$A$ 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! 


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. 

