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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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$m/m^k$ 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? $\endgroup$ – GraduateStudent Apr 9 '14 at 3:08
  • $\begingroup$ I think this answer is unnecessary complicated. $\endgroup$ – user26857 Apr 9 '14 at 8:07
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    $\begingroup$ @user26857 it is amusing you make this claim when the answer is actually shorter than yours. This answer is at the same level as the one you gave, except Nate opted to phrase the point about containment of the power in a different way. $\endgroup$ – rschwieb Apr 9 '14 at 10:44
  • $\begingroup$ @rschwieb it is amusing you make this claim: "This answer is at the same level as the one you gave". I don't know how I've missed that "The nilradical is the intersection of all prime ideals" is a trivial result! (Maybe because my knowledge in commutative algebra is so poor.) $\endgroup$ – user26857 Apr 9 '14 at 12:18

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