# Composition of Ring Homomorphisms

• July 12th 2011, 12:29 PM
dwsmith
Composition of Ring Homomorphisms
Let $f:R\to S$, $g:S\to T$ be functions. Let $g\circ f: R\to T$ be the composition of f and g. Show that if f and g are homomorphisms, then so is $g\circ f$.

Let $u,v\in R$.
$(g\circ f)(uv)=g(f(uv))=g(f(u)f(v))=g(f(u))g(f(v))$
$=(g\circ f)(u)(g\circ f)(v)$

$(g\circ f)(u+v)=g(f(u+v))=g(f(u)+f(v))=g(f(u))+g(f(v))$
$=(g\circ f)(u)+(g\circ f)(v)$

$(g\circ f)(1)=g(f(1))=g(1)=1$

Therefore, $g\circ f$ is a homomorphism. Correct?
• July 12th 2011, 12:42 PM
Drexel28
Re: Composition of Ring Homomorphisms
Quote:

Originally Posted by dwsmith
Let $f:R\to S$, $g:S\to T$ be functions. Let $g\circ f: R\to T$ be the composition of f and g. Show that if f and g are homomorphisms, then so is $g\circ f$.

Let $u,v\in R$.
$(g\circ f)(uv)=g(f(uv))=g(f(u)f(v))=g(f(u))g(f(v))$
$=(g\circ f)(u)(g\circ f)(v)$

$(g\circ f)(u+v)=g(f(u+v))=g(f(u)+f(v))=g(f(u))+g(f(v))$
$=(g\circ f)(u)+(g\circ f)(v)$

$(g\circ f)(1)=g(f(1))=g(1)=1$

Therefore, $g\circ f$ is a homomorphism. Correct?

Right.