# Composition of Ring Homomorphisms

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

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

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

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

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

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

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

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

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

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

Right.