Re: Hom_R(R,R) question, rings

From: Tony (Ttiger222_at_hotmail.com)
Date: 01/31/05 

Date: Sun, 30 Jan 2005 19:55:50 -0500

"Ron Sperber" <ronsperber@optonline.net> wrote in message
news:_ffLd.4848$2_2.2309@fe08.lga...
> 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).

??????

Multiplication in Hom_R (R,R) is defined by multiplication? I thought
composition.

I am quoting from Dummit and Foote Abstract Algebra :

"With addition defined as above and multiplication defined as function
composition, Hom_R (M,M) is a ring with 1."

This is in the module chapter with M an R-module. Addition, we know what it
is in Hom_R(M,M).

Also, are you a ring homomorphism must take 1 to 1? You showed that h(r) =
h(1)h(r), but this doesn't show
that h(1) = 1, since h(r) might not have an inverse under multiplication
since rings don't require multiplicative inverses.

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