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Proposition 5.1. The following are equivalent:\n\ni) \( x \in B \) is integral over \( A \) ;\n\nii) \( A\left\lbrack x\right\rbrack \) is a finitely generated \( A \) -module;\n\niii) \( A\left\lbrack x\right\rbrack \) is contained in a subring \( C \) of \( B \) such that \( C \) is a finitely generated A-module;\n\n... | Proof. i) \( \Rightarrow \) ii). From (1) we have\n\n\[ \n{x}^{n + r} = - \left( {{a}_{1}{x}^{n + r - 1} + \cdots + {a}_{n}{x}^{r}}\right)\n\]\n\nfor all \( r \geq 0 \) ; hence, by induction, all positive powers of \( x \) lie in the \( A \) -module generated by \( 1, x,\ldots ,{x}^{n - 1} \) . Hence \( A\left\lbrack x... | Yes |
Corollary 5.3. The set \( C \) of elements of \( B \) which are integral over \( A \) is a subring of \( B \) containing \( A \) . | Proof. If \( x, y \in C \) then \( A\left\lbrack {x, y}\right\rbrack \) is a finitely generated \( A \) -module by (5.2). Hence \( x \pm y \) and \( {xy} \) are integral over \( A \), by iii) of (5.1). ∎ | Yes |
Corollary 5.5. Let \( A \subseteq B \) be rings and let \( C \) be the integral closure of \( A \) in B. Then \( C \) is integrally closed in \( B \) . | Proof. Let \( x \in B \) be integral over \( C \) . By (5.4) \( x \) is integral over \( A \), hence \( x \in C \) . ∎ | Yes |
Proposition 5.6. Let \( A \subseteq B \) be rings, \( B \) integral over \( A \). i) If \( \mathfrak{b} \) is an ideal of \( B \) and \( \mathfrak{a} = {\mathfrak{b}}^{\mathfrak{c}} = A \cap \mathfrak{b} \), then \( B/\mathfrak{b} \) is integral over \( A/\mathfrak{a} \). ii) If \( S \) is a multiplicatively closed sub... | Proof. i) If \( x \in B \) we have, say, \( {x}^{n} + {a}_{1}{x}^{n - 1} + \cdots + {a}_{n} = 0 \), with \( {a}_{i} \in A \). Reduce this equation mod. b. ii) Let \( x/s \in {S}^{-1}B\left( {x \in B, s \in S}\right) \). Then the equation above gives \[ {\left( x/s\right) }^{n} + \left( {{a}_{1}/s}\right) {\left( x/s\ri... | No |
Proposition 5.7. Let \( A \subseteq B \) be integral domains, \( B \) integral over \( A \) . Then \( B \) is a field if and only if \( A \) is a field. | Proof. Suppose \( A \) is a field; let \( y \in B, y \neq 0 \) . Let\n\n\( {y}^{n} + {a}_{1}{y}^{n - 1} + \cdots + {a}_{n} = 0\;\left( {{a}_{1} \in A}\right) \)\n\nbe an equation of integral dependence for \( y \) of smallest possible degree. Since \( B \) is an integral domain we have \( {a}_{n} \neq 0 \), hence \( {y... | Yes |
Corollary 5.8. Let \( A \subseteq B \) be rings, \( B \) integral over \( A \) ; let \( \mathfrak{q} \) be a prime ideal of \( B \) and let \( \mathfrak{p} = {\mathfrak{q}}^{c} = \mathfrak{q} \cap A \) . Then \( \mathfrak{q} \) is maximal if and only if \( \mathfrak{p} \) is maximal. | Proof. By (5.6), \( B/\mathfrak{q} \) is integral over \( A/\mathfrak{p} \), and both these rings are integral | No |
Corollary 5.9. Let \( A \subseteq B \) be rings, \( B \) integral over \( A \) ; let \( \mathfrak{q},{\mathfrak{q}}^{\prime } \) be prime ideals of \( B \) such that \( \mathfrak{q} \subseteq {\mathfrak{q}}^{\prime } \) and \( {\mathfrak{q}}^{c} = {\mathfrak{q}}^{\prime c} = \mathfrak{p} \) say. Then \( \mathfrak{q} = ... | Proof. By (5.6), \( {B}_{\mathfrak{p}} \) is integral over \( {A}_{\mathfrak{p}} \) . Let \( \mathfrak{m} \) be the extension of \( \mathfrak{p} \) in \( {A}_{\mathfrak{p}} \) and let \( \mathfrak{n},{\mathfrak{n}}^{\prime } \) be the extensions of \( \mathfrak{q},{\mathfrak{q}}^{\prime } \) respectively in \( {B}_{\ma... | Yes |
Proposition 5.12. Let \( A \subseteq B \) be rings, \( C \) the integral closure of \( A \) in \( B \) . Let \( S \) be a multiplicatively closed subset of \( A \) . Then \( {S}^{-1}C \) is the integral closure of \( {S}^{-1}A \) in \( {S}^{-1}B \) . | Proof. By (5.6), \( {S}^{-1}C \) is integral over \( {S}^{-1}A \) . Conversely, if \( b/s \in {S}^{-1}B \) is integral over \( {S}^{-1}A \), then we have an equation of the form\n\n\[ \n{\left( b/s\right) }^{n} + \left( {{a}_{1}/{s}_{1}}\right) {\left( b/s\right) }^{n - 1} + \cdots + {a}_{n}/{s}_{n} = 0 \n\]\n\nwhere \... | Yes |
Proposition 5.13. Let \( A \) be an integral domain. Then the following are equivalent: i) \( A \) is integrally closed; ii) \( {A}_{\mathfrak{p}} \) is integrally closed, for each prime ideal \( \mathfrak{p} \) ; iii) \( {A}_{\mathfrak{m}} \) is integrally closed, for each maximal ideal \( \mathfrak{m} \) . | Proof. Let \( K \) be the field of fractions of \( A \), let \( C \) be the integral closure of \( A \) in \( K \) , and let \( f : A \rightarrow C \) be the identity mapping of \( A \) into \( C \) . Then \( A \) is integrally closed \( \Leftrightarrow f \) is surjective, and by (5.12) \( {A}_{\mathfrak{p}} \) (resp. ... | No |
Lemma 5.14. Let \( C \) be the integral closure of \( A \) in \( B \) and let \( {\mathfrak{a}}^{e} \) denote the extension of \( \mathfrak{a} \) in \( C \) . Then the integral closure of \( \mathfrak{a} \) in \( B \) is the radical of \( {\mathfrak{a}}^{e} \) | Proof. If \( x \in B \) is integral over \( \mathfrak{a} \), we have an equation of the form\n\n\[ \n{x}^{n} + {a}_{1}{x}^{n - 1} + \cdots + {a}_{n} = 0 \n\]\n\nwith \( {a}_{1},\ldots ,{a}_{n} \) in \( \mathfrak{a} \) . Hence \( x \in C \) and \( {x}^{n} \in {\mathfrak{a}}^{e} \), that is \( x \in r\left( {\mathfrak{a}... | Yes |
Proposition 5.15. Let \( A \subseteq B \) be integral domains, \( A \) integrally closed, and let \( x \in B \) be integral over an ideal \( \mathfrak{a} \) of \( A \). Then \( x \) is algebraic over the field of fractions \( K \) of \( A \), and if its minimal polynomial over \( K \) is \( {t}^{n} + {a}_{1}{t}^{n - 1}... | Proof. Clearly \( x \) is algebraic over \( K \). Let \( L \) be an extension field of \( K \) which contains all the conjugates \( {x}_{1},\ldots ,{x}_{n} \) of \( x \). Each \( {x}_{i} \) satisfies the same equation of integral dependence as \( x \) does, hence each \( {x}_{i} \) is integral over \( a \). The coeffic... | Yes |
Proposition 5.17. Let \( A \) be an integrally closed domain, \( K \) its field of fractions,\n\n\( L \) a finite separable algebraic extension of \( K, B \) the integral closure of \( A \) in \( L \) .\n\nThen there exists a basis \( {v}_{1},\ldots ,{v}_{n} \) of \( L \) over \( K \) such that \( B \subseteq \mathop{\... | Proof. If \( v \) is any element of \( L \), then \( v \) is algebraic over \( K \) and therefore satisfies an equation of the form\n\n\[ \n{a}_{0}{v}^{r} + {a}_{1}{v}^{r - 1} + \cdots + {a}_{n} = 0\left( {{a}_{i} \in A}\right) .\n\]\n\nMultiplying this equation by \( {a}_{0}^{n - 1} \), we see that \( {a}_{0}v = u \) ... | Yes |
Proposition 5.18. i) \( B \) is a local ring. | Proof. i) Let \( \mathfrak{m} \) be the set of non-units of \( B \), so that \( x \in \mathfrak{m} \Leftrightarrow \) either \( x = 0 \) or \( {x}^{-1} \notin B \) . If \( a \in B \) and \( x \in \mathfrak{m} \) we have \( {ax} \in \mathfrak{m} \), for otherwise \( {\left( ax\right) }^{-1} \in B \) and therefore \( {x}... | No |
Lemma 5.19. B is a local ring and \( \mathfrak{m} = \operatorname{Ker}\left( g\right) \) is its maximal ideal. | Proof. Since \( g\left( B\right) \) is a subring of a field and therefore an integral domain, the ideal \( \bar{g}\left( {b/s}\right) = g\left( b\right) /g\left( s\right) \) for all \( b \in B \) and all \( s \in B - m \), since \( g\left( s\right) \) will not be zero. Since the pair \( \left( {B, g}\right) \) is maxim... | No |
Theorem 5.21. Let \( \left( {B, g}\right) \) be a maximal element of \( \sum \) . Then \( B \) is a valuation ring of the field \( K \) . | Proof. We have to show that if \( x \neq 0 \) is an element of \( K \), then either \( x \in B \) or \( {x}^{-1} \in B \) . By (5.20) we may as well assume that \( m\left\lbrack x\right\rbrack \) is not the unit ideal of the ring \( {B}^{\prime } = B\left\lbrack x\right\rbrack \) . Then \( m\left\lbrack x\right\rbrack ... | Yes |
Corollary 5.22. Let \( A \) be a subring of a field \( K \) . Then the integral closure \( \bar{A} \) of \( A \) in \( K \) is the intersection of all the valuation rings of \( K \) which contain \( A \) | Proof. Let \( B \) be a valuation ring of \( K \) such that \( A \subseteq B \) . Since \( B \) is integrally closed, Conversely, let \( x \notin \bar{A} \) . Then \( x \) is not in the ring \( {A}^{\prime } = A\left\lbrack {x}^{-1}\right\rbrack \) . Hence \( {x}^{-1} \) is a non-unit in \( {A}^{\prime } \) and is ther... | No |
Corollary 5.24. Let \( k \) be a field and \( B \) a finitely generated \( k \) -algebra. If \( B \) is a | Proof. Take \( A = k, v = 1 \) and \( \Omega = \) algebraic closure of \( k \) . - | No |
Proposition 6.1. The following conditions on \( \sum \) are equivalent:\ni) Every increasing sequence \( {x}_{1} \leq {x}_{2} \leq \cdots \) in \( \sum \) is stationary (i.e., there\nii) Every non-empty subset of \( \sum \) has a maximal element. | Proof. i) \( \Rightarrow \) ii). If ii) is false there is a non-empty subset \( T \) of \( \sum \) with no maximal element, and we can construct inductively a non-terminating strictly increasing sequence in \( T \) .\n\nii) \( \Rightarrow \) i). The set \( {\left( {x}_{m}\right) }_{m \geq 1} \) has a maximal element, s... | No |
Proposition 6.3. Let \( 0 \rightarrow {M}^{\prime }\xrightarrow[]{\alpha }M\xrightarrow[]{\beta }{M}^{\prime \prime } \rightarrow 0 \) be an exact sequence of\ni) \( M \) is Noetherian \( \Leftrightarrow {M}^{\prime } \) and \( {M}^{\prime \prime } \) are Noetherian;\nii) \( M \) is Artinian \( \Leftrightarrow {M}^{\pr... | Proof. We shall prove i); the proof of ii) is similar.\n\n\( \Rightarrow \) : An ascending chain of submodules of \( {M}^{\prime } \) (or \( {M}^{\prime \prime } \) ) gives rise to a chain in \( M \), hence is stationary.\n\n\n\n\( \Leftarrow \) : Let \( {\left( {L}_{n}\right) }_{n \geq 1} \) be an ascending chain of s... | Yes |
Corollary 6.4. If \( {M}_{i}\left( {1 \leq i \leq n}\right) \) are Noetherian (resp. Artinian) A-modules, so is \( {\bigoplus }_{i = 1}^{n}{M}_{i} \) . | Proof. Apply induction and (6.3) to the exact sequence \[ 0 \rightarrow {M}_{n} \rightarrow {\bigoplus }_{i = 1}^{n}{M}_{i} \rightarrow {\bigoplus }_{i = 1}^{n - 1}{M}_{i} \rightarrow 0. \] | Yes |
Proposition 6.5. Let \( A \) be a Noetherian (resp. Artinian) ring, \( M \) a finitely-generated A-module. Then \( M \) is Noetherian (resp. Artinian). | Proof. \( M \) is a quotient of \( {A}^{n} \) for some \( n \) : apply (6.4) and (6.3). ∎ | No |
Proposition 6.7. Suppose that \( M \) has a composition series of length \( n \) . Then every composition series of \( M \) has length \( n \), and every chain in \( M \) can be extended to a composition series. | Proof. Let \( l\left( M\right) \) denote the least length of a composition series of a module \( M \) .\n\ni) \( N \subset M \Rightarrow l\left( N\right) < l\left( M\right) \) . Let \( \left( {M}_{i}\right) \) be a composition series of \( M \) of minimum length, and consider the submodules \( {N}_{t} = N \cap {M}_{i} ... | Yes |
Proposition 6.9. The length \( l\left( M\right) \) is an additive function on the class of all A-modules of finite length. | Proof. We have to show that if \( 0 \rightarrow {M}^{\prime }\xrightarrow[]{e}M\xrightarrow[]{g}{M}^{\prime \prime } \rightarrow 0 \) is an exact sequence, then \( l\left( M\right) = l\left( {M}^{\prime }\right) + l\left( {M}^{\prime \prime }\right) \) . Take the image under \( \alpha \) of any composition series of \(... | Yes |
Corollary 6.11. Let \( A \) be a ring in which the zero ideal is a product \( {\mathfrak{m}}_{1}\cdots {\mathfrak{m}}_{n} \) of (not necessarily distinct) maximal ideals. Then \( A \) is Noetherian if and only if \( A \) is Artinian. | Proof. Consider the chain of ideals \( A \supset {\mathfrak{m}}_{1} \supseteq {\mathfrak{m}}_{1}{\mathfrak{m}}_{2} \supseteq \cdots \supseteq {\mathfrak{m}}_{1}\cdots {\mathfrak{m}}_{n} = 0 \) . a.c.c. \( \Leftrightarrow \) d.c.c. for each factor. But a.c.c. (resp. d.c.c.) for each factor \( \Leftrightarrow \) a.c.c. (... | Yes |
Proposition 7.3. If \( A \) is Noetherian and \( S \) is any multiplicatively closed subset of \( A \), then \( {S}^{-1}A \) is Noetherian. | Proof. By (3.11-i) and (1.17-iii) the ideals of \( {S}^{-1}A \) are in one-to-one order-preserving correspondence with the contracted ideals of \( A \), hence satisfy the max-of generators, say \( {x}_{1},\ldots ,{x}_{n} \), and it is clear that \( {S}^{-1}a \) is generated by \( {x}_{1}/1,\ldots \) , \( \left. {{x}_{n... | No |
Theorem 7.5. (Hilbert’s Basis Theorem). If \( A \) is Noetherian, then the polynomial ring \( A\left\lbrack x\right\rbrack \) is Noetherian. | Proof. Let \( \mathfrak{a} \) be an ideal in \( A\left\lbrack x\right\rbrack \) . The leading coefficients of the polynomials in \( \mathfrak{a} \) form an ideal \( \mathfrak{l} \) in \( A \) . Since \( A \) is Noetherian, \( \mathfrak{l} \) is finitely generated, say by \( {f}_{i} = {a}_{i}{x}^{{r}_{i}} + \) (lower te... | Yes |
Proposition 7.8. Let \( A \subseteq B \subseteq C \) be rings. Suppose that \( A \) is Noetherian, that \( C \) is finitely generated as an A-algebra and that \( C \) is either (i) finitely generated as a B-module or (ii) integral over B. Then B is finitely generated as an \( A \) -algebra. | Proof. It follows from (5.1) and (5.2) that the conditions (i) and (ii) are equiva-\n\nLet \( {x}_{1},\ldots ,{x}_{m} \) generate \( C \) as an \( A \) -algebra, and let \( {y}_{1},\ldots ,{y}_{n} \) generate \( C \) as a \( B \) -module. Then there exist expressions of the form\n\n\[ \n{x}_{i} = \mathop{\sum }\limits_... | No |
Proposition 7.9. Let \( k \) be a field, \( E \) a finitely generated \( k \) -algebra. If \( E \) is a field then it is a finite algebraic extension of \( k \) . | Proof. Let \( E = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . If \( E \) is not algebraic over \( k \) then we can renumber the \( {x}_{i} \) so that \( {x}_{1},\ldots ,{x}_{r} \) are algebraically independent over \( k \), where \( r \geq 1 \), and each of \( {x}_{r + 1},\ldots ,{x}_{n} \) is algebraic o... | Yes |
Lemma 7.11. In a Noetherian ring \( A \) every ideal is a finite intersection of irreducible ideals. | Proof. Suppose not; then the set of ideals in \( A \) for which the lemma is false is not empty, hence has a maximal element \( a \) . Since \( a \) is reducible, we have irreducible ideals and therefore so is a: contradiction. - | No |
Lemma 7.12. In a Noetherian ring every irreducible ideal is primary. | Let \( {xy} = 0 \) with \( y \neq 0 \), and consider the chain of ideals \( \operatorname{Ann}\left( x\right) \subseteq \operatorname{Ann}\left( {x}^{2}\right) \subseteq \cdots \) . By the a.c.c., this chain is stationary, i.e., we for if \( a \in \left( y\right) \) then \( {ax} = 0 \), and if \( a \in \left( {x}^{n}\r... | No |
Proposition 7.14. In a Noetherian ring \( A \), every ideal \( \mathfrak{a} \) contains a power of | Proof. Let \( {x}_{1},\ldots ,{x}_{k} \) generate \( r\left( \alpha \right) \) : say \( {x}_{i}^{{n}_{i}} \in \alpha \left( {1 \leq i \leq k}\right) \) . Let \( m = \) \( \mathop{\sum }\limits_{{i = 1}}^{k}\left( {{n}_{i} - 1}\right) + 1 \) . Then \( r{\left( a\right) }^{m} \) is generated by the products \( {x}_{1}^{{... | No |
Corollary 7.16. Let \( A \) be a Noetherian ring, \( \mathfrak{m} \) a maximal ideal of \( A,\mathfrak{q} \) any ideal of \( A \) . Then the following are equivalent:\n\nii) \( r\left( q\right) = m \) ;\n\niii) \( {\mathfrak{m}}^{n} \subseteq \mathfrak{q} \subseteq \mathfrak{m} \) for some \( n > 0 \) . | Proof. i) \( \Rightarrow \) ii) is clear; ii) \( \Rightarrow \) i) from (4.2); ii) \( \Rightarrow \) iii) from (7.14); iii) \( \Rightarrow \) ii) by taking radicals: \( \mathfrak{m} = r\left( {\mathfrak{m}}^{n}\right) \subseteq r\left( \mathfrak{q}\right) \subseteq r\left( \mathfrak{m}\right) = \mathfrak{m} \) . ∎ | Yes |
Proposition 7.17. Let \( \mathfrak{a} \neq \left( 1\right) \) be an ideal in a Noetherian ring. Then the prime ideals which belong to a are precisely the prime ideals which occur in the set of ideals \( \left( {\alpha : x}\right) \left( {x \in A}\right) \). | Proof. By passing to \( A/a \) we may assume that \( a = 0 \) . Let \( \mathop{\bigcap }\limits_{{i = 1}}^{n}{q}_{i} = 0 \) be a minimal primary decomposition of the zero ideal, and let \( {\mathfrak{p}}_{i} \) be the radical of \( {q}_{i} \) . 84 NOETHERIAN RINGS\n\nLet \( {\mathfrak{a}}_{i} = \mathop{\bigcap }\limits... | No |
Proposition 8.3. An Artin ring has only a finite number of maximal ideals. | This set has a minimal element, say \( {m}_{1} \cap \cdots \cap {m}_{n} \) ; hence for any maximal ideal \( m \) we have \( m \cap {m}_{1} \cap \cdots \cap {m}_{n} = {m}_{1} \cap \cdots \cap {m}_{n} \), and therefore \( m \supseteq {m}_{1} \cap \cdots \cap {m}_{n} \) . By (1.11) \( m \supseteq {m}_{1} \) for some \( i ... | Yes |
Proposition 8.4. In an Artin ring the nilradical \( \mathfrak{N} \) is nilpotent. | Proof. By d.c.c. we have \( {\mathfrak{N}}^{k} = {\mathfrak{N}}^{k + 1} = \cdots = a \) say, for some \( k > 0 \) . Suppose empty, since \( \alpha \in \sum \) . Let \( \mathfrak{c} \) be a minimal element of \( \sum \) ; then there exists \( x \in \mathfrak{c} \) such that \( {xa} \neq 0 \) ; we have \( \left( x\right)... | No |
Theorem 8.5. \( A \) ring \( A \) is \( A \) rtin \( \Leftrightarrow A \) is Noetherian and \( \dim A = 0 \) . | Proof. \( \Rightarrow : \) By (8.1) we have \( \dim A = 0 \) . Let \( {\mathfrak{m}}_{i}\left( {1 \leq i \leq n}\right) \) be the distinct maximal ideals of \( A \) (8.3). Then \( \mathop{\prod }\limits_{{i = 1}}^{n}{m}_{i}^{k} \subseteq {\left( \mathop{\bigcap }\limits_{{i = 1}}^{n}{m}_{i}\right) }^{k} = {\Re }^{k} = ... | No |
Proposition 8.6. Let \( A \) be a Noetherian local ring, \( \mathfrak{m} \) its maximal ideal. Then exactly one of the following two statements is true: i) \( {\mathfrak{m}}^{n} \neq {\mathfrak{m}}^{n + 1} \) for all \( n \) ; ii) \( {\mathfrak{m}}^{n} = 0 \) for some \( n \), in which case \( A \) is an Artin local ri... | Proof. Suppose \( {m}^{n} = {m}^{n + 1} \) for some \( n \) . By Nakayama’s lemma (2.6) we have \( \mathfrak{m} = \mathfrak{p} \) . Hence \( \mathfrak{m} \) is the only prime ideal of \( A \) and therefore \( A \) is Artinian. | No |
Theorem 8.7. (structure theorem for Artin rings). An Artin ring \( A \) is uniquely (up to isomorphism) a finite direct product of Artin local rings. | Proof. Let \( {m}_{i}\left( {1 \leq i \leq n}\right) \) be the distinct maximal ideals of \( A \) . From the proof of (8.5) we have \( \mathop{\prod }\limits_{{i = 1}}^{n}{\mathfrak{m}}_{i}^{k} = 0 \) for some \( k > 0 \) . By (1.16) the ideals \( {\mathfrak{m}}_{i}^{k} \) are the natural mapping \( A \rightarrow \math... | Yes |
Proposition 8.8. Let \( A \) be an Artin local ring. Then the following are equivalent:\n\ni) every ideal in \( A \) is principal;\n\nii) the maximal ideal \( \mathfrak{m} \) is principal;\n\niii) \( {\dim }_{k}\left( {\mathfrak{m}/{\mathfrak{m}}^{2}}\right) \leq 1 \) . | Proof. i) \( \Rightarrow \) ii) \( \Rightarrow \) iii) is clear.\n\niii) \( \Rightarrow \) i): If \( {\dim }_{k}\left( {\mathfrak{m}/{\mathfrak{m}}^{2}}\right) = 0 \), then \( \mathfrak{m} = {\mathfrak{m}}^{2} \), hence \( \mathfrak{m} = 0 \) by Nakayama’s lemma (2.6), and therefore \( A \) is a field and there is noth... | No |
Proposition 9.2. Let \( A \) be a Noetherian local domain of dimension one, \( \mathfrak{m} \) its maximal ideal, \( k = A/\mathfrak{m} \) its residue field. Then the following are equivalent: i) \( A \) is a discrete valuation ring;\n\nii) \( A \) is integrally closed;\n\niii) \( m \) is a principal ideal;\n\niv) \( {... | Proof. Before we start going the rounds, we make two remarks:\n\n(A) If \( a \) is an ideal \( \neq 0,\left( 1\right) \), then \( a \) is \( m \) -primary and \( a \supseteq {m}^{n} \) for some \( n \) . For \( r\left( a\right) = m \), since \( m \) is the only non-zero prime ideal; now use (7.16).\n\ni) \( \Rightarrow... | Yes |
Theorem 9.3. Let \( A \) be a Noetherian domain of dimension one. Then the following are equivalent:\n\ni) \( A \) is integrally closed;\n\nii) Every primary ideal in \( A \) is a prime power;\n\niii) Every local ring \( {A}_{\mathfrak{p}}\left( {\mathfrak{p} \neq 0}\right) \) is a discrete valuation ring. | Proof. i) \( \Leftrightarrow \) iii) by (9.2) and (5.13). behave well under localization: (4.8), (3.11). | No |
Corollary 9.4. In a Dedekind domain every non-zero ideal has a unique factorization as a product of prime ideals. | Proof. (9.1) and (9.3). | No |
Proposition 9.6. For a fractional ideal \( M \), the following are equivalent:\ni) \( M \) is invertible;\niii) \( M \) is finitely generated and, for each maximal ideal \( \mathfrak{m},{M}_{\mathfrak{m}} \) is invertible. | ii) \( \Rightarrow \) iii) as usual. ideal \( m \) we have \( {a}_{m} = {M}_{m} \cdot \left( {{A}_{m} : {M}_{m}}\right) \) (by (3.11) and (3.15)) \( = {A}_{m} \) because \( {M}_{\mathfrak{m}} \) is invertible. Hence \( \mathfrak{a} \nsubseteq \mathfrak{m} \) . Consequently \( \mathfrak{a} = A \) and therefore \( M \) i... | No |
Proposition 9.7. Let \( A \) be a local domain. Then \( A \) is a discrete valuation ring \( \Leftrightarrow \) every non-zero fractional ideal of \( A \) is invertible. | be a fractional ideal. Then there exists \( y \in A \) such that \( {yM} \subseteq A \) : thus \( {yM} \) is an integral ideal, say \( \left( {x}^{r}\right) \), and therefore \( M = \left( {x}^{r - s}\right) \) where \( s = v\left( y\right) \).\n\n\( \Leftarrow \) : Every non-zero integral ideal is invertible and there... | No |
Theorem 9.8. Let \( A \) be an integral domain. Then \( A \) is a Dedekind domain \( \Leftrightarrow \) | Proof. \( \Rightarrow \) : Let \( M \neq 0 \) be a fractional ideal. Since \( A \) is Noetherian, \( M \) is finitely generated. For each prime ideal \( \mathfrak{p} \neq 0,{M}_{\mathfrak{p}} \) is a fractional ideal \( \neq 0 \) invertible, by (9.6).\n\n\( \Leftarrow \) : Every non-zero integral ideal is invertible, h... | No |
If \( A \) is a Dedekind domain, the non-zero fractional ideals of \( A \) form a group with respect to multiplication. | This group is called the group of ideals of \( A \) ; we denote it by \( I \) . In this terminology (9.4) says that \( I \) is a free (abelian) group, generated by the non-zero ideals: the quotient group \( H = I/P \) is called the ideal class group of \( A \) . The kernel \( U \) of \( \phi \) is the set of all \( u \... | Yes |
Lemma 10.1. Let \( H \) be the intersection of all neighborhoods of 0 in \( G \) . Then i) \( H \) is a subgroup. | Proof. i) follows from the continuity of the group operations. | No |
Proposition 10.2. If \( 0 \rightarrow \left\{ {A}_{n}\right\} \rightarrow \left\{ {B}_{n}\right\} \rightarrow \left\{ {C}_{n}\right\} \rightarrow 0 \) is an exact sequence of inverse systems then \[ 0 \rightarrow \mathop{\lim }\limits_{ \leftarrow }{A}_{n} \rightarrow \mathop{\lim }\limits_{ \leftarrow }{B}_{n} \righta... | Then Ker \( {d}^{A} \cong \mathop{\lim }\limits_{ \leftarrow }{A}_{n} \) . Define \( B, C \) and \( {d}^{B},{d}^{C} \) similarly. The exact sequence of inverse systems then defines a commutative diagram of exact sequences \[ 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 \] \[ d \downarrow d \Downarrow d \do... | Yes |
Corollary 10.3. Let \( 0 \rightarrow {G}^{\prime } \rightarrow G\xrightarrow[]{p}{G}^{\prime \prime } \rightarrow 0 \) be an exact sequence of groups. Let \( G \) have the topology defined by a sequence \( \left\{ {G}_{n}\right\} \) of subgroups, and give \( {G}^{\prime },{G}^{\prime \prime } \) the induced topologies,... | \[ 0 \rightarrow {\widehat{G}}^{\prime } \rightarrow \widehat{G} \rightarrow {\widehat{G}}^{\prime \prime } \rightarrow 0 \] is exact. Proof. Apply (10.2) to the exact sequences \[ 0 \rightarrow \frac{{G}^{\prime }}{{G}^{\prime } \cap {G}_{n}} \rightarrow \frac{G}{{G}_{n}} \rightarrow \frac{{G}^{\prime \prime }}{p{G}_{... | No |
Corollary 10.4. \( {\widehat{G}}_{n} \) is a subgroup of \( \widehat{G} \) | Taking inverse limits in (10.4) we deduce Proposition 10.5. \( \widehat{G} \cong \widehat{G} \) . ∎ | No |
Lemma 10.6. If \( \left( {M}_{n}\right) ,\left( {M}_{n}^{\prime }\right) \) are stable a-filtrations of \( M \), then they have bounded difference: that is, there exists an integer \( {n}_{0} \) such that \( {M}_{n + {n}_{0}} \subseteq {M}_{n} \) and \( {M}_{n + {n}_{0}}^{\prime } \subseteq {M}_{n} \) for all \( n \geq... | \( {a}^{n}M \subseteq {M}_{n} \) ; also \( a{M}_{n} = {M}_{n + 1} \) for all \( n \geq {n}_{0} \) say, hence \( {M}_{n + {n}_{0}} = {a}^{n}{M}_{{n}_{0}} \) \( \subseteq {\mathfrak{a}}^{n}M \) . ∎ | No |
Proposition 10.7. The following are equivalent, for a graded ring \( A \) :\ni) \( A \) is a Noetherian ring;\nii) \( {A}_{0} \) is Noetherian and \( A \) is finitely generated as an \( {A}_{0} \) -algebra. | Proof. i) \( \Rightarrow \) ii). \( {A}_{0} \cong A/{A}_{ + } \), hence is Noetherian. \( {A}_{ + } \) is an ideal in \( A \), hence is finitely generated, say by \( {x}_{1},\ldots ,{x}_{s} \), which we may take to be homogeneous elements of \( A \), of degrees \( {k}_{1},\ldots ,{k}_{s} \) say (all \( > 0 \) ). Let \(... | Yes |
Lemma 10.8. Let \( A \) be a Noetherian ring, \( M \) a finitely-generated \( A \) -module\n\n\( \left( {M}_{n}\right) \) an a-filtration of \( M \) . Then the following are equivalent:\n\n i) \( {M}^{ * } \) is a finitely-generated \( {A}^{ * } \) -module;\n\n ii) The filtration \( \left( {M}_{n}\right) \) is stable. | Proof. Each \( {M}_{n} \) is finitely generated, hence so is each \( {Q}_{n} = {\bigoplus }_{r = 0}^{n}{M}_{r} \) : this is a subgroup of \( {M}^{ * } \) but not (in general) an \( {A}^{ * } \) -submodule. However, it generates one, namely\n\n\( {M}_{n}^{ * } = {M}_{0} \oplus \cdots \oplus {M}_{n} \oplus \mathfrak{a}{M... | Yes |
Proposition 10.13. For any ring \( A \), if \( M \) is finitely-generated, \( \widehat{A}{ \otimes }_{A}M \rightarrow \widehat{M} \) morphism. | Proof. Using (10.3) or otherwise it is clear that a-adic completion commutes with finite direct sums. Hence if \( F \cong {A}^{n} \) we have \( \widehat{A}{ \otimes }_{A}F \cong \widehat{F} \) . Now assume \( M \) is finitely generated so that we have an exact sequence\n\n\[ 0 \rightarrow N \rightarrow F \rightarrow M ... | Yes |
Proposition 10.15. If \( A \) is Noetherian, \( \widehat{A} \) its \( \mathfrak{a} \) -adic completion, then\n\ni) \( \widehat{a} = \widehat{A}a \cong \widehat{A}{ \otimes }_{A}a \) ;\n\nii) \( {\left( {a}^{n}\right) }^{ \frown } = {\left( \widehat{a}\right) }^{n} \) ;\n\niii) \( {\alpha }^{n}/{\alpha }^{n + 1} \cong {... | \[ A{ \otimes }_{A}a \rightarrow \widehat{a} \] we deduce that\n\n\[ \begin{aligned} {\left( {a}^{n}\right) }^{ \frown } & = \widehat{A}{a}^{n} = {\left( \widehat{A}a\right) }^{n} \\ & = {\left( \widehat{a}\right) }^{n} \end{aligned} \] \n\nby (1.18)\n\nby i).\n\nApplying (10.4) we now deduce\n\n\[ A/{a}^{n} \cong \wid... | Yes |
Proposition 10.16. Let \( A \) be a Noetherian local ring, \( \mathfrak{m} \) its maximal ideal. Then the \( \mathfrak{m} \) -adic completion \( \widehat{A} \) of \( A \) is a local ring with maximal ideal \( \widehat{\mathfrak{m}} \) . | Proof. By (10.15) iii) we have \( \widehat{A}/\widehat{\mathfrak{m}} \cong A/\mathfrak{m} \), hence \( \widehat{A}/\widehat{\mathfrak{m}} \) is a field and so \( \widehat{\mathfrak{m}} \) is a maximal ideal. By (10.15) iv) it follows that \( \widehat{m} \) is the Jacobson radical of \( \widehat{A} \) and so is the uniq... | Yes |
Theorem 10.17. Let \( A \) be a Noetherian ring, \( \mathfrak{a} \) an ideal, \( M \) a finitely-generated \( A \) -module and \( \widehat{M} \) the a-completion of \( M \) . Then the kernel \( E = \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{\mathfrak{a}}^{n}M \) of \( M \rightarrow \widehat{M} \) consists of those \(... | Proof. Since \( E \) is the intersection of all neighborhoods of \( 0 \in M \), the topology induced topology on \( E \) coincides with its \( a \) -topology. Since \( {aE} \) is a neighborhood in the \( a \) -topology it follows that \( {aE} = E \) . Since \( M \) is finitely-generated and \( A \) is Noetherian, \( E ... | Yes |
Corollary 10.18. Let \( A \) be a Noetherian domain, \( \mathfrak{a} \neq \left( 1\right) \) an ideal of \( A \). Then \( \bigcap {a}^{n} = 0 \) . | Proof. \( 1 + a \) contains no zero-divisors. - | No |
Corollary 10.19. Let \( A \) be a Noetherian ring, \( \mathfrak{a} \) an ideal of \( A \) contained in the Jacobson radical and let \( M \) be a finitely-generated A-module. Then the \( \mathfrak{a} \) -topology of \( M \) is Hausdorff, i.e. \( \cap {\mathfrak{a}}^{n}M = 0 \) . | Proof. By (1.9) every element of \( 1 + a \) is a unit. - | No |
Let \( A \) be a Noetherian ring, \( \mathfrak{p} \) a prime ideal of \( A \) . Then the intersection of all \( \mathfrak{p} \) -primary ideals of \( A \) is the kernel of \( A \rightarrow {A}_{\mathfrak{p}} \) . | ∎ | No |
Proposition 10.22. Let \( A \) be a Noetherian ring, \( a \) an ideal of \( A \). Then\n\ni) \( {G}_{\mathfrak{a}}\left( A\right) \) is Noetherian;\n\nii) \( {G}_{a}\left( A\right) \) and \( {G}_{\widehat{a}}\left( \widehat{A}\right) \) are isomorphic as graded rings;\n\niii) if \( M \) is a finitely-generated \( A \) ... | Proof. i) Since \( A \) is Noetherian, \( a \) is finitely generated, say by \( {x}_{1},\ldots ,{x}_{n} \). Let Noetherian, \( G\left( A\right) \) is Noetherian by the Hilbert basis theorem.\n\nii) \( {a}^{n}/{a}^{n + 1} \cong {\widehat{a}}^{n}/{\widehat{a}}^{n + 1} \) by (10.15) iii).\n\niii) There exists \( {n}_{0} \... | Yes |
Lemma 10.23. Let \( \phi : A \rightarrow B \) be a homomorphism of filtered groups, i.e. morphisms of the associated graded and completed groups. Then\ni) \( G\left( \phi \right) \) injective \( \Rightarrow \widehat{\phi } \) injective; | Proof. Consider the commutative diagram of exact sequences\n\n\[ \begin{matrix} 0 \rightarrow {A}_{n}/{A}_{n + 1} \rightarrow A/{A}_{n + 1} \rightarrow A/{A}_{n} \rightarrow 0 \\ { \downarrow }^{{G}_{n}}{}^{\left( \phi \right) } \downarrow { \downarrow }^{{\alpha }_{n + 1}}{ \downarrow }^{{\alpha }_{n}} \end{matrix} \]... | Yes |
Proposition 10.24. Let \( A \) be a ring, \( a \) an ideal of \( A, M \) an \( A \) -module, \( \left( {M}_{n}\right) \) an Hausdorff in its filtration topology (i.e. that \( \left. {{ \cap }_{n}{M}_{n} = 0}\right) \) . Suppose also that \( G\left( M\right) \) is a finitely-generated \( G\left( A\right) \) -module. The... | Proof. Pick a finite set of generators of \( G\left( M\right) \), and split them up into their homogeneous components, say \( {\xi }_{i}\left( {1 \leq i \leq v}\right) \) where \( {\xi }_{i} \) has degree say \( n\left( i\right) \), and is therefore the image of say \( {x}_{i} \in {M}_{n\left( i\right) } \) . Let \( {F... | Yes |
Corollary 10.25. With the hypotheses of (10.24), if \( G\\left( M\\right) \) is a Noetherian \( G\\left( A\\right) \) -module, then \( M \) is a Noetherian \( A \) -module. | Proof. We have to show that every submodule \( {M}^{\prime } \) of \( M \) is finitely generated (6.2). Let \( {M}_{n}^{\prime } = {M}^{\prime } \\cap {M}_{n} \) ; then \( \\left( {M}_{n}^{\prime }\\right) \) is an a-filtration of \( {M}^{\prime } \), and the em- \( {M}_{n}/{M}_{n + 1} \), hence to an embedding of \( G... | Yes |
Theorem 10.26. If \( A \) is a Noetherian ring, a an ideal of \( A \), then the a- | Proof. By (10.22) we know that\n\n\[ \n{G}_{\mathrm{a}}\left( A\right) = {G}_{\mathrm{a}}\left( \widehat{A}\right) \]\n\nis Noetherian. Now apply (10.25) to the complete ring \( \widehat{A} \), taking \( M = \widehat{A} \) (filtered by \( {\widehat{a}}^{n} \), and so Hausdorff). - | No |
Corollary 10.27. If \( A \) is a Noetherian ring, the power series ring \( B = \)\n\n\( A\left\lbrack \left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \right\rbrack \) in \( n \) variables is Noetherian. In particular \( k\left\lbrack \left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \right\rbrack \) (k a field) i... | Proof. \( A\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) is Noetherian by the Hilbert basis theorem, and \( B \) is its | No |
Theorem 11.1. (Hilbert, Serre). \( P\left( {M, t}\right) \) is a rational function in \( t \) of the form\n\n\( f\left( t\right) /\mathop{\prod }\limits_{{i = 1}}^{s}\left( {1 - {t}^{{k}_{i}}}\right) \), where \( f\left( t\right) \in \mathbf{Z}\left\lbrack t\right\rbrack \) . | Proof. By induction on \( s \), the number of generators of \( A \) over \( {A}_{0} \) . Start with \( s = 0 \) ; this means that \( {A}_{n} = 0 \) for all \( n > 0 \), so that \( A = {A}_{0} \) and \( M \) is a finitely-in this case.\n\nNow suppose \( s > 0 \) and the theorem true for \( s - 1 \) . Multiplication by \... | Yes |
Corollary 11.2. If each \( {k}_{i} = 1 \), then for all sufficiently large \( n,\lambda \left( {M}_{n}\right) \) is a polynomial in \( n \) (with rational coefficients) of degree* \( d - 1 \) . | Proof. By (11.1) we have \( \lambda \left( {M}_{n}\right) = \) coefficient of \( {t}^{n} \) in \( f\left( t\right) \cdot {\left( 1 - t\right) }^{-s} \) . Canceling powers of \( \left( {1 - t}\right) \) we may assume \( s = d \) and \( f\left( 1\right) \neq 0 \) . Suppose \( f\left( t\right) = \) \( \mathop{\sum }\limit... | Yes |
Proposition 11.4. Let \( A \) be a Noetherian local ring, \( \mathfrak{m} \) its maximal ideal, \( \mathfrak{q} \) an m-primary ideal, \( M \) a finitely-generated \( A \) -module, \( \left( {M}_{n}\right) \) a stable \( \mathfrak{q} \) - filtration of \( M \) . Then\ni) \( M/{M}_{n} \) is of finite length, for each \(... | Proof. i) Let \( G\left( A\right) = {\bigoplus }_{n}{\mathfrak{q}}^{n}/{\mathfrak{q}}^{n + 1}, G\left( M\right) = {\bigoplus }_{n}{M}_{n}/{M}_{n + 1}.{G}_{0}\left( A\right) = A/\mathfrak{q} \) is an Artin local ring, say by (8.5); \( G\left( A\right) \) is Noetherian, and \( G\left( M\right) \) is a finitely-generated ... | Yes |
Proposition 11.6. If \( A,\mathfrak{m},\mathfrak{q} \) are as above\n\n\[ \deg {\chi }_{\mathfrak{q}}\left( n\right) = \deg {\chi }_{\mathfrak{m}}\left( n\right) \] | Proof. We have \( \mathfrak{m} \supseteq \mathfrak{q} \supseteq {\mathfrak{m}}^{r} \) for some \( r \) by (7.16), hence \( {\mathfrak{m}}^{n} \supseteq {\mathfrak{q}}^{n} \supseteq {\mathfrak{m}}^{m} \) and therefore\n\n\( {\chi }_{\mathfrak{m}}\left( n\right) \leq {\chi }_{\mathfrak{q}}\left( n\right) \leq {\chi }_{\m... | Yes |
Proposition 11.7. \( \delta \left( A\right) \geq d\left( A\right) \) | ∎ | No |
Proposition 11.8. Let \( A,\mathfrak{m},\mathfrak{q} \) be as before. Let \( M \) be a finitely-generated \( A \) -module, \( x \in A \) a non-zero-divisor in \( M \) and \( {M}^{\prime } = M/{xM} \) . Then\n\n\[ \deg {\chi }_{q}^{{M}^{\prime }} \leq \deg {\chi }_{q}^{M} - 1 \]\non \( x \) . Let \( {N}_{n} = N \cap {q}... | Proof. Put \( M = A \) in (11.8). ∎ | Yes |
Proposition 11.10. \( d\left( A\right) \geq \dim A \) . | Proof. By induction on \( d = d\left( A\right) \) . If \( d = 0 \) then \( l\left( {A/{\mathfrak{m}}^{n}}\right) \) is constant for all large \( n \), hence \( {\mathfrak{m}}^{n} = {\mathfrak{m}}^{n + 1} \) for some \( n \), hence \( {\mathfrak{m}}^{n} = 0 \) by Nakayama’s lemma (2.6). Thus \( A \) is an Artin ring and... | Yes |
Corollary 11.11. If \( A \) is a Noetherian local ring, \( \dim A \) is finite. | ∎ | No |
Proposition 11.13. Let \( A \) be a Noetherian local ring of dimension \( d \) . Then there exists an \( \mathfrak{m} \) -primary ideal in \( A \) generated by \( d \) elements \( {x}_{1},\ldots ,{x}_{d} \) , and therefore \( \dim A \geq \delta \left( A\right) \) . containing \( \left( {{x}_{1},\ldots ,{x}_{i}}\right) ... | Suppose \( i > 0 \) and \( {x}_{1},\ldots ,{x}_{i - 1} \) constructed. Let \( {\mathfrak{p}}_{j}\left( {1 \leq j \leq s}\right) \) be the minimal prime ideals (if any) of height \( \mathfrak{m} \), we have \( \mathfrak{m} \neq {\mathfrak{p}}_{j}\left( {1 \leq j \leq s}\right) \), hence \( \mathfrak{m} \neq \mathop{\big... | Yes |
Theorem 11.14. (Dimension theorem.) For any Noetherian local ring \( A \) the following three integers are equal: i) the maximum length of chains of prime ideals in \( A \) ; ii) the degree of the characteristic polynomial \( {\chi }_{\mathfrak{m}}\left( n\right) = l\left( {A/{\mathfrak{m}}^{n}}\right) \) ; iii) the le... | Proof. (11.7), (11.10), (11.13). | No |
Corollary 11.15. \( \dim A \leq {\dim }_{k}\left( {\mathfrak{m}/{\mathfrak{m}}^{2}}\right) \) . | Proof. If \( {x}_{i} \in \mathfrak{m}\left( {1 \leq i \leq s}\right) \) are such that their images in \( \mathfrak{m}/{\mathfrak{m}}^{2} \) form a basis of this vector space, then the \( {x}_{i} \) generate \( m \) by \( \left( {2.8}\right) \) ; hence \( {\dim }_{k}\left( {m/{m}^{2}}\right) = \) \( s \geq \dim A \) by ... | Yes |
Corollary 11.16. Let \( A \) be a Noetherian ring, \( {x}_{1},\ldots ,{x}_{r} \in A \) . Then every minimal ideal \( \mathfrak{p} \) belonging to \( \left( {{x}_{1},\ldots ,{x}_{r}}\right) \) has height \( \leq r \) . | Proof. In \( {A}_{\mathfrak{p}} \) the ideal \( \left( {{x}_{1},\ldots ,{x}_{r}}\right) \) becomes \( {\mathfrak{p}}^{e} \) -primary, hence \( r \geq \dim {A}_{\mathfrak{p}} = \) height \( \varphi \) . - | No |
Corollary 11.17. (Krull’s principal ideal theorem). Let \( A \) be a Noetherian ring and let \( x \) be an element of \( A \) which is neither a zero-divisor nor a unit. Then every minimal prime ideal \( \mathfrak{p} \) of \( \left( x\right) \) has height 1. | Proof. By (11.16), height \( \mathfrak{p} \leq 1 \) . If height \( \mathfrak{p} = 0 \), then \( \mathfrak{p} \) is a prime ideal belonging to 0, hence every element of \( \mathfrak{p} \) is a zero-divisor by (4.7): contra- | No |
Proposition 11.20. Let \( {x}_{1},\ldots ,{x}_{d} \) be a system of parameters for \( A \) and let \( q = \left( {{x}_{1},\ldots ,{x}_{d}}\right) \) be the m-primary ideal generated by them. Let \( f\left( {{t}_{1},\ldots ,{t}_{d}}\right) \) that \( f\left( {{x}_{1},\ldots ,{x}_{d}}\right) \in {\mathfrak{q}}^{s + 1} \)... | Proof. Consider the epimorphism of graded rings \( \alpha : \left( {A/\mathfrak{q}}\right) \left\lbrack {{t}_{1},\ldots ,{t}_{d}}\right\rbrack \rightarrow {G}_{\mathfrak{q}}\left( A\right) \) given by \( {t}_{i} \rightarrow {\bar{x}}_{i} \), where \( {t}_{i} \) are indeterminates and \( {\bar{x}}_{i} \) is \( {x}_{i}{\... | Yes |
Corollary 11.21. If \( k \subset A \) is a field mapping isomorphically onto \( A/\mathfrak{m} \) and if \( {x}_{1},\ldots ,{x}_{d} \) is a system of parameters, then \( {x}_{1},\ldots ,{x}_{d} \) are algebraically independent over \( k \) . | Proof. Assume \( f\left( {{x}_{1},\ldots ,{x}_{d}}\right) = 0 \) where \( f \) is a polynomial with coefficients in \( k \) . If \( f ≢ 0 \) we can write \( f = {f}_{s} + \) higher terms, where \( {f}_{s} \) is homogeneous of degree in \( \mathfrak{m} \) . Since \( {f}_{s} \) has coefficients in \( k \) this implies \(... | Yes |
Theorem 11.22. Let \( A \) be a Noetherian local ring of dimension \( d,\mathfrak{m} \) its maximal ideal, \( k = A/\mathfrak{m} \). Then the following are equivalent:\n\ni) \( {G}_{\mathrm{m}}\left( A\right) \cong k\left\lbrack {{t}_{1},\ldots ,{t}_{d}}\right\rbrack \) where the \( {t}_{\mathrm{i}} \) are independent ... | Proof. i) \( \Rightarrow \) ii) is clear. ii) \( \Rightarrow \) iii) by (2.8): see the proof of (11.15). iii) \( \Rightarrow \) i): isomorphism of graded rings. - | No |
Proposition 11.24. Let \( A \) be a Noetherian local ring. Then \( A \) is regular if | Proof. By (10.16),(10.26) and (11.19) we know that \( \widehat{A} \) is a Noetherian local ring of the same dimension as \( A \) and with \( \widehat{m} \) as maximal ideal. Now use (10.22) | No |
Lemma 11.26. Let \( B \subseteq A \) be integral domains with \( B \) integrally closed and \( A \) integral over \( B \) . Let \( \mathfrak{m} \) be a maximal ideal of \( A \), and let \( \mathfrak{n} = \mathfrak{m} \cap B \) . Then \( \mathfrak{n} \) is maximal and \( \dim {A}_{\mathfrak{m}} = \dim {B}_{\mathfrak{n}}... | Proof. This is an easy consequence of the results of Chapter 5. First \( n \) is\n\n\[ m \supset {q}_{1} \supset {q}_{2} \supset \cdots \supset {q}_{d} \]\n\n(1)\n\nis a strict chain of primes in \( A \), its intersection with \( B \) is by (5.9) a strict chain of primes\n\n\[ n \supset {\mathfrak{p}}_{1} \supset {\mat... | No |
Corollary 11.27. For every maximal ideal \( \mathfrak{m} \) of \( A\left( V\right) \) we have\n\n\[ \dim A\left( V\right) = \dim A{\left( V\right) }_{\mathfrak{m}}. \] | all \( A{\left( V\right) }_{\mathrm{m}} \) have the same dimension. - | No |
Describe the set \( \Gamma \left( {U,\mathcal{O}}\right) \) of all functions holomorphic in a given domain \( U \) . | For example, the Liouville theorem asserts that \( \Gamma \left( {\mathbb{C}\cup \{ \infty \} ,\mathcal{O}}\right) \) consists of constant functions only. | No |
Theorem 5.1 (Cayley). Every group is isomorphic to a collection of permutations. | Proof. We have just seen that from the columns of any group's multiplication table, we can create a permutation for each element of the group, as Figure 5.31 exemplifies. We can also make a multiplication table out of those permutations. This proof explains why such a multiplication table must behave the same as the or... | Yes |
Theorem 6.7. If \( H \) is a subgroup of \( G \), then each element of \( G \) belongs to exactly one left coset of \( H \) . | Proof. Suppose the element \( g \) of \( G \) appears to belong to two different left cosets, say \( {aH} \) and \( {bH} \) . In such a case, \( {aH} \) and \( {bH} \) must actually be two different names for the same coset (in the sense of Observation 6.5); here's why.\n\nSince \( g \) is in \( {aH} \), we can conclud... | Yes |
Theorem 6.8 (Lagrange’s Theorem). If \( H < G \), then the order \( \left| H\right| \) of the subgroup divides the order \( \left| G\right| \) of the larger group. | Proof. Theorem 6.7 proved that the group \( G \) is partitioned into copies of \( H \) . So the size of \( G \) can be determined just by counting how many copies of \( H \) there are and multiplying that number by the size of each one, the number \( \left| H\right| \) . So if \( n \) is the number of left cosets (incl... | Yes |
Theorem 7.6. If \( H < G \), then a quotient group \( \frac{G}{H} \) can be constructed just when \( H \vartriangleleft G \) . | Proof. The quotient process from Definition 7.5 succeeds just when the resulting diagram is a valid Cayley diagram. Most aspects of valid Cayley diagrams are guaranteed by the quotient process. For instance, because we begin with a diagram that has an arrow of every color exiting every node, our resulting diagram has t... | No |
Theorem 8.5 (Fundamental Homomorphism Theorem). If \( \phi : G \rightarrow H \) is a homomorphism, then \( \operatorname{Im}\left( \phi \right) \cong \frac{G}{\operatorname{Ker}\left( \phi \right) } \) . | Proof. For any homomorphism \( \phi : G \rightarrow H \), we know from Observations 8.2 to 8.4 that \( \operatorname{Ker}\left( \phi \right) \vartriangleleft G \), so we know that we can take a quotient \( \frac{G}{\operatorname{Ker}\left( \phi \right) } \) and obtain a group. Now I must explain why that group is isomo... | Yes |
Theorem 8.7. \( {C}_{n} \times {C}_{m} \cong {C}_{nm} \) if and only if \( n \) and \( m \) are relatively prime. | Proof. If \( {C}_{n} \times {C}_{m} \) is cyclic, it must be generated by one of its elements; let’s call it \( \left( {a, b}\right) \) because we do not know specifically which \( a \in {C}_{n} \) or \( b \in {C}_{m} \) it involves. Because the orbit of \( \left( {a, b}\right) \) includes every element in \( {C}_{n} \... | No |
Theorem 9.4 (Orbit-Stabilizer Theorem). The size of an element's orbit times the size of its stabilizer is the size of the group. | Proof. Because Stab \( \left( s\right) \) is a subgroup of \( G \), the definition of subgroup index (Definition 6.9) tells us that\n\n\[ \underset{\text{size of subgroup }}{\underbrace{\left| \operatorname{Stab}\left( S\right) \right| }} \cdot \underset{\text{number of cosets }}{\underbrace{\left\lbrack G : \operatorn... | Yes |
Theorem 9.5. If a group \( G \) of prime order \( p \) acts on a set \( S \), then the order of \( S \) and the number of stable elements in \( S \) are congruent mod \( p \) . | Proof. The Orbit-Stabilizer Theorem tells us that the size of each \( \operatorname{Orb}\left( s\right) \) is a factor of \( \left| G\right| \), so when \( \left| G\right| \) is a prime \( p \), all orbits have size 1 or \( p \) . The stable elements are in orbits of size 1 (each by itself) and the rest of \( S \) is p... | Yes |
Theorem 9.6 (Cauchy’s Theorem). If \( p \) is a prime number that divides \( \left| G\right| \), then \( G \) has an element \( g \) of order \( p \), and therefore a subgroup \( \langle g\rangle \) of order \( p \) . | Proof. Because \( p \) is prime, if I find some \( g \neq e \) satisfying \( {g}^{p} = e \), then \( g \) must have order \( p \) . Exercise 9.15 asks you to explain why this is so, but for now I use the fact without justification.\n\nIn most groups there are lots of ways to multiply \( p \) group elements together and... | No |
Theorem 9.8. If a p-group \( G \) acts on a set \( S \), then the order of \( S \) and the number of stable elements in \( S \) are congruent mod \( p \) . | Proof. As in the proof of Theorem 9.5, the size of each orbit must divide the order of \( G \) . In this case, only powers of \( p \) divide \( \left| G\right| \), so the orbits are therefore of various sizes including \( 1, p,{p}^{2},{p}^{3} \), up to at most \( {p}^{n} \) . So simply modify the illustration for Theor... | No |
Theorem 9.9. If \( H \) is a p-subgroup of \( G \), then \( \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack { \equiv }_{p}\left\lbrack {G : H}\right\rbrack \) . | Proof. Let \( S \) be the left cosets of \( H \) in \( G \) and consider the group \( H \) acting on \( S \) by the interpretation homomorphism \( \phi : G \rightarrow \operatorname{Perm}\left( S\right) \) defined by\n\n\[ \phi \left( h\right) = \text{the permutation that sends a coset}{gH}\text{to the coset}{hgH}\text... | No |
Theorem 9.10 (First Sylow Theorem). If \( G \) is a group and \( {p}^{n} \) is the highest power of \( p \) dividing \( \left| G\right| \), then there are subgroups of \( G \) of every order \( 1, p,{p}^{2},{p}^{3} \), up to \( {p}^{n} \) . Also, every p-subgroup with fewer than \( {p}^{n} \) elements is inside one of ... | Proof. It is easy to find a \( p \) -subgroup of order 1 (which is \( {p}^{0} \) ) because it is obviously \( \{ e\} \) . We also know that there is a \( p \) -subgroup of order \( p \) (which is \( {p}^{1} \) ) from Cauchy’s | No |
Theorem 9.12 (Second Sylow Theorem). Any two Sylow p-subgroups are conjugates. | Proof. I use again the strategy described after the statement of Theorem 9.9. Let \( S \) be the left cosets of some Sylow \( p \) -subgroup \( H < G \) and have another Sylow \( p \) -subgroup \( K \) act on \( S \) by left multiplication (as in the proof of Theorem 9.9).\n\nA stable element of this action is a left c... | Yes |
Theorem 9.13 (Third Sylow Theorem). The number \( n \) of Sylow p-subgroups of \( G \) obeys the following two restrictions.\n\n\( n \) divides \( \left| G\right| \)\n\[ n{ \equiv }_{p}1 \] | Proof. The first of the two restrictions is the easier to prove. Let \( H \) be one of the \( n \) Sylow \( p \) -subgroups of \( G \) . We know that the set of Sylow \( p \) -subgroups is the set of conjugates of \( H \) . Another way to say this is that if \( G \) acts on its subgroups by conjugation, then the set of... | Yes |
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