User pablo zadunaisky - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:29:58Z http://mathoverflow.net/feeds/user/17353 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73533/local-cohomology-and-maximal-cohen-macaulay-modules Local Cohomology and Maximal-Cohen-Macaulay modules Pablo Zadunaisky 2011-08-24T00:15:46Z 2013-04-21T15:39:49Z <p>Checking a recent article [<a href="http://arxiv.org/abs/1108.1552" rel="nofollow">this one</a>, specifically section 3.1] I found the following claim (I'm paraphrasing, of course): </p> <blockquote> <p>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.</p> </blockquote> <p>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?</p> http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/116973#116973 Answer by Pablo Zadunaisky for Awfully sophisticated proof for simple facts Pablo Zadunaisky 2012-12-21T14:28:42Z 2012-12-21T14:28:42Z <p>The skew-field of quaternions $\mathbb H$ is isomorphic to its opposite algebra. </p> <p>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 $\mathbb H \cong \mathbb H^\mathsf{opp}$.</p> <p>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.</p> http://mathoverflow.net/questions/116469/as-cohen-macaulay-algebras-and-dualizing-complexes AS Cohen Macaulay algebras and dualizing complexes Pablo Zadunaisky 2012-12-15T17:46:17Z 2012-12-15T22:17:12Z <p>Let $A$ be an $\mathbb N$-graded algebra such that $A_0 = k$ is a field. This are usually called graded connected algebras. </p> <p>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 $M$</p> <p>$$\Gamma_{\mathfrak m}(M) = \{m \in M | A_{\geq i}m = 0 \mbox{ for } i \gg 0 \} \cong \varinjlim_i Hom_A(A/A_{\geq i},M).$$</p> <p>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 $$H_\mathfrak m(-) \cong \varinjlim Ext_A^i(A/A_{\geq i}, -)$$</p> <p>We say $A$ is</p> <ul> <li><p>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)</p></li> <li><p>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.</p></li> </ul> <p>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.)</p> <p>We have the usual implication chain </p> <blockquote> <p>AS regular $\Rightarrow$ AS Gorenstein $\Rightarrow$ AS Cohen Macaulay.</p> </blockquote> <p>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 $i > n$.</p> <p><strong>Question 1</strong>: Is this result still true for noncommutative noetherian AS Cohen Macaulay algebras? </p> <p>The result is true for noetherian AS Gorenstein algebras, as explained in <a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=TI&amp;pg6=PC&amp;pg7=ALLF&amp;pg8=ET&amp;r=1&amp;review_format=html&amp;s4=Yekutieli&amp;s5=Serre&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq" rel="nofollow">this paper</a> 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</p> <p><strong>Question 2</strong>: Are there AS Cohen Macaulay algebras <em>without</em> (balanced or unbalanced) dualizing complexes?</p> <p>Thanks!</p> http://mathoverflow.net/questions/5751/can-infinity-shorten-proofs-a-lot/115389#115389 Answer by Pablo Zadunaisky for Can infinity shorten proofs a lot? Pablo Zadunaisky 2012-12-04T11:28:11Z 2012-12-04T11:28:11Z <p>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:</p> <blockquote> <p>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?</p> </blockquote> <p>The problem is easier to solve if we think of the grain as a fluid and apply, say, Pascal's principle.</p> http://mathoverflow.net/questions/114383/examples-where-adding-complexity-made-a-problem-simpler/114482#114482 Answer by Pablo Zadunaisky for Examples where adding complexity made a problem simpler Pablo Zadunaisky 2012-11-26T03:44:01Z 2012-11-26T03:44:01Z <p>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". </p> <p>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.</p> http://mathoverflow.net/questions/89453/injective-dimension-of-graded-injective-modules Injective dimension of graded-injective modules. Pablo Zadunaisky 2012-02-25T00:33:43Z 2012-04-23T06:57:26Z <p>In "<a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=TI&amp;pg6=PC&amp;pg7=ALLF&amp;pg8=ET&amp;r=1&amp;review_format=html&amp;s4=van%2520den%2520Bergh&amp;s5=Existence&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq" rel="nofollow">Existence theorems...</a>" Van den Bergh proposes the following "pleasant excercise in homological algebra":</p> <blockquote> <p>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.</p> </blockquote> <p>There is a proof of this fact for commutative graded algebras in <a href="http://www.mscand.dk/article.php?id=2222" rel="nofollow">this paper</a> 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.</p> <p>PS: In a <a href="http://www.math.bgu.ac.il/~amyekut/publications/aus-gor/journal.pdf" rel="nofollow">later paper</a>, 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...</p> http://mathoverflow.net/questions/87291/inverse-limit-of-spectral-sequences Inverse limit of spectral sequences Pablo Zadunaisky 2012-02-01T23:01:12Z 2012-02-02T08:05:06Z <p>I find myself in the following situation:</p> <p>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 <code>$E(n)_{*,*}^* \to E(n-1)_{*,*}^*$</code>, so we have an inverse directed system of spectral sequences.</p> <p>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?</p> http://mathoverflow.net/questions/74252/things-that-should-be-positive-integers-really/74275#74275 Answer by Pablo Zadunaisky for Things that should be positive integers...really? Pablo Zadunaisky 2011-09-01T17:59:44Z 2011-09-01T17:59:44Z <p>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.</p> <p>As a complementary answer... what would be a ring of characteristic $-\pi$?</p> http://mathoverflow.net/questions/73533/local-cohomology-and-maximal-cohen-macaulay-modules/73612#73612 Answer by Pablo Zadunaisky for Local Cohomology and Maximal-Cohen-Macaulay modules Pablo Zadunaisky 2011-08-24T21:54:12Z 2011-08-24T22:01:27Z <p>Well, I don't know if I'm supposed to, but since I found a solution, I'll write the general idea here.</p> <p>[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 $$E^{pq}_2 = \underline{Ext}^p_A(\underline{Ext}^q(M, \omega_A),\omega_A) \Rightarrow \begin{cases}M&amp;\mbox{ if p = q} \\ 0 &amp;\mbox{ otherwise}\end{cases}$$</p> <p>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 $i \leq d$)</p> http://mathoverflow.net/questions/117891/what-are-conjectures-that-are-true-for-primes-but-then-turned-out-to-be-false-for Comment by Pablo Zadunaisky Pablo Zadunaisky 2013-01-03T03:21:15Z 2013-01-03T03:21:15Z Does the classical theorem proved by physical induction, &quot;all odd numbers are prime&quot;, count as an example? http://mathoverflow.net/questions/117923/connectedness-of-the-complement-of-small-subsets-extended-question/117924#117924 Comment by Pablo Zadunaisky Pablo Zadunaisky 2013-01-03T03:14:05Z 2013-01-03T03:14:05Z You need $n&gt;1$ for this. http://mathoverflow.net/questions/96510/have-we-ever-lost-any-mathematics/96549#96549 Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-12-16T22:34:57Z 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. http://mathoverflow.net/questions/4994/fundamental-examples/5034#5034 Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-12-16T15:03:12Z 2012-12-16T15:03:12Z +1 for the parenthetical comment! http://mathoverflow.net/questions/116469/as-cohen-macaulay-algebras-and-dualizing-complexes Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-12-16T00:26:55Z 2012-12-16T00:26:55Z Thanks! I spent some time playing with it and couldn't get it right. http://mathoverflow.net/questions/109672/question-on-bigraded-modules Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-10-15T14:53:58Z 2012-10-15T14:53:58Z $M^{(n)}$ is not stable by multiplication by $T_j$, since this increases the &quot;$n$&quot;, 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. http://mathoverflow.net/questions/89453/injective-dimension-of-graded-injective-modules/94911#94911 Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-04-25T18:54:54Z 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. http://mathoverflow.net/questions/89453/injective-dimension-of-graded-injective-modules/94911#94911 Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-04-24T01:51:45Z 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 :). http://mathoverflow.net/questions/89453/injective-dimension-of-graded-injective-modules/92854#92854 Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-04-03T16:51:01Z 2012-04-03T16:51:01Z @Ralph, I meant finite dimensional over $k$. http://mathoverflow.net/questions/89453/injective-dimension-of-graded-injective-modules/92854#92854 Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-04-02T19:02:36Z 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 $k$. http://mathoverflow.net/questions/89453/injective-dimension-of-graded-injective-modules Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-02-26T18:52:08Z 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 $\bigoplus_{\mathbb Z} I \rightarrow \bigoplus_{\mathbb Z} E$. http://mathoverflow.net/questions/87291/inverse-limit-of-spectral-sequences Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-02-02T15:22:06Z 2012-02-02T15:22:06Z Thanks both Dylan Wilson and Leonid Positselski for their answers! http://mathoverflow.net/questions/87291/inverse-limit-of-spectral-sequences/87313#87313 Comment by Pablo Zadunaisky Pablo Zadunaisky 2012-02-02T15:19:11Z 2012-02-02T15:19:11Z Aside from answering the question, it is a crystal clear explanation. Thank you very much, Leonid!