# Let $S$ be a subring of a commutative Noetherian ring $R$. Then $R$ is finitely generated as an $S$-module? [closed]

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?

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

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

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 $S=\mathbb{R}$.
• 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. – Severin Schraven May 9 '16 at 15:15
• excluding the case (and similar ones involving polynomials rings) where $R=\mathbb{R}[x]$ and $S=\mathbb{R}$. – adrw_k May 9 '16 at 15:15
• 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? – adrw_k May 9 '16 at 15:17
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.