y}_{r}^{\prime }}\right\rbrack \) .
|
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 \( P\left( c\right) \neq 0 \), then there is a homomorphism\n\n\[ \varphi : k\left( x\right) \left\lbrack {u, P{\left( u\right) }^{-1}}\right\rbrack \rightarrow k\left( x\right) \]\n\nsuch that \( \varphi \left( u\right) = \left( c\right) \), and such that \( \varphi \) is the identity on \( k\left( x\right) \) . We can apply \( \varphi \) to an integral equation for \( {x}_{i} \) over \( {k}_{u}\left\lbrack y\right\rbrack \) to get an integral equation for \( {x}_{i} \) over \( k\left\lbrack {y}^{\prime }\right\rbrack \), thus concluding the proof.
|
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 \( k \) . Let \( \{ \lambda \} \) be a basis of \( L \) over \( E \) . Then \( \{ {\alpha \lambda }\} \) is a basis of \( L \) over \( k \) . If \( K \) and \( L \) are not linearly disjoint over \( k \), then there exists a relation\n\n\[ \mathop{\sum }\limits_{{\lambda ,\alpha }}\left( {\mathop{\sum }\limits_{\kappa }{c}_{\kappa \lambda \alpha }\kappa }\right) {\lambda \alpha } = 0\;\text{ with some }{c}_{\kappa \lambda \alpha } \neq 0,{c}_{\kappa \lambda \alpha } \in k.\]\n\nChanging the order of summation gives\n\n\[ \mathop{\sum }\limits_{\lambda }\left( {\mathop{\sum }\limits_{{\kappa ,\lambda }}{c}_{\kappa \lambda \alpha }{\kappa \alpha }}\right) \lambda = 0 \]\n\ncontradicting the linear disjointness of \( L \) and \( {KE} \) over \( E \) .\n\nConversely, assume that \( K \) and \( L \) are linearly disjoint over \( k \) . Then \( a \) fortiori, \( K \) and \( E \) are also linearly disjoint over \( k \), and the field \( {KE} \) is the quotient field of the ring \( E\left\lbrack K\right\rbrack \) generated over \( E \) by all elements of \( K \) . This ring is a vector space over \( E \), and a basis for \( K \) over \( k \) is also a basis for this ring \( E\left\lbrack K\right\rbrack \) over \( E \) . With this remark, and the criteria for linear disjointness, we see that it suffices to prove that the elements of such a basis remain linearly independent over \( L \) . At this point we see that the arguments given in the first part of the proof are reversible. We leave the formalism to the reader.
|
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 \( {y}_{\alpha } \) in \( L \) . This gives a linear relation among the \( {M}_{\alpha }\left( x\right) \) . But these are linearly independent over \( k \) because the \( x \) ’s are assumed algebraically independent over \( k \) . This is a contradiction.
|
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 \( k\left\lbrack u\right\rbrack \) over \( k \) . They must remain linearly independent over \( L \), because as we have seen, a linear relation gives an algebraic relation. This proves our proposition.
|
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 4.2, we may assume that \( K \) is finitely generated over \( k \) . Let \( \left( t\right) \) be a transcendence base of \( K \) over \( k \), so \( K \) is separable algebraic over \( k\left( t\right) \) . By hypothesis, \( \left( t\right) \) is a transcendence base of \( {KL} \) over \( L \), and since every element of \( K \) is separable algebraic over \( k\left( t\right) \), it is also separable over \( L\left( t\right) \) . Hence \( {KL} \) is separably generated over \( L \) . This proves the corollary.
|
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 4.2, we may assume that \( K \) is finitely generated over \( k \) . Let \( \left( t\right) \) be a transcendence base of \( K \) over \( k \), so \( K \) is separable algebraic over \( k\left( t\right) \) . By hypothesis, \( \left( t\right) \) is a transcendence base of \( {KL} \) over \( L \), and since every element of \( K \) is separable algebraic over \( k\left( t\right) \), it is also separable over \( L\left( t\right) \) . Hence \( {KL} \) is separably generated over \( L \) . This proves the corollary.
|
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 that \( {KL} \) is not separable over \( L \) . The converse is a special case of Corollary 4.5, taking into account that linearly disjoint fields are free.
|
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 to \( K \). Finally, if there exists an element \( t \) of \( K \) transcendental over \( k \), then \( k\left( {t}^{1/{p}^{m}}\right) \) has degree \( {p}^{m} \) over \( k\left( t\right) \), and hence there exists a \( t \) such that \( {\mathrm{t}}^{1/{p}^{m}} \) does not lie in \( K \). This proves our proposition.
|
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 to \( K \) . Finally, if there exists an element \( t \) of \( K \) transcendental over \( k \), then \( k\left( {t}^{1/{p}^{m}}\right) \) has degree \( {p}^{m} \) over \( k\left( t\right) \), and hence there exists a \( t \) such that \( {\mathrm{t}}^{1/{p}^{m}} \) does not lie in \( K \) . This proves our proposition.
|
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 \( \left\lbrack {k\left( x\right) : k}\right\rbrack = \left\lbrack {K\left( x\right) : K}\right\rbrack \) .
|
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 powers form a basis of \( K\left( x\right) \) over \( K \) . This proves the lemma.
|
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 \( k \) .\n\n(c) If \( k \) is algebraically closed, then every extension of \( k \) is regular.
|
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 \( {y}_{i} \in L \) . Let \( \varphi \) be a \( {k}^{a} \) -valued place of \( L \) over \( k \) . Let \( \left( t\right) \) be a transcendence base of \( K \) over \( k \) . By hypothesis, the elements of \( \left( t\right) \) remain algebraically independent over \( L \), and hence \( \varphi \) can be extended to a place of \( {KL} \) which is identity on \( k\left( t\right) \) . This place must then be an isomorphism of \( K \) on its image, because \( K \) is a finite algebraic extension of \( k\left( t\right) \) (remark at the end of Chapter VII, §3). After a suitable isomorphism, we may take a place equivalent to \( \varphi \) which is the identity on \( K \) . Say \( \varphi \left( {{y}_{i}/{y}_{n}}\right) \) is finite for all \( i \) (use Proposition 3.4 of Chapter VII). We divide the relation of linear dependence by \( {y}_{n} \) and apply \( \varphi \) to get \( \sum {x}_{i}\varphi \left( {{y}_{i}/{y}_{n}}\right) = 0 \), which gives a linear relation among the \( {x}_{i} \) with coefficients in \( {k}^{\mathrm{a}} \), contradicting the linear disjointness. This proves the theorem.
|
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 \( L \), and hence \( {KL} \) is regular over \( L \) .
|
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 \( f\left( x\right) = 0 \) . Let \( L \) be an extension of \( k \) , free from \( k\left( x\right) \) over \( k \) . Let \( {\mathfrak{p}}_{L} \) be the prime ideal in \( L\left\lbrack X\right\rbrack \) vanishing on \( \left( x\right) \) . Then \( {\mathfrak{p}}_{L} = \mathfrak{p}L\left\lbrack X\right\rbrack \), that is \( {\mathfrak{p}}_{L} \) is the ideal generated by \( \mathfrak{p} \) in \( L\left\lbrack X\right\rbrack \), and in particular, this ideal is prime.
|
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\n\[ 0 \rightarrow L{ \otimes }_{k}\mathfrak{p} \rightarrow L\left\lbrack X\right\rbrack \rightarrow L{ \otimes }_{k}k\left\lbrack x\right\rbrack \rightarrow 0. \]\n\nBy Corollary 4.15, we know that \( L{ \otimes }_{k}k\left\lbrack x\right\rbrack \approx L\left\lbrack x\right\rbrack \), and the image of \( L{ \otimes }_{k}\mathfrak{p} \) in \( L\left\lbrack X\right\rbrack \) is \( \mathfrak{p}L\left\lbrack X\right\rbrack \), so the lemma is proved.
|
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 \( \left( x\right) \) in \( K\left\lbrack X\right\rbrack \). Then, if \( \left( u\right) \) is any set of elements of \( K\left( x\right) \) satisfying the equations\n\n\[ 0 = {f}_{\alpha }^{D}\left( x\right) + \sum \left( {\partial {f}_{\alpha }/\partial {x}_{i}}\right) {u}_{i}, \]\n\nthere is one and only one derivation \( {D}^{ * } \) of \( K\left( x\right) \) coinciding with \( D \) on \( K \), and such that \( {D}^{ * }{x}_{i} = {u}_{i} \) for every \( i \).
|
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\[ {D}^{ * }g\left( x\right) = {g}^{D}\left( x\right) + \sum \frac{\partial g}{\partial {x}_{i}}{u}_{i} \]\n\n\[ {D}^{ * }\left( {g/h}\right) = \frac{h{D}^{ * }g - g{D}^{ * }h}{{h}^{2}} \]\n\nis well defined and is a derivation of \( K\left( x\right) \).
|
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 uppermost step of this latter type, one sees immediately how to construct a derivation trivial on the bottom and nontrivial on top of the tower.
|
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) \( \det \left( {\partial {f}_{i}/\partial {x}_{j}}\right) \neq 0 \) .\n\nThen \( \left( x\right) \) is separably algebraic over \( K \) .
|
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 \( D \) is trivial on \( K\left( x\right) \) . Hence \( K\left( x\right) \) is separable algebraic over \( K \) by Proposition 5.2.
|
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 \( {K}^{p}k \) such that \( {Dz} = 1 \) . This derivation is at first defined on the field \( {K}^{p}k\left( z\right) \) . One can extend it to \( K \) as follows. Suppose there is an element \( w \in K \) such that \( w \notin {K}^{p}k\left( z\right) \) . Then \( {w}^{p} \in {K}^{p}k \), and \( D \) vanishes on \( {w}^{p} \) . We can then again apply Case 3 to extend \( D \) from \( {K}^{p}k\left( z\right) \) to \( {K}^{p}k\left( {z, w}\right) \) . Proceeding stepwise, we finally reach \( K \), thus proving our proposition.
|
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 separating transcendence base of \( K \) over \( k \) if and only if \( d{t}_{1},\ldots, d{t}_{r} \) form a basis of the dual space of \( \mathfrak{D} \) over \( K \) .
|
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 \( {w}_{i} = D{t}_{i} \) . Then clearly \( D = \sum {w}_{i}{D}_{i} \), and so the \( {D}_{i} \) form a basis for \( \mathfrak{D} \) over \( K \), and the \( d{t}_{i} \) form the dual basis. Conversely, if \( d{t}_{1},\ldots, d{t}_{r} \) is a basis for \( \mathcal{F} \) over \( K \), and if \( K \) is not separably generated over \( k\left( t\right) \), then by Cases 2 and 3 we can find a derivation \( D \) which is trivial on \( k\left( t\right) \) but nontrivial on \( K \) . If \( {D}_{1},\ldots ,{D}_{r} \) is the dual basis of \( d{t}_{1},\ldots, d{t}_{r} \) (so \( {D}_{i}{t}_{j} = {\delta }_{ij} \) ) then \( D,{D}_{1},\ldots ,{D}_{r} \) would be linearly independent over \( K \), contradicting the first part of the theorem.
|
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, it follows that \( K \) will be separable over \( k\left( z\right) \) .
|
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 all except possibly one value of \( c \in k, K \) is a separable extension of \( k\left( {y + {cz}}\right) \). Furthermore, \( {K}_{u} \) is separable over \( {k}_{u}\left( {y + {uz}}\right) \). (b) Assume that \( K \) is regular over \( k \), and that its transcendence degree is at least 2. Then for all but a finite number of elements \( c \in k, K \) is a regular extension of \( k\left( {y + {cz}}\right) \). Furthermore, \( {K}_{u} \) is regular over \( {k}_{u}\left( {y + {uz}}\right) \).
|
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 \( k\left( w\right) \), by Corollary 5.6. Also by Corollary 5.6, there exists \( D \) such that \( {Dz} \neq 0 \). Then for all elements \( c \in k \), except possibly one, we have \( D\left( {y + {cz}}\right) = {Dy} + {cDz} \neq 0 \). Also we may extend \( D \) to \( {K}_{u} \) over \( {k}_{u} \) by putting \( {Du} = 0 \), and then one sees that \( D\left( {y + {uz}}\right) = {Dy} + {uDz} \neq 0 \), so \( K \) is separable over \( k\left( {y + {cz}}\right) \) except possibly for one value of \( c \), and \( {K}_{u} \) is separable over \( {k}_{u}\left( {y + {uz}}\right) \). In what follows, we assume that the constants \( {c}_{1},{c}_{2},\ldots \) are different from the exceptional constant, and hence that \( K \) is separable over \( k\left( {y + {c}_{i}z}\right) \) for \( i = 1,2 \). Assume next that \( K \) is regular over \( k \) and that the transcendence degree is at least 2. Let \( {E}_{i} = k\left( {y + {c}_{i}z}\right) \left( {i = 1,2}\right) \) and let \( {E}_{i}^{\prime } \) be the algebraic closure of \( {E}_{i} \) in \( K \). We must show that \( {E}_{i}^{\prime } = {E}_{i} \) for all but a finite number of constants. Note that \( k\left( {y, z}\right) = {E}_{1}{E}_{2} \) is the compositum of \( {E}_{1} \) and \( {E}_{2} \), and that \( k\left( {y, z}\right) \) has transcendence degree 2 over \( k \). Hence \( {E}_{1}^{\prime } \) and \( {E}_{2}^{\prime } \) are free over \( k \). Being subfields of a regular extension of \( k \), they are regular over \( k \), and are therefore linearly disjoint by Theorem 4.12.
|
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\nand let \( {k}_{u} = k\left( {{u}_{1},\ldots ,{u}_{n},{u}_{n + 1}}\right) \) . Then \( {k}_{u}\left( x\right) \) is separable over \( {k}_{u} \), and if the\ntranscendence degree of \( k\left( x\right) \) over \( k \) is \( \geqq 2 \), then \( {k}_{u}\left( x\right) \) is regular over \( {k}_{u} \) .
|
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 Theorem 5.7 to conclude the proof.
|
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 \( \varphi \) to a homomorphism of \( k\left\lbrack x\right\rbrack \) into \( L \) .
|
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 \( {\sigma k}\left\lbrack {\sigma {x}_{1},\ldots ,\sigma {x}_{n}}\right\rbrack \) is a field, and is in fact an extension field of \( {\sigma k} \) . If we can prove our theorem when the finitely generated ring is in fact a field, then we apply \( \varphi \circ {\sigma }^{-1} \) on \( {\sigma k} \) and extend this to a homomorphism of \( {\sigma k}\left\lbrack {\sigma {x}_{1},\ldots ,\sigma {x}_{n}}\right\rbrack \) into \( L \) to get what we want.\n\nWithout loss of generality, we therefore assume that \( k\left\lbrack x\right\rbrack \) is a field. If it is algebraic over \( k \), we are done (by the known result for algebraic extensions). Otherwise, let \( {t}_{1},\ldots ,{t}_{r} \) be a transcendence basis, \( r \geqq 1 \) . Without loss of generality, we may assume that \( \varphi \) is the identity on \( k \) . Each element \( {x}_{1},\ldots ,{x}_{n} \) is algebraic over \( k\left( {{t}_{1},\ldots ,{t}_{r}}\right) \) . If we multiply the irreducible polynomial \( \operatorname{Irr}\left( {{x}_{i}, k\left( t\right), X}\right) \) by a suitable non-zero element of \( k\left\lbrack t\right\rbrack \), then we get a polynomial all of whose coefficients lie in \( k\left\lbrack t\right\rbrack \) . Let \( {a}_{1}\left( t\right) ,\ldots ,{a}_{n}\left( t\right) \) be the set of the leading coefficients of these polynomials, and let \( a\left( t\right) \) be their product,\n\n\[ a\left( t\right) = {a}_{1}\left( t\right) \cdots {a}_{n}\left( t\right) \]\n\nSince \( a\left( t\right) \neq 0 \), there exist elements \( {t}_{1}^{\prime },\ldots ,{t}_{r}^{\prime } \in {k}^{\mathrm{a}} \) such that \( a\left( {t}^{\prime }\right) \neq 0 \), and hence \( {a}_{i}\left( {t}^{\prime }\right) \neq 0 \) for any \( i \) . Each \( {x}_{i} \) is integral over the ring\n\n\[ k\left\lbrack {{t}_{1},\ldots ,{t}_{r},\frac{1}{{a}_{1}\left( t\right) },\ldots ,\frac{1}{{a}_{r}\left( t\right) }}\right\rbrack . \]\n\nConsider the homomorphism\n\n\[ \varphi : k\left\lbrack {{t}_{1},\ldots ,{t}_{r}}\right\rbrack \rightarrow {k}^{\mathrm{a}} \]\n\nsuch that \( \varphi \) is the identity on \( k \), and \( \varphi \left( {t}_{j}\right) = {t}_{j}^{\prime } \) . Let \( \mathfrak{p} \) be its kernel. Then \( a\left( t\right) \notin \mathfrak{p} \).\n\nOur homomorphism \( \varphi \) extends uniquely to the local ring \( k{\left\lbrack t\right\rbrack }_{\mathfrak{p}} \) and by the preceding remarks, it extends to a homomorphism of\n\n\[ k{\left\lbrack t\right\rbrack }_{\mathfrak{p}}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \]\n\ninto \( {k}^{\mathrm{a}} \), using Proposition 3.1 of Chapter VII. This proves what we wanted.
|
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\nover \( k \) such that \( \psi \left( {y}_{j}\right) \neq 0 \) for all \( j = 1,\ldots, m \) .
|
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 \( {X}_{1},\ldots ,{X}_{n}{\;\operatorname{mod}\;m} \) . By Corollary 2.2, this field is algebraic over \( k \), and can therefore be embedded in the algebraic closure \( {k}^{\mathrm{a}} \) . The homomorphism on \( k\left\lbrack X\right\rbrack \) obtained by the composition of the canonical map mod \( m \), followed by this embedded gives the desired zero of \( a \), and concludes the proof of the theorem.
|
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 \( \mathfrak{a} \) in \( {k}^{\mathfrak{a}} \) . Then there exists an integer \( m > 0 \) such that \( {f}^{m} \in \mathfrak{a} \) .
|
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 \( {\mathfrak{a}}^{\prime } \) must be the whole polynomial ring \( k\left\lbrack {X, Y}\right\rbrack \), so there exist polynomials \( {g}_{i} \in k\left\lbrack {X, Y}\right\rbrack \) and \( {h}_{i} \in \mathfrak{a} \) such that\n\n\[ 1 = {g}_{0}\left( {1 - {Yf}}\right) + {g}_{1}{h}_{1} + \cdots + {g}_{r}{h}_{r} \]\n\nWe substitute \( {f}^{-1} \) for \( Y \) and multiply by an appropriate power \( {f}^{m} \) of \( f \) to clear denominators on the right-hand side. This concludes the proof.
|
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 \( \left( {\mathfrak{a},\mathfrak{b}}\right) \) .
|
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 polynomial \( f \in \mathfrak{a} \) such that \( f\left( x\right) \neq 0 \) . But \( \mathfrak{a}\mathfrak{b} \subset \mathfrak{a} \cap \mathfrak{b} \) and hence \( \left( {fg}\right) \left( x\right) = 0 \) for all \( g \in \mathfrak{b} \), whence \( g\left( x\right) = 0 \) for all \( g \in \mathfrak{b} \) . Hence \( \left( x\right) \) lies in \( B \), and \( A \cup B \) is an algebraic set of zeros of \( \mathfrak{a} \cap \mathfrak{b} \) .\n\nTo prove that \( A \cap B \) is an algebraic set, let \( \left( x\right) \in A \cap B \) . Then \( \left( x\right) \) is a zero of \( \left( {\mathfrak{a},\mathfrak{b}}\right) \) . Conversely, let \( \left( x\right) \) be a zero of \( \left( {\mathfrak{a},\mathfrak{b}}\right) \) . Then obviously \( \left( x\right) \in A \cap B \), as desired. This proves our theorem.
|
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 unique.\n\n(iii) Let \( W,{V}_{1},\ldots ,{V}_{r} \) be irreducible algebraic sets such that\n\n\[ W \subset {V}_{1} \cup \ldots \cup {V}_{r} \]\n\nThen \( W \subset {V}_{i} \) for some \( i \) .
|
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 \( B \neq V \) . Since each one of \( A, B \) is strictly smaller than \( V \), we can express \( A, B \) as finite unions of irreducible algebraic sets, and thus get an expression for \( V \), contradiction.\n\nThe uniqueness will follow from (iii), which we prove next. Let \( W \) be contained in the union \( {V}_{1} \cup \ldots \cup {V}_{r} \) . Then\n\n\[ W = \left( {W \cap {V}_{1}}\right) \cup \ldots \cup \left( {W \cap {V}_{r}}\right) \]\n\nSince each \( W \cap {V}_{i} \) is an algebraic set, by the irreducibility of \( W \) we must have \( W = W \cap {V}_{i} \) for some \( i \) . Hence \( W \subset {V}_{i} \) for some \( i \), thus proving (iii).\n\nNow to prove (ii), apply (iii) to each \( {W}_{j} \) . Then for each \( j \) there is some \( i \) such that \( {W}_{j} \subset {V}_{i} \) . Similarly for each \( i \) there exists \( \nu \) such that \( {V}_{i} \subset {W}_{\nu } \) . Since there is no inclusion relation among the \( {W}_{j} \) ’s, we must have \( {W}_{j} = {V}_{i} = {W}_{\nu } \) . This proves that each \( {W}_{j} \) appears among the \( {V}_{i} \) ’s and each \( {V}_{i} \) appears among the \( {W}_{j} \) ’s, and proves the uniqueness of the representation. It also concludes the proof of Theorem 2.2.
|
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 on \( \left( w\right) \) is principal.
|
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 \( R\left\lbrack W\right\rbrack \) . Let \( G\left( W\right) \in R\left\lbrack W\right\rbrack \) vanish on \( \left( w\right) \) . To prove that \( P \) divides \( G \), after selecting some irreducible factor of \( G \) vanishing on \( \left( w\right) \) if necessary, we may assume without loss of generality that \( G \) is a prime element in \( R\left\lbrack W\right\rbrack \) . One of the variables \( {W}_{i} \) occurs in \( P\left( W\right) \), say \( {W}_{m} \), so that \( {w}_{m} \) is algebraic over \( k\left( {{w}_{1},\ldots ,{w}_{m - 1}}\right) \) . Then \( \left( {{w}_{1},\ldots ,{w}_{m - 1}}\right) \) are algebraically independent, and hence \( {W}_{m} \) also occurs in \( G \) . Furthermore, \( P\left( {{w}_{1},\ldots ,{w}_{m - 1},{W}_{m}}\right) \) is irreducible as a polynomial in \( k\left( {{w}_{1},\ldots ,{w}_{m - 1}}\right) \left\lbrack {W}_{m}\right\rbrack \) by the Gauss lemma as in Chapter IV, Theorem 2.3. Hence there exists a polynomial \( H\left( {W}_{m}\right) \in k\left( {{w}_{1},\ldots ,{w}_{m - 1}}\right) \left\lbrack {W}_{m}\right\rbrack \) such that\n\n\[ G\left( W\right) = H\left( {W}_{m}\right) P\left( W\right) \]\n\nLet \( {R}^{\prime } = R\left\lbrack {{w}_{1},\ldots ,{w}_{m - 1}}\right\rbrack \) . Then \( P, G \) have content 1 as polynomials in \( {R}^{\prime }\left\lbrack {W}_{m}\right\rbrack \) . By Chapter IV Corollary 2.2 we conclude that \( H \in {R}^{\prime }\left\lbrack {W}_{m}\right\rbrack \approx R\left\lbrack W\right\rbrack \) , which proves Theorem 2.4.
|
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 \( i, d \) ) form a set of homogeneous generators. Conversely, let \( f \) be a homogeneous element in \( \mathfrak{a} \) and let \( g \in K\left\lbrack X\right\rbrack \) be arbitrary. For each \( d,{g}^{\left( d\right) }f \) lies in \( \mathfrak{a} \), and \( {g}^{\left( d\right) }f \) is homogeneous, so all the homogeneous components of \( {gf} \) also lie in a. Applying this remark to the case when \( f \) ranges over a set of homogeneous generators for \( \mathfrak{a} \) shows that \( \mathfrak{a} \) is homogeneous, and concludes the proof of the proposition.
|
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 \( i = 2,\ldots, r \) . By hypothesis, for \( t \) transcendental over \( k\left( x\right) \) it follows that \( \left( {tx}\right) \in \mathcal{Z} \) so \( \left( {tx}\right) \in {V}_{i} \) for some \( i \) . Specializing to \( t = 1 \), we conclude that \( \left( x\right) \in {V}_{i} \), so \( i = 1 \), which proves that \( {V}_{1} \) is homogeneous, as was to be shown.
|
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. \( \mathbf{Z}\left\lbrack {W, X}\right\rbrack \) ) and assume \( \mathfrak{p} \) is homogeneous in \( \left( X\right) \) . Let \( V \) be the corresponding irreducible algebraic space in \( {\mathbf{A}}^{m} \times {\mathbf{P}}^{n - 1} \) . Let \( {\mathfrak{p}}_{1} = \mathfrak{p} \cap k\left\lbrack W\right\rbrack \) (resp. \( \mathfrak{p} \cap \mathbf{Z}\left\lbrack W\right\rbrack \) ), and let \( {V}_{1} \) be the projection of \( V \) on the first factor. Then \( {V}_{1} \) is the algebraic space of zeros of \( {\mathfrak{p}}_{1} \) in \( {\mathbf{A}}^{m} \) .
|
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 \( \left( {x}^{\prime }\right) \) are equal to 0 . By assumption, not all the coordinates of ( \( x \) ) are equal to 0, since we viewed \( V \) as a subset of \( {\mathbf{A}}^{m} \times {\mathbf{P}}^{n - 1} \) . For definiteness, say we are dealing with the case of a field \( k \) . By Chapter VII, Proposition 3.3, the homomorphism \( k\left\lbrack w\right\rbrack \rightarrow k\left\lbrack {w}^{\prime }\right\rbrack \) can be extended to a place \( \varphi \) of \( k\left( {w, x}\right) \) . By Proposition 3.4 of Chapter VII, there is some coordinate \( {x}_{j} \) such that \( \varphi \left( {{x}_{i}/{x}_{j}}\right) \neq \infty \) for all \( i = 1,\ldots, n \) . We let \( {x}_{i}^{\prime } = \varphi \left( {{x}_{i}/{x}_{j}}\right) \) for all \( i \) to conclude the proof. The proof is similar when dealing with algebraic spaces over \( \mathbf{Z} \), replacing \( k \) by \( \mathbf{Z} \) .
|
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\[ \n{F}_{i}\left( {W, X}\right) = \sum {W}_{i,\left( \nu \right) }{M}_{\left( \nu \right) }\left( X\right) \n\] \n\nwhere \( {M}_{\left( \nu \right) }\left( X\right) \) ranges over all monomials in \( \left( X\right) \) of degree \( {d}_{i} \), so \( \left( W\right) = {\left( W\right) }_{F} \) . We call \( {F}_{1},\ldots ,{F}_{r} \) generic forms. Let \n\n\[ \n\mathfrak{a} = \text{ideal in}\mathbf{Z}\left\lbrack {W, X}\right\rbrack \text{generated by}{F}_{1},\ldots ,{F}_{r}\text{.} \n\] \n\nThen \( \mathfrak{a} \) is homogeneous in \( \left( X\right) \) . Thus we are in the situation of Theorem 3.1, with a defining an algebraic space \( \mathbf{Q} \) in \( {\mathbf{A}}^{m} \times {\mathbf{P}}^{n - 1} \) . Note that \( \left( w\right) \) is a specialization of \( \left( W\right) \), or, as we also say, \( \left( f\right) \) is a specialization of \( \left( F\right) \) . As in Theorem 3.1, let \( {\mathbf{Q}}_{1} \) be the projection of \( \mathbf{Q} \) on the first factor. Then directly from the definitions, \( \left( f\right) \) has a non-trivial zero if and only if \( {\left( w\right) }_{f} \) lies in \( {\mathbf{Q}}_{1} \), so Theorem 3.2 is a special case of Theorem 3.1.
|
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 transcendence degree of \( k\left( w\right) \) over \( k \) is \( m - 1 \) .
|
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 \( {z}_{i} = - {F}_{i}^{ * }\left( {y, u}\right) \), so\n\n\[ k\left( w\right) = k\left( {y, z}\right) \subset k\left( {y, u}\right) ,\]\n\nand so the transcendence degree of \( k\left( w\right) \) over \( k \) is \( \leqq m - 1 \) . We claim that this transcendence degree is \( m - 1 \) . It will suffice to prove that \( {u}_{1},\ldots ,{u}_{n - 1} \) are algebraic over \( k\left( w\right) = k\left( {y, z}\right) \) . Suppose this is not the case. Then there exists a place \( \varphi \) of \( k\left( {w, u}\right) \), which is the identity on \( k\left( w\right) \) and maps some \( {u}_{j} \) on \( \infty \) . Select an index \( q \) such that \( \varphi \left( {{u}_{i}/{u}_{q}}\right) \) is finite for all \( i = 1,\ldots, n - 1 \) . Let \( {v}_{i} = {u}_{i}/{u}_{q} \) and \( {v}_{i}^{\prime } = \varphi \left( {{u}_{i}/{u}_{q}}\right) \) . Denote by \( {Y}_{iq} \) the coefficient of \( {X}_{q}^{{d}_{i}} \) in \( {F}_{i} \) and let \( {Y}^{ * } \) denote the variables \( \left( Y\right) \) from which \( {Y}_{1q},\ldots ,{Y}_{nq} \) are deleted. By (3) we have for \( i = 1,\ldots, n \) :\n\n\[ 0 = {y}_{iq}{u}_{q}^{{d}_{i}} + {z}_{i} + {F}_{i}^{* * }\left( {{y}^{ * }, u}\right) \]\n\n\[ = {y}_{iq} + {z}_{i}/{u}_{q}^{{d}_{i}} + {F}_{i}^{* * }\left( {{y}^{ * }, u/{u}_{q}}\right) .\n\nApplying the place yields\n\n\[ 0 = {y}_{iq} + {F}_{i}^{* * }\left( {{y}^{ * },{v}^{\prime }}\right) \]\n\nIn particular, \( {y}_{iq} \in k\left( {{y}^{ * },{v}^{\prime }}\right) \) for each \( i = 1,\ldots, n \) . But the transcendence degree of \( k\left( {v}^{\prime }\right) \) over \( k \) is at most \( n - 1 \), while the elements \( \left( {{y}_{1q},\ldots ,{y}_{nq},{y}^{ * }}\right) \) are algebraically independent over \( k \), which gives a contradiction proving the theorem.
|
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\[ \left( \alpha \right) = \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \;\text{ and }\;\left( \beta \right) = \left( {{\beta }_{1},\ldots ,{\beta }_{n}}\right) \]\n\nwith \( \left| \alpha \right| = \left| \beta \right| = {d}_{i} \), we have\n\n\[ {X}_{n}^{s}\left( {{M}_{\left( \alpha \right) }\left( X\right) \frac{\partial R}{\partial {W}_{i,\left( \beta \right) }} - {M}_{\left( \beta \right) }\left( X\right) \frac{\partial R}{\partial {W}_{i,\left( \alpha \right) }}}\right) \equiv 0{\;\operatorname{mod}\;\left( {{F}_{1},\ldots ,{F}_{n}}\right) }.\]
|
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\[ {X}_{n}^{s}\frac{\partial R}{\partial {W}_{i,\left( \beta \right) }} \equiv {Q}_{i}{M}_{\left( \beta \right) }\left( X\right) {\;\operatorname{mod}\;\left( {{F}_{1},\ldots ,{F}_{n}}\right) }, \]\n\nand similarly\n\n\[ {X}_{n}^{s}\frac{\partial R}{\partial {W}_{i,\left( \alpha \right) }} \equiv {Q}_{i}{M}_{\left( \alpha \right) }\left( X\right) {\;\operatorname{mod}\;\left( {{F}_{1},\ldots ,{F}_{n}}\right) }. \]\n\nWe multiply the first congruence by \( {M}_{\left( \alpha \right) }\left( X\right) \) and the second by \( {M}_{\left( \beta \right) }\left( X\right) \), and we subtract to get our lemma.
|
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 \( G, H,{F}_{2},\ldots ,{F}_{n} \) respectively, and let \( \left( w\right) \) be the coefficients of \( {GH},{F}_{2},\ldots ,{F}_{n} \). Then \[ R\left( w\right) = R\left( {{GH},{F}_{2},\ldots ,{F}_{n}}\right) , \] and we obtain \[ {X}_{n}^{s}R\left( w\right) = {Q}_{1}\left( {w, X}\right) {GH} + {Q}_{2}\left( {w, X}\right) {F}_{2} + {Q}_{n}\left( {w, X}\right) {F}_{n}. \] Hence \( R\left( {{GH},{F}_{2},\ldots ,{F}_{n}}\right) \) belongs to the elimination ideal of \( G,{F}_{2},\ldots ,{F}_{n} \) in the ring \( \mathbf{Z}\left\lbrack {{W}_{G},{W}_{H},{W}_{{F}_{2}},\ldots ,{W}_{{F}_{n}}}\right\rbrack \), and similarly with \( H \) instead of \( G \). Since \( {W}_{H} \) is a family of independent variables over \( \mathbf{Z}\left\lbrack {{W}_{G},{W}_{{F}_{2}},\ldots ,{W}_{{F}_{n}}}\right\rbrack \), it follows that \( R\left( {G,{F}_{2},\ldots ,{F}_{n}}\right) \) divides \( R\left( {{GH},{F}_{2},\ldots ,{F}_{n}}\right) \) in that ring, and similarly for \( R\left( {H,{F}_{2},\ldots ,{F}_{n}}\right) \). But \( \left( {W}_{G}\right) \) and \( \left( {W}_{H}\right) \) are independent sets of variables, and so \( R\left( {G,{F}_{2},\ldots ,{F}_{n}}\right), R\left( {H,{F}_{2},\ldots ,{F}_{n}}\right) \) are distinct prime elements in that ring, so their product divides \( R\left( {{GH},{F}_{2},\ldots ,{F}_{n}}\right) \) as stated, thus proving the lemma.
|
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\[ \n\operatorname{Res}\left( {{gh},{f}_{2},\ldots ,{f}_{n}}\right) = \operatorname{Res}\left( {g,{f}_{2},\ldots ,{f}_{n}}\right) \operatorname{Res}\left( {h,{f}_{2},\ldots ,{f}_{n}}\right) .\n\]
|
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 \( n \geqq 2 \) . Then \[ \operatorname{Res}\left( {{F}_{1},\ldots ,{F}_{n - 1},{X}_{n}^{{d}_{n}}}\right) = \operatorname{Res}{\left( {\bar{F}}_{1},\ldots ,{\bar{F}}_{n - 1}\right) }^{{d}_{n}}. \]
|
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 \( {Q}_{j} \in \mathbf{Z}\left\lbrack {W, X}\right\rbrack \) (depending on the choice of \( i \) ). The left-hand side can be written as a polynomial in the coefficients of \( {F}_{1},\ldots ,{F}_{n - 1} \) with the notation \[ {X}_{i}^{s}R\left( {{W}_{{F}_{1}},\ldots ,{W}_{{F}_{n - 1}},{1}_{{X}_{n}}}\right) = {X}_{i}^{s}P\left( {{W}_{{F}_{1}},\ldots ,{W}_{{F}_{n - 1}}}\right) = {X}_{i}^{s}P\left( {W}^{\left( n - 1\right) }\right) \text{, say;} \] thus in the generic linear form in \( {X}_{1},\ldots ,{X}_{n} \) we have specialized all the coefficients to 0 except the coefficient of \( {X}_{n} \), which we have specialized to 1 . Substitute \( {X}_{n} = 0 \) in the right side of (*). By Theorem 3.4, we conclude that \( P\left( {W}^{\left( n - 1\right) }\right) \) lies in the resultant ideal of \( {\bar{F}}_{1},\ldots ,{\bar{F}}_{n - 1} \), and therefore \( \operatorname{Res}\left( {{\bar{F}}_{1},\ldots ,{\bar{F}}_{n - 1}}\right) \) divides \( P\left( {W}^{\left( n - 1\right) }\right) \) . By Theorem 3.8 we know that \( P\left( {W}^{\left( n - 1\right) }\right) \) has the same homogeneity degree in \( {W}_{{\bar{F}}_{i}}\left( {i = 1,\ldots, n - 1}\right) \) as \( \operatorname{Res}\left( {{\bar{F}}_{1},\ldots ,{\bar{F}}_{n - 1}}\right) \) . Hence there is \( c \in \mathbf{Z} \) such that \[ c\operatorname{Res}\left( {{\bar{F}}_{1},\ldots ,{\bar{F}}_{n - 1}}\right) = \operatorname{Res}\left( {{F}_{1},\ldots ,{F}_{n - 1},{X}_{n}}\right) . \] One finds \( c = 1 \) by specializing \( {\bar{F}}_{1},\ldots ,{\bar{F}}_{n - 1} \) to \( {X}_{1}^{{d}_{1}},\ldots ,{X}_{n - 1}^{{d}_{n - 1}} \) respectively, thus concluding the proof.
|
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 in \( A\left\lbrack X\right\rbrack \) . Then\n\n\[ \operatorname{Res}\left( {{f}_{1},\ldots ,{f}_{n}}\right) \text{divides}\operatorname{Res}\left( {{g}_{1},\ldots ,{g}_{n}}\right) \text{in}A\text{.} \]
|
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 Theorem 3.4, for each \( i \) we have a relation\n\n\[ {X}_{i}^{s}\operatorname{Res}\left( {{g}_{1},\ldots ,{g}_{n}}\right) = {Q}_{1}{g}_{1} + \cdots + {Q}_{n}{g}_{n}\text{with some}{Q}_{i} \in \mathbf{Z}\left\lbrack {{W}_{H},{W}_{F}}\right\rbrack \text{,} \]\n\n where \( {W}_{H},{W}_{F} \) denote the independent variable coefficients of the polynomials \( {H}_{ij} \) and \( {F}_{j} \) respectively. In particular,\n\n(*)\n\n\[ {X}_{i}^{s}\operatorname{Res}\left( {{g}_{1},\ldots ,{g}_{n}}\right) \equiv 0{\;\operatorname{mod}\;\left( {{F}_{1},\ldots ,{F}_{n}}\right) }\mathbf{Z}\left\lbrack {{W}_{H},{W}_{F}, X}\right\rbrack . \]\n\n Note that \( \operatorname{Res}\left( {{g}_{1},\ldots ,{g}_{n}}\right) = P\left( {{W}_{H},{W}_{F}}\right) \in \mathbf{Z}\left\lbrack {{W}_{H},{W}_{F}}\right\rbrack \) is a polynomial with integer coefficients. If \( \left( {w}_{F}\right) \) is a generic point of the resultant variety \( {\mathbf{Q}}_{1} \) over \( \mathbf{Z} \), then \( P\left( {{W}_{H},{w}_{F}}\right) = 0 \) by \( \left( *\right) \) . Hence \( \operatorname{Res}\left( {{F}_{1},\ldots ,{F}_{n}}\right) \) divides \( P\left( {{W}_{H},{W}_{F}}\right) \), thus proving the lemma.
|
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\[ \n\operatorname{Res}\left( {{f}_{1},\ldots ,{f}_{n - 1},{f}_{n} + \mathop{\sum }\limits_{{j = 1}}^{{n - 1}}{h}_{j}{f}_{j}}\right) = \operatorname{Res}\left( {{f}_{1},\ldots ,{f}_{n}}\right) \text{ in }A.\n\]
|
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 respective polynomials. We note that the ideals \( \left( {{F}_{1},\ldots ,{F}_{n}}\right) \) and \( \left( {{F}_{1},\ldots ,{F}_{n} + \mathop{\sum }\limits_{{\mathrm{j} \neq n}}{H}_{j}{F}_{j}}\right) \) are equal. From Lemma 3.12 we conclude that the two resultants in the statement of the theorem differ by a factor of 1 or -1 . We may now specialize \( {H}_{ij} \) to 0 to determine that the factor is +1, thus concluding the proof.
|
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\[ \operatorname{Res}\left( {{F}_{\pi \left( 1\right) },\ldots ,{F}_{\pi \left( n\right) }}\right) = \varepsilon {\left( \pi \right) }^{{d}_{1}\ldots {d}_{n}}\operatorname{Res}\left( {{F}_{1},\ldots ,{F}_{n}}\right) . \]
|
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 determine this sign by specializing \( {F}_{i} \) to \( {X}_{i}^{{d}_{i}} \), and using the multiplicativity of Theorem 3.10. We are then reduced to the case when \( {F}_{i} = {X}_{i} \), so a linear form; and we can apply Corollary 3.14 to conclude the proof.
|
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 \( j \) -th minor determinant of the coefficient matrix of the forms \( \left( {{L}_{1},\ldots ,{L}_{n - 1}}\right) \) . Then\n\n\[ \n\operatorname{Res}\left( {{L}_{1},\ldots ,{L}_{n - 1}, F}\right) = F\left( {{\Delta }_{1},\ldots ,{\Delta }_{n}}\right) .\n\]
|
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 coefficients of the forms \( {L}_{1},\ldots ,{L}_{n - 1}, F \) . To see this, we consider the system of linear equations\n\n\[ \n{W}_{11}{X}_{1} + \cdots + {W}_{1, n - 1}{X}_{n - 1} = {L}_{1}\left( {W, X}\right) - {W}_{1, n}{X}_{n} \n\]\n\n\[ \n{W}_{n - 1,1}{X}_{1} + \cdots + {W}_{n - 1, n - 1}{X}_{n - 1} = {L}_{n - 1}\left( {W, X}\right) - {W}_{n - 1, n}{X}_{n}. \n\]\n\nIf \( C = \left( {{C}^{1},\ldots ,{C}^{n - 1}}\right) \) is a square matrix with columns \( {\mathrm{C}}^{j} \), then a solution of a system of linear equations \( {CX} = {C}^{n} \) satisfies Cramer’s rule\n\n\[ \n{X}_{j}\det \left( {{C}^{1},\ldots ,{C}^{n - 1}}\right) = \det \left( {{C}^{1},\ldots ,{C}^{n},\ldots ,{C}^{n - 1}}\right) . \n\]\n\nUsing the fact that the determinant is linear in each column, (*) falls out.\n\nThen from the congruence (*) it follows that\n\n\[ \n{X}_{n}^{d}F\left( {{\Delta }_{1},\ldots ,{\Delta }_{n}}\right) \equiv {\Delta }_{n}^{d}F\left( {{X}_{1},\ldots ,{X}_{n}}\right) {\;\operatorname{mod}\;\left( {{L}_{1},\ldots ,{L}_{n - 1}}\right) }\mathbf{Z}\left\lbrack {W, X}\right\rbrack , \n\]\n\nwhence\n\n\[ \n{X}_{n}^{d}F\left( {{\Delta }_{1},\ldots ,{\Delta }_{n}}\right) \equiv 0{\;\operatorname{mod}\;\left( {{L}_{1},\ldots ,{L}_{n - 1}, F}\right) }. \n\]\n\nHence by Theorem 3.4 and the fact that \( \operatorname{Res}\left( {{L}_{1},\ldots ,{L}_{n - 1}, F}\right) = R\left( W\right) \) generates the elimination ideal, it follows that there exists \( c \in \mathbf{Z}\left\lbrack W\right\rbrack \) such that\n\n\[ \nF\left( {{\Delta }_{1},\ldots ,{\Delta }_{n}}\right) = c\operatorname{Res}\left( {{L}_{1},\ldots ,{L}_{n - 1}, F}\right) . \n\]\n\nSince the left side is homogeneous of degree 1 in the coefficients \( {W}_{F} \) and homogeneous of degree \( d \) in the coefficients \( {W}_{{L}_{i}} \) for each \( i = 1,\ldots, n - 1 \), it follows from Theorem 3.8 that \( c \in \mathbf{Z} \) . Specializing \( {L}_{i} \) to \( {X}_{i} \) and \( F \) to \( {X}_{n}^{d} \) makes \( {\Delta }_{i} \) specialize to 0 if \( j \neq n \) and \( {\Delta }_{n} \) specializes to 1 . Hence the left side specializes to 1, and so does the right side, whence \( c = 1 \) . This concludes the proof.
|
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 determinable finite number of polynomials \( {R}_{\rho }\left( W\right) \in \mathbf{Z}\left\lbrack W\right\rbrack \) having the following property. Let \( \left( f\right) \) be as in (1), a system of forms of the given degrees with coefficients (w) in some field \( k \) . Then \( \left( f\right) \) has a non-trivial common zero if and only if \( {R}_{\rho }\left( w\right) = 0 \) for all \( \rho \) .
|
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 a paper of Kapferer (Über Resultanten und Resultantensysteme, Sitzungsber. Bayer. Akad. München 1929, pp. 179-200). The family of polynomials \( \left\{ {{R}_{\rho }\left( W\right) }\right\} \) is called a resultant system, because of the way they are constructed. They form a set of generators for an ideal \( {b}_{1} \) such that the arithmetic variety \( {Q}_{1} \) is the set of zeros of \( {b}_{1} \) . I don’t know how close the system constructed below is to being a set of generators for the prime ideal \( {\mathfrak{p}}_{1} \) in \( \mathbf{Z}\left\lbrack W\right\rbrack \) associated with \( {\mathbf{Q}}_{1} \) . Actually we shall not need the whole theory of Chapter IV, \( \$ {10} \) ; we need only one of the characterizing properties of resultants.
|
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 to be finite).
|
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 \( \left( {{\alpha }_{i} - {\beta }_{j}}\right) \) the proposition follows at once.
|
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 \( {Q}_{\mu }\left( w\right) = 0 \) for all \( \mu \) . Conversely, suppose that \( {Q}_{\mu }\left( w\right) = 0 \) for all \( \mu \) . Let \( {f}_{i}\left( X\right) = {F}_{i}\left( {w, X}\right) \) . We want to show that \( {f}_{i}\left( X\right) \) for \( i = 1,\ldots, r \) have a common non-trivial zero in some extension of\n\n\( k \) . If all \( {f}_{i} \) are 0 in \( k\left\lbrack {{X}_{1},{X}_{2}}\right\rbrack \) then they have a common non-trivial zero. If, say, \( {f}_{1} \neq 0 \) in \( k\left\lbrack X\right\rbrack \), then specializing \( {T}_{2},\ldots ,{T}_{r} \) to 0 and \( {T}_{1} \) to 1 in the resultant \( \operatorname{Res}\left( {f, g}\right) \), we see that\n\n\[ \operatorname{Res}\left( {{f}_{1},{f}_{2}{U}_{2} + \cdots + {f}_{r}{U}_{r}}\right) = 0 \]\n\n\nas a polynomial in \( k\left\lbrack {{U}_{2},\ldots ,{U}_{r}}\right\rbrack \) . After making a finite extension of \( k \) if necessary, we may assume that \( {f}_{1}\left( X\right) \) splits into linear factors. Let \( \left\{ {\alpha }_{i}\right\} \) be the roots of \( {f}_{1}\left( {{X}_{1},1}\right) \) . Then some \( \left( {{\alpha }_{i},1}\right) \) must also be a zero of \( {f}_{2}{U}_{2} + \cdots + {f}_{r}{U}_{r} \) , which implies that \( \left( {{\alpha }_{i},1}\right) \) is a common zero of \( {f}_{1},\ldots ,{f}_{r} \) since \( {U}_{2},\ldots ,{U}_{r} \) are algebraically independent over \( k \) . This proves Proposition 4.3.
|
Yes
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.