0
$\begingroup$

Let $S$ be a subring of a commutative Noetherian ring $R$. Then how can I show that $R$ is finitely generated as an $S$-module?

$\endgroup$

closed as off-topic by Carl Mummert, Shailesh, choco_addicted, Claude Leibovici, Watson May 10 '16 at 7:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Shailesh, choco_addicted, Claude Leibovici, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The paper imposes the "finitely generated" condition as a hypothesis. It need not hold in general; consider $S=\mathbb{Z}$, $R=\mathbb{Z}[x]$. $\endgroup$ – vadim123 May 9 '16 at 15:09
1
$\begingroup$

They assume that $R$ is finitely generated over $S$. In general this is not true. An example where this is not the case would be $R=\mathbb{R}[x]$ and $S=\mathbb{R}$.

$\endgroup$
  • $\begingroup$ I see -- thanks! I've edited the question and was wondering if this is true... $\endgroup$ – adrw_k May 9 '16 at 15:12
  • $\begingroup$ Well, you still need to assume that it is finitely generated. My counterexample is still valid, as $\mathbb{R}[x]$ is noetherian by the Hilbert basis theorem. $\endgroup$ – Severin Schraven May 9 '16 at 15:15
  • $\begingroup$ excluding the case (and similar ones involving polynomials rings) where $R=\mathbb{R}[x]$ and $S=\mathbb{R}$. $\endgroup$ – adrw_k May 9 '16 at 15:15
  • $\begingroup$ What is the precise statement, you'd like to prove? $\endgroup$ – Severin Schraven May 9 '16 at 15:17
  • $\begingroup$ Let's say $S$ is a subring of $\mathbb{R}[x]$ such that $\mathbb{R}\subsetneq S$. Then can I show that $\mathbb{R}[x]$ is finitely generated as an $S$-module? $\endgroup$ – adrw_k May 9 '16 at 15:17
0
$\begingroup$

It is not possible to show. For example, $\Bbb Q$ is a subring of the Noetherian ring $\Bbb R$ but of course $\Bbb R_\Bbb Q$ is not finitely generated.

$\endgroup$

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