# Canonical Ring Homomorphism

1. ### 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. ### aPhilosopher

244
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. ### Dragonfall

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

4. ### aPhilosopher

244
Yeah, that's right

5. ### Dragonfall

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

6. ### aPhilosopher

244
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. ### 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. ### aPhilosopher

244
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. ### Dragonfall

Ok, I think I got it. Thanks!