Maximal Ideals under Canonical Ring Homomorphism: A/I to A

[Dragonfall]

Homework Statement

f:A->A/I is a ring homomorphism. Does f^-1 take maximal ideas of A/I to maximal ideals of A?

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.

 

[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.

 

[Dragonfall]

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

 

[aPhilosopher]

Yeah, that's right

 

[Dragonfall]

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

 

[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.

 

[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)?

 

[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.

 

[Dragonfall]

Ok, I think I got it. Thanks!