21
$\begingroup$

A common example showing that a subring of a Noetherian ring is not necessarily Noetherian is to take a polynomial ring over a field $k$ in infinitely many indeterminates, $k[x_1,x_2,\dots]$. The quotient field is then obviously Noetherian, but the subring $k[x_1,x_2,\dots]$ is not since there is an infinite ascending chain of ideals which never stabilizes.

Is there an instance of a finitely generated Noetherian ring over some ground ring $R$, that has an intermediate ring which is not finitely generated over $R$, and hence not Noetherian either?

$\endgroup$
5
  • $\begingroup$ This is answered by Mariano's answer here. That question did not have a Noetherian condition, but it's clearly satisfied in the example. $\endgroup$ Feb 5 '12 at 22:23
  • $\begingroup$ Oh yeah, Hilbert's basis theorem shows the Noetherian condition. Should I delete this somehow? $\endgroup$
    – Emilia
    Feb 5 '12 at 22:32
  • $\begingroup$ I actually don't know what the interface for asking questions is like, but if you want to then that seems fine: no one has written an answer, or anything. $\endgroup$ Feb 5 '12 at 22:46
  • 1
    $\begingroup$ In last sentence, I think that "hence not Noetherian either " doesn't make sense ,,because Noetherian property should be checked over itself ring. (not over $R$) $\endgroup$
    – hew
    Mar 11 '20 at 13:40
  • $\begingroup$ By Eakin - Nagata theorem, converse is true. $\endgroup$
    – hew
    Mar 11 '20 at 13:58
28
$\begingroup$

To get this question off the unanswered list, I copy Mariano's example from here.

The polynomial ring $k[x,y]$ is Noetherian (by Hilbert's basis theorem), but the subring generated by $\{xy^i:i\geq0\}$ is not finitely generated over $k$.

$\endgroup$
1
  • $\begingroup$ Nice, simple example. $\endgroup$
    – anomaly
    Dec 15 '15 at 20:09

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