$A$ 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?
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$\begingroup$ @kwan: If $M$ has full depth, then a regular element in $R$ is also regular on $M$. $\endgroup$ – Hailong Dao Aug 16 '10 at 15:54

1$\begingroup$ @kwan: I actually meant depth(M) =depth(R), that why we need to pass to a high syzygy. $\endgroup$ – Hailong Dao Aug 16 '10 at 16:47

1$\begingroup$ @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 Rreg element has to be outside all ass. primes of R, so also outside all ass. primes of M. $\endgroup$ – Hailong Dao Aug 17 '10 at 5:09

1$\begingroup$ No, but if $M$ is maximal CohenMacaulay, then $M_p$ also maximal CohenMacaulay. $\endgroup$ – 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 nonzerodivisor $a\in A$, 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.

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