TITLE: Basic counterexample re: preimages of ideals
QUESTION [18 upvotes]: I'm trying to think of an example of a homomorphism of commutative rings $f:A\rightarrow B$ and ideals $I,J$ of $B$ such that $f^{-1}(I)+f^{-1}(J)$ is not a preimage of any ideal of $B$.  I can't seem to come up with one... anyone know one?
Edit: To clear up some basic facts / head off some mistakes:
As Arturo points out, we can assume $f$ is an inclusion.  Perhaps I should have written the question in terms of inclusions in the first place, but, eh.
No, $f^{-1}(I)+f^{-1}(J)$ is not equal to $f^{-1}(I+J)$ in general.  A counterexample would be the inclusion of $\mathbb{C}$ in $\mathbb{C}[x]$; consider $(x)$ and $(1-x)$.
To show an ideal $K\subseteq A$ is not a preimage of any ideal of $B$, it suffices to show that it's not equal to $f^{-1}(Bf(K))$.

REPLY [3 votes]: Let $k$ be field of caracteristic $\neq 2$, $A=k[x,y], B=k[x,y,x^{-1},y^{-1}]$ and take $f:A\hookrightarrow B$ be inclusion. Example for what you want is $I=(x+y)B, J=(x-y)B$. 
Then $f^{-1}I=(x+y)A, f^{-1}J=(x-y)A$ and $f^{-1}I+f^{-1}J=(x,y)A$ . Since $(x,y)B=B$,  sum 
$f^{-1}I+f^{-1}J$ can not be $f^{-1}$ of ideal in $B$