Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

It is easy to show that a PID must be noetherian. My question is:

Does UFD imply noetherian? If not, is there an easy counterexample?

I apologize if this turns out to be a simple question. Thanks in advance!

share|cite|improve this question
What about $\prod_{n=1}^\infty\mathbb{Z}$? – Clayton Dec 9 '12 at 3:38
Over a field $\,k\,$ , the polynomial ring $\,k[X_1, X_2,\ldots ]\,$ is a UFD but not Noetherian – DonAntonio Dec 9 '12 at 3:39 (see Properties and also Equivalent conditions for a ring to be a UFD* – Amzoti Dec 9 '12 at 3:39
@Clayton: that isn't a UFD (because it isn't a domain). – Qiaochu Yuan Dec 9 '12 at 4:01
Thanks @JulianKuelshammer, done. I used to go into unanswered questions from time to time and tried to answer some, but haven't done that for a while. I shall in the near future, hopefully. I don't know though how to post the answer in the chat room... – DonAntonio Jun 20 '13 at 8:54

Since any $\,f\in k[X_1,X_2,\ldots]\;$ is a polynomial in a finite number of indeterminates $\,X_{i_1},\ldots, X_{i_n}\,$ , then in fact $\,f\in k[X_{i_1},\ldots,X_{i_n}]\;$ and this last is a UFD whenever $\,k\,$ is (in fact, this is an iff claim).

Clearly though, $\,k[X_1,\ldots]\;$ is not Noetherian since the proper ideal $\,\langle X_1,\ldots\rangle\;$ isn't finitely generated.

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.