Csdrhrt

Let $R$ and $S$ be rings and $R'$ an integral domain. $f: R times S to R'$ is a ring homomorphism. I have to prove that there exists a ring homomorphism $g: R to R'$ such that $f(x,y) = g(x) forall x in R, y in S$. OR there exists a ring homomorphism $h: S to R'$ such that $f(x,y) = h(y) forall x in R, y in S$.

What I have so far:
$$f(x_1+x_2,y_1+y_2) = f(x_1,y_1)+f(x_2,y_2) forall x_1,x_2 in R, y_1,y_2 in S$$
$$f(x_1,y_1)=f(x_1,0)+f(0,y_1)$$
$$f(0,0) = f(x_1,0)cdot f(0,y_1)$$
$R'$ has no zero divisors, so $f(0,0) notin R'$

Now I am stuck. Can somebody help me?

$$f(0,0)=f((1, 0)(0,1))=f(1,0)f(0,1)=0$$
By the integral domain property, either $f(1,0)=0$ or $f(0,1)=0$. Assume the latter is true. Then
$$f(0,s)=f(0,s)f(0,1)=0$$
for all $s$. Thus
$$f(r, s) =f(r, 0)$$
for all $(r, s) $. Take
$$g(x)=f(x, 0)$$
and you're done.

From $f(0,0)=f(x,0)f(0,y)$ you cannot conclude that $f(0,0)notin R'$, but rather that

either $f(x,0)=0$ or $f(0,y)=0$

since $R'$ is an integral domain.

In particular, this is true for $x=1$ and $y=1$ (the unities in $R$ and $S$ respectively). Suppose $f(0,1)=0$; then $f(0,y)=0$, for every $yin S$.

Can you finish?