Posted by Jason Polak on 05. January 2015 · Write a comment · Categories: commutative-algebra · Tags:
Problem. Let $R$ be a (commutative!) Noetherian local ring, $M\subset R$ its maximal ideal, and $A$ a finitely generated $R$-module. If ${\rm Ext}^1(A,R/M) = 0$ then $A$ is a free $R$-module.

This problem will be a stepping stone to showing that a Noetherian local ring is regular if and only if the injective dimension of $R/M$ is finite. A closely related variation of this statement is that a Noetherian local ring is regular if and only if it has finite global dimension. This characterisation will then allow us to prove swiftly that a regular local ring is a unique factorisation domain.

Solution. Fix an exact sequence $0\to K\to F\to A\to 0$ with the rank of $F$ minimal. We will show that $K = 0$, which will finish the proof.

To do this, apply the functor ${\rm Hom}_R(-,R/M)$ to this exact sequence. The hypothesis that ${\rm Ext}^1(A,R/M) = 0$ shows that every homomorphism $f:K\to R/M$ can be factored as $K\subseteq F\to R/M$. Since we chose the rank of $F$ to be minimal, $K\subseteq MF$, and hence ${\rm Hom}_R(K,R/M) = 0$.

Now, $K/MK$ is an $R/M$ vector space, so if $K/MK$ is nonzero, there exists a nontrivial $R/M$-module homomorphism $\varphi:K/MK\to R/M$, which is also an $R$-module homomorphism since the action of $R$ on both is through $R\to R/M$. Composing with $K\to K/MK$ gives a nontrivial homomorphism $K\to R/M$, which is a contradiction. Hence $K/MK = 0$, or equivalently, $K = MK$. Since $R$ is Noetherian, $K$ is finitely generated and so by Nakayama’s lemma, $K = 0$.

And, the short version of this proof: applying ${\rm Hom}_R(-,R/M)$ to a minimal short exact sequence $0\to K\to F\to A\to 0$ with $F$ free shows that ${\rm Hom}_R(K,R/M) = 0$ and hence $K = MK$ whence $K = 0$.

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