16
$\begingroup$

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!

$\endgroup$
17
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.