# Canonical Ring Homomorphism

1. Sep 24, 2009

### Dragonfall

1. The problem statement, all variables and given/known data
f:A->A/I is a ring homomorphism. Does f^-1 take maximal ideas of A/I to maximal ideals of A?

3. The attempt at a solution

I think it does, since there is a bijection between A and A/I preserving subsets-ordering. But f might not be that bijection.

2. Sep 24, 2009

### aPhilosopher

There's no bijection between R and R/I unless I is {0} ;)

Are you supposed to prove this or do you just want to know?

EDIT: Unless of course, R is infinite *blushes*

And even then, it won't always be the case.

Last edited: Sep 24, 2009
3. Sep 24, 2009

### Dragonfall

I mean bijection between ideals of A/I and those of A containing I.

4. Sep 24, 2009

### aPhilosopher

Yeah, that's right

5. Sep 24, 2009

### Dragonfall

So how do I use that to show that f^-1 takes maximal ideas of A/I to those of A?

6. Sep 24, 2009

### aPhilosopher

Well if you're allowed to use the fact that f induces a bijection between the ideals of A containing I and the ideals of A/I that preserves inclusion, then that should be easy. Think about a maximal ideal $$M_{A}$$ containing $$f^{-1}(N_{A/I})$$ where $$N_{A/I}$$ is maximal in A/I.

7. Sep 24, 2009

### Dragonfall

Do you mean that f *is* a bijection between the ideals of A containing I and the ideas of A/I that preserves inclusion? Not sure what you meant by "induces". Do you mean defining g which acts on the power set of A, and g(x) is the image f(x)?

8. Sep 24, 2009

### aPhilosopher

No because f isn't a map on the ideals! It's a map on elements of A onto cosets of I. I guess we were being a little sloppy earlier. You can think of it like that but formally, they are two distinct maps. It's common to abuse notation and write them the same however.

9. Sep 24, 2009

### Dragonfall

Ok, I think I got it. Thanks!