# $R$ is normal. Are $R[x]$ and $R[[x]]$ normal?

Studying about normalizations I've bumped in the following theorem:

Theorem. Let $R$ be a normal (integrally closed) domain, then $R[x]$ is a normal domain.

How to prove (elegantly, if possible) it?

Is true that if $R$ a normal domain, then $R[[x]]$ is a normal domain? How to prove it?

Thank you for help.

EDIT Now it's clear that, if $R$ is normal $R[[x]]$ is not necessary normal. In the answer, Martin Brandenburg cited the condition of be "completely integrally closed" and a counter example with DVR of dimension equal to 2.

What about if $R$ is a PID or in general a ring of dimension 1?

• Which definition of "normal" are you using, integrally closed? – Bill Dubuque Jun 26 '15 at 3:10
• Regarding the edit: If $R$ is a PID, then $R$ is noetherian and integrally closed, and therefore $R[[x]]$ is integrally closed. The question is more interesting for other integral domains of dimension $1$. – Martin Brandenburg Jun 26 '15 at 10:31
• If $R$ is a PID then $R[[X]]$ is a UFD. – user26857 Jun 26 '15 at 11:16
• @user26857 can you link some reference? – Sabino Di Trani Jun 26 '15 at 11:33
• Kaplansky, CR, Theorem 72. – user26857 Jun 26 '15 at 11:35

Yes for $R[x]$. No for $R[[x]]$.

If $R$ is a valuation ring of dimension $\geq 2$, then $R[[x]]$ is not integrally closed (see math.SE/202203; this is exercise 10.4 in Matsumura's Commutative ring theory). However, if $R$ is completely integrally closed (see Wikipedia), then $R[[x]]$ is completely integrally closed, too. It follows, in particular, that if $R$ is a noetherian integrally closed domain, then $R[[x]]$ is a (noetherian) integrally closed domain.

A reference for the statement about $R[x]$ is:

N. Bourbaki, Commutative Algebra: Chapters 1-7, Ch. V, §1.3, Corollary 2 to Proposition 13.

More generally, if $R \to S$ is a homomorphism of commutative rings, and $R \to R' \to S$ is its integral closure, then $R[x] \to R'[x] \to S[x]$ is the integral closure of $R[x] \to S[x]$ (Proposition 13).

• PS: In a paper by Craig Huneke (a famous commutative algebraist), it is claimed that $R[[x]]$ is integrally closed when $R$ is. Now I'm confused! – Martin Brandenburg Jun 26 '15 at 8:59
• Would you mind giving the information for this paper? Sounds interesting. – CPM Jun 26 '15 at 10:00
• @Martin Can you provide a link to that paper? (Or, if you found something that explains it, like Huneke only considering Noetherian rings in that paper, maybe delete the comment.) – Daniel Fischer Jul 2 '15 at 9:08