$\surd (I^e)=(\surd I)^e$ I'm trying to solve this question: =(\surd I)^e$ in the part (ii). I'm trying a lot proving the inclusions $\subset$ and $\supset$ without any success, I really need help. Thanks in advance

Let $r \in \sqrt{I}$. That means $r^n \in I$ for some $n > 0$. But then

$$f(r)^n = f(r^n) \in f(I) = I^e,$$

so $f(r) \in \sqrt{I^e}$, thus $f(\sqrt{I}) = (\sqrt{I})^e \subset \sqrt{I^e}$.

Conversely, let $s \in \sqrt{I^e}$ and $r \in f^{-1}(\\{s\\})$ (since $f$ is surjective, such an $r$ exists). Let $n > 0$ such that $s^n \in I^e$. Then $f(r^n) = f(r)^n = s^n \in I^e$, so $r^n \in f^{-1}(I^e) = I$, hence $r \in \sqrt{I}$ and $s \in f(\sqrt{I}) = (\sqrt{I})^e$, hence $\sqrt{I^e} \subset (\sqrt{I})^e$.