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.
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.
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 [tex]M_{A}[/tex] containing [tex]f^{-1}(N_{A/I})[/tex] where [tex]N_{A/I}[/tex] is maximal in A/I.
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)?
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.