# Thread: Composition of Ring Homomorphisms

1. ## 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?

2. ## Re: Composition of Ring Homomorphisms

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.