Let $R$ be a ring and $I$ be a finitely generated nilpotent ideal. If $R/I$ is noetherian (resp. Artinian) then $R$ is so.

Let $R$ be a ring and $I$ be a finitely generated nilpotent ideal. If $R/I$ is noetherian (resp. Artinian) then $R$ is so.

In between step is $I^j/I^{j+1}$ is noetherian (artinian) $\forall j$. I am not getting this too.

Can anyone help?

• May be you should give reference of the book/notes where you have seen this statement.
– user312648
Commented Apr 4, 2016 at 7:24
• Sorry $I$ should be finitely generated is the statement still false? Commented Apr 4, 2016 at 7:32
• Commented Apr 13, 2016 at 9:12

Since $I^n=0$ the ideal $I^{n-1}$ is a finitely generated $R/I$-module, so it is noetherian (respectively, artinian).
Now $I^{n-2}/I^{n-1}$ is also a finitely generated $R/I$-module, so it is noetherian (respectively, artinian).
From the exact sequence $$0\to I^{n-1}\to I^{n-2}\to I^{n-2}/I^{n-1}\to 0$$ we get that $I^{n-2}$ is a noetherian (respectively, artinian) $R$-module.

Step by step, we get that $I^j$ is a noetherian (respectively, artinian) $R$-module for $j=n-1,n-2,\dots,1,0$, so $R$ is noetherian (respectively, artinian).

• One should also cite the following basic fact to avoid confusion while switching between $R$- and $R/I$-modules: If $R \to S$ is a ring homomorphism, any $S$-module can be regarded as a $R$-module. If $R \to S$ is surjective, then noetherianness/artinianness will not be lost during this process, since any $R$-submodule is also an $S$-submodule in this case.
– MooS
Commented Apr 6, 2016 at 13:07
• I just have a question: from $I^n = 0$ how do you conclude that $I^{n-1}$ is finitely generated? (this is relevant to a question I just asked a couple of hours ago, so it might help me arrive at a solution - link to question: math.stackexchange.com/questions/1785889/…)
– user290300
Commented May 15, 2016 at 9:58
• If M is a R-module and I, an ideal of R is contained in Annihilator of M then M is a R/I-module. Here, M is taken as $I^{n-1}$ Commented Dec 2, 2018 at 16:59