Canonical Ring Homomorphism


by Dragonfall
Tags: canonical, homomorphism, ring
Dragonfall
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#1
Sep24-09, 08:02 PM
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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.
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aPhilosopher
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#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
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#3
Sep24-09, 08:33 PM
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I mean bijection between ideals of A/I and those of A containing I.

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

Canonical Ring Homomorphism


Yeah, that's right
Dragonfall
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#5
Sep24-09, 08:48 PM
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So how do I use that to show that f^-1 takes maximal ideas of A/I to those of A?
aPhilosopher
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#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
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#7
Sep24-09, 09:07 PM
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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
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#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
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#9
Sep24-09, 09:12 PM
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Ok, I think I got it. Thanks!


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