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 $f^{-1}(Bf(K))$.

  • $\begingroup$ Not an answer yet, but some comments: $f$ would not be a surjection; passing through $A/\mathrm{ker}(f)$, I think you can assume that $A$ is a subring of $B$, so that you are looking at $(A\cap I)+(A\cap J)$, and you want it to not equal $A\cap K$ for any ideal $K$. $\endgroup$ – Arturo Magidin Oct 28 '10 at 3:32

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....

  • $\begingroup$ No, your initial assertion is false. $f^{-1}(I+J)$ is not equal to $f^{-1}(I)+f^{-1}(J)$ in general; for a simple counterexample, consider the inclusion of $\mathbb{C}$ in $\mathbb{C}[x]$ and consider the ideals $(x)$ and $(1-x)$. $\endgroup$ – Harry Altman Oct 28 '10 at 7:44
  • $\begingroup$ you are right. forgot to pay attention to the fact that f is not surjective (as Arturo pointed above) $\endgroup$ – Ofir Oct 28 '10 at 8:04
  • $\begingroup$ Ah, thank you! I feel silly for not thinking of this one myself. I'll accept this one rather than the other one as it was earlier... $\endgroup$ – Harry Altman Oct 28 '10 at 20:30

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 $B$

  • $\begingroup$ The general construction on the basis of your example and that of Prometheus seems to be the following: take a domain $A$ and a multiplicative subset $S$ of $A$ that contains at least one non-unit $s$ of $A$. Consider the ring extension $B := S^{-1}A / A$. Chose a non-unit $x\in A$, then the ideals $xB$ and $(x+s)B$ yield the desired example, provided that $x+s$ is a non-unit in $A$. $\endgroup$ – Hagen Knaf Oct 28 '10 at 14:47
  • $\begingroup$ @Hagen: Does that work in general? It's not obvious to me that $xB \cap A$ should be $xA$ in general, e.g. $\endgroup$ – Harry Altman Oct 31 '10 at 5:40

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