Register to reply

Canonical Ring Homomorphism

by Dragonfall
Tags: canonical, homomorphism, ring
Share this thread:
Dragonfall
#1
Sep24-09, 08:02 PM
Dragonfall's Avatar
P: 994
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.
Phys.Org News Partner Science news on Phys.org
Scientists discover RNA modifications in some unexpected places
Scientists discover tropical tree microbiome in Panama
'Squid skin' metamaterials project yields vivid color display
aPhilosopher
#2
Sep24-09, 08:19 PM
P: 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.
Dragonfall
#3
Sep24-09, 08:33 PM
Dragonfall's Avatar
P: 994
I mean bijection between ideals of A/I and those of A containing I.

aPhilosopher
#4
Sep24-09, 08:36 PM
P: 244
Canonical Ring Homomorphism

Yeah, that's right
Dragonfall
#5
Sep24-09, 08:48 PM
Dragonfall's Avatar
P: 994
So how do I use that to show that f^-1 takes maximal ideas of A/I to those of A?
aPhilosopher
#6
Sep24-09, 09:02 PM
P: 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 [tex]M_{A}[/tex] containing [tex]f^{-1}(N_{A/I})[/tex] where [tex]N_{A/I}[/tex] is maximal in A/I.
Dragonfall
#7
Sep24-09, 09:07 PM
Dragonfall's Avatar
P: 994
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
#8
Sep24-09, 09:09 PM
P: 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.
Dragonfall
#9
Sep24-09, 09:12 PM
Dragonfall's Avatar
P: 994
Ok, I think I got it. Thanks!


Register to reply

Related Discussions
Ring homomorphism Calculus & Beyond Homework 10
Ring homomorphism Calculus & Beyond Homework 0
Ring homomorphism Calculus & Beyond Homework 3
Ring homomorphism Calculus & Beyond Homework 7
Ring Homomorphism Question Linear & Abstract Algebra 3