# A Case of No Positive Finite Projective Dimension

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 $R$?