Let $A$ be a ring, and $I$ and $J$ are distinct maximal ideals of $A$. Then, $I$ never contains $J^n$ for every positive integer $n$

Question

Let $A$ be a ring, and $I$ and $J$ are distinct maximal ideals of $A$. Then, $I$ never contains $J^n$ for every positive integer $n$? I want to figure out contradiction using $J$'s maximality. But I cannot say $J^n$ contains $J$, so I'm having trouble.

Thank you in advance.

Answer 1

Assume that $J^n \subseteq I$. Taking radicals, we get $r(J^n) \subseteq r(I)$. Then we have $J \subseteq r(I)$. We also have $I \subseteq r(I)$, which is just a property of the radical. But, since $I$ and $J$ are maximal, this implies that $I = r(I) = J$, which is a contradiction.