UFD implies noetherian?

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!

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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
en.wikipedia.org/wiki/Unique_factorization_domain (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.