# Solution: Kaplansky’s Commutative Rings 4.1.2

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 $K = 0$.

# Projective Modules over Local Rings are Free

Posted by Jason Polak on 06. May 2012 · 2 comments · Categories: commutative-algebra, modules · Tags: , , , ,

Conventions and definitions: Rings are unital and not necessarily commutative. Modules over rings are left modules. A local ring is a ring in which the set of nonunits form an ideal. A module is called projective if it is a direct summand of a free module.

Today I shall share with you the wonderful result that any projective module over a local ring is free. We shall follow Kaplansky (reference given below), who first proved this result.

Now modules are in fact my favourite mathematical objects. They are like vector spaces, except that they are interesting. Of course, this "interesting" can be irksome if one has to solve a problem and these interesting properties throw a wrench in the works. However, by themselves modules are certainly curious creatures worthy of intense and gruelling analysis!

Of course, when the idea of a module was first conceived, mathematicians attempted to port all kinds of ideas from vector spaces into the world of modules. Some, like the direct sum construction, worked flawlessly. Other concepts such as rank, fortunately or unfortunately depending on your perspective, did not turn out so well (think about it: if everything worked well with modules then there'd be much less interesting math).
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