Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Example 2. Suppose \( \mathrm{H} \) is a degree \( \mathrm{d} \) hypersurface in \( {\mathbb{P}}^{n} \) . Then from the closed subscheme exact sequence\n\n\[ 0 \rightarrow {\mathcal{O}}_{{\mathbb{P}}^{n}}\left( {-\mathrm{d}}\right) \rightarrow {\mathcal{O}}_{{\mathbb{P}}^{n}} \rightarrow {\mathcal{O}}_{\mathrm{H}} \rig... | (Implicit in this argument is the fact that if \( i : H \hookrightarrow {\mathbb{P}}^{n} \) is the closed embedding, then \( \left( {{i}_{ * }{\mathcal{O}}_{\mathbb{H}}}\right) \otimes {\mathcal{O}}_{{\mathbb{P}}^{n}}\left( m\right) \cong {i}_{ * }\left( {{\mathcal{O}}_{\mathbb{H}} \otimes {i}^{ * }{\mathcal{O}}_{{\mat... | No |
Proof of the Semicontinuity Theorem 28.1.1. The result is local on \( Y \), so we may assume \( Y \) is affine. Let \( {K}^{ \bullet } \) be a complex as in Key Theorem [28,2,1]. Then for \( q \in Y \), | \[ \begin{array}{lll} {\dim }_{\kappa \left( q\right) }{H}^{p}\left( {{X}_{q},\mathcal{F}{|}_{{X}_{q}}}\right) & = & {\dim }_{\kappa \left( q\right) }\ker \left( {{\delta }^{p}{ \otimes }_{B}\kappa \left( q\right) }\right) - {\dim }_{\kappa \left( q\right) }\operatorname{im}\left( {{\delta }^{p - 1}{ \otimes }_{B}\kapp... | Yes |
(a) If \( A \) is an integral domain and algebraic over \( K \), then \( A \) is a field. | Proof. (a) We need to show that every \( a \in A \smallsetminus \{ 0\} \) is invertible in \( A \) . For this, it suffices to show that \( K\left\lbrack a\right\rbrack \) is a field. We may therefore assume that \( A = K\left\lbrack a\right\rbrack \) . With \( x \) an indeterminate, let \( I \subseteq K\left\lbrack x\r... | Yes |
Proposition 1.2 (Preimages of maximal ideals). Let \( \varphi : A \rightarrow B \) be a homomorphism of algebras over a field \( K \), and let \( \mathfrak{m} \subset B \) be a maximal ideal. If \( B \) is finitely generated, then the preimage \( {\varphi }^{-1}\left( \mathfrak{m}\right) \subseteq A \) is also a maxima... | Proof. The map \( A \rightarrow B/\mathfrak{m}, a \mapsto \varphi \left( a\right) + \mathfrak{m} \), has kernel \( {\varphi }^{-1}\left( \mathfrak{m}\right) = : \mathfrak{n} \) . So \( A/\mathfrak{n} \) is isomorphic to a subalgebra of \( B/\mathfrak{m} \) . By Lemma 1.1(b), \( B/\mathfrak{m} \) is algebraic over \( K ... | Yes |
Lemma 1.4. Let \( K \) be a field and \( P = \left( {{\xi }_{1},\ldots ,{\xi }_{n}}\right) \in {K}^{n} \) a point in \( {K}^{n} \) . Then the ideal\n\n\[ \n{\mathfrak{m}}_{P} \mathrel{\text{:=}} \left( {{x}_{1} - {\xi }_{1},\ldots ,{x}_{n} - {\xi }_{n}}\right) \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbr... | Proof. It is clear from the definition of \( {\mathfrak{m}}_{P} \) that every polynomial \( f \in \) \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) is congruent to \( f\left( {{\xi }_{1},\ldots ,{\xi }_{n}}\right) \) modulo \( {\mathfrak{m}}_{P} \) . It follows that \( {\mathfrak{m}}_{P} \) is the kernel of... | Yes |
Proposition 1.5 (Maximal ideals in a polynomial ring). Let \( K \) be an algebraically closed field, and let \( \mathfrak{m} \subset K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be a maximal ideal in a polynomial ring over \( K \) . Then there exists a point \( P = \left( {{\xi }_{1},\ldots ,{\xi }_{n}}\righ... | Proof. By Proposition 1.2, the intersection \( K\left\lbrack {x}_{i}\right\rbrack \cap \mathfrak{m} \) is a maximal ideal in \( K\left\lbrack {x}_{i}\right\rbrack \) for each \( i = 1,\ldots, n \) . Since \( K\left\lbrack {x}_{i}\right\rbrack \) is a principal ideal domain, \( K\left\lbrack {x}_{i}\right\rbrack \cap \m... | Yes |
Theorem 1.7 (Correspondence points-maximal ideals). Let \( K \) be an algebraically closed field and \( S \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) a set of polynomials. Let \( {\mathcal{M}}_{S} \) be the set of all maximal ideals \( \mathfrak{m} \subset K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\r... | Proof. Let \( P \mathrel{\text{:=}} \left( {{\xi }_{1},\ldots ,{\xi }_{n}}\right) \in \mathcal{V}\left( S\right) \) . Then \( \Phi \left( P\right) \) is a maximal ideal by Lemma 1.4. All \( f \in S \) satisfy \( f\left( P\right) = 0 \), so \( f \in \Phi \left( P\right) \) . It follows that \( \Phi \left( P\right) \in {... | Yes |
Corollary 1.8 (Hilbert’s Nullstellensatz, first version). Let \( K \) be an algebraically closed field and let \( I \subsetneqq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be a proper ideal in a polynomial ring. Then \[ \mathcal{V}\left( I\right) \neq \varnothing \text{.} \] | Proof. Consider the set of all proper ideals \( J \subsetneqq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) containing \( I \) . Using Zorn's lemma, we conclude that this set contains a maximal element \( \mathfrak{m} \) . (Instead of Zorn’s lemma, we could also use the fact that \( K\left\lbrack {{x}_{1},\ld... | Yes |
Lemma 1.10. Let \( R \) be a ring, \( I \subseteq R \) an ideal, and \( \mathcal{M} \subseteq \operatorname{Spec}\left( R\right) \) a subset. Then \n\nIf there exist no \( P \in \mathcal{M} \) with \( I \subseteq P \),... | Proof. Let \( a \in \sqrt{I} \), so \( {a}^{k} \in I \) for some \( k \) . Let \( P \in \mathcal{M} \) with \( I \subseteq P \) . Then \( {a}^{k} \in P \) . Since \( P \) is a prime ideal, it follows that \( a \in P \) . | Yes |
Proposition 1.11 (The raison d’être of the Rabinowitsch spectrum). Let \( I \) \( \subseteq R \) be an ideal in a ring. Then \n\nIf there exist no \( P \in {\operatorname{Spec}}_{\mathrm{{rab}}}\left( R\right) \) with ... | Proof. The inclusion \ | No |
Corollary 1.12 (Intersecting prime ideals). Let \( R \) be a ring and \( I \subseteq R \) an ideal. Then \n\nIf there exist no \( P \in \operatorname{Spec}\left( R\right) \) with \( I \subseteq P \), the intersection i... | Proof. This follows from Lemma 1.10 and Proposition 1.11. | No |
Theorem 1.13 (Intersecting maximal ideals). Let \( A \) be an affine algebra and \( I \subseteq A \) an ideal. Then\n\n\[ \sqrt{I} = \mathop{\bigcap }\limits_{\substack{{\mathfrak{m} \in {\operatorname{Spec}}_{\max }\left( A\right) ,} \\ {I \subseteq \mathfrak{m}} }}\mathfrak{m} \]\n\nIf there exist no \( \mathfrak{m} ... | Proof. The inclusion \ | No |
Theorem 1.17 (Hilbert’s Nullstellensatz, second version). Let \( K \) be an algebraically closed field and let \( I \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an ideal in a polynomial ring. Then \[ \mathcal{I}\left( {\mathcal{V}\left( I\right) }\right) = \sqrt{I} \] | Proof. We start by showing the inclusion \ | No |
Lemma 1.18. Let \( K \) be a field and \( X \subseteq {K}^{n} \) an affine variety. Then\n\n\[ \mathcal{V}\left( {\mathcal{I}\left( X\right) }\right) = X \] | Proof. By assumption, \( X = \mathcal{V}\left( S\right) \) with \( S \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . So \( S \subseteq \mathcal{I}\left( X\right) \), and applying \( \mathcal{V} \) yields\n\n\[ \mathcal{V}\left( {\mathcal{I}\left( X\right) }\right) \subseteq \mathcal{V}\left( S\right... | Yes |
Lemma 1.22 (Ideals in quotient rings). Let \( R \) be a ring and let \( I \subseteq R \) be an ideal. Consider the sets\n\n\[ \mathcal{A} \mathrel{\text{:=}} \{ J \subseteq R \mid J\text{ is an ideal and }I \subseteq J\} \]\n\nand\n\n\[ \mathcal{B} \mathrel{\text{:=}} \{ \mathcal{J} \subseteq R/I \mid \mathcal{J}\text{... | Proof. It is easy to check that \( \Phi \) and \( \Psi \) are inclusion-preserving maps and that \( \Psi \circ \Phi = {\operatorname{id}}_{\mathcal{A}} \) and \( \Phi \circ \Psi = {\operatorname{id}}_{\mathcal{B}} \) . The isomorphism (1.4) follows since \( \Phi \left( J\right) \) is the kernel of the epimorphism \( R/... | Yes |
Theorem 1.23 (Correspondence subvarieties-radical ideals). Let \( X \) be an affine variety over an algebraically closed field \( K \) . Then there is an inclusion-reversing bijection between the set of subvarieties \( Y \subseteq X \) and the set of radical ideals \( J \subseteq K\left\lbrack X\right\rbrack \) . The b... | Proof. All claims are shown by putting Corollary 1.19 and Lemma 1.22 together. | No |
The ring \( \mathbb{Z} \) of integers is Noetherian. | since ascending chains of ideals correspond to chains of integers \( {a}_{1},{a}_{2},\ldots \) with \( {a}_{i + 1} \) a divisor of \( {a}_{i} \) . So the well-ordering of the natural numbers yields the result. | Yes |
Lemma 2.6 (Ideal powers and radical ideals). Let \( R \) be a ring and \( I, J \subseteq R \) ideals. If \( I \) is finitely generated, then\n\n\[ I \subseteq \sqrt{J}\; \Leftrightarrow \;\text{ there exists }\;k \in {\mathbb{N}}_{0}\;\text{ such that }\;{I}^{k} \subseteq J. \] | Proof. We have \( I = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) . Suppose that \( I \subseteq \sqrt{J} \) . Then there exists \( m > 0 \) with \( {a}_{i}^{m} \in J \) for \( i = 1,\ldots, n \) . Set \( k \mathrel{\text{:=}} n \cdot \left( {m - 1}\right) + 1 \) . We need to show that the product of \( k \) arbitrary el... | Yes |
Lemma 2.7. Let \( R \) be a ring and \( {\mathfrak{m}}_{1},\ldots ,{\mathfrak{m}}_{n} \in {\operatorname{Spec}}_{\max }\left( R\right) \) maximal ideals (which are not assumed to be distinct) such that the ideal product \( {\mathfrak{m}}_{1}\cdots {\mathfrak{m}}_{n} \) is zero. Then \( R \) is Artinian if and only if i... | Proof. Setting\n\n\[ {I}_{i} \mathrel{\text{:=}} {\mathfrak{m}}_{1}\cdots {\mathfrak{m}}_{i} \]\n\nwe get a chain\n\n\[ \{ 0\} = {I}_{n} \subseteq {I}_{n - 1} \subseteq \cdots \subseteq {I}_{2} \subseteq {I}_{1} \subseteq {I}_{0} \mathrel{\text{:=}} R \]\n\nof ideals. Applying Proposition 2.4 repeatedly, we see that \(... | Yes |
Theorem 2.8 (Artinian and Noetherian rings). Let \( R \) be a ring. Then the following statements are equivalent:\n\n(a) \( R \) is Artinian.\n\n(b) \( R \) is Noetherian and every prime ideal of \( R \) is maximal. | Proof of \ | No |
Theorem 2.9 (Alternative definition of Noetherian modules). Let \( R \) be a ring and \( M \) an \( R \)-module. The following statements are equivalent:\n\n(a) \( M \) is Noetherian.\n\n(b) For every subset \( S \subseteq M \) there exist finitely many elements \( {m}_{1},\ldots ,{m}_{k} \in \) \( S \) such that\n\n\[... | Proof. Assume that \( M \) is Noetherian, but there exists \( S \subseteq M \) that does not satisfy (b). We define finite subsets \( {S}_{i} \subseteq S\left( {i = 1,2,\ldots }\right) \) recursively, starting with \( {S}_{1} = \varnothing \). Suppose \( {S}_{i} \) has been defined. Since \( S \) does not satisfy (b), ... | Yes |
Theorem 2.10 (Noetherian modules and finite generation). Let \( R \) be a Noetherian ring and \( M \) an \( R \)-module. Then the following statements are equivalent:\n\n(a) \( M \) is Noetherian.\n\n(b) \( M \) is finitely generated.\n\nIn particular, every submodule of a finitely generated \( R \)-module is also fini... | Proof. We need to show only that (b) implies (a), since the converse implication is a consequence of Theorem 2.9. So let \( M = {\left( {m}_{1},\ldots ,{m}_{k}\right) }_{R} \). We use induction on \( k \). There is nothing to show for \( k = 0 \), so assume \( k > 0 \). Consider the submodule\n\n\[ N \mathrel{\text{:=}... | Yes |
Theorem 2.11 (Polynomial rings over Noetherian rings). Let \( R \) be a Noetherian ring. Then the polynomial ring \( R\left\lbrack x\right\rbrack \) is Noetherian, too. | Proof. Let \( I \subseteq R\left\lbrack x\right\rbrack \) be an ideal. By Theorem 2.9, we need to show that \( I \) is finitely generated. For a nonnegative integer \( i \), set\n\n\[ \n{J}_{i} \mathrel{\text{:=}} \left\{ {{a}_{i} \in R \mid \text{ there exist }{a}_{0},\ldots ,{a}_{i - 1} \in R\text{ such that }\mathop... | Yes |
Proposition 3.1 (Unions and intersections of affine varieties). Let \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be a polynomial ring over a field \( K \) .\n\n(a) Let \( I, J \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be ideals. Then\n\n\[ \mathcal{V}\left( I\right) \cup \mathcal{V... | Proof. We first prove (a). It is clear that \( \mathcal{V}\left( I\right) \cup \mathcal{V}\left( J\right) \subseteq \mathcal{V}\left( {I \cap J}\right) \) . To prove the reverse inclusion, let \( P \in \mathcal{V}\left( {I \cap J}\right) \) . Assume \( P \notin \mathcal{V}\left( I\right) \), so there exists\n\n\( f \in... | No |
Proposition 3.6 (Properties of \( {\mathcal{V}}_{\operatorname{Spec}\left( R\right) } \) and \( {\mathcal{I}}_{R} \) ). Let \( R \) be a ring.\n\n(a) Let \( S, T \subseteq R \) be subsets. Then\n\n\[ \n{\mathcal{V}}_{\operatorname{Spec}\left( R\right) }\left( S\right) \cup {\mathcal{V}}_{\operatorname{Spec}\left( R\rig... | Proof. (a) If \( P \in {\mathcal{V}}_{\operatorname{Spec}\left( R\right) }\left( S\right) \), then \( S \subseteq P \), so also \( {\left( S\right) }_{R} \subseteq P \) and \( {\left( S\right) }_{R} \cap \) \( {\left( T\right) }_{R} \subseteq P \) . The same follows if \( P \in {\mathcal{V}}_{\operatorname{Spec}\left( ... | Yes |
Theorem 3.9 (Noether property of the Zariski topology). (a) Let \( K \) be a field and \( X \subseteq {K}^{n} \) a set of points, equipped with the Zariski topology. Then \( X \) is Noetherian. (b) Let \( R \) be a Noetherian ring and \( X \subseteq \operatorname{Spec}\left( R\right) \) a set of prime ideals, equipped ... | Proof. First observe that if \( X \) is any Noetherian topological space and \( Y \subseteq X \) is a subset equipped with the subset topology, then \( Y \) is also Noetherian. So we may assume \( X = {K}^{n} \) in part (a), and \( X = \operatorname{Spec}\left( R\right) \) in part (b). To prove (a), let \( {Y}_{1},{Y}_... | Yes |
Theorem 3.10 (Irreducible subsets of \( {K}^{n} \) and \( \operatorname{Spec}\left( R\right) \) ).\n\n(a) Let \( K \) be a field and \( X \subseteq {K}^{n} \) a set of points, equipped with the Zariski topology. Then \( X \) is irreducible if and only if \( {\mathcal{I}}_{K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\r... | Proof. (a) First assume that \( X \) is irreducible. Then \( I \mathrel{\text{:=}} {\mathcal{I}}_{K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }\left( X\right) \subsetneqq \) \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), since \( X \neq \varnothing \) . To show that \( I \) is a prime ideal, let \... | Yes |
Theorem 3.11 (Decomposition into irreducibles). Let \( X \) be a Noetherian topological space.\n\n(a) There exist a nonnegative integer \( n \) and closed, irreducible subsets \( {Z}_{1},\ldots ,{Z}_{n} \subseteq X \) such that\n\n\[ X = {Z}_{1} \cup \cdots \cup {Z}_{n}\;\text{ and }\;{Z}_{i} \nsubseteq {Z}_{j}\;\text{... | Proof. First observe that every nonempty set of closed subsets of \( X \) has a minimal element, since otherwise it would contain an infinite strictly descending chain. Assume that there exists a nonempty closed subset \( Y \subseteq X \) that is not a finite union of closed, irreducible subsets. Then we may assume \( ... | Yes |
Corollary 3.14 (Minimal prime ideals). Let \( R \) be a Noetherian ring.\n\n(a) There exist only finitely many minimal prime ideals \( {P}_{1},\ldots ,{P}_{n} \) of \( R \) .\n\n(b) Every prime ideal of \( R \) contains at least one of the \( {P}_{i} \) .\n\n(c) The nilradical is the intersection of the \( {P}_{i} \) :... | Proof. By Proposition 3.6(e) and by Theorem 3.10(b), the (maximal) closed, irreducible subsets of \( X \mathrel{\text{:=}} \operatorname{Spec}\left( R\right) \) correspond to (minimal) prime ideals of \( R \) . So for (a) and (b), we need to show that \( X \) has only finitely many maximal closed, irreducible subsets, ... | Yes |
Theorem 5.5 (Dimension of algebras, upper bound). Let \( A \) be a (not necessarily finitely generated) algebra over a field \( K \) . Then\n\n\[ \dim \left( A\right) \leq \operatorname{trdeg}\left( A\right) \] | Proof. This is the special case \( S = A \) of the following lemma. | No |
Lemma 5.6. Let \( A \) be an algebra over a field \( K \), and let \( S \subseteq A \) be a subset that generates \( A \) as an algebra. Then\n\n\[ \dim \left( A\right) \leq \sup \{ \left| T\right| \mid T \subseteq S\text{ is finite and algebraically independent }\} . \] | Proof. Let \( n \) be the supremum on the right-hand side of the claimed inequality. There is nothing to show if \( n = \infty \), and the lemma is correct if \( n = - 1 \) . So assume \( n \in {\mathbb{N}}_{0} \) . We need to show that \( \dim \left( {A/P}\right) \leq n \) for all \( P \in \operatorname{Spec}\left( A\... | Yes |
Corollary 5.7 (Dimension of a polynomial ring). If \( K \) is a field, then\n\n\[ \dim \left( {K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }\right) = n. \] | Proof. With \( S \mathrel{\text{:=}} \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \), Lemma 5.6 yields \( \dim \left( {K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }\right) \leq n \) . Since we have the chain\n\n\[ \{ 0\} \subsetneqq \left( {x}_{1}\right) \subsetneqq \left( {{x}_{1},{x}_{2}}\right) \subsetneqq \cd... | Yes |
Theorem 5.11 (0-dimensional affine algebras). Let \( A \neq \{ 0\} \) be an affine \( K \) -algebra. Then the following statements are equivalent:\n\n(a) \( \dim \left( A\right) = 0 \) .\n\n(b) \( A \) is algebraic over \( K \) .\n\n(c) \( {\dim }_{K}\left( A\right) < \infty \) .\n\n(d) \( A \) is Artinian.\n\n(e) \( \... | Proof. If \( \dim \left( A\right) = 0 \), then \( A \) is algebraic by Theorem 5.9. Assume that \( A \) is algebraic. We can write \( A = K\left\lbrack {{a}_{1},\ldots ,{a}_{n}}\right\rbrack \), so there exist nonzero polynomials \( {g}_{i} \in K\left\lbrack x\right\rbrack \) with \( {g}_{i}\left( {a}_{i}\right) = 0 \)... | Yes |
Proposition 5.12 (0-dimensional sets). Let \( K \) be a field and \( X \subseteq {K}^{n} \) nonempty. Then \( \dim \left( X\right) = 0 \) if and only if \( X \) is finite. | Proof. Assume \( \dim \left( X\right) = 0 \) . Since \( X \) is a subset of a Noetherian space, \( X \) is Noetherian, too. By Theorem 3.11(a), \( X \) is a finite union of closed, irreducible subsets \( {Z}_{i} \) . Choose \( {x}_{i} \in {Z}_{i} \) . Then \( \left\{ {x}_{i}\right\} \subseteq {Z}_{i} \) is a chain of c... | Yes |
Lemma 5.14 (Height-one prime ideals in a factorial ring). Let \( R \) be a factorial ring and let \( P \in \operatorname{Spec}\left( R\right) \) be prime ideal that is minimal among all nonzero prime ideals. (According to Definition 6.10, this means that \( P \) has height 1.) Then \( P = \left( a\right) \) with \( a \... | Proof. Let \( a \in P \smallsetminus \{ 0\} \) . Since \( P \) is a prime ideal, at least one factor of a factorization of \( a \) into prime elements also lies in \( P \), so we may assume \( a \) to be a prime element. Then \( \left( a\right) \) is a prime ideal and \( \{ 0\} \subsetneqq \left( a\right) \subseteq P \... | Yes |
Theorem 5.15 (Dimension of a product variety). Let \( X \subseteq {K}^{n} \) and \( Y \subseteq {K}^{m} \) be nonempty affine varieties over an algebraically closed field \( K \) . Then the product variety \( X \times Y \subseteq {K}^{n + m} \) satisfies\n\n\[ \dim \left( {X \times Y}\right) = \dim \left( X\right) + \d... | Proof. The proof is very easy and straightforward, even if it takes some space to write it down.\n\nWrite \( d = \dim \left( X\right) = \dim \left( {K\left\lbrack X\right\rbrack }\right) \) and \( e = \dim \left( Y\right) = \dim \left( {K\left\lbrack Y\right\rbrack }\right) \) . By Theorem 5.9 and Proposition 5.10, \( ... | Yes |
Theorem 6.5 (The spectrum of a localized ring). Let \( R \) be a ring and \( U \subseteq R \) a multiplicative subset. Let \( \varepsilon : R \rightarrow {U}^{-1}R \) be the canonical map and\n\n\[ \n\mathcal{A} \mathrel{\text{:=}} \{ Q \in \operatorname{Spec}\left( R\right) \mid U \cap Q = \varnothing \} .\n\]\n\nThen... | Proof. Since preimages of prime ideals under ring homomorphisms are always prime ideals, \( {\varepsilon }^{-1}\left( \mathfrak{Q}\right) \in \operatorname{Spec}\left( R\right) \) for \( \mathfrak{Q} \in \operatorname{Spec}\left( {{U}^{-1}R}\right) \) . Moreover, \( U \cap \) \( {\varepsilon }^{-1}\left( \mathfrak{Q}\r... | Yes |
Lemma 7.1 (Adjugate matrix over rings). Let \( A = {\left( {a}_{i, j}\right) }_{1 \leq i, j \leq n} \in {R}^{n \times n} \) be a square matrix with entries in a ring \( R \) . For \( i, k \in \{ 1,\ldots, n\} \), let \( {c}_{i, k} \in R \) be the determinant of the matrix obtained from \( A \) by deleting the ith row a... | Proof. If determinant theory is developed over a ring, this is the standard result on the adjugate matrix. For readers who are familiar with determinants only over a field, we present a proof by reduction to the field case.\n\nFor \( i, j \in \{ 1,\ldots, n\} \), let \( {x}_{i, j} \) be an indeterminate over \( \mathbb... | Yes |
Lemma 7.2. Let \( R \) be a ring, \( M = {\left( {m}_{1},\ldots ,{m}_{n}\right) }_{R} \) a finitely generated \( R \) -module, and \( {a}_{i, j} \in R \) ring elements \( \left( {i, j \in \{ 1,\ldots, n\} }\right) \) with\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{i, j}{m}_{j} = 0\;\text{ for }\;i \in \{ 1,\ldots, n... | Proof. Let \( A \mathrel{\text{:=}} \left( {a}_{i, j}\right) \in {R}^{n \times n} \), and let \( {c}_{i, k} \in R \) be as in Lemma 7.1. For every \( k \in \{ 1,\ldots, n\} \), it follows from Lemma 7.1 that\n\n\[ \det \left( A\right) \cdot {m}_{k} = \mathop{\sum }\limits_{{j = 1}}^{n}{\delta }_{j, k}\det \left( A\righ... | Yes |
Theorem 7.3 (Nakayama's lemma). Let \( R \) be a ring with Jacobson radical \( J \), and let \( M \) be a finitely generated \( R \) -module. If\n\n\[ J \cdot M = M \]\n\nthen \( M = \{ 0\} \) . | Proof. Write \( M = {\left( {m}_{1},\ldots ,{m}_{n}\right) }_{R} \) . By hypothesis, \( {m}_{i} = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{i, j}{m}_{j} \) with \( {a}_{i, j} \in J \) . By Lemma 7.2,\n\n\[ d \mathrel{\text{:=}} \det {\left( {\delta }_{i, j} - {a}_{i, j}\right) }_{1 \leq i, j \leq n} \in \operatorname{Ann... | Yes |
Theorem 7.4 (Principal ideal theorem, first version). Let \( R \) be a Noetherian ring and \( P \in \operatorname{Spec}\left( R\right) \) a prime ideal that is minimal over a principal ideal \( \left( a\right) \subseteq R \) . Then \[ \operatorname{ht}\left( P\right) \leq 1\text{.} \] In particular, a proper principal ... | Proof. Let \( {R}_{P} \) be the localization at \( P \) . Using Theorem 6.5, we see that \( {P}_{P} \) is a prime ideal that is minimal over \( {\left( \frac{a}{1}\right) }_{{R}_{P}} \), and \( \operatorname{ht}\left( {P}_{P}\right) = \operatorname{ht}\left( P\right) \) . So by replacing \( R \) with \( {R}_{P} \), we ... | Yes |
Theorem 7.5 (Principal ideal theorem, generalized version). Let \( R \) be a Noetherian ring and \( P \in \operatorname{Spec}\left( R\right) \) a prime ideal that is minimal over an ideal \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \subseteq R \) generated by \( n \) elements. Then\n\n\[\n\operatorname{ht}\left( P\right... | Proof. We use induction on \( n \) . The result is correct for \( n = 0 \), so assume \( n > 0 \) . As in the proof of Theorem 7.4, we may assume that \( R \) is a local ring with maximal ideal \( P \) . Let \( Q \subsetneqq P \) be a prime ideal such that no other prime ideals lie between \( Q \) and \( P \) . We need... | Yes |
Lemma 7.7 (Prime avoidance). Let \( R \) be a ring and \( I,{P}_{1},\ldots ,{P}_{n} \subseteq R \) ideals, with \( n \) a positive integer. Assume that \( {P}_{i} \) is a prime ideal for \( i > 2 \) . Then\n\n\[ I \subseteq \mathop{\bigcup }\limits_{{i = 1}}^{n}{P}_{i} \]\n\nimplies that there exists an \( i \) with \(... | Proof. The proof is by induction on \( n \) . There is nothing to show for \( n = 1 \), so assume \( n > 1 \) . By way of contradiction, assume that for each \( i \in \{ 1,\ldots, n\} \) there exists\n\n\[ {x}_{i} \in I \smallsetminus \mathop{\bigcup }\limits_{{j \neq i}}{P}_{j} \]\n\nSo by assumption, \( {x}_{i} \in {... | Yes |
Theorem 7.8 (A converse of the principal ideal theorem). Let \( R \) be a Noetherian ring and \( P \in \operatorname{Spec}\left( R\right) \) a prime ideal of height \( n \) . Then there exist \( {a}_{1},\ldots ,{a}_{n} \in R \) such that \( P \) is minimal over \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) . | Proof. We will show that there exist \( {a}_{1},\ldots ,{a}_{n} \in P \) with\n\n\[ \operatorname{ht}\left( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \right) = k\;\text{ for all }\;k \leq n.\]\n\nAssume that \( {a}_{1},\ldots ,{a}_{k - 1} \) have been found. Let \( \mathcal{M} \subseteq \operatorname{Spec}\left( R\right)... | Yes |
Corollary 7.9 (Systems of parameters). Let \( R \) be a Noetherian local ring with maximal ideal \( \mathfrak{m} \) . Then \( \dim \left( R\right) \) is the least number \( n \) such that there exist \( {a}_{1},\ldots ,{a}_{n} \in \mathfrak{m} \) with\n\n\[ \mathfrak{m} = \sqrt{\left( {a}_{1},\ldots ,{a}_{n}\right) } \... | Proof. Using Corollary 1.12, we see that (7.2) is equivalent to the condition that \( \mathfrak{m} \) is minimal over \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) . The existence of \( {a}_{1},\ldots ,{a}_{n} \) with \( n = \) \( \operatorname{ht}\left( \mathfrak{m}\right) = \dim \left( R\right) \) is guaranteed by Th... | Yes |
Lemma 7.10 (Fiber dimension, lower bound, local case). Let \( R \) and \( S \) be Noetherian local rings with maximal ideals \( \mathfrak{m} \) and \( \mathfrak{n} \), respectively. Let \( \varphi : R \rightarrow \) \( S \) be a homomorphism with \( \varphi \left( \mathfrak{m}\right) \subseteq \mathfrak{n} \), and let ... | Proof. Let \( {a}_{1},\ldots ,{a}_{m} \in \mathfrak{m} \) be a system of parameters of \( R \), so \( m = \dim \left( R\right) \) by Corollary 7.9. By Lemma 2.6, there exists a nonnegative integer \( k \) with \( {\mathfrak{m}}^{k} \subseteq {\left( {a}_{1},\ldots ,{a}_{m}\right) }_{R} \) . It is easy to check that thi... | Yes |
Proposition 7.11. In the above situation, the map\n\n\\[ \n\\Phi : \\operatorname{Spec}\\left( {S}_{\\left\\lbrack P\\right\\rbrack }\\right) \\rightarrow {f}^{-1}\\left( {\\{ P\\}}\\right) ,\\mathfrak{Q} \\mapsto {\\pi }^{-1}\\left( {{\\varepsilon }^{-1}\\left( \\mathfrak{Q}\\right) }\\right) ,\n\\]\n\nis an inclusion... | Proof. By Lemma 1.22 and Theorem \\( {6.5},\\Phi \\) is an inclusion-preserving injection \\( \\operatorname{Spec}\\left( {S}_{\\left\\lbrack P\\right\\rbrack }\\right) \\rightarrow \\operatorname{Spec}\\left( S\\right) \\), and its image is\n\n\\[ \n\\operatorname{im}\\left( \\Phi \\right) = \\{ Q \\in \\operatorname{... | Yes |
Theorem 7.12 (Fiber dimension, lower bound). Let \( \varphi : R \rightarrow S \) be a homomorphism of Noetherian rings. Moreover, let \( Q \in \operatorname{Spec}\left( S\right) \) and \( P \mathrel{\text{:=}} {\varphi }^{-1}\left( Q\right) \) . Then\n\n\[ \dim \left( {S}_{\left\lbrack P\right\rbrack }\right) \geq \ope... | Proof. The second inequality (7.7) implies the first, so we only need to prove (7.7). We do this by reduction to the local case. We have a (well-defined) homomorphism\n\n\[ \psi : {R}_{P} \rightarrow {S}_{Q},\frac{a}{b} \mapsto \frac{\varphi \left( a\right) }{\varphi \left( b\right) } \]\n\nmapping \( {P}_{P} \) into \... | Yes |
Corollary 7.13 (Dimension of polynomial rings). Let \( R \neq \{ 0\} \) be a Noetherian ring and \( R\left\lbrack x\right\rbrack \) the polynomial ring in one indeterminate. Then\n\n\[ \dim \left( {R\left\lbrack x\right\rbrack }\right) = \dim \left( R\right) + 1. \] | Proof. Let \( \varphi : R \rightarrow R\left\lbrack x\right\rbrack = : S \) be the natural embedding. For an ideal \( I \subseteq R \) we have \( S/{\left( \varphi \left( I\right) \right) }_{S} \cong \left( {R/I}\right) \left\lbrack x\right\rbrack \) and \( {\varphi }^{-1}\left( {\left( \varphi \left( I\right) \right) ... | Yes |
Lemma 7.15 (Going down and fiber dimension). In the situation of Theorem 7.12, let \( U \mathrel{\text{:=}} \varphi \left( {R \smallsetminus P}\right) \) . If going down holds for the homomorphism \( {R}_{P} \rightarrow {U}^{-1}S \) induced by \( \varphi \), then equality holds in (7.7). | Proof. By Proposition 7.11, \( \operatorname{ht}\left( \mathfrak{Q}\right) \) is the maximal length of a chain\n\n\[ \n{Q}_{0} \subsetneqq {Q}_{1} \subsetneqq \cdots \subsetneqq {Q}_{m} = Q \n\]\n\n(7.9)\n\nof prime ideals \( {Q}_{i} \in \operatorname{Spec}\left( S\right) \) with \( {\varphi }^{-1}\left( {Q}_{i}\right)... | Yes |
Lemma 7.16 (Freeness implies going down). Let \( \varphi : R \rightarrow S \) be a ring homomorphism with \( S \) Noetherian.\n\n(a) If \( S \) is free as an \( R \) -module, then going down holds for \( \varphi \) . | Proof. For the proof of (a), let \( P \in \operatorname{Spec}\left( R\right) \) and \( {Q}^{\prime } \in \operatorname{Spec}\left( S\right) \) with \( \varphi \left( P\right) \subseteq {Q}^{\prime } \) . Set \( I \mathrel{\text{:=}} {\left( \varphi \left( P\right) \right) }_{S} \), and let \( Q \in \operatorname{Spec}\... | Yes |
Lemma 8.3 (Integral elements and finite modules). Let \( S \) be a ring, \( R \subseteq S \) a subring, and \( s \in S \) . Then the following statements are equivalent:\n\n(a) The element \( s \) is integral over \( R \) .\n\n(b) The subalgebra \( R\left\lbrack s\right\rbrack \subseteq S \) generated by \( s \) is fin... | Proof. Assume that \( s \) is integral over \( R \), so we have an integral equation \( {x}^{n} + {a}_{1}{x}^{n - 1} + \cdots + {a}_{n - 1}x + {a}_{n} \in R\left\lbrack x\right\rbrack \) for \( s \) . We claim that \( R\left\lbrack s\right\rbrack \) is generated by the \( {s}^{i}, i \in \{ 0,\ldots, n - 1\} \), i.e.,\n... | Yes |
Theorem 8.4 (Generated by integral elements implies integral). Let \( S \) be a ring and \( R \subseteq S \) a subring such that \( S = R\left\lbrack {{a}_{1},\ldots ,{a}_{n}}\right\rbrack \) is finitely generated as an R-algebra. Then the following statements are equivalent:\n\n(a) All \( {a}_{i} \) are integral over ... | Proof. Clearly (b) implies (a). We use induction on \( n \) to show that (a) implies (c). We may assume \( n > 0 \) . By induction, \( {S}^{\prime } \mathrel{\text{:=}} R\left\lbrack {{a}_{1},\ldots ,{a}_{n - 1}}\right\rbrack \) is finitely generated as an \( R \) -module, so \( {S}^{\prime } = {\left( {m}_{1},\ldots ,... | Yes |
Corollary 8.5 (Integral elements form a subalgebra). Let \( S \) be a ring and \( R \subseteq S \) a subring. Then the set \[ {S}^{\prime } \mathrel{\text{:=}} \{ s \in S \mid s\text{ is integral over }R\} \subseteq S \] is an \( R \) -subalgebra. | Proof. Clearly all elements from \( R \) lie in \( {S}^{\prime } \) . So all we need to show is that if \( a, b \in {S}^{\prime } \), then also \( a + b \in {S}^{\prime } \) and \( a \cdot b \in {S}^{\prime } \) . But this follows since \( R\left\lbrack {a, b}\right\rbrack \) is integral over \( R \) by Theorem 8.4. | Yes |
Corollary 8.6 (Towers of integral extensions). Let \( T \) be a ring and \( R \subseteq \) \( S \subseteq T \) subrings. If \( T \) is integral over \( S \) and \( S \) is integral over \( R \), then \( T \) is integral over \( R \) . | Proof. For every \( t \in T \) we have an integral equation\n\n\[ {t}^{n} + {s}_{1}{t}^{n - 1} + \cdots + {s}_{n - 1}t + {s}_{n} = 0 \]\n\nwith \( {s}_{i} \in S \) . So \( t \) is integral over \( {S}^{\prime } \mathrel{\text{:=}} R\left\lbrack {{s}_{1},\ldots ,{s}_{n}}\right\rbrack \subseteq S \) . By Lemma 8.3, \( {S... | Yes |
Proposition 8.8. Every factorial ring is normal. | Proof. Let \( R \) be a factorial ring, and let \( a/b \in \operatorname{Quot}\left( R\right) \) be integral over \( R \) with \( a, b \in R \) coprime. So we have\n\n\[ \frac{{a}^{n}}{{b}^{n}} + {a}_{1}\frac{{a}^{n - 1}}{{b}^{n - 1}} + \cdots + {a}_{n - 1}\frac{a}{b} + {a}_{n} = 0 \]\n\nwith \( {a}_{i} \in R \) . Mult... | Yes |
(2) By Example \( {8.2}\left( 3\right), R \mathrel{\text{:=}} \mathbb{Z}\left\lbrack \sqrt{5}\right\rbrack \) is not normal. In fact, the normalization is \[ \widetilde{R} = \mathbb{Z}\left\lbrack {\left( {1 + \sqrt{5}}\right) /2}\right\rbrack = : S \] | To see this, let \( a + b\sqrt{5} \in \mathbb{Q}\left\lbrack \sqrt{5}\right\rbrack = \operatorname{Quot}\left( S\right) \) (with \( a, b \in \mathbb{Q} \) ) be integral over \( S \) . Since \( S \) is integral over \( \mathbb{Z} \) by Theorem 8.4, \( a + b\sqrt{5} \) is integral over \( \mathbb{Z} \) by Corollary 8.6, ... | Yes |
Proposition 8.10 (Normal rings and localization). For an integral domain \( R \), the following statements are equivalent:\n\n(a) \( R \) is normal.\n\n(b) For every multiplicative subset \( U \subset R \) with \( 0 \notin U \), the localization \( {U}^{-1}R \) is normal.\n\n(c) For every maximal ideal \( \mathfrak{m} ... | Proof. Let \( K = \operatorname{Quot}\left( R\right) \) be the field of fractions. Assume that \( R \) is normal, and let \( U \subset R \) be a multiplicative subset with \( 0 \notin U \) . We have \( {U}^{-1}R \subseteq K \) and \( \operatorname{Quot}\left( {{U}^{-1}R}\right) = K \) . To show that \( {U}^{-1}R \) is ... | Yes |
Lemma 8.11 (Almost integral elements). In the above setting, if \( s \) is integral, then it is almost integral. If \( R \) is Noetherian, the converse holds. | Proof. By Lemma 8.3, \( s \) is integral if and only if \( R\left\lbrack s\right\rbrack \subseteq \operatorname{Quot}\left( R\right) \) is finitely generated as an \( R \) -module. In this case there exists \( c \in R \smallsetminus \{ 0\} \) such that \( {cf} \in R \) for all \( f \in R\left\lbrack s\right\rbrack \) .... | Yes |
Theorem 8.12 (Lying over and going up). Let \( R \subseteq S \) be an integral extension of rings, \( P \in \operatorname{Spec}\left( R\right) \) a prime ideal, and \( I \subseteq S \) an ideal with \( R \cap I \subseteq P \) . (Notice that the zero ideal always satisfies the condition on I.) Set\n\n\[ \mathcal{M} \mat... | Proof of Theorem 8.12. With \( {S}^{\prime } \mathrel{\text{:=}} S/I,{R}^{\prime } \mathrel{\text{:=}} R/\left( {R \cap I}\right) \), and \( {P}^{\prime } \mathrel{\text{:=}} \) \( P/\left( {R \cap I}\right) \), we have an integral extension \( {R}^{\prime } \subseteq {S}^{\prime } \), and Lemma 1.22 yields an inclusio... | Yes |
Lemma 8.15 (Elements fixed by field automorphisms). Let \( N \) be a field of characteristic \( p \geq 0 \) and let \( K \subseteq N \) be a subfield such that \( N \) is finite and normal over \( K \) (see Lang [33, Chapter VII, Theorem 3.3] for the definition of a normal field extension). Let \( G \mathrel{\text{:=}}... | Proof. In the separable case, the lemma follows directly from Galois theory. The proof we give works for the separable case, too.\n\nLet \( g = \operatorname{irr}\left( {\alpha, K}\right) \in K\left\lbrack x\right\rbrack \) be the minimal polynomial of \( \alpha \) over \( K \) . Let \( \bar{N} \) be the algebraic clos... | Yes |
Lemma 8.16. Let \( N \) be a field and \( K \subseteq N \) a subfield such that \( N \) is finite and normal over \( K \) . Let \( R \subseteq K \) be a subring that is integrally closed in \( K \) , and let \( S \subseteq N \) be the integral closure of \( R \) in \( N \) . Then for two prime ideals \( Q,\widetilde{Q}... | Proof. Let \( a \in \widetilde{Q} \) . Then the product \( \mathop{\prod }\limits_{{\sigma \in G}}\sigma \left( a\right) \) lies in \( {N}^{G} \), so by Lemma 8.15 there exists \( n \in {\mathbb{N}}_{0} \) with\n\n\[ b \mathrel{\text{:=}} \mathop{\prod }\limits_{{\sigma \in G}}\sigma {\left( a\right) }^{{p}^{n}} \in K ... | Yes |
Theorem 8.17 (Going down for integral extensions of normal rings). Let \( S \) be a ring and \( R \subseteq S \) a subring such that\n\n(1) \( S \) is an integral domain,\n\n(2) \( R \) is normal,\n\n(3) \( S \) is integral over \( R \), and\n\n(4) \( S \) is finitely generated as an \( R \) -algebra.\n\nThen going dow... | Proof. The proof is not difficult but a bit involved. Fig. 8.3 shows what is going on. Given prime ideals \( P \in \operatorname{Spec}\left( R\right) \) and \( {Q}^{\prime } \in \operatorname{Spec}\left( S\right) \) with \( P \subseteq {Q}^{\prime } \), we need to produce \( Q \in \operatorname{Spec}\left( S\right) \) ... | Yes |
Proposition 8.18 (Geometric properties of normalization). Let \( R \) be an integral domain with normalization \( \widetilde{R} \), and consider the morphism \( f : \operatorname{Spec}\left( \widetilde{R}\right) \rightarrow \operatorname{Spec}\left( R\right) \) induced from the inclusion \( R \subseteq \widetilde{R} \)... | Proof. Parts (a) and (b) follow from Corollary 8.13 and Theorem 8.12(a).\n\nTo prove (c), take \( P \in \operatorname{Spec}\left( R\right) \) with \( {R}_{P} \) normal. Both \( {R}_{P} \) and \( \widetilde{R} \) are contained in \( \operatorname{Quot}\left( R\right) \). With \( U \mathrel{\text{:=}} R \smallsetminus P ... | Yes |
Theorem 8.19 (Noether normalization). Let \( A \neq \{ 0\} \) be an affine \( K \) -algebra. Then there exist algebraically independent elements \( {c}_{1},\ldots ,{c}_{n} \in A \) (with \( n \in {\mathbb{N}}_{0} \) ) such that \( A \) is integral over the subalgebra \( C \mathrel{\text{:=}} K\left\lbrack {{c}_{1},\ldo... | Proof. Write \( A \) as a quotient ring of a polynomial ring: \( A = K\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack /I \) . We use induction on \( m \) for proving the first statement. There is nothing to show for \( m = 0 \) . If \( I = \{ 0\} \), we can set \( {c}_{i} = {x}_{i} + I \), and again there is nothin... | Yes |
Consider the affine variety \( X = {\mathcal{V}}_{{K}^{2}}\left( {{x}_{1}{x}_{2} - 1}\right) \), which is a hyperbola as shown in Fig. 8.4. We write \( {\bar{x}}_{i} \) for the image of \( {x}_{i} \) in the coordinate ring \( K\left\lbrack X\right\rbrack = K\left\lbrack {{x}_{1},{x}_{2}}\right\rbrack /\left( {{x}_{1}{x... | \[ 0 = {\bar{x}}_{1}{\bar{x}}_{2} - 1 = {\bar{x}}_{1}^{2} - {\bar{x}}_{1}c - 1 \] so \( K\left\lbrack X\right\rbrack \) is integral over \( C \mathrel{\text{:=}} K\left\lbrack c\right\rbrack \) . The morphism induced by the embedding \( C \hookrightarrow K\left\lbrack X\right\rbrack \) is \( f : X \rightarrow {K}^{1},\... | Yes |
Theorem 8.22 (Chains of prime ideals in an affine algebra). Let \( A \) be an affine algebra and let\n\n\[ \n{P}_{0} \subsetneqq {P}_{1} \subsetneqq \cdots \subsetneqq {P}_{n} \n\]\n\n(8.6)\n\nbe a maximal chain of prime ideals \( {P}_{i} \in \operatorname{Spec}\left( A\right) \) . Then\n\n\[ \n n = \dim \left( {A/{P}_... | Proof. We use induction on \( n \) . Substituting \( A \) by \( A/{P}_{0} \), we may assume that \( A \) is an affine domain and \( {P}_{0} = \{ 0\} \) . If \( n = 0 \), then \( {P}_{0} \) is a maximal ideal, so \( A \) is a field and we are done. So we may assume \( n > 0 \) . Applying Lemma 1.22 yields a maximal chai... | Yes |
Corollary 8.23 (Dimension and height). Let \( A \) be an affine domain or, more generally, an equidimensional affine algebra. If \( I \subseteq A \) is an ideal, then\n\n\[ \operatorname{ht}\left( I\right) = \dim \left( A\right) - \dim \left( {A/I}\right) . \] | Proof. If \( I \) is a prime ideal, there exists a maximal chain \( \mathcal{C} \subseteq \operatorname{Spec}\left( A\right) \) with \( I \in \mathcal{C} \), so the result follows from Theorem 8.22, Lemma 1.22, and Definition 6.10(a). For \( I = A \), it follows from Definition 6.10(b). For all other \( I \) , Definiti... | Yes |
Corollary 8.24 (Height of maximal ideals). Let \( A \) be an affine algebra with minimal prime ideals \( {P}_{1},\ldots ,{P}_{n} \) . (There are finitely many \( {P}_{i} \) by Corollaries 2.12 and 3.14(a).) If \( \mathfrak{m} \in {\operatorname{Spec}}_{\max }\left( A\right) \) is a maximal ideal, then\n\n\[ \operatorna... | Proof. This is an immediate consequence of Theorem 8.22. | Yes |
Theorem 8.25 (Principal ideal theorem for affine domains). Let \( A \) be an affine domain or, more generally, an equidimensional affine algebra, and let \( I = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \subseteq A \) be an ideal generated by \( n \) elements. Then every prime ideal \( P \in \operatorname{Spec}\left( A\r... | Proof. By Theorem 7.5, every \( P \in \operatorname{Spec}\left( A\right) \) that is minimal over \( I \) satisfies \( \operatorname{ht}\left( P\right) \leq n \), so by Corollary 8.23. | No |
Theorem 8.26. Let \( A \) be an affine domain. Then the normalization \( \widetilde{A} \) of \( A \) is an affine domain, too. | Proof. By Noether normalization (Theorem 8.19), we have a subalgebra \( R \subseteq A \) which is isomorphic to a polynomial algebra, such that \( A \) is integral over \( R \) . In particular, \( N \mathrel{\text{:=}} \operatorname{Quot}\left( A\right) \) is a finite field extension of \( \operatorname{Quot}\left( R\r... | No |
Corollary 8.28 (Normalization of an affine variety). Let \( X \) be an irreducible affine variety over an algebraically closed field \( K \) . Then there exists a normal affine variety \( \widetilde{X} \) with a surjective morphism \( f : \widetilde{X} \rightarrow X \) such that:\n\n(a) \( \dim \left( \widetilde{X}\rig... | Proof. By Theorem 8.26, the normalization \( \widetilde{A} \) of the coordinate ring \( A \mathrel{\text{:=}} \) \( K\left\lbrack X\right\rbrack \) is an affine domain, so by Theorem 1.25(b) there exists an affine variety \( \widetilde{X} \) with \( K\left\lbrack \widetilde{X}\right\rbrack \cong \widetilde{A} \) . The ... | Yes |
We give some examples of monomial orderings. Let \( t = \) \( {x}_{1}^{{e}_{1}}\cdots {x}_{n}^{{e}_{n}} \) and \( {t}^{\prime } = {x}_{1}^{{e}_{1}^{\prime }}\cdots {x}_{n}^{{e}_{n}^{\prime }} \) be monomials. | (1) The lexicographic ordering is given by saying \( t \leq {t}^{\prime } \) if \( t = {t}^{\prime } \) or \( {e}_{i} < {e}_{i}^{\prime } \) for the smallest index \( i \) with \( {e}_{i} \neq {e}_{i}^{\prime } \) . As we will see on page 131, the lexicographic ordering is useful for solving systems of polynomial equat... | No |
Example 9.7. Let \( S = \left\{ {{x}_{1},{x}_{1} + 1}\right\} \), as in Example 9.5. Then 1 is congruent to 0 modulo \( \left( S\right) \), but 0 is not a normal form of 1 . Moreover, \( f = {x}_{1} \) has two normal forms: 0 and -1 . So in general, normal forms are not uniquely determined. | Observe that the set \( S \) from the above example is not a Gröbner basis. We will see that normal forms with respect to a Gröbner basis are unique (Theorem 9.9). But first we present an algorithm for computing a normal form, thereby also proving its existence. To actually run the algorithm on a computer, we need to a... | Yes |
Theorem 9.9 (The normal form map). Let \( G \) be a Gröbner basis of an ideal \( I \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . (a) Every \( f \in K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) has precisely one normal form with respect to G. So there is a map \( {\mathrm{{NF}}}_{G} : K\... | Proof. We prove (a) and (c) together. To this end, let \( {f}^{ * } \) and \( \widetilde{f} \) be normal forms of \( f \) with respect to \( G \) and \( \widetilde{G} \), respectively. It follows from (9.3) that \( {f}^{ * } - \widetilde{f} \in I \), so \[ \operatorname{LM}\left( {{f}^{ * } - \widetilde{f}}\right) \in ... | Yes |
Corollary 9.10 (Gröbner bases are ideal bases). Let \( G \) be a Gröbner basis of an ideal \( I \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then \( I = {\left( G\right) }_{K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack } \) . | Proof. By definition, \( G \subseteq I \), so \( \left( G\right) \subseteq I \) . Conversely, for \( f \in I \) we have \( {\mathrm{{NF}}}_{G}\left( f\right) = 0 \) by Theorem 9.9(b), so \( f \in \left( G\right) \) by (9.3). | Yes |
Proposition 9.17 (Kernel of a homomorphism of affine algebras). Let\n\n\\[ \n\\varphi : K\\left\\lbrack {{x}_{1},\\ldots ,{x}_{n}}\\right\\rbrack \\rightarrow A \\mathrel{\\text{:=}} K\\left\\lbrack {{y}_{1},\\ldots ,{y}_{m}}\\right\\rbrack /I \n\\]\n\nbe a homomorphism of \\( K \\) -algebras, given by \\( \\varphi \\l... | Proof. It follows from the definition of \\( J \\) that for every \\( f \\in K\\left\\lbrack {{x}_{1},\\ldots ,{x}_{n}}\\right\\rbrack \\) we have\n\n\\[ \nf\\left( {{g}_{1},\\ldots ,{g}_{n}}\\right) - f \\in J. \n\\]\n\n(9.9)\n\nAssume \\( f \\in \\ker \\left( \\varphi \\right) \\) . Then \\( f\\left( {{g}_{1},\\ldots... | Yes |
Lemma 9.19. In the situation of Proposition 9.18(c), let \( f \in K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right. \) , \( \left. {{y}_{1},\ldots ,{y}_{m}}\right\rbrack \) be such that there exists no \( g \in {G}_{x} \) with \( \operatorname{LM}\left( g\right) \) dividing \( \operatorname{LM}\left( f\right) \) . Write \... | Proof. We may assume \( f \neq 0 \) . Since \ | No |
Lemma 10.1 (Generic freeness, constructive version). Let \( R \subseteq S \) be a finitely generated ring extension, so that there is an epimorphism\n\n\[ \n\psi : R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow S \n\]\n\n(with \( {x}_{i} \) indeterminates). Let \( G \subseteq R\left\lbrack {{x}_{1},\l... | Proof of Lemma 10.1. Let \( B \subset R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be the set of all monomials that are not divisible by any leading monomial \( \operatorname{LM}\left( g\right) \) with \( g \in G \) . Since \( \psi \) is injective on \( R \), we have \( 1 \in B \) . Moreover, \( \psi \left( ... | Yes |
Corollary 10.2 (Generic freeness lemma). Let \( R \) be an integral domain and let \( S \) be a ring extension of \( R \) that is finitely generated as an \( R \) -algebra. Then there exists a nonzero element \( a \in R \) such that for every multiplicative subset \( U \subseteq R \) with \( a \in U \), the localizatio... | Proof. We have an epimorphism \( \psi : R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow S \) . Let \( I \mathrel{\text{:=}} \ker \left( \psi \right) \) , \( K \mathrel{\text{:=}} \operatorname{Quot}\left( R\right) \), and \( J \mathrel{\text{:=}} {\left( I\right) }_{K\left\lbrack {{x}_{1},\ldots ,{x}_{... | Yes |
Theorem 10.4. Algorithm 10.3 terminates after finitely many steps and calculates the image of \( {\varphi }^{ * } \) and its closure correctly. | Proof. We use the notation from the algorithm. By way of contradiction, assume that there exists an ideal \( I \subseteq K\left\lbrack {{y}_{1},\ldots ,{y}_{m}}\right\rbrack \) such that the algorithm applied to \( I \) does not terminate after finitely many steps. By Hilbert’s basis theorem (Corollary 2.13), we may as... | Yes |
Theorem 10.5. Let \( \varphi : R \rightarrow S \) be a ring homomorphism such that\n\n(1) \( R \) is a Noetherian integral domain,\n\n(2) \( S \) is finitely generated as an \( R \) -algebra, and\n\n(3) \( \varphi \) is injective.\n\nThen there exists a nonzero \( a \in R \) such that for all \( P \in \operatorname{Spe... | Proof. Corollary 10.2 yields \( a \in R \smallsetminus \{ 0\} \) such that for \( P \in \operatorname{Spec}\left( R\right) \) with \( a \notin P \) the localization \( {U}^{-1}S \) (with \( U \mathrel{\text{:=}} R \smallsetminus P \) ) is a free \( {R}_{P} \) -module with 1 contained in a basis. Moreover, \( S \) and \... | Yes |
Let \( f : X \rightarrow Y \) be a morphism of equidimensional affine varieties over an algebraically closed field. For a point \( y \in Y \), every irreducible component \( Z \subseteq {f}^{-1}\left( {\{ y\} }\right) \) of the fiber has dimension \[ \dim \left( Z\right) \geq \dim \left( X\right) - \dim \left( Y\right)... | Proof. Let \( P \in {\operatorname{Spec}}_{\max }\left( {K\left\lbrack Y\right\rbrack }\right) \) be the maximal ideal corresponding to a point \( y \in Y \) . The fiber over \( y \) is an affine variety, so by Corollary 8.24, the inequality (10.8) follows if we can show that every maximal ideal in the coordinate ring ... | Yes |
Corollary 10.8 (Chevalley's theorem on images of morphisms).\n\nLet \( \varphi : R \rightarrow S \) be a homomorphism of Noetherian rings making \( S \) into a finitely generated \( R \) -algebra. Then the image \( \operatorname{im}\left( {\varphi }^{ * }\right) \) of the induced map \( {\varphi }^{ * } : \operatorname... | Proof. The proof technique we use here is sometimes called Noetherian induction. This works as follows. We assume that the assertion is false. Since \( S \) is Noetherian, there exists an ideal \( I \subseteq S \) that is maximal with the property that\n\n\[ Y\left( I\right) \mathrel{\text{:=}} {\varphi }^{ * }\left( {... | Yes |
(1) Let \( I = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) . Then \( {h}_{I}\left( d\right) = 1 \) for all \( d \), so | \[ {H}_{I}\left( t\right) = \mathop{\sum }\limits_{{d = 0}}^{\infty }{t}^{d} = \frac{1}{1 - t}. \] | Yes |
Proposition 11.4 (The Hilbert series of a principal ideal). If \( I = \left( f\right) \subseteq \) \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) is a principal ideal, then\n\n\[ \n{H}_{I}\left( t\right) = \frac{1 - {t}^{\deg \left( f\right) }}{{\left( 1 - t\right) }^{n + 1}}\;\text{ if }\;f \neq 0 \n\]\n\n... | Proof. We start with the case \( f = 0 \) . Since the Hilbert function and Hilbert series of the zero ideal depend on the number \( n \) of indeterminates, we will write them in this proof as \( {h}_{n}\left( d\right) \) and \( {H}_{n}\left( t\right) \), respectively. We use induction on \( n \), starting with \( n = 0... | Yes |
Theorem 11.6 (Hilbert series and leading ideal). Suppose that the polynomial ring \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) is equipped with a total degree ordering, and let \( I \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an ideal. Then\n\n\[ \n{H}_{I}\left( t\right) = {H}_{L\... | Proof. Set \( A \mathrel{\text{:=}} K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /I \) . By Theorem 9.9, the normal form map \( {\mathrm{{NF}}}_{G} \), given by a Gröbner basis \( G \) of \( I \), induces an injective linear map \( \varphi : A \rightarrow K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) ... | Yes |
Lemma 11.7 (Hilbert series of the sum and intersection of ideals). Let \( I, J \) \( \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be homogeneous ideals. Then\n\n\[ \n{H}_{I + J}\left( t\right) + {H}_{I \cap J}\left( t\right) = {H}_{I}\left( t\right) + {H}_{J}\left( t\right) .\n\] | Proof. Let \( d \) be a nonnegative integer. For an ideal \( L \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) we write \( {L}_{ \leq d} \mathrel{\text{:=}} \{ f \in L \mid \deg \left( f\right) \leq d\} \) . It follows from the hypothesis that \( I + J \) is generated by homogeneous polynomials \( {g}... | Yes |
Theorem 11.9. Algorithm 11.8 terminates after finitely many steps and calculates \( {H}_{I}\left( t\right) \) correctly. | Proof. With each recursive call of the algorithm, the number \( r \) decreases strictly. This guarantees termination.\n\nLet \( \widetilde{I} \mathrel{\text{:=}} \left( {{m}_{1},\ldots ,{m}_{r}}\right) = L\left( I\right) \) . By Theorem 11.6, we need to show that steps (2) through (5) calculate \( {H}_{\widetilde{I}}\l... | Yes |
Corollary 11.10 (Hilbert-Serre theorem). Let \( I \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an ideal. Then the Hilbert series has the form\n\n\[ \n{H}_{I}\left( t\right) = \frac{{a}_{0} + {a}_{1}t + \cdots + {a}_{k}{t}^{k}}{{\left( 1 - t\right) }^{n + 1}} \n\]\n\n(11.3)\n\nwith \( k \in {\mat... | Proof. Induction on the recursion depth in Algorithm 11.8 immediately yields (11.3). By Remark 11.5, we can write \( \frac{1}{{\left( 1 - t\right) }^{n + 1}} = \mathop{\sum }\limits_{{d = 0}}^{\infty }\left( \begin{matrix} d + n \\ n \end{matrix}\right) {t}^{d} \), so\n\n\[ \n{H}_{I}\left( t\right) = \mathop{\sum }\lim... | Yes |
Lemma 11.12 (The degree of the Hilbert polynomial is an invariant). Let \( I \subseteq K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and \( J \subseteq K\left\lbrack {{y}_{1},\ldots ,{y}_{m}}\right\rbrack \) be ideals in polynomial rings such that the \( K \) -algebras \( A \mathrel{\text{:=}} K\left\lbrack {... | Proof. We have an isomorphism \( \varphi : A \rightarrow B \) of \( K \) -algebras, so there exist polynomials \( {g}_{1},\ldots ,{g}_{m} \in K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) such that \( \varphi \left( {{g}_{i} + I}\right) = {y}_{i} + J \) . Set \( k \mathrel{\text{:=}} \max \left\{ {\deg \left(... | Yes |
Corollary 11.14 (Computing dimension via the leading ideal). Let \( I \subseteq \) \( K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an ideal, and let \( L\left( I\right) \) be its leading ideal with respect to a total degree ordering. Then\n\n\[ \dim \left( {K\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rb... | Proof. This follows from Theorems 11.6 and 11.13. | Yes |
Theorem 12.3 (Basic facts about length). Let \( M \) be a module over a ring \( R \) .\n\n(a) If \( M \) has a finite maximal chain \( {M}_{0} \subsetneqq {M}_{1} \subsetneqq \cdots \subsetneqq {M}_{n} \) of submodules, then \( \operatorname{length}\left( M\right) = n \) . So in particular, all maximal chains have the ... | Proof. (a) We use induction on \( n \) . If \( n = 0 \), then \( M = \{ 0\} \) and so \( \operatorname{length}\left( M\right) = 0 \) . Therefore we may assume \( n > 0 \) . Let \( N \subsetneqq M \) be a proper submodule, and set \( {N}_{i} \mathrel{\text{:=}} N \cap {M}_{i} \) . The \( {N}_{i} \) need not be distinct,... | Yes |
Lemma 12.6 (Artin-Rees lemma). Let \( I \subseteq R \) be an ideal. Then there exists a nonnegative integer \( r \) such that \[ I \cap {\mathfrak{m}}^{n} = {\mathfrak{m}}^{n - r} \cdot \left( {I \cap {\mathfrak{m}}^{r}}\right) \] for all \( n \geq r \) . | Proof. Let \( {J}_{d} \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 0}}^{d}{R}^{ * }\left( {I \cap {\mathfrak{m}}^{i}}\right) {t}^{i} \) be the ideal in \( {R}^{ * } \) generated by the \( \left( {I \cap {\mathfrak{m}}^{i}}\right) {t}^{i} \) with \( i \leq d \) . Since \( {R}^{ * } \) is Noetherian, there exists a no... | Yes |
Lemma 12.7. Let \( a \in \mathfrak{m} \) . If \( a \) is not a zero divisor, then\n\n\[ \dim \left( {\operatorname{gr}\left( {R/{Ra}}\right) }\right) < \dim \left( {\operatorname{gr}\left( R\right) }\right) . \] | Proof. By Proposition 12.5, we need to show that \( \deg \left( {p}_{R/{Ra}}\right) < \deg \left( {p}_{R}\right) \) . So we need to compare the Hilbert-Samuel functions \( {h}_{R/{Ra}} \) and \( {h}_{R} \) . Since \( \mathfrak{m}/{Ra} \) is the maximal ideal of \( R/{Ra},{h}_{R/{Ra}}\left( d\right) \) is the length of ... | Yes |
Theorem 12.8 (The dimensions of \( R \) and \( \operatorname{gr}\left( R\right) \) ). Let \( R \) be a Noetherian local ring and let \( \operatorname{gr}\left( R\right) \) be its associated graded ring. Then\n\n\[ \dim \left( R\right) = \dim \left( {\operatorname{gr}\left( R\right) }\right) \]\n\nEquivalently, the Hilb... | Proof. From (12.7) we know that \( \dim \left( {\operatorname{gr}\left( R\right) }\right) \leq \dim \left( R\right) \) . For the reverse inequality we use induction on \( \dim \left( {\operatorname{gr}\left( R\right) }\right) \) . We first reduce to the case that \( R \) is an integral domain. We need to prove that \( ... | Yes |
Theorem 12.9 (Krull's intersection theorem). If \( R \) is a Noetherian local ring with maximal ideal \( \mathfrak{m} \) (as always in this section), then\n\n\[ \mathop{\bigcap }\limits_{{n \in \mathbb{N}}}{\mathfrak{m}}^{n} = \{ 0\} \] | Proof. Set \( I \mathrel{\text{:=}} \mathop{\bigcap }\limits_{{n \in \mathbb{N}}}{\mathfrak{m}}^{n} \), and let \( r \) be the integer given by the Artin-Rees lemma (Lemma 12.6). Then \( I \cap {\mathfrak{m}}^{r + 1} = \mathfrak{m} \cdot \left( {I \cap {\mathfrak{m}}^{r}}\right) \) . By the definition of \( I \) , this... | Yes |
Theorem 12.10 (Properties passing from \( \operatorname{gr}\left( R\right) \) to \( R \) ). Let \( R \) be a Noetherian local ring and let \( \operatorname{gr}\left( R\right) \) be its associated graded ring.\n\n(a) If \( \operatorname{gr}\left( R\right) \) is an integral domain, then the same is true for \( R \) . | Proof. (a) \( R \) is not the zero ring since it is local. Let \( a, b \in R \) be nonzero elements of orders \( d \) and \( e \), respectively. By hypothesis, \( \operatorname{gr}\left( a\right) \cdot \operatorname{gr}\left( b\right) \neq 0 \) , so \( {ab} \notin {\mathfrak{m}}^{d + e + 1} \) by the discussion precedi... | Yes |
Lemma 13.1 (Generating modules over a local ring). In the above setting, assume \( M \) to be finitely generated. Let \( {m}_{1},\ldots ,{m}_{n} \in M \) . Then the following statements are equivalent:\n\n(a) \( M \) is generated by \( {m}_{1},\ldots ,{m}_{n} \) as an \( R \) -module.\n\n(b) \( M/\mathfrak{m}M \) is ge... | Proof. It is clear that (a) implies (b). Conversely, assume (b) and set \( N \mathrel{\text{:=}} \) \( \left( {{m}_{1},\ldots ,{m}_{n}}\right) \subseteq M \) . Then (b) implies \( M \subseteq N + \mathfrak{m}M \), so \( M/N \subseteq \mathfrak{m} \cdot M/N \) . By Nakayama’s lemma (Theorem 7.3), this implies \( M/N = \... | Yes |
Theorem 13.4 (Associated graded ring and regularity). The local ring \( R \) is regular if and only if the associated graded ring \( \operatorname{gr}\left( R\right) \) is isomorphic to a polynomial ring over \( K \) . | Proof. Write \( A \mathrel{\text{:=}} \operatorname{gr}\left( R\right) \) . By Theorem 12.8 we have \( \dim \left( A\right) = \dim \left( R\right) = : n \) .\n\nFirst assume that \( R \) is regular, so the maximal ideal \( \mathfrak{m} \) is generated by \( n \) elements. By the discussion preceding (12.5) on page 172,... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.