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Let $R$ be a commutative ring with identity. Suppose $\text{nilrad}(R)$ is a finitely generated ideal of $R$ and $R/\text{nilrad}(R)$ is isomorphic to finite product of fields. I want to show $R$ is Noetherian.

I can see that $R/\text{nilrad}(R)$ is Noetherian directly from the assumption above, but not why $R$ is Noetherian too. Thank you for your help.

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    $\begingroup$ One hint, although it might be an overkill approach: Use the fact that for a commutative ring to be Noetherian, it suffices that all the prime ideals are finitely generated --- see here. $\endgroup$ – I I Jun 6 '18 at 15:38
  • $\begingroup$ Let $I$ be the nilradical. Show that $R/I^n$ is Noetherian by induction and $I^n=0$ for large $n$, using the finite generation of $I$. $\endgroup$ – Mohan Jun 6 '18 at 17:11
  • $\begingroup$ @Mohan: I can see how $I$ is nilpotent, but how would you show $R/I^n$ is Noetherian? I tried the third isomorphism theorem but that doesn't quite give what is needed here I think? $\endgroup$ – dstivd Jun 6 '18 at 17:26
  • $\begingroup$ @Jeroen I can see how that result is utilised here. As you indicated, would be nice to see a proof without using that result as well. $\endgroup$ – dstivd Jun 6 '18 at 18:02
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    $\begingroup$ Here is the case when $I^2=0$. Then, you have $I$ a finitely generated module over $R/I$ and hence Noetherian. Since $R/I$ and $I$ are Noetherian, so is $R$. Induction should be clear. $\endgroup$ – Mohan Jun 6 '18 at 19:08
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This might be similar to what @Mohan proposed in the comments, but works different in detail.

Let $\mathfrak{n} \subseteq R$ be the nilradical. As $\mathfrak{n}$ is finitely generated, every $\mathfrak{n}^k$ for $k \geq 0$ is finitely generated and there exists $n \geq 0$ such that $\mathfrak{n}^n = 0$. Further, for any $k \geq 0$, the $R$-module $\mathfrak{n}^{k}/\mathfrak{n}^{k+1}$ is an $R/\mathfrak{n}R$-module in a natural way. As we have such observed, it is finitely generated as such, and because $R/\mathfrak{n}R$ is a noetherian ring, the module $\mathfrak{n}^{k}/\mathfrak{n}^{k+1}$ is noetherian (as $R/\mathfrak{n}R$-module and hence also as $R$-module).

But there is a canonical injection of $R$-modules

\begin{align*} R \hookrightarrow \prod_{k=0}^n \mathfrak{n}^k/\mathfrak{n}^{k-1} \end{align*}

which exhibits $R$ as an $R$-submodule of the noetherian $R$-module $\prod_{k=0}^n \mathfrak{n}^k/\mathfrak{n}^{k-1}$.

Observe the following: we only needed that $R/\mathfrak{n}R$ was noetherian and $\mathfrak{n} \subseteq R$ was finitely generated. In your case, one would even obtain that $R$ is Artinian as an extension of Artinian modules. I don't know how to show that $\mathfrak{n}$ is Artinian without showing it is noetherian first though.

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