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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?

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$$f(0,0)=f((1, 0)(0,1))=f(1,0)f(0,1)=0$$ 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.

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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 $y\in S$.

Can you finish?

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