Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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 $\dim R[1/a]=1$.

share|cite|improve this answer

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 $a$....

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.