Let $R = 2\mathbb{Z}$. Then $R[x]$ is not a noetherian ring.

I do not understand why this is so, because Hilbert's basis theorem says: If R Noetherian ring, then R[X] a is Noetherian ring (from wiki).

I suppose that $2\mathbb{Z}$ is principal ideal ring:
Let (2), (4), (6), ... are the ideals, therefore all elements are generated by one ideal, so $2\mathbb{Z}$ is principal ideal ring. And we conclude that $2\mathbb{Z}$ is a noetherian ring. Why can't use Hilbert's basis theorem for $R[x]$?

  • $\begingroup$ Does the definition of "ring" in the wikipedia article require the existence of $1$? And are you certain that $R[x]$ isn't Noetherian? $\endgroup$ – Arthur Mar 11 '18 at 13:40
  • $\begingroup$ I think is because $2\mathbb{Z}$ has not unity. $\endgroup$ – Gödel Mar 11 '18 at 13:42
  • $\begingroup$ @Arthur I believe Hilbert's theorem holds for non unity rings $\endgroup$ – krirkrirk Mar 11 '18 at 13:43
  • $\begingroup$ @Arthur I think, that by definition of Noetherian ring, Hilbert's theorem should be require the existence of 1 (or I'm wrong?). $\endgroup$ – Pennywise Mar 11 '18 at 13:54
  • $\begingroup$ I don’t know if the Hilbert basis theorem works in rings without identity or not, but it still seems like $R[x]$ is noetherian. A bigger question is why $R[x]$ looks like when $R$ doesn’t have identity. Where do you see the claim? $\endgroup$ – rschwieb Mar 11 '18 at 14:18

The Hilbert basis theorem only applies to unital rings (this is often not stated explicitly since unital is often included in the definition of "ring"). Since $2\mathbb{Z}$ is not unital, the Hilbert basis theorem does not apply in this case.

An example of a non-finitely generated ideal in $2\mathbb{Z}[x]$ is the entire ring $2\mathbb{Z}[x]$ itself. (Note that for this to be true, $2\mathbb{Z}[x]$ must be defined as the set of polynomials all of whose coefficients are in $2\mathbb{Z}$, rather than the ring obtained from $2\mathbb{Z}$ by freely adjoining a central element $x$. This distinction does not make a difference for unital rings.)


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.