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I'm studying on these notes. I have a question about page 64, the remark.

A local ring is Gorenstein if and only if the Gorenstein dimension of the residue field is finite.

Of course if the ring is Gorenstein then the Gorenstein dimension of the residue field is finite. Could you explain why the converse is true?

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up vote 1 down vote accepted

The hint from your notes leads easily to a proof. In particular, it says that for a local ring $(A,m,k)$ with $\text{id}_A(A)=\infty$ we should have $\mu_i(m)>0$ for all $i\ge \dim A$. Or if $\text{Gdim}(k)<\infty$, then $\operatorname{Ext}_A^i(k,A)=0$ for all $i>\text{Gdim}(k)$ (to show this take a $G$-resolution of $k$ and use the long exact homology sequence for $\operatorname{Ext}$), that is, $\mu_i(m)=0$ for all $i>\text{Gdim}(k)$. Conclusion: $\text{id}_A(A)<\infty$.

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@Chris: It’s water under the bridge, but I was in the process of rejecting the edit when someone preempted me. I don’t consider it appropriate to do more than correct obvious typos without the explicit consent of the poster; a comment would have been preferable. – Brian M. Scott Nov 3 '12 at 21:02

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