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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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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1R is **not** necessarily commutative – 2014-08-02