Re: Hom_R(R,R) question, rings

From: Ron Sperber (ronsperber_at_optonline.net)
Date: 01/31/05 

Date: Sun, 30 Jan 2005 19:42:20 -0500

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).

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