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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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    $\begingroup$ Can you tell where the exercise is taken from ? $\endgroup$
    – tj_
    Mar 12, 2013 at 13:50
  • $\begingroup$ 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 ${\rm pd} N = {\rm gldim} R < \infty$ $\endgroup$ Mar 13, 2013 at 11:49

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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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    $\begingroup$ Hi: this is a long time later, but I'm fascinated by this piece that $\prod_{i=1}^\infty \mathbb C$ having global dimension $2$. I'm surprised it depends on something such as the CH. Is it not known more generally for an infinite number of copies of a single field? Do you know where I can read more about that claim? $\endgroup$
    – rschwieb
    May 4, 2018 at 13:41
  • $\begingroup$ @rschwieb This can be found in Barbara Osofsky's book Homological dimensions of modules. This statement is part of Theorem 2.51: If $R = \prod_{i\in I}R_i$ is a product of division rings and $|2^I| = \aleph_k$, then the global dimension of $R$ equals $k+1$. $\endgroup$
    – Ben
    May 8, 2018 at 7:25

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