Regular local ring implies integral domain

Statement
Any regular local ring is an integral domain. In other words, if a ring has a unique maximal ideal and that ideal is generated by a set whose size is the Krull dimension of the ring, then the ring is an integral domain.

Results used

 * Nakayama's lemma
 * Prime avoidance lemma

Proof
Let $$R$$ be a regular local ring and $$M$$ be its unique maximal ideal. We now prove the result by induction on the Krull dimension of $$R$$.

Base case for induction
If the dimension of $$R$$ is zero, then, by the definition of regular local ring, the maximal ideal must be trivial and hence, the ring must actually be a field, and hence an integral domain.

Induction step
Suppose the result is true for dimensions up to $$d - 1$$. We need to prove that the result is true for $$R$$ of Krull dimension $$d$$.

We know the following:


 * By Nakayama's lemma, $$M^2 \ne M$$
 * The set of minimal prime ideals of $$R$$ is finite

Now, suppose $$M$$ were contained in the union of $$M^2$$ and the minimal prime ideals. Then, by the prime avoidance lemma, $$M$$ must be contained either in $$M^2$$ or in one of the minimal prime ideals. $$M^2 \ne M$$ thus forces $$M$$ to be a minimal prime ideal, which would make the Krull dimension zero, contradicting our assumption that the Krull dimension is at least 1.

Thus, there exists an element $$x$$ in $$M$$ which is outside the union of $$M^2$$ and all the minimal prime ideals. Let $$S = R/(x)$$ and $$N = MS$$. Clearly $$N$$ is the unique maximal ideal in $$S$$. By the choice of $$S$$, $$dim(S) < dim(R)$$, and in fact we can conclude that $$dim(S) = d - 1$$.

Now $$N/N^2$$ is a proper homomorphic image of $$M/M^2$$ so it can be generated by $$(d-1)$$ elements. By Nakayama's lemma, $$N$$ can also be generated by $$(d-1)$$ elements.

Thus $$S$$ is a regular local ring, and hence, by the induction step assumption, $$S$$ is an integral domain. Hence $$x$$ is a prime ideal of $$R$$. But since $$x$$ lies outside every minimal prime ideal, there is a minimal prime ideal properly contained inside $$x$$. Call this minimal prime ideal $$Q$$.

If $$y \in Q$$ is any element, then we may write $$y = ax$$ for some $$a \in R$$. But then since $$x \notin Q$$, $$a \in Q$$. Thus, $$Q = xQ$$. Nakayama's lemma now yields $$Q = 0$$, and hence $$R$$ is an integral domain.