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$A$ a Gorenstein ring, $M\neq 0$ a finite $A$-module with finite injective dimension. According to Bruns, this implies that $M$ has finite projective dimension. How do I see that?

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up vote 3 down vote accepted

Here is a proof which may not be the best but demonstrates some standard techniques:

Since $R$ has finite inj. dim. one can replace $M$ by a high syzygy, so one can assume $M$ has full depth. Thus one can kill a full regular sequence for both $M$ and $R$ and (as finiteness of proj. or inj. dim are not affected) assume $R$ is Artinian. Now, your last question shows that you already know in this case $M$ must be injective.

Map a free module onto $M$ and look at the exact sequence:

$$ 0 \to N \to R^n \to M \to 0$$

As $R$ is Gorenstein, $N$ also has fin. inj. dim., so injective. But then $\text{Ext}_R^1(M,N)=0$, hence the sequence splits, and $M$ must be free.

PS: The name is Bruns. You said in your profile that you are a graduate student interested in commutative algebra. If that is indeed the case, then perhaps taking a serious course and talking to the experts at your institution would be more effective then learning it on MO (-: Good luck!

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+1 for the last paragraph. – Yemon Choi Aug 8 '10 at 5:34
@kwan: If $M$ has full depth, then a regular element in $R$ is also regular on $M$. – Hailong Dao Aug 16 '10 at 15:54
@kwan: I actually meant depth(M) =depth(R), that why we need to pass to a high syzygy. – Hailong Dao Aug 16 '10 at 16:47
@kwan: the set of associated primes of M is a subset of the set of associated primes of R (depth M_p=0 implies depth R_p=0). A R-reg element has to be outside all ass. primes of R, so also outside all ass. primes of M. – Hailong Dao Aug 17 '10 at 5:09
No, but if $M$ is maximal Cohen-Macaulay, then $M_p$ also maximal Cohen-Macaulay. – Hailong Dao Sep 6 '10 at 15:40

I found this proof in Kaplansky's Commutative Rings:

Induction on $\mbox{dim }A$.

$\mbox{dim }A =0 \ $:

Suppose $M\neq 0$. $\mbox{id}(M)=\mbox{depth}(A)=0$ so $M$ is injective, and hence is a direct sum of $\mbox{E}(A/\mathfrak{m})$. Since $A$ is Artin Gorenstein, $\mbox{E}(A/\mathfrak{m})\simeq A$ so $M$ is free.

$\mbox{dim }A \geq 1\ $:

$ 0\leftarrow M\leftarrow A^{n}\leftarrow K\leftarrow 0$

Then $\mbox{id}(K)<\infty$. Since $\mbox{dim }A\geq 1$ there is a non-zero-divisor $a\in A$, which is $K$-regular. Then $\mbox{id} _{A/(a)}(K/aK)\leq\mbox{id} _{A}(K)-1<\infty$ by one of the change of rings formulae. Now $A/(a)$ is Gorenstein and $\mbox{dim }A/(a)<\mbox{dim }A$ so by the induction hybothesis $\mbox{pd} _{A/(a)}(K/aK)<\infty$. By another change of rings formula, $\mbox{pd} _{A}(K)=\mbox{pd} _{A/(a)}(K/aK)<\infty $ so $\mbox{pd} _{A}(M)<\infty$ from the above short exact sequence.

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This is the same proof as Hailong's, with induction replacing the (more straightforward) reduction to the Artinian case. – Graham Leuschke Aug 16 '10 at 17:03

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