Basic counterexample re: preimages of ideals 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 $f^{-1}(Bf(K))$.
 A: Unless I'm missing something, this is the preimage of I+J (I missed something... this is not correct)
If $I,J$ are ideals in $B$ then $f^{-1}(I),f^{-1}(J)\subseteq f^{-1}(I+J)$ so $f^{-1}(I)+f^{-1}(J)\subseteq f^{-1}(I+J)$.
On the other hand, if $x\in f^{-1}(I+J)$ then $f(x)=i+j,\; i\in I,j\in J$. Let $a\in f^{-1}(I)$ such that $f(a)=i$, and $b\in f^{-1}(J),\;f(b)=j$ then $f(a+b)=i+j=f(x)$ so $a+b=x+k$ where $k \in ker(f)$. $ker(f)\subseteq f^{-1}(I)+f^{-1}(J)$ and $a,b\in  f^{-1}(I)+f^{-1}(J)$ so also $x \in f^{-1}(I)+f^{-1}(J)$ and we get that $f^{-1}(I+J)\subseteq f^{-1}(I)+f^{-1}(J)$

ok, another try
let $f:\mathbb{Z}[x]\rightarrow \mathbb{Q}[x]$ be the inclusion function. If $I=(x-3)\mathbb{Q}[x],\; J=x \mathbb{Q}[x]$ then their preimage is $(x-3)\mathbb{Z}[x], \; x\mathbb{Z}[x]$ (I used here Gauss lemma). The sum of the preimages is not all of $\mathbb{Z}[x]$ and it contains 3. If it is in itself a preimage of K then it  $3\in K$ in $\mathbb{Q}[x]$ which is invertible and so this is all of $\mathbb{Q}[x]$ and we get a contradiction.
Hope this one is ok....
A: 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 $B$
