If a proper ideal contains some power of a maximal ideal then the maximal ideal is the only prime ideal that contains the ideal.

Question

Let $R$ be a commutative ring with $1$ and $\mathfrak{m}\subset R$ be a maximal ideal. Show that if $I\subset R$ is a proper ideal containing $\mathfrak{m}^n$ for some $n\geq 1$, then $\mathfrak{m}$ is the unique prime ideal containing $I$.

This question was asked in a recent exam that i appeared, but could not solve it. Any hints would be highly appreciated. Thank you.

What role does $n$ play?

Answer 1

Let $p$ be a prime ideal containing $I$. Since $m^n \subseteq I$, $m^n \subseteq p$. This implies $m \subseteq p$. Since $m$ is maximal $m = p$.