Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is problem 3.1.25 (page 97) in Cohen-Macaulay Rings by Bruns and Herzog. The direction I am interested in is the following.

Let $R$ be a Gorenstein local ring and $M$ a finite $R$-module. If the injective dimension of $M$ is finite, then prove that the projective dimension of $M$ is finite.

I am interested in knowing how to do this exercise by using results proved upto this point in the book.

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

Use induction on $d=\dim R$.

If $d=0$, then $\text{inj}\dim M=0$, so $M$ is injective. Take a short exact sequence $$0\longrightarrow N\longrightarrow F\longrightarrow M\longrightarrow 0$$ with $F$ free of finite rank, notice that $F$ is injective and then $\text{inj}\dim N<\infty$. Furthermore, $\text{inj}\dim N=0$ and then the short exact sequence is split. It follows that $M$ is projective.

If $d>0$, then choose $x\in m$ ($m$ is the maximal ideal of $R$) which is nonzerodivisor on $R$ and on $N$, where $N$ (in this case) is the $d$th syzygy of $M$. (One can do this, since from B&H, exercise 2.1.26, one knows that $N$ is $0$ or maximal CM, so $m$ cannot be contained in the set of zerodivisors of $N$, otherwise $\text{depth } N=0$ which implies $d=0$, contradiction.) Notice again that $\text{inj}\dim N<\infty$ and now specialize all by $x$ and apply the induction hypothesis.

share|improve this answer
    
Thanks, that was great. –  messi Jun 5 '12 at 22:18
    
Can you indicate how one could do exercise 2.1.26? –  messi Jun 6 '12 at 12:26
    
I just saw this comment from you, so i will post this exercise as a different problem. The above answer from you is very good, and i would like to vote it up but i dont know how to do that. –  messi Jun 17 '12 at 8:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.