If in a commutative ring $R$ a maximal ideal is nilpotent, then $R$ is local

Question

Let $R$ a commutative ring, and $M$ a maximal ideal of $R$.

If there exists $n\in \mathbb{N}$ such that $M^n=0$, then $R$ is local.

In general, I have proved that $R/M^n$ is local with a unique maximal ideal $M/M^n$. But I don't see how to prove that $R$ is local.

But if $M^n=0$, then $R\simeq R/M^n$ so it seems you're done, or not? — Torsten Schoeneberg, May 11 '20 at 00:09

score 0 · Accepted Answer · answered May 11 '20 at 00:01

$R$ is local is equivalent to say that $R$ has a unique maximal ideal. Let $N$ be a maximal ideal, $x\in M$, $x^n=0\in N$ implies $x\in N$ or $x^{n-1}\in N$ since $N$ is prime, we deduce recursively that $x\in N, M\subset N$ since $M$ is maximal, $M=N$.

If in a commutative ring $R$ a maximal ideal is nilpotent, then $R$ is local

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