User ashpool - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T01:54:18Z http://mathoverflow.net/feeds/user/5292 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54065/about-injective-hull About injective hull ashpool 2011-02-02T05:22:25Z 2012-12-01T21:38:28Z <p>Let $M$ be an $A$-module. Is its injective hull affected by whether I regard $M$ as an $A$-module or $A/\mbox{Ann}(M)$-module ?</p> http://mathoverflow.net/questions/23261/length-of-a-resolution Length of a resolution ashpool 2010-05-02T15:37:38Z 2012-11-20T14:46:13Z <p>Is a (say, projective) resolution (of a module) consisting entirely of zero modules considered to have a length (of zero) at all? I think this possibility causes problems in some books.</p> http://mathoverflow.net/questions/37740/projective-dimension-of-zero-module Projective dimension of zero module ashpool 2010-09-04T17:18:22Z 2012-11-20T08:27:10Z <p>Is there any consensus on what the projective dimension of the zero module should be? Here are three statements one commonly encounters in textbooks, sometimes with or without the condition $M\neq 0$:</p> <p>(1) $\mbox{pd}(M)\leq n$ iff $\mbox{Ext}^{n+1}(M,-)=0$</p> <p>(2) $\mbox{pd}(M)=0$ iff $M$ is projective</p> <p>(3) $\mbox{grade}(M):=\infty$ if $M=0$</p> <p>If one attempts to define $\mbox{pd}((0))$ by extending one of these results, (1), (2), (3) suggest $\mbox{pd}=-1, 0, \infty$, respectively.</p> http://mathoverflow.net/questions/25687/projective-module projective module ashpool 2010-05-23T16:36:01Z 2012-11-07T16:40:39Z <p>Is it true that if $\mbox{Ext}^{1}_{A}(P,A/I)=0 \ \forall I$ then P is projective? Similar statements are true for flat and injective modules, but I'm beginning to suspect that projective modules cannot be characterized soley by ideals.</p> http://mathoverflow.net/questions/34704/projective-injective-dimensions Projective & injective dimensions ashpool 2010-08-06T00:05:23Z 2012-05-02T13:09:23Z <p>$A$ a Noetherian local ring, $M\neq 0$ a finite $A$-module. I'm not quite sure about the relation between finiteness of projective and injective dimensions of $M$. Does the finiteness (or infiniteness) of one necessarily imply the finiteness (or infiniteness) of another?</p> http://mathoverflow.net/questions/90391/technical-question-about-height-of-minimal-associated-primes Technical question about height of minimal associated primes ashpool 2012-03-06T18:56:13Z 2012-03-06T19:11:11Z <p>Let $A$ be a Noetherian ring, $\mathfrak{p}\subset A$ a prime ideal of height $p$, $N$ an $A_{\mathfrak{p}}$-module of finite length, $M,M'\subset N$ finitely generated $A$-submodules such that $M\varsubsetneq M'$ and $M_{\mathfrak{p}}=M'_{\mathfrak{p}}=N$. Then is it true that every minimal associated prime of $M'/M$ has height $p+1$? I could show that they must have height $\geq p+1$.</p> http://mathoverflow.net/questions/34702/depth-and-dimension Depth and dimension ashpool 2010-08-05T23:45:23Z 2012-01-11T18:47:56Z <p>$A$ a Noetherian local ring, $M\neq 0$ a finite $A$-module. Then is it true that $\mbox{depth }M\le\mbox{depth }A$ just like $\mbox{dim }M\le\mbox{dim }A$? I don't see any relation between an $M$-sequence and an $A$-sequence. At least I know it is true when $\mbox{inj.dim }M&lt;\infty$, from the relation $\mbox{depth }M\leq\mbox{dim }M\leq\mbox{inj. dim }M=\mbox{depth }A\leq\mbox{dim }A$. But what happens when $\mbox{inj.dim }M=\infty$? Another inequality I'm not quite sure about when $\mbox{inj.dim }M=\infty\$: is it true that $\mbox{dim }M\leq\mbox{depth }A$?</p> http://mathoverflow.net/questions/84695/why-are-canonical-modules-supported-everywhere Why are canonical modules supported everywhere? ashpool 2012-01-01T19:24:47Z 2012-01-02T03:01:44Z <p>Let $A$ be a local CM ring, and $\omega$ a canonical module of $A$. Here are two properties of $\omega$ from Bruns &amp; Herzog:</p> <ul> <li>$\omega_{\mathfrak{p}}$ is a canonical module of $A_{\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$.</li> <li>$\mu_i(\mathfrak{p},\omega)=\delta_{i}^{\operatorname{ht}\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$, where $\mu$ denotes the Bass number.</li> </ul> <p>These properties seem to imply that $\operatorname{Supp}\omega=\operatorname{Spec}A$. As <a href="http://mathoverflow.net/questions/84685/are-maximal-cohen-macaulay-modules-supported-everywhere/84688#84688" rel="nofollow">Graham Leuschke</a> pointed out, this is not a property of maximal CM modules. Why, then, are canonical modules supported everywhere?</p> http://mathoverflow.net/questions/84685/are-maximal-cohen-macaulay-modules-supported-everywhere Are maximal Cohen-Macaulay modules supported everywhere? ashpool 2012-01-01T15:43:37Z 2012-01-01T16:58:32Z <p>Let $A$ be a local CM ring, and $M$ a maximal CM $A$-module. Is it true that $\operatorname{Supp}M=\operatorname{Spec}A$ ? This suspicion stems from such statements as:</p> <ul> <li>If $\omega$ is a canonical module of $A$, then $\omega_{\mathfrak{p}}$ is a canonical module of $A_{\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$.</li> <li>If $\omega$ is a canonical module of $A$, then $\mu_i(\mathfrak{p},\omega)=\delta_{i}^{\operatorname{ht}\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$, where $\mu$ denotes the Bass number.</li> </ul> <p>And am I correct in understanding that maximal CM module is by definition nonzero?</p> http://mathoverflow.net/questions/33522/flatness-and-local-freeness Flatness and local freeness ashpool 2010-07-27T14:50:58Z 2011-07-20T16:33:56Z <p>The following statement is well-known:</p> <p>$A$ a commutative Noetherian ring, $M$ a finitely generated $A$-module. Than $M$ is flat if and only if $M_{\mathfrak{p}}$ is free for all $\mathfrak{p}$.</p> <p>My question is: do we need the assumption that $A$ is Noetherian? I have a proof (from Matsumura) which doesn't require that assumption, but the fact that other references (e.g. Atiyah, Wikipedia) are including this assumption makes me rather uneasy.</p> http://mathoverflow.net/questions/65138/metrizability-of-mathfraka-adic-topology Metrizability of $\mathfrak{a}$-adic topology ashpool 2011-05-16T13:59:07Z 2011-05-16T14:45:42Z <p>Let $A$ be a ring, $\mathfrak{a}\subset A$ an ideal. Then is the $\mathfrak{a}$-adic topology on $A$ necessarily a metric space? I can see that it is true when $A$ is a DVR, but is it true in general?</p> http://mathoverflow.net/questions/64399/does-completion-commute-with-localization Does completion commute with localization? ashpool 2011-05-09T15:12:50Z 2011-05-09T16:26:54Z <p>Suppose $A$ is a Noetherian (not necessarily local) ring and $\mathfrak{m}\subset A$ a maximal ideal. Then is it true that $$\hat{A}_{\hat{\mathfrak{m}}}=\widehat{A _{\mathfrak{m}}},$$ where hats denote completion and subscripts denote localization? If one uses superscripts to denote completion it would be</p> <p>$$(A^{\mathfrak{m}})_{\mathfrak{m^{\mathfrak{m}}}}=(A _{\mathfrak{m}})^{\mathfrak{m} _{\mathfrak{m}}}.$$</p> http://mathoverflow.net/questions/53906/primary-decomposition-for-modules Primary decomposition for modules ashpool 2011-01-31T19:15:50Z 2011-01-31T19:15:50Z <p>It is well known that the associated primes of a module over a commutative ring (those primes associated to primary decomposition of the zero submodule, provided such decomposition exists) are precisely the radicals of the annihilators of elements of the module that are prime. It is easy to show that the word "radical" can be omitted if the ring is Noetherian. Apparently it can also be omitted if the module (and not the ring) is Noetherian. The only proof I know constructs a huge theory of injective modules, and I'm curious to know if there is a more elementary proof.</p> http://mathoverflow.net/questions/52341/related-to-fractional-ideals Related to fractional ideals ashpool 2011-01-17T19:03:02Z 2011-01-25T21:55:12Z <p>$K$ a field, $A\subset K$ a subring, $M\subset K$ an $A$-submodule. Define $$(A:_{K}M):= \lbrace s\in K|sM\subset A\rbrace$$ Then it is easy to see that </p> <p>$$M\subset A\Longleftrightarrow A\subset (A:_{K}M),$$</p> <p>$$A\subset M\Longrightarrow (A:_{K}M)\subset A$$</p> <p>But I couldn't show the reverse implication of the second. It is true if $M$ is invertible, and I'm guessing that it is true only if $M$ is invertible. Any ideas?</p> http://mathoverflow.net/questions/49187/projectively-splitting-module Projectively splitting module ashpool 2010-12-12T23:34:32Z 2010-12-12T23:45:22Z <p>Is there a name for such class of modules $M$ such that $M\rightarrow N\rightarrow 0$ splits for every $N$ ?</p> http://mathoverflow.net/questions/47534/unit-ideal-in-non-commutative-rings Unit ideal in non-commutative rings ashpool 2010-11-27T20:35:14Z 2010-11-27T21:13:01Z <p>In a non-commutative ring (with identity), is it possible for an element which does not possess left or right inverses to generate the entire ring? i.e. $(r)=R$, where (r) is the two-sided ideal generated by $r\in R$ ?</p> http://mathoverflow.net/questions/46742/cm-module-is-height-unmixed CM module is height-unmixed? ashpool 2010-11-20T15:42:36Z 2010-11-20T23:38:48Z <p>$A$ a Cohen-Macaulay ring (not necessarily local), $M$ a Cohen-Macaulay $A$-module. Then does it necessarily follow that $\mbox{ann}(M)$ is height-unmixed?</p> http://mathoverflow.net/questions/37497/torsion-submodule Torsion submodule ashpool 2010-09-02T13:42:36Z 2010-09-02T15:16:11Z <p>$A$ a commutative Noetherian domain, $M$ a finitely generated $A$-module. How can I show that the kernel of the natural map $M\rightarrow M^{**}$, where $M^{ * *}$ is the double dual (with respect to $A$), is <em>the</em> torsion submodule of $M$?</p> <p>I do know that in this situation torsionlessness coincides with torsion-freeness. According to Auslander this result is well-know'' but I can't seem to prove it or find any reference on this.</p> http://mathoverflow.net/questions/24031/dimension-of-module Dimension of module ashpool 2010-05-09T16:57:42Z 2010-08-17T06:00:55Z <p>Does dimension of a module (say, dimension of its support) have anything to do with the supremum length of chains of prime submodules like rings? Let's restrict to finitely generated modules over Noetherian ring. Prime submodules are defined analogously to primary submodules: a submodule P in M is prime if P$\neq$M and $M/P$ has no zero divisors, i.e. $am\in P$ implies $m\in P$ or $a \in \mbox{Ann}(M/P)$.</p> http://mathoverflow.net/questions/34784/finiteness-of-injective-hull-of-residue-field-for-artin-local-ring Finiteness of injective hull of residue field for Artin local ring ashpool 2010-08-06T16:24:57Z 2010-08-16T18:11:45Z <p>$(A,\mathfrak{m})$ an Artin local ring, $E(A/\mathfrak{m})$ the injective hull of $A/\mathfrak{m}$. How do I see that $E(A/\mathfrak{m})$ is a finite $A$-module?</p> http://mathoverflow.net/questions/34784/finiteness-of-injective-hull-of-residue-field-for-artin-local-ring/35783#35783 Answer by ashpool for Finiteness of injective hull of residue field for Artin local ring ashpool 2010-08-16T18:11:45Z 2010-08-16T18:11:45Z <p>This is a proof of $\ell(M)=\ell(\mbox{Hom}(M,\mbox{E}(A/\mathfrak{m}))$ suggested by Mariano:</p> <p>Induction on $\ell(M)\$: </p> <p>If $\ell(M)=0$, $M=0$ so obviously true. Suppose $\ell(M)=n\geq 1$. From a composition series of $M$ choose the submodule N right beneath M so that $\ell(N)=n-1$ and $M/N\simeq A/\mathfrak{m}$. $0\rightarrow N\rightarrow M \rightarrow A/\mathfrak{m}\rightarrow 0$ induces $0\leftarrow \mbox{Hom}(N,E(A/\mathfrak{m}))\leftarrow \mbox{Hom}(M,E(A/\mathfrak{m}))\leftarrow \mbox{Hom}(A/\mathfrak{m},E(A/\mathfrak{m}))\leftarrow 0$.</p> <p>Now $\mbox{Hom}(A/\mathfrak{m},E(A/\mathfrak{m}))\simeq A/\mathfrak{m}$ since $E(A/\mathfrak{m})$ is an essential extension of $A/\mathfrak{m}$, and $\ell(\mbox{Hom}(N,E(A/\mathfrak{m})))=\ell(N)=n-1$ by the induction hypothesis. $\ell(A/\mathfrak{m})=1$ so $\ell(\mbox{Hom}(M,E(A/\mathfrak{m}))=(n-1)+1=n$</p> http://mathoverflow.net/questions/34879/modules-over-a-gorenstein-ring Modules over a Gorenstein ring ashpool 2010-08-08T02:13:56Z 2010-08-16T16:01:39Z <p>$A$ a Gorenstein ring, $M\neq 0$ a finite $A$-module with finite injective dimension. According to Bruns, this implies that $M$ has finite projective dimension. How do I see that?</p> http://mathoverflow.net/questions/34879/modules-over-a-gorenstein-ring/35769#35769 Answer by ashpool for Modules over a Gorenstein ring ashpool 2010-08-16T14:50:03Z 2010-08-16T16:01:39Z <p>I found this proof in Kaplansky's Commutative Rings:</p> <p>Induction on $\mbox{dim }A$.</p> <p>$\mbox{dim }A =0 \$:</p> <p>Suppose $M\neq 0$. $\mbox{id}(M)=\mbox{depth}(A)=0$ so $M$ is injective, and hence is a direct sum of $\mbox{E}(A/\mathfrak{m})$. Since $A$ is Artin Gorenstein, $\mbox{E}(A/\mathfrak{m})\simeq A$ so $M$ is free.</p> <p>$\mbox{dim }A \geq 1\$:</p> <p>$0\leftarrow M\leftarrow A^{n}\leftarrow K\leftarrow 0$</p> <p>Then $\mbox{id}(K)&lt;\infty$. Since $\mbox{dim }A\geq 1$ there is a non-zero-divisor $a\in A$, which is $K$-regular. Then $\mbox{id} _{A/(a)}(K/aK)\leq\mbox{id} _{A}(K)-1&lt;\infty$ by one of the change of rings formulae. Now $A/(a)$ is Gorenstein and $\mbox{dim }A/(a)&lt;\mbox{dim }A$ so by the induction hybothesis $\mbox{pd} _{A/(a)}(K/aK)&lt;\infty$. By another change of rings formula, $\mbox{pd} _{A}(K)=\mbox{pd} _{A/(a)}(K/aK)&lt;\infty$ so $\mbox{pd} _{A}(M)&lt;\infty$ from the above short exact sequence.</p> http://mathoverflow.net/questions/33205/tor-and-projective-dimension Tor and projective dimension ashpool 2010-07-24T15:14:30Z 2010-08-14T02:12:37Z <p>Is it possible that $\mbox{Tor }^{r+1}(M,N)=0 \ \ \forall N$ yet $\mbox{proj. dim }M>r$?</p> <p>What I do know is that if $(A,\mathfrak{m})$ is Noetherian local and $M$ is finitely generated over $A$ then $\mbox{Tor }^{r+1}(M,A/\mathfrak{m})=0$ if and only if $\mbox{proj. dim }M\leq r$.</p> <p>Generally speaking, is $\mbox{Tor }$ functor as good a tool to measure projective dimension as $\mbox{Ext }$ even when the ring/module is not Noetherian or local?</p> <p>I suspect we can use $\mbox{Tor }$ to measure projective dimension when ring is Neotherian local and module is finitely generated because flatness and projectivity coincide in such case.</p> http://mathoverflow.net/questions/34098/is-the-isomorphism-class-of-a-fixed-cardinality-a-set Is the isomorphism class of a fixed cardinality a set? ashpool 2010-08-01T10:50:55Z 2010-08-09T15:14:48Z <p>Is the isomorphism class of a fixed cardinality a set(not a proper class)? Or a fixed ordinality for that matter? By "isomorphism" I mean just bijection for cardinals and order preserving bijection for ordinals, in the category of sets.</p> http://mathoverflow.net/questions/34771/do-gorenstein-rings-necessarily-have-a-finite-projective-dimension-as-a-module-o Do Gorenstein rings necessarily have a finite projective dimension (as a module over itself)? ashpool 2010-08-06T15:04:02Z 2010-08-06T15:53:39Z <p>Do Gorenstein rings necessarily have finite projective dimensions?</p> http://mathoverflow.net/questions/33784/extension-problem Extension problem ashpool 2010-07-29T13:01:57Z 2010-07-29T14:21:17Z <p>As I understand, if $0\rightarrow A\rightarrow X\rightarrow B\rightarrow 0$ is a short exact sequence of abelian groups, $\mbox{Ext }_{\mathbb{Z}}^{1}(B,A)$ gives all the isomorphism classes of what can come in as $X$. But when I consider $0\rightarrow \mathbb{Z}\rightarrow X\rightarrow \mathbb{Z}/(3)\rightarrow 0$,$\ \ \$ $\mbox{Ext } _{\mathbb{Z}}(\mathbb{Z}/(3),\mathbb{Z})=\mathbb{Z}/(3)\ \$ but all I can think of for $X$ are only two, $\mathbb{Z}$ and $\mathbb{Z}\oplus\mathbb{Z}/(3)$. Am I missing something or am I not understanding the result of extension problem correctly?</p> http://mathoverflow.net/questions/33736/homological-dimensions-of-module Homological dimensions of module ashpool 2010-07-29T00:07:35Z 2010-07-29T12:50:59Z <p>$(A,\mathfrak{m})$ a Noetherian local ring, $M\neq 0$ a finitely generated $A$-module. As I understand, $\mbox{Ext }^{j}(A/\mathfrak{m}, M) = 0$ for $j&lt;\mbox{depth }(M)$ and for $j>\mbox{inj. dim }(M)$, while $\mbox{Ext }^{j}(A/\mathfrak{m}, M) \neq 0$ for $j=\mbox{depth }(M)$ and $j=\mbox{inj. dim }(M)$. And I cannot help but wonder if $\mbox{Ext }^{j}(A/\mathfrak{m}, M) \neq 0$ for every $j$ between $\mbox{depth }(M)$ and $\mbox{inj. dim }(M)$ ?</p> http://mathoverflow.net/questions/33513/non-finite-version-of-nakayamas-lemma Non-finite version of Nakayama's lemma? ashpool 2010-07-27T14:07:03Z 2010-07-27T22:01:45Z <p>Let $A$ be a local ring with nilpotent maximal ideal $\mathfrak{m}$ (i.e., some power of $\mathfrak{m}$ vanishes), and $M$ an $A$-module (not necessarily finitely generated). Let $\bar{S}\subset M/\mathfrak{m}M$ be a set of generators and $S$ a set of representatives of $\bar{S}$ in $M$. Then is it true that $S$ is a set of generators of $M$? This is a common form of Nakayama's lemma with the assumption of finite generation of $M$ replacing the nilpotence of $\mathfrak{m}$. A passage in Matsumura's book "Commutative Ring Theory" (see Theorem 7.10) seems to imply this result, and I can't figure out why.</p> http://mathoverflow.net/questions/33540/existence-of-a-minimal-generating-set-of-a-module Existence of a minimal generating set of a module ashpool 2010-07-27T16:14:49Z 2010-07-27T16:27:06Z <p>Does a module (over a commutative ring) always possess a minimal generating set? When the module is not finitely generated, the typical Zorn's lemma type argument doesn't seem to work. More precisely, if $(S_{\alpha})_{\alpha}$ is a chain of generating subsets of a module $M$, $M=\cap _ {\alpha}(S _ {\alpha})\supset (\cap_{\alpha}S_{\alpha})$ is not in general (I think) equality (the parentheses in the last equation indicate submodule generated by).</p> http://mathoverflow.net/questions/84695/why-are-canonical-modules-supported-everywhere/84700#84700 Comment by ashpool ashpool 2012-01-01T23:38:25Z 2012-01-01T23:38:25Z @ Mahdi Majidi-Zolbanin $\operatorname{Ass}\omega=\operatorname{Ass}A$ from (1.7) also follows from the isomorphism $\operatorname{Hom}(\omega,\omega)=A$. Thanks for the reference! http://mathoverflow.net/questions/84695/why-are-canonical-modules-supported-everywhere/84700#84700 Comment by ashpool ashpool 2012-01-01T23:09:53Z 2012-01-01T23:09:53Z I just realized that $\operatorname{Supp}\omega=\operatorname{Spec}A$ can be deduced from the isomorphism $\operatorname{Hom}(\omega,\omega)\simeq A$! http://mathoverflow.net/questions/84685/are-maximal-cohen-macaulay-modules-supported-everywhere/84688#84688 Comment by ashpool ashpool 2012-01-01T19:14:05Z 2012-01-01T19:14:05Z I guess the question is, then, Why are canonical modules supported everywhere? http://mathoverflow.net/questions/84685/are-maximal-cohen-macaulay-modules-supported-everywhere/84688#84688 Comment by ashpool ashpool 2012-01-01T18:20:35Z 2012-01-01T18:20:35Z I think Graham's example is a good counter-example to the common phrase &quot;MCM modules localize.&quot; http://mathoverflow.net/questions/84685/are-maximal-cohen-macaulay-modules-supported-everywhere/84688#84688 Comment by ashpool ashpool 2012-01-01T17:57:06Z 2012-01-01T17:57:06Z Thanks! So if the statements about canonical module $\omega$ are true, then the reason why $\operatorname{Supp}\omega=\operatorname{Spec}A$ does not come from MCM property... Then I guess it comes from type=1 condition? http://mathoverflow.net/questions/64399/does-completion-commute-with-localization Comment by ashpool ashpool 2011-05-16T14:01:32Z 2011-05-16T14:01:32Z @Martin Brandenburg: Sorry, I realized it was a simple problem after I posted it. http://mathoverflow.net/questions/52341/related-to-fractional-ideals/53287#53287 Comment by ashpool ashpool 2011-02-18T15:33:01Z 2011-02-18T15:33:01Z It was very helpful. Thanks! http://mathoverflow.net/questions/53906/primary-decomposition-for-modules Comment by ashpool ashpool 2011-02-05T05:48:15Z 2011-02-05T05:48:15Z @Greg Marks: I can't figure out your farcical exercise. Is there any hint? http://mathoverflow.net/questions/53906/primary-decomposition-for-modules Comment by ashpool ashpool 2011-02-01T03:13:46Z 2011-02-01T03:13:46Z @Manny Reyes: Thanks! Can you point to a reference where I can find the result about the ring with a faithful notherian module being itself Noetherian? http://mathoverflow.net/questions/53906/primary-decomposition-for-modules Comment by ashpool ashpool 2011-01-31T21:34:56Z 2011-01-31T21:34:56Z @Manny Reyes: Sorry, I meant the radical is a prime ideal. http://mathoverflow.net/questions/53906/primary-decomposition-for-modules Comment by ashpool ashpool 2011-01-31T19:52:16Z 2011-01-31T19:52:16Z No, I'm asking for the proof in the case of a Noetherian module over a non-Noetherian (commutative) ring. http://mathoverflow.net/questions/46742/cm-module-is-height-unmixed/46783#46783 Comment by ashpool ashpool 2010-11-21T01:45:58Z 2010-11-21T01:45:58Z Thanks! I was aware of the local result, but spent an inordinate amount of time today trying to prove the non-local version. Your example puts an end to that. Greatly appreciated! http://mathoverflow.net/questions/34879/modules-over-a-gorenstein-ring/34888#34888 Comment by ashpool ashpool 2010-09-08T22:32:01Z 2010-09-08T22:32:01Z @Hailong: I see. Thanks! http://mathoverflow.net/questions/34879/modules-over-a-gorenstein-ring/34888#34888 Comment by ashpool ashpool 2010-09-08T17:40:20Z 2010-09-08T17:40:20Z @Hailong: I couldn't find it in Bruns-Herzog but I found a nice proof in Eisenbud which also covers A-regularity implying M-regularity. Thanks! By the way, in your comment when you mentioned Ass(M) is a subset of Ass(A) (Aug 17 5:09), did you mean that the union of primes in Ass(M) is contained in the union of primes in Ass(A)? Your argument doesn't seem to go beyond that. http://mathoverflow.net/questions/34879/modules-over-a-gorenstein-ring/34888#34888 Comment by ashpool ashpool 2010-09-07T19:39:47Z 2010-09-07T19:39:47Z @Hailong: Sorry, I can't figure out why maximal Cohen-Macaulay modules localize. Is it a general fact or are you assuming $R$ to be Gorenstein?