When are dual modules free? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:58:01Z http://mathoverflow.net/feeds/question/4590 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4590/when-are-dual-modules-free When are dual modules free? David Speyer 2009-11-08T04:15:25Z 2010-01-08T20:52:29Z <p>Let $A$ be a commutative integral domain, with fraction field $K$. Let $T$ be a torsion-free finitely generated $A$ module, so $T \otimes_A K$ is a finite dimensional vector space $V$. Let $T^*$ be the set of $y$ in the dual vector space, $V^*$, such that $\langle x, y \rangle \in A$ for every $x \in T$.</p> <p>Under what hypotheses on $A$ can I conclude that $T^*$ is a free $A$-module? My current conjecture is that this holds whenever $A$ is a <a href="http://en.wikipedia.org/wiki/Unique%5Ffactorization%5Fdomain" rel="nofollow">UFD</a>. (Of course, it trivially holds if $A$ is a PID.)</p> <p>Here are a few ideas of mine. Define the rank of $T$ to be $\dim_K T \otimes_A K$. I can show that, if $A$ is a UFD, then $T^*$ is free for $T$ of rank $1$. For any $T$, we can make a short exact sequence $$0 \to S \to T \to U \to 0$$ where $S$ is rank $1$ and $U$ is torsion free with rank one less than $T$. So we have $$0 \to U^* \to T^* \to S^* \to \mathrm{Ext}^1(U,A) \to \cdots$$. This looks like a good start, but I don't know how to control that Ext group. I suspect that one of you does!</p> <p>This is motivated by Kevin Buzzard's <a href="http://mathoverflow.net/questions/3270/which-rings-are-subrings-of-matrix-rings" rel="nofollow">question</a> about matrix rings.</p> http://mathoverflow.net/questions/4590/when-are-dual-modules-free/4608#4608 Answer by Greg Muller for When are dual modules free? Greg Muller 2009-11-08T05:23:49Z 2010-01-08T20:52:29Z <p>The dual module of a finitely generated module is <em>reflexive</em>, that is, $M^{**}=M$, and reflexives are awfully close to projectives. Specifically, if $R$ is a Noetherian domain, then a module is projective if $Ext^i(M,R)=0$ for all $i>0$, and its reflexive if $Ext^i(M,R)=0$ for $i=1,2$.</p> <p>It is also worth noting that <strong>every</strong> reflexive is the dual of some module, specifically of $M^*$. Therefore, your question amounts to "for what rings is every reflexive module free?" In this light, its very similar to the question of when every projective module is free.</p> <p>From the above Ext criterion, its clear that if the global dimension of $R$ is less than or equal to 2, that being reflexive is the same as being projective. I would go so far as to conjecture the converse is true: that if gldim of R is 3 or more, that there is a non-projective module which is reflexive (and hence it is non-free).</p> <p>If this conjecture is true, then the answer to your question is "rings with global dimension 2 or less, such that every projective is free". Of course, its not immediately clear what these are, but its a start.</p> http://mathoverflow.net/questions/4590/when-are-dual-modules-free/4615#4615 Answer by Greg Stevenson for When are dual modules free? Greg Stevenson 2009-11-08T06:01:36Z 2009-11-08T06:01:36Z <p>In the case of regular local rings a criterion for a reflexive module to be free is given <a href="http://www.springerlink.com/content/j2406t088q58n710/fulltext.pdf" rel="nofollow"> here </a> (academic access required). The result is as follows </p> <p>Let <img src="http://latex.mathoverflow.net/png?A" alt="R" title="" /> be a regular local ring and <img src="http://latex.mathoverflow.net/png?M" alt="M" title="" /> a finitely generated reflexive <img src="http://latex.mathoverflow.net/png?A" alt="R" title="" />-module. Then if<br /> <img src="http://latex.mathoverflow.net/png?Ext%5FA%5E1%28Hom%28M%2CM%29%2CA%29%20%3D%200" alt="Ext\sb R^1(Hom(M,M),R) = 0" title="" /><br /> M is free over <img src="http://latex.mathoverflow.net/png?A" alt="R" title="" />. This is not great though since we have just asked for cohomological vanishing elsewhere.</p> <p>A criterion is also given in terms of unmixed ideals of height <img src="http://latex.mathoverflow.net/png?2" alt="2" title="" /> (more vanishing though). This I guess is not terribly surprising since the obstruction in your construction lives in codimension <img src="http://latex.mathoverflow.net/png?%5Cgeq%202" alt="\geq 2" title="" />. Indeed, with the hypothesis that <img src="http://latex.mathoverflow.net/png?A" alt="A" title="" /> be a UFD one has<br /> <img src="http://latex.mathoverflow.net/png?codim%5C%3B%20Supp%5FA%28Ext%5FA%5E1%28U%2CA%29%29" alt="codim\; Supp\sb A(Ext\sb A^1(U,A))" title="" /> <img src="http://latex.mathoverflow.net/png?%5Cgeq%202" alt="\geq 2" title="" /></p> <p>I've been trying to do a bit better (or find a counterexample) by assuming that <img src="http://latex.mathoverflow.net/png?A" alt="A" title="" /> is at least Gorenstein of finite Krull dimension and using the fact that one can write down the injective resolution for <img src="http://latex.mathoverflow.net/png?A" alt="A" title="" /> but I haven't had any luck yet.</p> http://mathoverflow.net/questions/4590/when-are-dual-modules-free/4646#4646 Answer by David Speyer for When are dual modules free? David Speyer 2009-11-08T16:23:16Z 2009-11-08T16:23:16Z <p>OK, I now have a counter-example. Thanks to the previous answers for showing me where to look.</p> <p>Let $A = k[x,y,z]$. Let $M$ be the kernel of the map $(x,y,z) : A^3 \to A$ and $N$ the co-kernel of the map $(x,y,z)^T: A \to A^3$. I claim that $M = N^{*}$ but $M$ is not free.</p> <p>To see that $M=N^{*}$, consider the defining sequence $$0 \to A \to A^3 \to N \to 0.$$ This gives rise to $$0 \to N^{*} \to A^3 \to A.$$ The kernel of the right hand map is $M$ by definition.</p> <p>Now, let's see that $M$ is not free. We have a graded short exact sequence $$0 \to M \to A^3 \to A[1] \to k[1] \to 0.$$ So the Hilbert series of $M$ is $$\frac{3}{(1-t)^3} - \frac{t^{-1}}{(1-t)^3} + t^{-1} = \frac{(1-t)^3 - 1 + 3t}{t(1-t)^3} = \frac{3t-t^2}{(1-t)^3}.$$ If $M$ were a free module, its Hilbert series would look like $(t^a+t^b)/(1-t)^3$.</p> <p>$N$ is also not free; I have not figured out whether $N$ is reflexive.</p> http://mathoverflow.net/questions/4590/when-are-dual-modules-free/4870#4870 Answer by Graham Leuschke for When are dual modules free? Graham Leuschke 2009-11-10T14:49:38Z 2009-11-10T14:49:38Z <p>Every dual $T^*$, where $T$ is torsionfree -- and hence every reflexive module -- is a second syzygy, as displayed by dualizing a projective presentation of $T$. On the other hand, it follows from Auslander-Bridger (or see a slightly more readable presentation <a href="http://arxiv1.library.cornell.edu/abs/math/9809121v2" rel="nofollow">by Masek (last Corollary in this paper</a>) that if a ring $R$ satisfies S1 and is Gorenstein at the minimal primes, then every second syzygy is reflexive. </p> <p>Therefore, for reduced rings being a dual of a torsionfree is equivalent to being reflexive is equivalent to being a second syzygy. In particular, as long as the global dimension is at least 3, there are duals that are not projective. </p> <p>Back to the original question: when can you conclude the dual of $T$ is free? (I'm going to talk only about local rings, so ignore the distinction between free and projective.) Assume $A$ is a regular local ring. If there is a module $N$ such that $Ext(T,N)=0$ for $i = 1, ..., depth N-2$, then the dual of $T$ is free. In particular, if $N$ has depth less than or equal to $3$ and $Ext_R^1(T,N)=0$, then the dual of $T$ is free. This is in a recent paper by Jothilingam, but is not hard to prove directly. It's not a condition solely on $A$, but maybe it's useful.</p>