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Proposition 2.9. i) Let\n\n\\[ \n{M}^{\prime }\xrightarrow[]{u}M\xrightarrow[]{v}{M}^{\prime \prime } \rightarrow 0 \n\\] \n\n(4) \n\nbe a sequence of \\( A \\) -modules and homomorphisms. Then the sequence (4) is exact \\( \\Leftrightarrow \\) for all \\( A \\) -modules \\( N \\), the sequence \n\n\\[ \n0 \rightarrow ... | All four parts of this proposition are easy exercises. For example, suppose that \\( \\left( {4}^{\prime }\\right) \\) is exact for all \\( N \\) . First of all, since \\( \\bar{v} \\) is injective for all \\( N \\) it follows that \\( v \\) is surjective. Next, we have \\( \\bar{u} \\circ \\bar{v} = 0 \\), that is \\(... | No |
Proposition 2.10. Let\n\n\[ \n0 \rightarrow {M}^{\prime }\xrightarrow[]{u}M\xrightarrow[]{v}{M}^{\prime \prime } \rightarrow 0 \]\n\n\[ \n\begin{array}{l} {f}^{\prime } \downarrow \;{f}_{ \downarrow }\;{ \downarrow }^{{f}^{\prime \prime }} \\ \end{array} \]\n\n\[ \n0 \rightarrow {N}^{\prime }\underset{{u}^{\prime }}{ \... | The boundary homomorphism \( d \) is defined as follows: if \( {x}^{\prime \prime } \in \operatorname{Ker}\left( {f}^{\prime \prime }\right) \), we have \( {x}^{\prime \prime } = v\left( x\right) \) for some \( x \in M \), and \( {v}^{\prime }\left( {f\left( x\right) }\right) = {f}^{\prime \prime }\left( {v\left( x\rig... | No |
Proposition 2.11. Let \( 0 \rightarrow {M}_{0} \rightarrow {M}_{1} \rightarrow \cdots \rightarrow {M}_{n} \rightarrow 0 \) be an exact sequence of \( A \) -modules in which all the modules \( {M}_{i} \) and the kernels of all the homomorphisms belong to \( C \) . Then for any additive function \( \lambda \) on \( C \) ... | Proof. Split up the sequence into short exact sequences\n\n\[ 0 \rightarrow {N}_{i} \rightarrow {M}_{i} \rightarrow {N}_{i + 1} \rightarrow 0 \]\n\n\( \left( {{N}_{0} = {N}_{n + 1} = 0}\right) \) . Then we have \( \lambda \left( {M}_{i}\right) = \lambda \left( {N}_{i}\right) + \lambda \left( {N}_{i + 1}\right) \) . Now... | Yes |
Proposition 2.12. Let \( M, N \) be \( A \) -modules. Then there exists a pair \( \left( {T, g}\right) \) consisting of an A-module \( T \) and an \( A \) -bilinear mapping \( g : M \times N \rightarrow T \), with the following property:\n\nGiven any A-module \( P \) and any A-bilinear mapping \( f : M \times N \righta... | Proof. i) Uniqueness. Replacing \( \left( {P, f}\right) \) by \( \left( {{T}^{\prime },{g}^{\prime }}\right) \) we get a unique \( j : T \rightarrow {T}^{\prime } \) such that \( {g}^{\prime } = j \circ g \) . Interchanging the roles of \( T \) and \( {T}^{\prime } \), we get \( {j}^{\prime } : {T}^{\prime } \rightarro... | Yes |
Corollary 2.13. Let \( {x}_{i} \in M,{y}_{i} \in N \) be such that \( \sum {x}_{i} \otimes {y}_{i} = 0 \) in \( M \otimes N \) . Then there exist finitely generated submodules \( {M}_{0} \) of \( M \) and \( {N}_{0} \) of \( N \) such that \( \sum {x}_{i} \otimes {y}_{i} = 0 \) in \( {M}_{0} \otimes {N}_{0}. \) | Proof. If \( \sum {x}_{i} \otimes {y}_{i} = 0 \) in \( M \otimes N \), then in the notation of the proof of (2.12) we have \( \sum \left( {{x}_{i},{y}_{i}}\right) \in D \), and therefore \( \sum \left( {{x}_{i},{y}_{i}}\right) \) is a finite sum of generators of \( D \) . Let \( {M}_{0} \) be the submodule of \( M \) g... | Yes |
Proposition 2.14. Let \( M, N, P \) be \( A \) -modules. Then there exist unique isomorphisms\ni)\( M \otimes N \rightarrow N \otimes M \)\nii)\( \left( {M \otimes N}\right) \otimes P \rightarrow M \otimes \left( {N \otimes P}\right) \rightarrow M \otimes N \otimes P \)\niii)\( \left( {M \oplus N}\right) \otimes P \rig... | Proof. In each case the point is to show that the mappings so described are well defined. The technique is to construct suitable bilinear or multilinear mappings, and use the defining property (2.12) or (2.12*) to infer the existence of homomorphisms of tensor products. We shall prove half of ii) as an example of the m... | No |
Proposition 2.16. Suppose \( N \) is finitely generated as a B-module and that \( B \) is finitely generated as an A-module. Then \( N \) is finitely generated as an \( A \) -module. | Proof. Let \( {y}_{1},\ldots ,{y}_{n} \) generate \( N \) over \( B \), and let \( {x}_{1},\ldots ,{x}_{m} \) generate \( B \) as an \( A \) -module. Then the \( {mn} \) products \( {x}_{i}{y}_{j} \) generate \( N \) over \( A \) . | Yes |
Proposition 2.17. If \( M \) is finitely generated as an \( A \) -module, then \( {M}_{B} \) is finitely generated as a B-module. | Proof. If \( {x}_{1},\ldots ,{x}_{m} \) generate \( M \) over \( A \), then the \( 1 \otimes {x}_{i} \) generate \( {M}_{B} \) over \( B \) . | Yes |
Proposition 2.18. Let\n\n\[ \n{M}^{\prime }\xrightarrow[]{f}M\xrightarrow[]{g}{M}^{\prime \prime } \rightarrow 0 \]\n\n(2)\n\nbe an exact sequence of \( A \) -modules and homomorphisms, and let \( N \) be any\n\nA-module. Then the sequence\n\n\[ \n{M}^{\prime } \otimes N\xrightarrow[]{f \otimes 1}M \otimes N\xrightarro... | Proof. Let \( E \) denote the sequence (2), and let \( E \otimes N \) denote the sequence (3). Let \( P \) be any \( A \) -module. Since (2) is exact, the sequence Hom \( \left( {E,\operatorname{Hom}\left( {N, P}\right) }\right) \) is exact by (2.9); hence by (1) the sequence Hom \( \left( {E \otimes N, P}\right) \) is... | Yes |
Proposition 2.19. The following are equivalent, for an A-module \( N \) :\ni) \( N \) is flat.\nii) If \( 0 \rightarrow {M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \rightarrow 0 \) is any exact sequence of \( A \) -modules, the\ntensored sequence \( 0 \rightarrow {M}^{\prime } \otimes N \rightarrow M \... | Proof. i) \( \Leftrightarrow \) ii) by splitting up a long exact sequence into short exact sequences.\nii) \( \Leftrightarrow \) iii) by (2.18).\niii) \( \cdot \Rightarrow \) iv): clear.\niv) \( \Rightarrow \) iii). Let \( f : {M}^{\prime } \rightarrow M \) be injective and let \( u = \sum {x}_{i} \otimes {y}_{i} \in \... | Yes |
Proposition 3.1. Let \( g : A \rightarrow B \) be a ring homomorphism such that \( g\left( s\right) \) is a unit in \( B \) for all \( s \in S \) . Then there exists a unique ring homomorphism \( h : {S}^{-1}A \rightarrow B \) such that \( g = h \circ f \) . | Proof. i) Uniqueness. If \( h \) satisfies the conditions, then \( h\left( {a/1}\right) = {hf}\left( a\right) = g\left( a\right) \) for all \( a \in A \) ; hence, if \( s \in S \), \[ h\left( {1/s}\right) = h\left( {\left( s/1\right) }^{-1}\right) = h{\left( s/1\right) }^{-1} = g{\left( s\right) }^{-1} \] and therefore... | Yes |
Corollary 3.2. If \( g : A \rightarrow B \) is a ring homomorphism such that\n\ni) \( s \in S \Rightarrow g\left( s\right) \) is a unit in \( B \) ;\n\nii) \( g\left( a\right) = 0 \Rightarrow {as} = 0 \) for some \( s \in S \) ;\n\niii) Every element of \( B \) is of the form \( g\left( a\right) g{\left( s\right) }^{-1... | Proof. By (3.1) we have to show that \( h : {S}^{-1}A \rightarrow B \), defined by\n\n\[ h\left( {a/s}\right) = g\left( a\right) g{\left( s\right) }^{-1} \]\n\n(this definition uses i)) is an isomorphism. By iii), \( h \) is surjective. To show \( h \) is injective, look at the kernel of \( h \) : if \( h\left( {a/s}\r... | Yes |
Proposition 3.3. The operation \( {S}^{-1} \) is exact, i.e., if \( {M}^{\prime }\xrightarrow[]{f}M\xrightarrow[]{g}{M}^{\prime \prime } \) is exact at \( M \), then \( {S}^{-1}{M}^{\prime }\xrightarrow[]{s - {1f}}{S}^{-1}M\xrightarrow[]{s - {1g}}{S}^{-1}{M}^{\prime \prime } \) is exact at \( {S}^{-1}M \) . | Proof. We have \( g \circ f = 0 \), hence \( {S}^{-1}g \circ {S}^{-1}f = {S}^{-1}\left( 0\right) = 0 \), hence \( \operatorname{Im}\left( {{S}^{-1}f}\right) \) \( \subseteq \operatorname{Ker}\left( {{S}^{-1}g}\right) \) . To prove the reverse inclusion, let \( m/s \in \operatorname{Ker}\left( {{S}^{-1}g}\right) \), the... | Yes |
Corollary 3.4. Formation of fractions commutes with formation of finite sums, finite intersections and quotients. Precisely, if \( N, P \) are submodules of an \( A \) -module \( M \), then\ni) \( {S}^{-1}\left( {N + P}\right) = {S}^{-1}\left( N\right) + {S}^{-1}\left( P\right) \)\nii) \( {S}^{-1}\left( {N \cap P}\righ... | Proof. i) follows readily from the definitions and ii) is easy to verify: if \( y/s = z/t\left( {y \in N, z \in P, s, t \in S}\right) \) then \( u\left( {{ty} - {sz}}\right) = 0 \) for some \( u \in S \), hence \( w = {uty} = {usz} \in N \cap P \) and therefore \( y/s = w/{stu} \in {S}^{-1}\left( {N \cap P}\right) \) .... | Yes |
Let \( M \) be an \( A \) -module. Then the \( {S}^{-1}A \) modules \( {S}^{-1}M \) and \( {S}^{-1}A{ \otimes }_{A}M \) are isomorphic; more precisely, there exists a unique isomorphism \( f : {S}^{-1}A{ \otimes }_{A}M \rightarrow {S}^{-1}M \) for which\n\n\[ f\left( {\left( {a/s}\right) \otimes m}\right) = {am}/s\text... | Proof. The mapping \( {S}^{-1}A \times M \rightarrow {S}^{-1}M \) defined by\n\n\[ \left( {a/s, m}\right) \mapsto {am}/s \]\n\n is \( A \) -bilinear, and therefore by the universal property (2.12) of the tensor product induces an \( A \) -homomorphism\n\n\[ f : {S}^{-1}A{ \otimes }_{A}M \rightarrow {S}^{-1}M \]\n\nsati... | Yes |
Corollary 3.6. \( {S}^{-1}A \) is a flat \( A \) -module. | Proof. (3.3), (3.5). | No |
Proposition 3.7. If \( M, N \) are \( A \) -modules, there is a unique isomorphism of \( {S}^{-1}A \) -modules \( f : {S}^{-1}M{ \otimes }_{{S}^{-1}A}{S}^{-1}N \rightarrow {S}^{-1}\left( {M{ \otimes }_{A}N}\right) \) such that \[ f\left( {\left( {m/s}\right) \otimes \left( {n/t}\right) }\right) = \left( {m \otimes n}\r... | Proof. Use (3.5) and the canonical isomorphisms of Chapter 2. | No |
Proposition 3.8. Let \( M \) be an \( A \) -module. Then the following are equivalent:\ni) \( M = 0 \) ;\nii) \( {M}_{\mathfrak{p}} = 0 \) for all prime ideals \( \mathfrak{p} \) of \( A \) ;\niii) \( {M}_{\mathfrak{m}} = 0 \) for all maximal ideals \( \mathfrak{m} \) of \( A \) . | Proof. Clearly i) \( \Rightarrow \) ii) \( \Rightarrow \) iii). Suppose iii) satisfied and \( M \neq 0 \) . Let \( x \) be a non-zero element of \( M \), and let \( \mathfrak{a} = \operatorname{Ann}\left( x\right) \) ; \( \mathfrak{a} \) is an ideal \( \neq \left( 1\right) \), hence is contained in a maximal ideal \( m... | Yes |
Proposition 3.10. For any A-module \( M \), the following statements are equivalent:\n\ni) \( M \) is a flat \( A \) -module:\n\nii) \( {M}_{\mathfrak{p}} \) is a flat \( {A}_{\mathfrak{p}} \) -module for each prime ideal \( \mathfrak{p} \) ;\n\niii) \( {M}_{\mathfrak{m}} \) is a flat \( {A}_{\mathfrak{m}} \) -module f... | Proof. i) \( \Rightarrow \) ii) by (3.5) and (2.20).\n\nii) \( \Rightarrow \) iii) O.K.\n\niii) \( \Rightarrow \) i). If \( N \rightarrow P \) is a homomorphism of \( A \) -modules, and \( \mathfrak{m} \) is any maximal ideal of \( A \), then\n\n\[ N \rightarrow P\text{injective} \Rightarrow {N}_{\mathfrak{m}} \rightar... | Yes |
Proposition 3.11. i) Every ideal in \( {S}^{-1}A \) is an extended ideal. | Proof. i) Let \( \mathfrak{b} \) be an ideal in \( {S}^{-1}A \), and let \( x/s \in \mathfrak{b} \) . Then \( x/1 \in \mathfrak{b} \), hence \( x \in {\mathfrak{b}}^{c} \) and therefore \( x/s \in {\mathfrak{b}}^{ce} \) . Since \( \mathfrak{b} \supseteq {\mathfrak{b}}^{ce} \) in any case (1.17), it follows that \( \mat... | Yes |
Corollary 3.13. If \( \mathfrak{p} \) is a prime ideal of \( A \), the prime ideals of the local ring \( {A}_{\mathfrak{p}} \) are in one-to-one correspondence with the prime ideals of \( A \) contained in \( \mathfrak{p} \) . | Proof. Take \( S = A - \mathfrak{p} \) in (3.11) (iv). | Yes |
Proposition 3.14. Let \( M \) be a finitely generated \( A \) -module, \( S \) a multiplicatively closed subset of \( A \) . Then \( {S}^{-1}\left( {\operatorname{Ann}\left( M\right) }\right) = \operatorname{Ann}\left( {{S}^{-1}M}\right) \) . | Proof. If this is true for two \( A \) -modules, \( M, N \), it is true for \( M + N \) :\n\n\[ \n{S}^{-1}\left( {\operatorname{Ann}\left( {M + N}\right) }\right) = {S}^{-1}\left( {\operatorname{Ann}\left( M\right) \cap \operatorname{Ann}\left( N\right) }\right) \text{by (2.2)} \n\]\n\n\[ \n= {S}^{-1}\left( {\operatorn... | Yes |
Corollary 3.15. If \( N, P \) are submodules of an \( A \) -module \( M \) and if \( P \) is finitely generated, then \( {S}^{-1}\left( {N : P}\right) = \left( {{S}^{-1}N : {S}^{-1}P}\right) \) . | Proof. \( \left( {N : P}\right) = \operatorname{Ann}\left( {\left( {N + P}\right) /N}\right) \) by (2.2); now apply (3.14). | No |
Proposition 3.16. Let \( A \rightarrow B \) be a ring homomorphism and let \( \mathfrak{p} \) be a prime ideal of \( A \) . Then \( \mathfrak{p} \) is the contraction of a prime ideal of \( B \) if and only if \( {\mathfrak{p}}^{ec} = \mathfrak{p} \) . | Proof. If \( \mathfrak{p} = {\mathfrak{q}}^{c} \) then \( {\mathfrak{p}}^{ec} = \mathfrak{p} \) by (1.17). Conversely, if \( {\mathfrak{p}}^{ec} = \mathfrak{p} \), let \( S \) be the image of \( A - \mathfrak{p} \) in \( B \) . Then \( {\mathfrak{p}}^{e} \) does not meet \( S \), therefore by (3.11) its extension in \(... | Yes |
Proposition 4.1. Let \( \mathfrak{q} \) be a primary ideal in a ring \( A \) . Then \( r\left( \mathfrak{q}\right) \) is the smallest prime ideal containing \( \mathfrak{q} \) . | Proof. By (1.8) it is enough to show that \( \mathfrak{p} = r\left( \mathfrak{q}\right) \) is prime. Let \( {xy} \in r\left( \mathfrak{q}\right) \), then \( {\left( xy\right) }^{m} \in \mathfrak{q} \) for some \( m > 0 \), and therefore either \( {x}^{m} \in \mathfrak{q} \) or \( {y}^{mn} \in \mathfrak{q} \) for some \... | Yes |
Proposition 4.2. If \( r\left( \alpha \right) \) is maximal, then \( \alpha \) is primary. In particular, the powers of a maximal ideal \( \mathfrak{m} \) are \( \mathfrak{m} -primary. | Proof. Let \( r\left( a\right) = \mathfrak{m} \) . The image of \( \mathfrak{m} \) in \( A/a \) is the nilradical of \( A/a \), hence \( A/a \) has only one prime ideal, by (1.8). Hence every element of \( A/\alpha \) is either a unit or nilpotent, and so every zero-divisor in \( A/a \) is nilpotent. | Yes |
Lemma 4.3. If \( {\mathfrak{q}}_{i}\left( {1 \leq i \leq n}\right) \) are \( \mathfrak{p} \) -primary, then \( \mathfrak{q} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i} \) is \( \mathfrak{p} \) -primary. | Proof. \( r\left( \mathfrak{q}\right) = r\left( {\mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i}}\right) = \bigcap r\left( {\mathfrak{q}}_{i}\right) = \mathfrak{p} \) . Let \( {xy} \in \mathfrak{q}, y \notin \mathfrak{q} \) . Then for some \( i \) we have \( {xy} \in {\mathfrak{q}}_{i} \) and \( y \notin {\mat... | Yes |
Lemma 4.4. Let \( \mathfrak{q} \) be a \( \mathfrak{p} \) -primary ideal, \( x \) an element of \( A \) . Then\ni) if \( x \in \mathfrak{q} \) then \( \left( {\mathfrak{q} : x}\right) = \left( 1\right) \) ;\nii) if \( x \notin \mathfrak{q} \) then \( \left( {\mathfrak{q};x}\right) \) is \( \mathfrak{p} \) -primary, and... | Proof. i) and iii) follow immediately from the definitions.\nii): if \( y \in \left( {\mathfrak{q} : x}\right) \) then \( {xy} \in \mathfrak{q} \), hence (as \( x \notin \mathfrak{q} \) ) we have \( y \in \mathfrak{p} \) . Hence \( \mathfrak{q} \subseteq \) \( \left( {\mathfrak{q} : x}\right) \subseteq \mathfrak{p} \) ... | Yes |
Theorem 4.5. (1st uniqueness theorem). Let \( \mathfrak{a} \) be a decomposable ideal and let \( \mathfrak{a} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i} \) be a minimal primary decomposition of \( \mathfrak{a} \) . Let \( {\mathfrak{p}}_{i} = r\left( {\mathfrak{q}}_{i}\right) \) \( \left( {1 \leq i \leq... | Proof. For any \( x \in A \) we have \( \left( {\mathfrak{a} : x}\right) = \left( {\cap {\mathfrak{q}}_{i} : x}\right) = \cap \left( {{\mathfrak{q}}_{i} : x}\right) \), hence \( r\left( {\mathfrak{a} : x}\right) = \) \( \mathop{\bigcap }\limits_{{i = 1}}^{n}r\left( {{\mathfrak{q}}_{i} : x}\right) = \mathop{\bigcap }\li... | Yes |
Proposition 4.6. Let \( \mathfrak{a} \) be a decomposable ideal. Then any prime ideal \( \mathfrak{p} \supseteq \mathfrak{a} \) contains a minimal prime ideal belonging to \( \mathfrak{a} \), and thus the minimal prime ideals of \( \mathfrak{a} \) are precisely the minimal elements in the set of all prime ideals contai... | Proof. If \( \mathfrak{p} \supseteq \mathfrak{a} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i} \), then \( \mathfrak{p} = r\left( \mathfrak{p}\right) \supseteq \bigcap r\left( {\mathfrak{q}}_{i}\right) = \bigcap {\mathfrak{p}}_{i} \) . Hence by (1.11) we have \( \mathfrak{p} \supseteq {\mathfrak{p}}_{i} \)... | Yes |
Proposition 4.7. Let \( \mathfrak{a} \) be a decomposable ideal, let \( \mathfrak{a} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i} \) be a minimal primary decomposition, and let \( r\left( {\mathfrak{q}}_{i}\right) = {\mathfrak{p}}_{i} \) . Then\n\n\[ \mathop{\bigcup }\limits_{{i = 1}}^{n}{\mathfrak{p}}_{i... | Proof. If \( \mathfrak{a} \) is decomposable, then 0 is decomposable in \( A/\mathfrak{a} \) : namely \( 0 = \bigcap {\widetilde{\mathfrak{q}}}_{t} \) where \( {\overline{\mathfrak{q}}}_{i} \) is the image of \( {\mathfrak{q}}_{i} \) in \( A/a \), and is primary. Hence it is enough to prove the last statement of (4.7).... | Yes |
Proposition 4.8. Let \( S \) be a multiplicatively closed subset of \( A \), and let \( \mathfrak{q} \) be a \( \mathfrak{p} \)-primary ideal.\n\ni) If \( S \cap \mathfrak{p} \neq \varnothing \), then \( {S}^{-1}\mathfrak{q} = {S}^{-1}A \).\n\nii) If \( S \cap \mathfrak{p} = \varnothing \), then \( {S}^{-1}\mathfrak{q}... | Proof. i) If \( s \in S \cap \mathfrak{p} \), then \( {s}^{n} \in S \cap \mathfrak{q} \) for some \( n > 0 \) ; hence \( {S}^{-1}\mathfrak{q} \) contains \( {s}^{n}/1 \), which is a unit in \( {S}^{-1}A \).\n\nii) If \( S \cap \mathfrak{p} = \varnothing \), then \( s \in S \) and \( {as} \in \mathfrak{q} \) imply \( a ... | Yes |
Proposition 4.9. Let \( S \) be a multiplicatively closed subset of \( A \) and let \( \mathfrak{a} \) be a decomposable ideal. Let \( \mathfrak{a} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i} \) be a minimal primary decomposition of \( \mathfrak{a} \). Let \( {\mathfrak{p}}_{i} = r\left( {\mathfrak{q}}_{... | Proof. \( {S}^{-1}a = \mathop{\bigcap }\limits_{{i = 1}}^{n}{S}^{-1}{q}_{i} \) by (3.11) \( = \mathop{\bigcap }\limits_{{i = 1}}^{m}{S}^{-1}{q}_{i} \) by (4.8), and \( {S}^{-1}{q}_{i} \) is \( {S}^{-1}{\mathfrak{p}}_{i} \) -primary for \( i = 1,\ldots, m \). Since the \( {\mathfrak{p}}_{i} \) are distinct, so are the \... | Yes |
Corollary 4.11. The isolated primary components (i.e., the primary components \( {\mathfrak{q}}_{i} \) corresponding to minimal prime ideals \( {\mathfrak{p}}_{i} \) ) are uniquely determined by \( \mathfrak{a} \) . | Proof of (4.10). We have \( {\mathfrak{q}}_{{i}_{1}} \cap \cdots \cap {\mathfrak{q}}_{{i}_{m}} = S\left( \mathfrak{a}\right) \) where \( S = A - {\mathfrak{p}}_{{i}_{1}} \cup \cdots \cup {\mathfrak{p}}_{{i}_{m}} \) , hence depends only on \( \alpha \) (since the \( {\mathfrak{p}}_{i} \) depend only on \( \alpha \) ). ∎ | No |
If a rational number \( x = r/s \) is integral over \( \mathbf{Z} \), where \( r, s \) have no common factor, we have from (1) | \[ {r}^{n} + {a}_{1}{r}^{n - 1}s + \cdots + {a}_{n}{s}^{n} = 0 \] the \( {a}_{i} \) being rational integers. Hence \( s \) divides \( {r}^{n} \), hence \( s = \pm 1 \), hence \( x \in \mathbf{Z} \). | Yes |
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... | Yes |
Corollary 5.2. Let \( {x}_{i}\left( {1 \leq i \leq n}\right) \) be elements of \( B \), each integral over \( A \) . Then the ring \( A\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) is a finitely-generated \( A \) -module. | Proof. By induction on \( n \) . The case \( n = 1 \) is part of (5.1). Assume \( n > 1 \), let \( {A}_{r} = A\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \) ; then by the inductive hypothesis \( {A}_{n - 1} \) is a finitely generated \( A \) -module. \( {A}_{n} = {A}_{n - 1}\left\lbrack {x}_{n}\right\rbrack \) ... | 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 |
If \( A \subseteq B \subseteq C \) are rings and if \( B \) is integral over \( A \), and \( C \) is integral over \( B \), then \( C \) is integral over \( A \) (transitivity of integral dependence). | Proof. Let \( x \in C \), then we have an equation\n\n\[ {x}^{n} + {b}_{1}{x}^{n - 1} + \cdots + {b}_{n} = 0\;\left( {{b}_{i} \in B}\right) .\n\]\n\nThe ring \( {B}^{\prime } = A\left\lbrack {{b}_{1},\ldots ,{b}_{n}}\right\rbrack \) is a finitely generated \( A \) -module by (5.2), and \( {B}^{\prime }\left\lbrack x\ri... | 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 \( \mathfrak{a} = {\mathfrak{b}}^{c} = A \cap \mathfrak{b} \), then \( B/\mathfrak{b} \) is integral over \( A/\mathfrak{a} \). ii) If \( S \) is a multiplicatively closed subset of \( A... | 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\[ \n{y}^{n} + {a}_{1}{y}^{n - 1} + \cdots + {a}_{n} = 0\;\left( {{a}_{i} \in A}\right) \n\] \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 ... | 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 domains. Now use (5.7). | Yes |
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 |
Theorem 5.10. Let \( A \subseteq B \) be rings, \( B \) integral over \( A \), and let \( \mathfrak{p} \) be a prime ideal of \( A \) . Then there exists a prime ideal \( \mathfrak{q} \) of \( B \) such that \( \mathfrak{q} \cap A = \mathfrak{p} \) . | Proof. By (5.6), \( {B}_{\mathfrak{p}} \) is integral over \( {A}_{\mathfrak{p}} \), and the diagram\n\n\[ \nA \rightarrow B \n\]\n\n\[ \n\alpha \downarrow \beta \n\]\n\n\[ \n{A}_{\mathfrak{p}} \rightarrow {B}_{\mathfrak{p}} \n\]\n\n(in which the horizontal arrows are injections) is commutative. Let \( n \) be a maxima... | 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\[{\left( b/s\right) }^{n} + \left( {{a}_{1}/{s}_{1}}\right) {\left( b/s\right) }^{n - 1} + \cdots + {a}_{n}/{s}_{n} = 0\]\n\nwhere \( {a}_... | Yes |
Proposition 5.13. Let \( A \) be an integral domain. Then the following are equivalent:\n\ni) \( A \) is integrally closed;\n\nii) \( {A}_{\mathfrak{p}} \) is integrally closed, for each prime ideal \( \mathfrak{p} \) ;\n\niii) \( {A}_{\mathfrak{m}} \) 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. ... | Yes |
Lemma 5.14. Let \( C \) be the integral closure of \( A \) in \( B \) and let \( {\mathfrak{a}}^{a} \) denote the extension of \( \mathfrak{a} \) in \( C \) . Then the integral closure of \( \mathfrak{a} \) in \( B \) is the radical of \( {\mathfrak{a}}^{e} \) (and is therefore closed under addition and multiplication)... | Proof. If \( x \in B \) is integral over \( a \), 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 a. Hence \( x \in C \) and \( {x}^{n} \in {a}^{e} \), that is \( x \in r\left( {a}^{e}\right) \) . Conversely, if \( x \in r\left( {... | 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 coef... | Yes |
Proposition 5.17. Let \( A \) be an integrally closed domain, \( K \) its field of fractions, \( L \) a finite separable algebraic extension of \( K, B \) the integral closure of \( A \) in \( L \) . Then there exists a basis \( {v}_{1},\ldots ,{v}_{n} \) of \( L \) over \( K \) such that \( B \subseteq \mathop{\sum }\... | Proof. If \( v \) is any element of \( L \), then \( v \) is algebraic over \( K \) and therefore satisfies an equation of the form \[ {a}_{0}{v}^{r} + {a}_{1}{v}^{r - 1} + \cdots + {a}_{n} = 0\left( {{a}_{i} \in A}\right) . \] Multiplying this equation by \( {a}_{0}^{r - 1} \), we see that \( {a}_{0}v = u \) is integr... | Yes |
Proposition 5.18. i) \( B \) is a local ring. | Proof. i) Let \( m \) be the set of non-units of \( B \), so that \( x \in m \Leftrightarrow \) 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}^{-1} = a \cdot {\left... | Yes |
Lemma 5.19. B is a local ring and \( \mathrm{m} = \operatorname{Ker}\left( g\right) \) is its maximal ideal. | Proof. Since \( g\left( B\right) \) is a subring of a field and therefore an integral domain, the ideal \( \mathfrak{m} = \operatorname{Ker}\left( g\right) \) is prime. We can extend \( g \) to a homomorphism \( \widetilde{g} : {B}_{\mathfrak{m}} \rightarrow \Omega \) by putting \( \bar{g}\left( {b/s}\right) = g\left( ... | Yes |
Lemma 5.20. Let \( x \) be a non-zero element of \( K \) . Let \( B\left\lbrack x\right\rbrack \) be the subring of \( K \) generated by \( x \) over \( B \), and let \( \mathfrak{m}\left\lbrack x\right\rbrack \) be the extension of \( \mathfrak{m} \) in \( B\left\lbrack x\right\rbrack \) . Then either \( \mathfrak{m}\... | Proof. Suppose that \( m\left\lbrack x\right\rbrack = B\left\lbrack x\right\rbrack \) and \( m\left\lbrack {x}^{-1}\right\rbrack = B\left\lbrack {x}^{-1}\right\rbrack \) . Then we shall have equations\n\n\[ \n{u}_{0} + {u}_{1}x + \cdots + {u}_{m}{x}^{m} = 1\;\left( {{u}_{i} \in \mathfrak{m}}\right) \n\]\n\n(1)\n\n\[ \n... | Yes |
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 \( \\mathfrak{m}\\left\\lbrack x\\right\\rbrack \) is not the unit ideal of the ring \( {B}^{\\prime } = B\\left\\lbrack x\\right\\rbrack \) . Then \( \\mat... | 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, by (5.18) iii), it follows that \( \bar{A} \subseteq B \) . Conversely, let \( x \notin \bar{A} \) . Then \( x \) is not in the ring \( {A}^{\prime } = A\left\lbrack {x}^{-1}\right\rbrack \) . Hence \( ... | Yes |
Proposition 5.23. Let \( A \subseteq B \) be integral domains, \( B \) finitely generated over \( A \). Let \( v \) be a non-zero element of \( B \). Then there exists \( u \neq 0 \) in \( A \) with the following property: any homomorphism \( f \) of \( A \) into an algebraically closed field \( \Omega \) such that \( ... | Proof. By induction on the number of generators of \( B \) over \( A \) we reduce immediately to the case where \( B \) is generated over \( A \) by a single element \( x \). i) Suppose \( x \) is transcendental over \( A \), i.e., that no non-zero polynomial with coefficients in \( A \) has \( x \) as a root. Let \( v... | Yes |
Corollary 5.24. Let \( k \) be a field and \( B \) a finitely generated \( k \) -algebra. If \( B \) is a field then it is a finite algebraic extension of \( k \) . | Proof. Take \( A = k, v = 1 \) and \( \Omega = \) algebraic closure of \( k \) . ∎ | No |
Proposition 6.1. The following conditions on \( \sum \) are equivalent:\n\ni) Every increasing sequence \( {x}_{1} \leq {x}_{2} \leq \cdots \) in \( \sum \) is stationary (i.e., there exists \( n \) such that \( {x}_{n} = {x}_{n + 1} = \cdots \) ).\n\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, sa... | Yes |
Proposition 6.2. \( M \) is a Noetherian A-module \( \Leftrightarrow \) every submodule of \( M \) is finitely generated. | Proof. \( \Rightarrow \) : Let \( N \) be a submodule of \( M \), and let \( \sum \) be the set of all finitely generated submodules of \( N \) . Then \( \sum \) is not empty (since \( 0 \in \sum \) ) and therefore has a maximal element, say \( {N}_{0} \) . If \( {N}_{0} \neq N \), consider the submodule \( {N}_{0} + {... | Yes |
Proposition 6.3. Let \( 0 \rightarrow {M}^{\prime }\xrightarrow[]{\alpha }M\xrightarrow[]{\beta }{M}^{\prime \prime } \rightarrow 0 \) be an exact sequence of A-modules. Then\ni) \( M \) is Noetherian \( \Leftrightarrow {M}^{\prime } \) and \( {M}^{\prime \prime } \) are Noetherian;\nii) \( M \) is Artinian \( \Leftrig... | 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\( \Leftarrow : \) Let \( {\left( {L}_{n}\right) }_{n \geq 1} \) be an ascending chain of submo... | Yes |
Corollary 6.4. If \( {M}_{1}\left( {1 \leq i \leq n}\right) \) are Noetherian (resp. Artinian) A-modules, so is \( {\bigoplus }_{i = 1}^{n}{M}_{i} \) . | Proof. Apply induction and (6.3) to the exact sequence\n\n\[ 0 \rightarrow {M}_{n} \rightarrow {\bigoplus }_{i = 1}^{n}{M}_{i} \rightarrow {\bigoplus }_{i = 1}^{n - 1}{M}_{i} \rightarrow 0. \] | 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 6.6. Let \( A \) be Noetherian (resp. Artinian), a an ideal of \( A \) .\n\nThen \( A/\mathfrak{a} \) is a Noetherian (resp. Artinian) ring. | Proof. By (6.3) \( A/a \) is Noetherian (resp. Artinian) as an \( A \) -module, hence also as an \( A/\mathfrak{a} \) -module. | Yes |
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 \( M \) . \( (l\left( M\right) = + \infty \) if \( M \) has no composition series.)\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 ... | Yes |
Proposition 6.8. \( M \) has a composition series \( \Leftrightarrow M \) satisfies both chain conditions. | Proof. \( \Rightarrow \) : All chains in \( M \) are of bounded length, hence both a.c.c. and d.c.c. hold.\n\n\( \Leftarrow \) : Construct a composition series of \( M \) as follows. Since \( M = {M}_{0} \) satisfies the maximum condition by (6.1), it has a maximal submodule \( {M}_{1} \subset {M}_{0} \). Similarly \( ... | 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 \( 0 \rightarrow {M}^{\prime }\xrightarrow[]{\alpha }M\xrightarrow[]{\beta }{M}^{\prime \prime } \rightarrow 0 \) 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 s... | Yes |
Proposition 6.10. For \( k \) -vector spaces \( V \) the following conditions are equivalent:\ni) finite dimension;\nii) finite length;\niii) a.c.c.;\niv) d.c.c.\nMoreover, if these conditions are satisfied, length \( = \) dimension. | Proof. i) \( \Rightarrow \) ii) is elementary; ii) \( \Rightarrow \) iii), ii) \( \Rightarrow \) iv) from (6.8). Remains to prove iii) \( \Rightarrow \) i) and iv) \( \Rightarrow \) i). Suppose i) is false, then there exists an infinite sequence \( {\left( {x}_{n}\right) }_{n > 1} \) of linearly independent elements of... | Yes |
Corollary 6.11. Let \( A \) be a ring in which the zero ideal is a product \( {m}_{1}\cdots {m}_{n} \) of (not necessarily distinct) maximal ideals. Then \( A \) is Noetherian if and only if \( A \) is Artinian. | Proof. Consider the chain of ideals \( A \supset {\mathfrak{m}}_{1} \supseteq {\mathfrak{m}}_{1}{\mathfrak{m}}_{2} \supseteq \cdots \supseteq {\mathfrak{m}}_{1}\cdots {\mathfrak{m}}_{n} = 0 \) . Each factor \( {m}_{1}\cdots {m}_{i - 1}/{m}_{1}\cdots {m}_{i} \) is a vector space over the field \( A/{m}_{i} \) . Hence a.... | Yes |
Proposition 7.1. If \( A \) is Noetherian and \( \phi \) is a homomorphism of \( A \) onto a ring \( B \), then \( B \) is Noetherian. | Proof. This follows from (6.6), since \( B \cong A/\mathfrak{a} \), where \( \mathfrak{a} = \operatorname{Ker}\left( \phi \right) \). | Yes |
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 a form an ideal \( 1 \) in \( A \) . Since \( A \) is Noetherian, \( 1 \) is finitely generated, say by \( {a}_{1},\ldots ,{a}_{n} \) . For each \( i = 1,\ldots, n \) there is a polynomial \(... | Yes |
Corollary 7.6. If \( A \) is Noetherian so is \( A\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . | Proof. By induction on \( n \) from (7.5). | No |
Corollary 7.7. Let \( B \) be a finitely-generated \( A \) -algebra. If \( A \) is Noetherian, then so is \( B \) . | Proof. \( B \) is a homomorphic image of a polynomial ring \( A\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), which is Noetherian by (7.6). | 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 equivalent in this situation. So we may concentrate on (i).\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 ... | Yes |
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 |
Corollary 7.10. Let \( k \) be a field, \( A \) a finitely generated \( k \) -algebra. Let \( \mathfrak{m} \) be a maximal ideal of \( A \) . Then the field \( A/\mathfrak{m} \) is a finite algebraic extension of \( k \) . In particular, if \( k \) is algebraically closed then \( A/\mathfrak{m} \cong k \) . | Proof. Take \( E = A/\mathfrak{m} \) in (7.9). | No |
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 \( \alpha \) . Since \( \alpha \) is reducible, we have \( \mathfrak{a} = \mathfrak{b} \cap \mathfrak{c} \) where \( \mathfrak{b} \supset \mathfrak{a} \) and \( \mathfrak{c} \supset \mathfrak{a} ... | Yes |
Lemma 7.12. In a Noetherian ring every irreducible ideal is primary. | Proof. By passing to the quotient ring, it is enough to show that if the zero ideal is irreducible then it 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... | Yes |
Proposition 7.14. In a Noetherian ring \( A \), every ideal \( \mathfrak{a} \) contains a power of its radical. | Proof. Let \( {x}_{1},\ldots ,{x}_{k} \) generate \( r\left( \mathfrak{a}\right) \) : say \( {x}_{i}^{{n}_{i}} \in \mathfrak{a}\left( {1 \leq i \leq k}\right) \) . Let \( m = \) \( \mathop{\sum }\limits_{{i = 1}}^{k}\left( {{n}_{i} - 1}\right) + 1 \) . Then \( r{\left( \mathfrak{a}\right) }^{m} \) is generated by the p... | Yes |
Corollary 7.15. In a Noetherian ring the nilradical is nilpotent. | Proof. Take \( \mathfrak{a} = \left( 0\right) \) in (7.14). | 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:\ni) \( q \) is \( \mathfrak{m} \) -primary;\nii) \( r\left( q\right) = m \);\niii) \( {\mathfrak{m}}^{n} \subseteq \mathfrak{q} \subseteq \mathfrak{m} \)... | 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 \( \left( {\mathfrak{a} : x}\right) \left( {x \in A}\right) \) . | Proof. By passing to \( A/a \) we may assume that \( \mathfrak{a} = 0 \) . Let \( \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i} = 0 \) be a minimal primary decomposition of the zero ideal, and let \( {\mathfrak{p}}_{i} \) be the radical of \( {\mathfrak{q}}_{i} \) . Let \( {\mathfrak{a}}_{i} = \mathop{\bigca... | Yes |
Proposition 8.1. In an Artin ring A every prime ideal is maximal. | Proof. Let \( \mathfrak{p} \) be a prime ideal of \( A \) . Then \( B = A/\mathfrak{p} \) is an Artinian integral domain. Let \( x \in B, x \neq 0 \) . By the d.c.c. we have \( \left( {x}^{n}\right) = \left( {x}^{n + 1}\right) \) for some \( n \) , hence \( {x}^{n} = {x}^{n + 1}y \) for some \( y \in B \) . Since \( B ... | Yes |
Proposition 8.3. An Artin ring has only a finite number of maximal ideals. | Proof. Consider the set of all finite intersections \( {m}_{1} \cap \cdots \cap {m}_{r} \), where the \( {m}_{i} \) are 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 \cdot... | Yes |
Proposition 8.4. In an Artin ring the nilradical \( \mathfrak{R} \) is nilpotent. | Proof. By d.c.c. we have \( {\Re }^{k} = {\Re }^{k + 1} = \cdots = \) a say, for some \( k > 0 \) . Suppose \( \mathfrak{a} \neq 0 \), and let \( \sum \) denote the set of all ideals \( \mathfrak{b} \) such that \( \mathfrak{a}\mathfrak{b} \neq 0 \) . Then \( \sum \) is not empty, since \( a \in \sum \) . Let \( c \) b... | Yes |
Theorem 8.5. A ring \( A \) is Artin \( \Leftrightarrow A \) is Noetherian and \( \dim A = 0 \) . | Proof. \( \Rightarrow \) : By (8.1) we have \( \dim A = 0 \) . Let \( {\mathfrak{m}}_{1}\left( {1 \leq i \leq n}\right) \) be the distinct maximal ideals of \( A \) (8.3). Then \( \mathop{\prod }\limits_{{i = 1}}^{n}{\mathfrak{m}}_{i}^{k} \subseteq {\left( \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{m}}_{i}\right)... | Yes |
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:\ni) \( {\mathfrak{m}}^{n} \neq {\mathfrak{m}}^{n + 1} \) for all \( n \) ;\nii) \( {\mathfrak{m}}^{n} = 0 \) for some \( n \), in which case \( A \) is an Artin local ... | Proof. Suppose \( {\mathfrak{m}}^{n} = {\mathfrak{m}}^{n + 1} \) for some \( n \) . By Nakayama’s lemma (2.6) we have \( {\mathfrak{m}}^{n} = 0 \) . Let \( \mathfrak{p} \) be any prime ideal of \( A \) . Then \( {\mathfrak{m}}^{n} \subseteq \mathfrak{p} \), hence (taking radicals) \( \mathfrak{m} = \mathfrak{p} \) . He... | Yes |
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 coprime in pairs, hence \( \cap {\mathfrak... | 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... | Yes |
Proposition 9.1. Let \( A \) be a Noetherian domain of dimension 1. Then every non-zero ideal \( \mathfrak{a} \) in \( A \) can be uniquely expressed as a product of primary ideals whose radicals are all distinct. | Proof. Since \( A \) is Noetherian, \( a \) has a minimal primary decomposition \( \mathfrak{a} = \mathop{\bigcap }\limits_{{i = 1}}^{n}{\mathfrak{q}}_{i} \) by (7.13), where each \( {\mathfrak{q}}_{i} \) is say \( {\mathfrak{p}}_{i} \) -primary. Since \( \dim A = 1 \) and \( A \) is an integral domain, each non-zero p... | Yes |
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:\n\ni) \( A \) is a discrete valuation ring;\n\nii) \( A \) is integrally closed;\n\niii) \( \mathfrak{m} \) is a principal idea... | Proof. Before we start going the rounds, we make two remarks:\n\n(A) If \( \mathfrak{a} \) is an ideal \( \neq 0,\left( 1\right) \), then \( \mathfrak{a} \) is \( \mathfrak{m} \) -primary and \( \mathfrak{a} \supseteq {\mathfrak{m}}^{n} \) for some \( n \) . For \( r\left( a\right) = m \), since \( m \) is the only non... | 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).\n\nii) \( \Leftrightarrow \) iii). Use (9.2) and the fact that primary ideals and powers of ideals 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 |
Theorem 9.5. The ring of integers in an algebraic number field \( K \) is a Dedekind domain. | Proof. \( K \) is a separable extension of \( \mathbf{Q} \) (because the characteristic is zero), hence by (5.17) there is a basis \( {v}_{1},\ldots ,{v}_{n} \) of \( K \) over \( \mathbf{Q} \) such that \( A \subseteq \sum \mathbf{Z}{v}_{J} \) . Hence \( A \) is finitely generated as a \( \mathbf{Z} \) -module and the... | Yes |
Proposition 9.6. For a fractional ideal \( M \), the following are equivalent:\n\ni) \( M \) is invertible;\n\nii) \( M \) is finitely generated and, for each prime ideal \( \mathfrak{p},{M}_{\mathfrak{p}} \) is invertible:\n\niii) \( M \) is finitely generated and, for each maximal ideal \( \mathrm{m},{M}_{\mathrm{m}}... | Proof. i) \( \Rightarrow \) ii): \( {A}_{\mathfrak{p}} = {\left( M \cdot \left( A : M\right) \right) }_{\mathfrak{p}} = {M}_{\mathfrak{p}} \cdot \left( {{A}_{\mathfrak{p}} : {M}_{\mathfrak{p}}}\right) \) by (3.11) and (3.15) (for \( M \) is finitely generated, because invertible).\n\nii) \( \Rightarrow \) iii) as usual... | 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. | Proof. \( \Rightarrow \) . Let \( x \) be a generator of the maximal ideal \( \mathfrak{m} \) of \( A \), and let \( M \neq 0 \) 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... | Yes |
Theorem 9.8. Let \( A \) be an integral domain. Then \( A \) is a Dedekind domain \( \Leftrightarrow \) every non-zero fractional ideal of \( A \) is invertible. | 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 \( \neq 0 \) of the discrete valuation ring \( {A}_{\mathfrak{p}} \), hence is invertible by (9.7). Hen... | Yes |
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 prime ideals of \( A \) . Let \( {K}^{ * } \) denote the multiplicative group of the field of fractions \( K \) of \( A \) . Each \( u \in {K}^... | Yes |
Lemma 10.1. Let \( H \) be the intersection of all neighborhoods of 0 in \( G \) . Then i) \( H \) is a subgroup.\n\nii) \( H \) is the closure of \( \{ 0\} \) .\n\niii) \( G/H \) is Hausdorff.\n\niv) \( G \) is Hausdorff \( \Leftrightarrow H = 0 \) . | Proof. i) follows from the continuity of the group operations. For ii) we have:\n\n\[ x \in H \Leftrightarrow 0 \in x - U\text{for all neighborhoods}U\text{of 0} \]\n\n\[ \Leftrightarrow x \in \overline{\{ 0\} }\text{. } \]\n\nii) implies that the cosets of \( H \) are all closed; thus points are closed in \( G/H \) an... | Yes |
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... | Proof. Let \( A = \mathop{\prod }\limits_{{n = 1}}^{\infty }{A}_{n} \) and define \( {d}^{A} : A \rightarrow A \) by \( {d}^{A}\left( {a}_{n}\right) = {a}_{n} - {\theta }_{n + 1}\left( {a}_{n + 1}\right) \) . Then \( \operatorname{Ker}{d}^{A} \cong \mathop{\lim }\limits_{ \leftarrow }{A}_{n} \) . Define \( B, C \) and ... | 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,... | Proof. Apply (10.2) to the exact sequences \[ 0 \rightarrow \frac{{G}^{\prime }}{{G}^{\prime } \cap {G}_{n}} \rightarrow \frac{G}{{G}_{n}} \rightarrow \frac{{G}^{\prime \prime }}{p{G}_{n}} \rightarrow 0. \] | No |
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