Re: Hom_R(R,R) question, rings
From: Ron Sperber (ronsperber_at_optonline.net)
Date: 01/31/05
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Date: Sun, 30 Jan 2005 20:01:19 -0500
Ron Sperber wrote:
> Tony wrote:
>
>> I am trying to prove that Hom_R (R,R) is isomorphic to R as a ring
>> isomorphism.
>>
>> Clearly if f : R ---> R is an R-module homomorphism, then f is
>> determined by where f sends 1, since f(r) = f(r*1) = rf(1).
>>
>> So, I want to make the ring map :
>>
>> h : Hom_R (R,R) ----> R
>>
>> h(f) = h(1).
>>
>> But I can't show part of the ring homomorphism property :
>>
>> Specifically,
>>
>> h(fg) = (f o g)(1) = f(g(1))
>>
>> h(f)h(g) = f(1)g(1).
>>
>> So don't we need to assume that a ring homomorphism takes 1 to 1?
>>
>> When can I not assume that a ring homomorphism takes 1 to 1 in general?
>>
>> Thanks for your help,
>>
>> Tony
>>
> A ring homomorphism MUST take 1 to 1, since if h:R->R is a ring
> homomorphism, then h(r)=h(1r)=h(1)h(r), so h(1)=1. But you are looking
> at Hom_R(R,R) the set of R-module homomorphisms. I think your trouble is
> that multiplication in Hom_R(R,R) is defined not by composition but by
> multiplication. i.e. h(fg)=fg(1)=f(1)*g(1) not, f o g(1).
Ugh, I really shouldn't post that quickly. That can't be the
multiplication because the "identity" would have to be a constant
function which is not a R-module homomorphism. However, having said
that, let's go back to what you have.
you computed that h(fg)=f(g(1))
and h(f)*h(g)=f(1)*g(1).
f and g are R-module homomorphisms. In particular f is an R-module
homomorphism, so f(g(1))=f(g(1)*1)=g(1)*f(1). Clearly this suffices if R
is commutative and either it's false for non-commutative rings or I'm
still screwed up. (I suspect the latter is more likely given my silly
statement the previous post)
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