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I'm studying on these notes, my question in about a proof on page 63. Basically this is my question:

Suppose $R$ local noetherian of positive depth, $M$ a module of finite gorenstein dimension $n\geq1$ and of depth 0. Take an exact sequence $0\rightarrow N\rightarrow P\rightarrow M\rightarrow0$ where $P$ is totally reflexive (we can take it projective) and the Gorenstein dimension of $N$ is $n-1$. The notes claim that $\mathrm{depth}\;N=1$, why this is true? (it would be ok also if you could tell me why $\mathrm{depth}\;N\geq1$).

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Since $P$ is projective, it is a summand, and so a subobject, of a free module. Thus $N$ is a subobject of a free module. It is also non-zero, since $M$ is not projective (being of depth $0$, while $R$ is of positive depth).

Since $R$ is of positive depth, there is a non-unit $r \in R$ which is a non-zero divisor. This element $r$ is then a non-zero divisor on any submodule of a free module, in particular on $N$. Thus depth $N \geq 1$.

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is it true that $\mathrm{depth}\;N=1?$ –  Chris Nov 15 '12 at 3:19
    
I think it is true because $\mathrm{depth}\;P>0$ because the depth of the ring is positive. By depth lemma I have $\mathrm{depth}\;M\geq\mathrm{min}\{\mathrm{depth}\;P,\mathrm{depth}\;N-1\}$ but the depth of $M$ is zero and so necessarily the depth of $N$ is less or equal to 1 and so it is 1, am I right? –  Chris Nov 15 '12 at 3:31
    
@Chris Yes, you are right. (I think you should reformulate your question in simpler terms, by eliminating those useless references to G-dim(M), and then can post your comment as an answer cause Matt E's answer is not complete.) –  user26857 Nov 15 '12 at 10:58
    
I'll accept his answer since it was helpfull –  Chris Nov 15 '12 at 22:16
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