TITLE: When is the maximal ideal of a zero-dimensional local non-noetherian commutative ring nilpotent?
QUESTION [7 upvotes]: Let $R$ be a non-Noetherian local commutative ring with identity such that it is of Krull-dimension zero. I am wondering if there are conditions which will force the maximal ideal to be nilpotent.

REPLY [7 votes]: This is a broad question... 
A necessary condition for the maximal ideal $\mathfrak m$ to be nilpotent is $R$ is separated for the $\mathfrak m$-adic topology: 
$$ \cap_{n\ge 1} \mathfrak m^n=0.$$ 
Now suppose moreover that $\mathfrak m/\mathfrak m^2$ is finite dimensional over $R/\mathfrak m$, then $\mathfrak m$ is nilpotent. 
Proof: let $I\subseteq \mathfrak m$ be a finitely generated ideal which generates $\mathfrak m$ modulo $\mathfrak m^2$. Let $N>0$ be such that $I^N=0$ ($N$ exists because $I$ is finitely generated). For any $n>0$, we have $\mathfrak m=I+ \mathfrak m^n$ and 
$$\mathfrak m^{N}\subseteq I^N+\mathfrak m^n=\mathfrak m^n.$$ 
Therefore $\mathfrak m^{N}\subseteq \cap_n \mathfrak m^n=0$. 
Of course in general, $ \mathfrak m/\mathfrak m^2$ is not finite dimensional. I don't know whether the above condition is reasonable for you.