User ashpool - MathOverflow

About injective hull

2011-02-02T05:22:25Z

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 ?

Length of a resolution

2010-05-02T15:37:38Z

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.

Projective dimension of zero module

2010-09-04T17:18:22Z

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 $M\neq 0$:

(1) $\mbox{pd}(M)\leq n$ iff $\mbox{Ext}^{n+1}(M,-)=0$

(2) $\mbox{pd}(M)=0$ iff $M$ is projective

(3) $\mbox{grade}(M):=\infty$ if $M=0$

If one attempts to define $\mbox{pd}((0))$ by extending one of these results, (1), (2), (3) suggest $\mbox{pd}=-1, 0, \infty$, respectively.

projective module

2010-05-23T16:36:01Z

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.

Projective & injective dimensions

2010-08-06T00:05:23Z

$A$ 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?

Technical question about height of minimal associated primes

2012-03-06T18:56:13Z

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 $\geq p+1$.

Depth and dimension

2010-08-05T23:45:23Z

$A$ 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 $\mbox{inj.dim }M<\infty$, 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 $\mbox{dim }M\leq\mbox{depth }A$?

Why are canonical modules supported everywhere?

2012-01-01T19:24:47Z

Let $A$ be a local CM ring, and $\omega$ a canonical module of $A$. Here are two properties of $\omega$ from Bruns & Herzog:

$\omega_{\mathfrak{p}}$ is a canonical module of $A_{\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$.
$\mu_i(\mathfrak{p},\omega)=\delta_{i}^{\operatorname{ht}\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$, where $\mu$ denotes the Bass number.

These properties seem to imply that $\operatorname{Supp}\omega=\operatorname{Spec}A$. As Graham Leuschke pointed out, this is not a property of maximal CM modules. Why, then, are canonical modules supported everywhere?

Are maximal Cohen-Macaulay modules supported everywhere?

2012-01-01T15:43:37Z

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:

If $\omega$ is a canonical module of $A$, then $\omega_{\mathfrak{p}}$ is a canonical module of $A_{\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$.
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.

And am I correct in understanding that maximal CM module is by definition nonzero?

Flatness and local freeness

2010-07-27T14:50:58Z

The following statement is well-known:

$A$ 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 $\mathfrak{p}$.

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.

Metrizability of $\mathfrak{a}$-adic topology

2011-05-16T13:59:07Z

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?

Does completion commute with localization?

2011-05-09T15:12:50Z

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

$$(A^{\mathfrak{m}})_{\mathfrak{m^{\mathfrak{m}}}}=(A _{\mathfrak{m}})^{\mathfrak{m} _{\mathfrak{m}}}.$$

Primary decomposition for modules

2011-01-31T19:15:50Z

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.

Related to fractional ideals

2011-01-17T19:03:02Z

$K$ 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

$$M\subset A\Longleftrightarrow A\subset (A:_{K}M),$$

$$A\subset M\Longrightarrow (A:_{K}M)\subset A$$

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?

Projectively splitting module

2010-12-12T23:34:32Z

Is there a name for such class of modules $M$ such that $M\rightarrow N\rightarrow 0$ splits for every $N$ ?

Unit ideal in non-commutative rings

2010-11-27T20:35:14Z

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 $r\in R$ ?

CM module is height-unmixed?

2010-11-20T15:42:36Z

$A$ 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?

Torsion submodule

2010-09-02T13:42:36Z

$A$ 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 the torsion submodule of $M$?

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.

Dimension of module

2010-05-09T16:57:42Z

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 $a \in \mbox{Ann}(M/P)$.

Finiteness of injective hull of residue field for Artin local ring

2010-08-06T16:24:57Z

$(A,\mathfrak{m})$ 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?

Answer by ashpool for Finiteness of injective hull of residue field for Artin local ring

2010-08-16T18:11:45Z

This is a proof of $\ell(M)=\ell(\mbox{Hom}(M,\mbox{E}(A/\mathfrak{m}))$ suggested by Mariano:

Induction on $\ell(M)\ $:

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 $0\leftarrow \mbox{Hom}(N,E(A/\mathfrak{m}))\leftarrow \mbox{Hom}(M,E(A/\mathfrak{m}))\leftarrow \mbox{Hom}(A/\mathfrak{m},E(A/\mathfrak{m}))\leftarrow 0$.

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 $\ell(\mbox{Hom}(M,E(A/\mathfrak{m}))=(n-1)+1=n$

Modules over a Gorenstein ring

2010-08-08T02:13:56Z

$A$ 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?

Answer by ashpool for Modules over a Gorenstein ring

2010-08-16T14:50:03Z

I found this proof in Kaplansky's Commutative Rings:

Induction on $\mbox{dim }A$.

$\mbox{dim }A =0 \ $:

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.

$\mbox{dim }A \geq 1\ $:

$ 0\leftarrow M\leftarrow A^{n}\leftarrow K\leftarrow 0$

Then $\mbox{id}(K)<\infty$. Since $\mbox{dim }A\geq 1$ there is a non-zero-divisor $a\in A$, which is $K$-regular. Then $\mbox{id} _{A/(a)}(K/aK)\leq\mbox{id} _{A}(K)-1<\infty$ by one of the change of rings formulae. Now $A/(a)$ is Gorenstein and $\mbox{dim }A/(a)<\mbox{dim }A$ so by the induction hybothesis $\mbox{pd} _{A/(a)}(K/aK)<\infty$. By another change of rings formula, $\mbox{pd} _{A}(K)=\mbox{pd} _{A/(a)}(K/aK)<\infty $ so $\mbox{pd} _{A}(M)<\infty$ from the above short exact sequence.

Tor and projective dimension

2010-07-24T15:14:30Z

Is it possible that $\mbox{Tor }^{r+1}(M,N)=0 \ \ \forall N$ yet $\mbox{proj. dim }M>r$?

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 $\mbox{proj. dim }M\leq r$.

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?

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.

Is the isomorphism class of a fixed cardinality a set?

2010-08-01T10:50:55Z

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.

Do Gorenstein rings necessarily have a finite projective dimension (as a module over itself)?

2010-08-06T15:04:02Z

Do Gorenstein rings necessarily have finite projective dimensions?

Extension problem

2010-07-29T13:01:57Z

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?

Homological dimensions of module

2010-07-29T00:07:35Z

$(A,\mathfrak{m})$ a Noetherian local ring, $M\neq 0$ a finitely generated $A$-module. As I understand, $\mbox{Ext }^{j}(A/\mathfrak{m}, M) = 0$ for $j<\mbox{depth }(M)$ 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 $\mbox{inj. dim }(M)$ ?

Non-finite version of Nakayama's lemma?

2010-07-27T14:07:03Z

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.

Existence of a minimal generating set of a module

2010-07-27T16:14:49Z

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).

Comment by ashpool

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!

Comment by ashpool

2012-01-01T23:09:53Z

I just realized that $\operatorname{Supp}\omega=\operatorname{Spec}A$ can be deduced from the isomorphism $\operatorname{Hom}(\omega,\omega)\simeq A$!

Comment by ashpool

2012-01-01T19:14:05Z

I guess the question is, then, Why are canonical modules supported everywhere?

Comment by ashpool

2012-01-01T18:20:35Z

I think Graham's example is a good counter-example to the common phrase "MCM modules localize."

Comment by ashpool

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?

Comment by ashpool

2011-05-16T14:01:32Z

@Martin Brandenburg: Sorry, I realized it was a simple problem after I posted it.

Comment by ashpool

2011-02-18T15:33:01Z

It was very helpful. Thanks!

Comment by ashpool

2011-02-05T05:48:15Z

@Greg Marks: I can't figure out your farcical exercise. Is there any hint?

Comment by ashpool

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?

Comment by ashpool

2011-01-31T21:34:56Z

@Manny Reyes: Sorry, I meant the radical is a prime ideal.

Comment by ashpool

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.

Comment by ashpool

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!

Comment by ashpool

2010-09-08T22:32:01Z

@Hailong: I see. Thanks!

Comment by ashpool

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.

Comment by ashpool

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?