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Say I have a commutative ring $R$ with a maximal ideal $m$. Then $m/m^k$ is a maximal ideal in $R/m^k$ for any $k$. Is it the only maximal ideal, i.e. is $R/m^k$ a local ring?

This is a well known result for $k = 1$, as $R/m$ is a field. It seems to be true in other cases, e.g. $p\mathbb{Z}\subset \mathbb{Z}$ for prime $p$ and for $(x,y) \subset \mathbb{F}[x,y]$, the maximal ideal generated by $x$ and $y$ in a polynomial ring over the field $\mathbb{F}$.

Equivalently, if an element $x\notin m$, is $x + m^k$ an invertible element in $R/m^k$?

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The ideals of a quotient ring $R/I$ are of the form $J/I$ with $I\subseteq J\subseteq R$, $J$ ideal of $R$. In particular, the prime ideals are of the form $P/I$ with $I\subseteq P$, $P$ prime ideal of $R$. Now let $I=M^k$ with $M$ maximal. A prime ideal $P$ of $R$ containing $M^k$ contains $M$ (by the definition of prime ideals), so equals $M$.
Conclusion: $R/M^k$ is not only local, it has only one prime ideal, namely $M/M^k$.

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  • $\begingroup$ This shows even more: $R/M^k$ has only one radical ideal, $M/M^k$. $\endgroup$
    – Jose Brox
    Jul 1 at 11:00
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$m/m^k$ consists of nilpotent elements in $R/m^k$. The nilradical is the intersection of all prime ideals, so $m/m^k$ is contained in any prime ideal and hence any maximal ideal. But it itself is a maximal ideal by construction, so it must be the unique one.

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  • $\begingroup$ Should we worry about the case where $m$ is not finitely generated, and so $m/m^k$ may not be nilpotent? $\endgroup$ Apr 9 '14 at 3:08

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