User pablo zadunaisky - MathOverflow

Local Cohomology and Maximal-Cohen-Macaulay modules

2011-08-24T00:15:46Z

Checking a recent article [this one, specifically section 3.1] I found the following claim (I'm paraphrasing, of course):

Let $A$ be a graded connected noetherian algebra (not necessarily commutative), and suppose it is AS-Cohen-Macaulay of depth $d$. If $M$ is a finitely generated graded module over $A$, and it is Maximal Cohen Macaulay (MCM, ie, its only non-zero local cohomology module is precisely the $d$-th), then its first syzygy is also MCM.

I have a proof for this in the commutative ungraded case, but it deppends on the fact that $\lbrace i|H^i_{\mathfrak m}(M) \neq 0 \rbrace$ is non-empty and contained in the interval $[0,d]$ (consider the short exact sequence involving $M$ and its first syzygy and look at the long exact sequence of local cohomology). I found results regarding the non-vanishing of this groups in the non-commutative case, but they demand much more strict conditions than in the paper (finite GK-dimension, enough normal elements, etc.). Any idea on how to prove this in this more general context?

Answer by Pablo Zadunaisky for Awfully sophisticated proof for simple facts

2012-12-21T14:28:42Z

The skew-field of quaternions $\mathbb H$ is isomorphic to its opposite algebra.

Indeed, by a theorem of Frobenius, division algebras over the reals are isomorphic to either $\mathbb R, \mathbb C$ or $\mathbb H$. Since $\mathbb H^\mathsf{opp}$ is again a division algebra, it must be isomorphic to one of these. There are several ways to conclude: since it is four dimensional, or since it is not commutative, or since it has more than two square roots of $-1$, etc., we conclude that the only possibility is $\mathbb H \cong \mathbb H^\mathsf{opp}$.

If you are only interested in Morita equivalence between these two algebras, you can do better: the Brauer group of $\mathbb R$ is isomorphic to $\mathbb Z_2$, and so all elements are of order $2$. This implies that the class of $\mathbb H$ coincides with its inverse, which is the class of $\mathbb H^{\mathsf{opp}}$. Thus $\mathbb H$ and $\mathbb H^\mathsf{opp}$ are Morita equivalent.

AS Cohen Macaulay algebras and dualizing complexes

2012-12-15T17:46:17Z

Let $A$ be an $\mathbb N$-graded algebra such that $A_0 = k$ is a field. This are usually called graded connected algebras.

One can define a torsion functor with respect to the ideal $\mathfrak m = \bigoplus_{i \geq 1} A_i$, setting for any graded $A$ module $M$

$$\Gamma_{\mathfrak m}(M) = \{m \in M | A_{\geq i}m = 0 \mbox{ for } i \gg 0 \} \cong \varinjlim_i Hom_A(A/A_{\geq i},M).$$

The derived functors of $\Gamma_{\mathfrak m}$ are the local cohomology functors with respect to $\mathfrak m$, and are denoted by $H^i_\mathfrak m$. As in the commutative case, there is a natural isomorphism $$H_\mathfrak m(-) \cong \varinjlim Ext_A^i(A/A_{\geq i}, -)$$

We say $A$ is

AS Cohen Macaulay if there is a natural number $n$ such that $H_\mathfrak m^i (A) = H_\mathfrak m(A^{op}) = 0$. Let us call $n$ the local dimension of the module $A$ (I'm not sure this is standard notation)
AS Gorenstein if it has finite injective dimension $n$ both as a right and left module, and furthermore $Ext_A^n(k,A) = k$, once again on both sides.

This are generalizations of ye olde condition of regularity for graded connected algebras introduced by Artin and Schelter, hence the AS. If $A$ is commutative and noetherian, then they are equivalent to their AS-less counterparts. (Maybe you can drop the noetherian hypothesis on this, but I'm not sure.)

We have the usual implication chain

AS regular $\Rightarrow$ AS Gorenstein $\Rightarrow$ AS Cohen Macaulay.

By Groethendick's vanishing theorem, if $A$ is a noetherian Cohen Macaulay algebra of local dimension $n$, then $H^i_\mathfrak m \equiv 0$ for $i > n$.

Question 1: Is this result still true for noncommutative noetherian AS Cohen Macaulay algebras?

The result is true for noetherian AS Gorenstein algebras, as explained in this paper by Yekutieli and Zhang, Corollary 4.3. The argument comes basically from the fact that AS Gorenstein algebras have balanced dualizing complexes, almost by definition. This brings me to my

Question 2: Are there AS Cohen Macaulay algebras without (balanced or unbalanced) dualizing complexes?

Thanks!

Answer by Pablo Zadunaisky for Can infinity shorten proofs a lot?

2012-12-04T11:28:11Z

I am surprised no one has mentioned any examples from physics. Fluid mechanics for example obtains some beautiful results by assuming that a fluid is continuous, while at the molecular level it is a discrete object. An example from my physics highschool textbook:

Suppose you have a circular grain storage, with a diameter of 2 meters and 10 meters high. The grain weights about $X \ kg/m^3$. What is the pressure at two meters depth? at four meters depth?

The problem is easier to solve if we think of the grain as a fluid and apply, say, Pascal's principle.

Answer by Pablo Zadunaisky for Examples where adding complexity made a problem simpler

2012-11-26T03:44:01Z

In Oxtoby's "Measure and [Baire] Category" it is mentioned that facts such as "Elements with property X form a set of measure zero/first category/countable" can be seen as existence results, as in the proof that there are many irrational numbers, since the set of rational numbers is "small" in all three senses, while the set of real numbers is "big".

From that point of view this three "complicated" ideas, [the Baire Category theorem, naive set theory and Lebesgue measure] introduce very simple ways of proving the existence of potentially complicated objects.

Injective dimension of graded-injective modules.

2012-02-25T00:33:43Z

In "Existence theorems..." Van den Bergh proposes the following "pleasant excercise in homological algebra":

Let $A$ be a connected graded noetherian $k$-algebra (that is, $\mathbb N$-graded with $A_0 = k$ a field). A graded module $I$ which is injective in the category of left graded modules has injective dimension at most $1$ in the category of left modules.

There is a proof of this fact for commutative graded algebras in this paper by Fossum and Foxby, but I don't really see how to transfer this to the non-commutative setting. Can anyone provide any pointers or a reference to a proof? Thanks in advance.

PS: In a later paper, Yekutieli and Zhang state that the only proof they know of this fact is "quite involved", which eased my anxiety at being unable to solve a pleasant excercise in the area I'm supposed to be PhD-ing in...

Inverse limit of spectral sequences

2012-02-01T23:01:12Z

I find myself in the following situation:

I have a sequence of first quadrant spectral sequences, let's call them $ E(n)_{p,q}^* $, each convergent to $E(n)_{p,q}^\infty$, with spectral sequence morphisms $E(n)_{*,*}^* \to E(n-1)_{*,*}^*$ , so we have an inverse directed system of spectral sequences.

Each module of each page is a locally finite graded vector space, so if you define $E_{p,q}^* = \varprojlim_n E(n)_{p,q}^* $, $E_{p,q}^* $ turns out to be a spectral sequence (differentials are the limit of differentials in the original sequences, etc.) by an argument found in a paper by John Carter. However, in this same paper he states that you can't say that $E_{p,q}^*$ converges to $\varprojlim E(n)_{p,q}^\infty$, and his counterexample rests on the fact that his spectral sequences can be non-bounded... Does anyone know if this "convergence" result is true for bounded (or, say, first quadrant) sequences?

Answer by Pablo Zadunaisky for Things that should be positive integers...really?

2011-09-01T17:59:44Z

It's interesting to note that this has happened with several notions of "dimension". Krull dimension of rings has been extended to notions as GK-dimension, for example.

As a complementary answer... what would be a ring of characteristic $-\pi$?

Answer by Pablo Zadunaisky for Local Cohomology and Maximal-Cohen-Macaulay modules

2011-08-24T21:54:12Z

Well, I don't know if I'm supposed to, but since I found a solution, I'll write the general idea here.

[This is from an unpublished manuscript by P. Smith, the first author of the paper]: If $A$ is CM, let $\omega_A = H^d_\mathfrak m(A)^*$ be its dualizing module. Then there is a spectral sequence $$ E^{pq}_2 = \underline{Ext}^p_A(\underline{Ext}^q(M, \omega_A),\omega_A) \Rightarrow \begin{cases}M&\mbox{ if p = q} \\ 0 &\mbox{ otherwise}\end{cases}$$

Since $H_\mathfrak{m}^i(M)^* \cong \underline{Ext}_A^i(M,\omega_A)$, the convergence of this SS to a non-zero result guarantees that there must be a non-zero local cohomology module (and in fact, that there is a non-zero one with $i \leq d$)

Comment by Pablo Zadunaisky

2013-01-03T03:21:15Z

Does the classical theorem proved by physical induction, "all odd numbers are prime", count as an example?

Comment by Pablo Zadunaisky

2013-01-03T03:14:05Z

You need $n>1$ for this.

Comment by Pablo Zadunaisky

2012-12-16T22:34:57Z

I always understood Russell's nightmare as reflecting insecurity in the lasting importance of his work, rather than the intelligence of its judges.

Comment by Pablo Zadunaisky

2012-12-16T15:03:12Z

+1 for the parenthetical comment!

Comment by Pablo Zadunaisky

2012-12-16T00:26:55Z

Thanks! I spent some time playing with it and couldn't get it right.

Comment by Pablo Zadunaisky

2012-10-15T14:53:58Z

$M^{(n)}$ is not stable by multiplication by $T_j$, since this increases the "$n$", so it only has the structure of an $R$-module. On the other hand, this assignation must be exact because a complex of $S$-modules is exact if and only if it is exact when we look at each component separately.

Comment by Pablo Zadunaisky

2012-04-25T18:54:54Z

The best I can tell you is that I sent an email to vdB but received no answer. I did not try to contact Yekutieli or Zhang.

Comment by Pablo Zadunaisky

2012-04-24T01:51:45Z

Wow! Thanks again Ralph. I will look into the details tomorrow, but on a first reading everything seems to work :).

Comment by Pablo Zadunaisky

2012-04-03T16:51:01Z

@Ralph, I meant finite dimensional over $k$.

Comment by Pablo Zadunaisky

2012-04-02T19:02:36Z

This s a very interesting answer... off the top of my head, I would even be optimistic of improving the result to $\mathbb Z$-graded injective modules, by making a finer analysis of the case when the filtration is indeed exhaustive, separated and the associated graded module is locally finite, but not bounded. Let me point out, however, that by Matlis duality an injective module is the dual of a flat one, and in the local case, flat modules are free. The dual of a free module is unbounded unless the algebra itself is finite over $k$.

Comment by Pablo Zadunaisky

2012-02-26T18:52:08Z

I don't see what the graded module structure on $\oplus_{\mathbb Z} E$ is, or how to make it compatible with the obvious morphism $\bigoplus_{\mathbb Z} I \rightarrow \bigoplus_{\mathbb Z} E$.

Comment by Pablo Zadunaisky

2012-02-02T15:22:06Z

Thanks both Dylan Wilson and Leonid Positselski for their answers!

Comment by Pablo Zadunaisky

2012-02-02T15:19:11Z

Aside from answering the question, it is a crystal clear explanation. Thank you very much, Leonid!