Posted by Jason Polak on 11. January 2017 · Write a comment · Categories: commutative-algebra

Wolmer Vasconcelos [1] gave the following classification theorem about commutative local rings of global dimension two:

Theorem. Let $A$ be a commutative local ring of global dimension two with maximal ideal $M$. If $M$ is principal or not finitely generated, then $A$ is a valuation domain. Otherwise $M$ is generated by a regular sequence of two elements, and $A$ will be Noetherian if and only if it is completely integrally closed.

In this post we shall prove a small part of this theorem: that if $A$ is a commutative local ring of global dimension two and $M$ is a principal ideal, then $A$ is a valuation domain (i.e. for all $a,b\in A$ either $a | b$ or $b | a). As always, we use the term local ring to mean a commutative ring with a unique maximal ideal.

Before we prove this theorem, let us warm up by discussing the simpler cases of global dimension zero and one.

Theorem. If a commutative local ring $A$ has global dimension zero, then it is a field. In particular it is a valuation domain.
Proof. By Wedderburn’s theorem, $A$ is a product of fields. Since it is local, it must therefore be a field.

Rings of (left) global dimension one are called (left) hereditary. (Left) hereditary is the same as saying that every (left) ideal is projective. If a commutative local ring has global dimension one, then it must be a domain. Indeed, annihilator ideals are projective, and hence free. And, it is well-known that the commutative domains of global dimension one are exactly the Dedekind domains. Therefore, if $A$ is a commutative local ring of global dimension one, then it is not only a valuation domain but a discrete valuation domain.

Theorem. Let $A$ be a commutative local ring with maximal ideal $M$. If $A$ has global dimension two and $M$ is principal, then $A$ is a valuation domain.
Proof. The global dimension may be computed as the supremum over projective dimensions of modules of the form $A/I$ where $I$ is an ideal of $A$. Therefore, ideals have projective dimension at most one as $A$-modules. Consider $a\in A$, and let $I$ be the annihilator of $a$. Thne there is an exact sequence $0\to I\to A\to (a)\to 0$. Since ideals have projective dimension at most one, the annihilator $I$ of $a$ is projective, and hence free. So $I = (0)$.

We claim that $A$ is a GCD-domain. For suppose $a,b\in A$ are not both zero. We will calculate the GCD of $a$ and $b$. Consider the map $A^2\to (a,b)$ given by $(x,y)\mapsto ax – by$. Let $K$ be the kernel of this map so that we have an exact sequence
$$0 \to K\to A^2\to (a,b).$$
Then $K$ is a free module of rank one generated by an element $(\alpha,\beta)\in A^2$. Since $(b,-a)\in K$ we can write $(b,-a) = \delta (\alpha,\beta)$. Now it is easy to see that $\delta$ is in fact the GCD of $a$ and $b$.

Let $a,b\in A$ and let $\delta$ be their GCD. Write $a = d\alpha$ and $b = d\beta$ for some $\alpha,\beta\in A$. If either $\alpha$ or $\beta$ is a unit, then we are done, since this implies that either $a | b$ or $b | a$. If this were not true, then it would be the case that $d | \alpha$ and $d | \beta$, showing that $d\delta | a$ and $d\delta | b$. But $d\delta$ does not divide $\delta$ because $A$ is a domain, and this contradicts the fact that $\delta$ is the GCD of $a$ and $b$. QED

As we mentioned in the beginning, the conclusion also holds if $M$ is not finitely generated, but requires a little more work that would have made this post too long. But, if $M$ is finitely generated, Vansconcelos’ theorem says that $M$ must be generated by a regular sequence of two elements $a,b\in M$. This means that $A$ cannot be a valuation domain! Indeed, $a | b$ would imply that $a$ would be a zero divisor on $R/b$.

For Noetherian local rings, this result becomes much easier because such rings of finite global dimension are automatically regular, and thus their maximal ideals must be generated by a regular sequence of two elements. However, this shows that Vasconcelos’ theorem is also cool in that even if $A$ is not Noetherian, if $M$ is finitely generated and not principal, then it has a two-element generating set.

[1] Vasconcelos, Wolmer V. The local rings of global dimension two. Proc. Amer. Math. Soc. 35 (1972), 381-386.

Leave a Reply

Your email address will not be published. Required fields are marked *