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I am reading an article in algebraic geometry and am having trouble understanding a particular point that reduces to a problem in commutative algebra. I'm not familiar with the concepts involved so am having a hard time understanding what's going on. I think that the following result should be true and relatively simple to prove:

Let $R$ be a Noetherian local ring of depth $1$ and $M$ a finitely generated $R-$module with finite free resolution $0\rightarrow F_n\rightarrow\dots\rightarrow F_0\rightarrow M$. Let $N$ be the kernel of the map $F_0\rightarrow M$. Then we have a short exact sequence of finitely generated $R-$modules $0 \rightarrow N \rightarrow F_0 \rightarrow M \rightarrow 0$. Then $N$ is free.

The article quotes the result of Auslander and Buchsbaum, that if $R$ is a Noetherian local ring and $M$ is a non-zero finitely generated $R-$module of finite projective dimension, then $\mathrm{pd}(M)+\mathrm{depth}(M) = \mathrm{depth}(R)$.

I see that we can apply this theorem in this instance to get that the projective dimension of $M$ is $0$ or $1$. I don't really see how that helps. I know that every projective module over a local ring is free, but I don't think that helps either.

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I looked up some more facts about projective dimension which ultimately lead to me solving part of the problem. It turns out that the fact that $F_0$ and $N$ have the same rank was a geometric implication, not an algebraic one so that probably can't be proved from the question as it is.

(Simple) Fact: If $pd(M)\leq n$ then given any exact sequence $0\rightarrow K_{n-1}\rightarrow P_{n-1}\rightarrow\dots\rightarrow P_0 \rightarrow M \rightarrow 0$ with the $P_i$ all projective, $K_{n-1}$ is projective.

Thus, since $\mathrm{pd}(M)\leq \mathrm{depth}(R)$ and in this case $\mathrm{depth}(R)$ is $1$, $N$ is projective. Since $R$ is local it is then free.

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