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, $M^{2}\neq M$
• The set of minimal prime ideals of $R$ is finite (Template:Justify)

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. $M^{2}\neq M$ 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. Fill this in later