# Help with identity for ideals

1. Aug 6, 2007

### learningphysics

I'm trying to prove this problem out of Allan Clark's Elements of abstract algebra.

Given an epimorphism $$\phi$$ from R -> R'
Prove that:

$$\phi^{-1}$$(a'b') = ($$\phi^{-1}$$a')($$\phi^{-1}$$b')

where a' and b' are ideals of R'

I had no trouble showing that ($$\phi^{-1}$$a')($$\phi^{-1}$$b') is a subset of $$\phi^{-1}$$(a'b'). But I'm having trouble with the forward direction. I'd appreciate any help/hints. Thanks.

2. Aug 6, 2007

### Kummer

What does the juxaposition of the inverse image ideals mean?

Again, I do not understand the juxtaposed notation. But I can tell you this immediately. In your first part of proof did you use the fact that $$\phi$$ was an epimorphism? I think not. Now for the reverse direction you need to use that fact.

3. Aug 6, 2007

### learningphysics

The juxtaposed notation is the direct product of sets.

Last edited: Aug 6, 2007
4. Aug 6, 2007

### mathwonk

it is nit the direct product is it? it shoukld be the product of the dieals, which means the ideal generated by the set of all pairwise products of elements. and have you used the hypothesis of surjectivity?

5. Aug 6, 2007

### learningphysics

Sorry, yes you're right it's not the direct product... it's the pairwise product as you said.

Do you mean the fact that it is an epimorphism? No, I didn't use it. Apologies to Kummer for not answering this in his post... I thought at first I used it for the reverse direction, but I actually didn't use it... I think you and Kummer are hinting at the same thing, but I'm not able to see it.

I'm not seeing how to use the fact that it is an epimorphism... If we let a = $$\phi^{-1}(a')$$ and let b=$$\phi^{-1}(b')$$ I'm not able to see why we couldn't have an element x outside of ab such that $$\phi(x)$$ belongs to a'b'... I know that it is related to the fact that $$\phi$$ is an epimorphism...

Last edited: Aug 6, 2007