Let $R$ be a Gorenstein (not necessarily commutative) ring and let $I$ be an injective finitely generated module over $R$. Is it true that if $\operatorname{Ext}_R^i(I, R)=0$ for $i > 0$, then $I$ is projective?

  • $\begingroup$ What definition of Gorenstein are you using? In the non-comm. case there are dozens... :P $\endgroup$ – Mariano Suárez-Álvarez Feb 21 '12 at 3:43
  • $\begingroup$ In my definition $S$ is left and right noetherian and have finite injective dimension as left or right module over itself. $\endgroup$ – Alex Feb 21 '12 at 11:58

If $R$ is commutative, then the answer is yes and we don't need the vanishing of those Ext's.

Actually we prove the following

Proposition. Let $R$ be a commutative Gorenstein ring and $I$ a finitely generated injective $R$-module. Then $I$ is projective.

We can consider $R$ local. The existence of a nonzero finitely generated injective $R$-module implies that $R$ is artinian; see Bruns and Herzog, 3.1.23 or look here. On the other side, over Gorenstein local rings finitely generated modules of finite injective dimension have also finite projective dimension; Bruns and Herzog, 3.1.25 or look here. Now simply apply the Auslander-Buchsbaum formula and deduce that the projective dimension of $I$ is $0$, i.e. $I$ is free.

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    $\begingroup$ In the non-commutative case, though, there do exist, for example, finite dimensional algebras of finite global dimension (on both sides), and therefore Gorenstein in the sense of the OP, with finitely generated injective modules which are not projective. The ring $T$ of $2\times 2$ upper triangular matrices is a concrete example. $\endgroup$ – Mariano Suárez-Álvarez Nov 5 '12 at 21:00
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    $\begingroup$ R is not necessarily commutative $\endgroup$ – user147308 Aug 2 '14 at 16:12

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