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Suppose R has finite global dimension n, N is a f.g. module, F is a free module and Ext^n(N,F) is not equal to zero, then Ext^n(N, R) is not trival either.

Note, HERE R is not Noetherian necessarily.

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Can you tell where the exercise is taken from ? –  tj_ Mar 12 '13 at 13:50
    
I'm mildly sceptical. If $R$ is not noetherian, how can it be relevant that $N$ is finitely generated? Also, the other conditions on $N$ are just a roundabout way of saying ${\rm pd} N = {\rm gldim} R < \infty$ –  Dag Oskar Madsen Mar 13 '13 at 11:49

1 Answer 1

This cannot be right if $R$ is injective as an $R$-module and the global dimension is non-zero but finite.

Let for instance $R$ be the ring $R = \prod_{i=1}^\infty \mathbb{C}$ and assume the Continuum Hypothesis so that ${\rm gldim}(R)=2$. Then by a theorem of Auslander there is a cyclic module $N$ such that ${\rm pd}(N)=2$. Take a free resolution of $N$ and obtain a free module $F$ such that ${\rm Ext}^2(N,F) \neq 0$. But ${\rm Ext}^2(N,R)=0$ since $R$ is injective.

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