Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $K$. – Arturo Magidin Oct 28 '10 at 3:32
up vote 4 down vote accepted

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....

share|cite|improve this answer
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 $(1-x)$. – Harry Altman Oct 28 '10 at 7:44
you are right. forgot to pay attention to the fact that f is not surjective (as Arturo pointed above) – Prometheus Oct 28 '10 at 8:04
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... – 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 $B$

share|cite|improve this answer
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 $A$. – Hagen Knaf Oct 28 '10 at 14:47
@Hagen: Does that work in general? It's not obvious to me that $xB \cap A$ should be $xA$ in general, e.g. – Harry Altman Oct 31 '10 at 5:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.