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?
Injective Maximal Cohen-Macaulay modules
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homological-algebra
projective-module
injective-module
cohen-macaulay
gorenstein
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0What definition of Gorenstein are you using? In the non-comm. case there are dozens... :P – 2012-02-21
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0In my definition $S$ is left and right noetherian and have finite injective dimension as left or right module over itself. – 2012-02-21
1 Answers
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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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1In 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. – 2012-11-05
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1R is **not** necessarily commutative – 2014-08-02