# Does UFD imply 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!

## 1 Answer

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.