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Let $M$ be a finitely generated module over a commutative ring $R$. The first order deformation of module $M$ is parametrized by $Ext^{1}(M,M)$ and the obstruction is parametrized by $Ext^{2}(M,M)$. Is there a similar story for noncommutative $R$? I don't expect this to be true for any noncommutative rings but wonder if this is still true for some $good$ ones. I would appreciate any reference suggestion, comments, and ideas.

Edit I would like naively to compute the tangent space of the moduli space $X$ of module with some data if it exists. At $M\in X$, the tangent space can be understood as the set of extensions of the $R$-module $M$ to some $R\otimes_{k} k[\epsilon]/(\epsilon^2)$-module. The obstruction is defined in the same manner.

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  • $\begingroup$ What is $E$ here? $\endgroup$
    – Qiaochu Yuan
    Commented Sep 10, 2012 at 16:52
  • $\begingroup$ $E$ must be $M$. Thank you for pointing this out. $\endgroup$
    – user2013
    Commented Sep 11, 2012 at 4:51
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    $\begingroup$ (I am asking because the simplest arguments I know to show the claims you make about $Ext^1$ and $Ext^2$ work excatly the same for commutative and non-commutative rings... so there must be something different) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 11, 2012 at 5:18
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    $\begingroup$ The references I know for deformations of modules are work of Waldmann and Burstyn, and also some previous work of Weinstein (as collaborator to other authors), and there indeed there was no need that the base ring be commutative. As M. Suarez-Alvarez said though, this is quite similar to standard things (like first explained by Gerstenhaber I think), so you should really explain if you're using a particular definition of deformation. $\endgroup$
    – Amin
    Commented Sep 11, 2012 at 6:57
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    $\begingroup$ I am with Mariano on this one. Commutativity is not really an issue if the deformations are usual. $\endgroup$
    – Bugs Bunny
    Commented Sep 11, 2012 at 13:30

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