Show ring is Noetherian 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. 
 A: 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.
