## Ring Homomorphisms

A mapping $f : R \rightarrow S$ where $R, S$ are rings is a ring homomorphism if

1. $f(1) = 1$

2. For all $x,y\in R$ we have

$\array { f(x+y)&=&f(x) +f(y) \\ f(x\cdot y) &=& f(x)\cdot f(y) }$

It is easily verified that if $f$ is a ring homomorphism, then:

1. $f(0)=0$

2. $f(-x) = -f(x)$ for all $x \in R$

3. The image of $f$, $f(R) = \{f(x)|x\in R\}$ is a subring of $S$

Also it is clear that the composition of ring homomorphisms is also a ring homomorphism.

An isomorphism is a bijective homomorphism. If $f:R\rightarrow S$ is an isomorphism we write $R \cong S$. Note that isomorphism is an equivalence relation.