Extension of the radical of an ideal

Question

Let $I$ be an ideal of a ring $R$ and let $f: R\rightarrow S$ be a homomorphism. Let $$I^e = \{\sum_{j=1}^n s_j f(r_j) : s_j \in S, r_j \in I\}$$ be the extension of $I$. Is it true that the radical $\sqrt{I^e}$ equals $(\sqrt I)^e$?

I have been able to show that $(\sqrt I)^e \subset \sqrt{I^e}$ by raising the expression for $s \in (\sqrt I)^e $ to a sufficiently large power that each term in its expansion has a factor belonging to $I$. But, I am unsure about the reverse inclusion. Does anyone have a proof or a counterexample?

Answer 1

The inclusion $(\sqrt I)^e \subset \sqrt{I^e}$ can be strict. For example, let $R=\mathbb{Z}$, let $S=\mathbb{Z}[x]/(x^2)$, and let $f:R\to S$ be the standard inclusion. Let $I=(0)$. Then $$(\sqrt{I})^e=(0)^e=(0)\subset S,$$ but $$\sqrt{I^e}=\sqrt{(0)}=(\overline{x})\subset S.$$