Let $A$ be a commutative Noetherian ring. We say it is regular if its localization at every prime ideal is a regular local ring.

If this is the case, is it true that $A[X]$ is regular?


1 Answer 1


Yes. Let $\mathfrak q$ be a prime of $A[X]$. Then $\mathfrak q$ lies over a prime $\mathfrak p$ of $A$, and the formation of $(A[X])_{\mathfrak q}$ can be broken up into two steps: first localize at $\mathfrak p$, and then localize this at the prime ideal generated by $\mathfrak q$.

The first step gives a ring which is canonically isomorphic to $A_{\mathfrak p}[X]$, and so the second step involves localizing at a prime of this ring which contains the maximal ideal of $A_{\mathfrak p}.$

Since $A$ was assumed regular, we have reduced to the following situation: let $A$ be a regular local ring with maximal ideal $\mathfrak m$, and let $\mathfrak q$ be a prime ideal of $A[X]$ which contains $\mathfrak m.$ We have to show that $A[X]_{\mathfrak q}$ is regular.

Can you see how to handle this special case? (Hint: think first about the case when $A$ is actually a field, and then try to reduce to this case.)

  • 1
    $\begingroup$ Thank you! As it is no harm to assume $\mathfrak{q}$ is maximal, I then apply the following lemma and the fact that k[x] is a PID to complete the remaining: In that case one has $ht(\mathfrak{q})=ht(\mathfrak{m})+1$. $\endgroup$
    – Ch Zh
    Jan 30, 2012 at 22:13
  • $\begingroup$ @Matt E: I guess if $X$ is NOT an indeterminate over $A$, but algebraic over $A$, then $A[X]$ is not necessarily regular? (assuming $A[X]$ is noetherian). $\endgroup$
    – user237522
    Jun 11, 2015 at 0:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .