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Homomorphisms, isomorphisms and the first isomorphism theorem for rings

Let $R$\ and $S$\ be two rings.
A mapping $f\st R\longrightarrow S$\ ...
... for all $x,y\in R$\ we have
$f(x+y)=f(x)+f(y)$\ and $f(xy)=f(x)f(y)$.

The mapping $f\::\: {\mathbb{Z}}\longrightarrow {\mathbb{Z}}_n$defined by $f(x)=x\pmod{n}$\ is a homomorphism.

The mapping $f\::\: {\mathbb{Z}}\longrightarrow{\mathbb{Z}}$defined b...
$f(x+y)=2(x+y)=2x+2y=f(x)+f(y)$, we have
$f(xy)=2xy\neq 4xy=f(x)f(y)$.

For any two rings $R$\ and $S$\ the mapping $f\::\: R\longrightarrow S$defined by $f(x)=0$\ (the zero of $S$) is a homomorphism.

Let $f\::\: R\longrightarrow S$\ be a homomorphism of rings.
The {\em...
...\in R\::\: f(x)=0\},\\
\im(f)=\{ f(x)\::\: x\in R\}.

Next we prove that kernels of homomorphisms are precisely the ideals. (The corresponding results for groups are Theorems 13.8 and 14.4.)

{\rm (I)}
If $f\::\: R\longrightarrow S$\ is a homomorphism of rings...
...ghtarrow R/I$defined by $f(x)=x+I$,
is a homomorphism with kernel $I$.

% latex2html id marker 4814(I)
That $\ker(f)$\ is a subgroup of ...
... x+I=0+I\Leftrightarrow x\in I,
\end{displaymath}and so $\ker(f)=I$.

Prove that if $f\::\: R\longrightarrow S$\ is a ring homomorphism
then $\im(f)$\ is a subring of $S$.

A bijective homomorphism of rings is called
an {\em isomorphism}.
($R\cong S$) if there is an isomorphism $f\::\: R\longrightarrow S$.

If $f\::\: R\longrightarrow S$\ is a homomorphism of groups then
R/\ker(f)\cong \im(f).

Denote $\ker(f)$\ by $I$\ and $\im(f)$\ by $K$. Let $\phi\::\: R/I...
\end{displaymath}and hence $\phi$\ is a ring isomorphism.

Edmund F Robertson

11 September 2006