A commutative Noetherian local ring $ R$ with maximal ideal $ M$ is called a regular local ring if the Krull dimension of $ R$ is the same as the dimension of $ M/M^2$ as a $ R/M$-vector space.

In studying regular local rings one often uses the following lemma in inductive arguments: if $ R$ is an arbitrary commutative Noetherian local ring with maximal ideal $ M$, and $ M$ consists entirely of zero divisors, then the projective dimension $ \mathrm{pd}_R(A)$ is either zero or infinity for any finitely-generated $ R$-module $ A$. In other words, the only $ R$-modules of finite projective dimension are the projective (hence free) modules.

This is a neat little result that has a fun application for some basic examples in ring theory. One example of a Noetherian local ring is $ \mathbb{Z}/4$: its maximal ideal is $ \{0,2,4\}$. Now consider $ \mathbb{Z}/2$ as a $ \mathbb{Z}/4$-module via reduction modulo two. By considering the short exact sequence $ 0\to \mathbb{Z}/2\to\mathbb{Z}/4\to\mathbb{Z}/2\to 0$ and applying $ \mathrm{Hom}_{\mathbb{Z}/4}(\mathbb{Z}/2,-)$, we see that $ \mathbb{Z}/2$ is not projective. Hence $ \mathbb{Z}/2$ as a $ \mathbb{Z}/4$ module has infinite projective dimension! Less trivially, this lemma is used in at least one proof of the Auslander-Buchsbaum formula: if $ A$ is a finitely-generated $ R$-module with finite projective dimension, then $ G(R) = G(A) + d(A)$, where $ G(X)$ is the grade of $ X$ (the length of a maximal regular sequence on $ X$ contained in $ M$). In the proof, the lemma is used in the 'grade zero' base case.

So how does one prove this lemma? We will give a complete proof, following Kaplansky in his book 'Rings and Fields'. Before we do this, we recall a very useful theorem in commutative algebra that I have seen used tons of times (see Kaplansky, 'Commutative Rings', Theorem 82):

**Theorem.**Let $ R$ be a commutative Noetherian ring, $ A$ a finitely-generated nonzero $ R$-module, and $ S$ an ideal of $ Z(A)$ where $ Z(A)$ are all the zero divisors of $ R$ on $ A$. Then there is a nonzero $ a\in A$ such that $ Sa = 0$.

### The Proof

**Theorem.**Let $ R$ be a Noetherian local ring and $ M$ its maximal ideal. If $ M$ consists entirely of zero divisors then for any finitely generated $ R$-module $ A$, we have $ \mathrm{pd}_R(A)$ either zero or infinity.

*Proof*. If not, then there exists a finitely generated $ A$ with $ \mathrm{pd}_R(A) = 1$. Indeed, if $ X$ is a finitely generated $ R$-module with $ \mathrm{pd}_R(X) = n > 1$ then in a short exact sequence $ 0\to K\to F\to X\to 0$ with $ F$ a finitely generated free module shows that $ \mathrm{pd}_R(K) = n-1$, so by repeatedly resolving, we can obtain $ A$ (these kernels are still finitely generated since $ R$ is Noetherian).

Choose a set of minimal generators $ a_1,\dots,a_r$ of $ A$. Let $ F$ be a free $ R$-module with basis $ u_1,\dots,u_r$ and let $ F\to A$ be the map $ u_i\to a_i$. Let $ K$ be the kernel, giving a short exact sequence $ 0\to K\to F\to A\to 0$. Now, $ K\subseteq MF$. Indeed, if $ \sum c_iu_i$ is in the kernel and not in $ MF$, then one of the coefficients in this sum will be a unit, and applying the homomorphism $ F\to A$ contradicts that $ a_i$ was a minimal generating set to begin with.

Now, we use the theorem we talked about earlier: since $ M$ consists entirely of zero divisors, and $ M$ is finitely generated, there exists an $ r\in R$ such that $ rM = 0$. That is, $ M$ annihilates the single $ r\in R$! However, we have seen $ K\subseteq MF$ and so $ r$ also annihilates $ K$, which is impossible because $ K$ is free, as it is projective over a local ring.

**Challenge**: What happens to the above theorem if we drop that $ A$ be finitely generated? Can you find an example of an $ R$-module $ A$ that has positive finite projective dimension over $ R$?