2
$\begingroup$

Given a definition of $R=2\Bbb Z$ prove that ring $R[x]$ is not noether.

I assume that the proof should be based on the fact that a ring is noether if and only if any ideal is finitely generated. However, I'm struggling with constructing a non-finitely generated ideal, because only ideals I can think of are "root-based" (i.e. polynomials that have certain roots), that are, of cource, finitely generated.

I would be very grateful for a description of such ideal or a piece of work that can explain the structure of polynomial ring ideals in detail.

$\endgroup$
4
$\begingroup$

It's unconventional to treat rings without multiplicative identity, but here goes.

Let $R=2\Bbb Z$. There is an easy example of a non-finitely generated $R[X]$-ideal, namely $R[X]$ itself. To see an infinite ascending chain of $R[X]$-ideals, let $I_n$ be the ideal generated by $2X$, $2X^2,\ldots,2X^n$.

$\endgroup$
  • $\begingroup$ but I think I can use the same reasoning for $R = \Bbb Z$ so $R[x]$ is noether due to Hilbert's basis theorem. What is the difference between these two cases? $\endgroup$ – youtur Mar 11 '18 at 17:11
  • 2
    $\begingroup$ As I pointed out, $R[x]$ is not finitely generated as a module over itself, and $R[x]$ has a nontrivial increasing sequence of ideals. If you study the proof of the Hilbert basis theorem carefully, you will see that it relies on $R$ being a unital ring. @youtur $\endgroup$ – Lord Shark the Unknown Mar 11 '18 at 17:14

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.