Show that $R$ is a Noetherian Ring

Question

$R$ is a ring and $I$ is a finitely generated nilpotent ideal. If $R/I$ is noetherian show that $R$ is noetherian.

Answer 1

Since $I$ is nilpotent and finitely generated, there exists $n$ such that $I^n = 0$. It thus suffices to show that $R/I^k$ is a Noetherian $R$-module for every $k$, which can be done by induction.

For $k = 1$, note that $R/I$ is Noetherian as an $R/I$-module, hence also as an $R$-module (the module structures are the same). Next, notice that for every $k$, $I^k/I^{k+1}$ is finitely generated and has $R$-annihilator equal to $I$, and is thus a Noetherian $R$-module. Finally, there is a short exact sequence of $R$-modules

$$0 \to I^k/I^{k+1} \to R/I^{k+1} \to R/I^k \to 0$$

The first term is always Noetherian, and the last term is Noetherian by induction, so the middle term is Noetherian as well.

Answer 2

By Cohen's theorem, $R$ is Noetherian iff all its prime ideals are finitely generated.

Since $I$ is nilpotent, the prime ideals of $R/I$ correspond to those of $R$ via the projection $R\to R/I$, and are of the form $P/I$ for various primes $P$. We make use of this last fact to show that all the $P$ are finitely generated $R$ ideals.

Let $p$ be in prime $P$. Since $P/I$ is finitely generated $R/I$ module, there is a set of $\{x_i+I\mid 1\leq i\leq n\}\subseteq P/I$ such that $p+I=\sum (x_i+I)(r_i+I)$ mod $I$, for some $r_i\in R$, i.e. $p-\sum x_ir_i\in I$.

Using a finite generating set $\{y_j\mid1\leq j\leq m\}$ for $I$ as an $R$ module, we then have that $p-\sum x_ir_i=\sum y_js_j\in I$ for some $s_j\in R$. Rearranging, we get that $p=\sum x_ir_i + \sum y_js_j$. This demonstrates that the $x_i$s and the $y_j$s together generate $P$ as an $R$-module.

By Cohen's theorem, we're done.