Adjoining an inverse to a local UFD of Krull dimension 2 gives a PID

Question

Let $R$ be a local UFD of Krull dimension 2. Let $a\in R$ be a nonzero, non-unit. I am trying to show that the ring $R[1/a]$ is a principal ideal domain. Does anyone have any suggestions as to how this can be done?

Answer 1

From this topic it follows that you have to prove that the Krull dimension of $R[1/a]$ is $1$. But the maximal ideal of $R$ is the only ideal of height $2$ and this explodes in $R[1/a]$ (since it contains $a$). The prime ideals of $R$ not containing the element $a$ remain prime in $R[1/a]$ and have height at most $1$. This shows that $\dim R[1/a]=1$.

Answer 2

The set of prime ideals of $R$ consists of

• A single ideal of height 0: $(0)$
• One or more ideals of height 1, all principal
• Exactly one ideal of height 2

The set of prime ideals of $R[1/a]$ consists of the (extensions of) prime ideals of $R$ that do not contain $a$....