0
$\begingroup$

If R is a Noetherian ring, is $R[X]/((X-1)^2X)$ Noetherian? Artinian? So first I have to understand if it is Noetherian or Artinian, and then prove it or find chains of ideals that don't stabilize. So often I am a bit lost because here it is not even clear to me if these things are Noetherian or Artinian in the first place. I know that the ideals from the chains will be in the form $J/((X-1)^2X)$ where $J$ contains $((X-1)^2X)$, but then...

$\endgroup$
2
$\begingroup$

It is noetherian, because it's a quotient of a noetherian ring.

But it is not Artinian in general. For instance if $R=\Bbb Z$ there's a chain of prime ideals $(2,\bar X)\supset (\bar X)$ so that $\dim R[X]/((X−1)^2X) >0$, whereas Artin rings have Krull dimension $0$.

$\endgroup$
  • $\begingroup$ For Noetherian part, it is clear, but is it possible to see this without the help of Krull dimension, which we didn't see yet? $\endgroup$ – roi_saumon Oct 28 '18 at 21:53
  • 1
    $\begingroup$ @roi_saumon: well, without invoking explicitely the Krull dimension, you should nonetheless know that in an Artin ring all prime ideals are maximal. $\endgroup$ – Andrea Mori Oct 28 '18 at 22:14

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.