Inverse image of anhilator ideals Let $f$ be a ring homomorphism from $R\rightarrow S$ and $J$ be annihilator of some ideal in $S$. Under what conditions on $R$ and kernel of $f$ , $f^{-1} (J) $ is annihilator of some ideal of $R$.
This question seems hard to me although its elementary and it gives me hard time in solving this. I think if kernel of $f$ is annihilator of some ideal in $R$ then it comes. But I still in process of proving that. Is there any sufficient/nesscasry conditions for this.

Let $f:R\to S$ be a ring homomorphism and $I$ an ideal of $S$ with annihilator $J$. Then $f^{-1}(I)$ is an ideal. If $j\in f^{-1}(J)$ and $a\in f^{-1}(I)$, then $f(ja)=f(j)f(a)=0$. So for $j$ to annihilate every element of $f^{-1}(I)$ it is necessary that the kernel of $f$ intersects $jf^{-1}(I)$ trivially. Therefore, for $f^{-1}(J)$ to annihilate $f^{-1}(I)$ we need that the kernel of $f$ intersects $f^{-1}(J)f^{-1}(I)$ trivially (watch out because this is not $f^{-1}(IJ)$ in general).
This condition is also sufficient because if $r\in R$ annihilates $f^{-1}(I)$, then $f(r)$ annihilates $I$, so $f(r)\in J$, meaning that $r\in f^{-1}(J)$.
