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Canonical Ring Homomorphism 
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#1
Sep2409, 08:02 PM

P: 995

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 subsetsordering. But f might not be that bijection. 


#2
Sep2409, 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. 


#4
Sep2409, 08:36 PM

P: 244

Canonical Ring Homomorphism
Yeah, that's right



#5
Sep2409, 08:48 PM

P: 995

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



#6
Sep2409, 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.



#7
Sep2409, 09:07 PM

P: 995

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
Sep2409, 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.



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