Is quotient of a ring by a power of a maximal ideal local?

Solution 1:

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 $M/M^k$.

Solution 2:

$m/m^k$ 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.