Let $R_1, R_2$ be rings with 1 and let $R$ be a domain. Let $f : R_1 \times R_2 \to R$ be a ring homomorphism. Prove that one of the following statements holds:
(a) there exists a ring homomorphism $g : R_1 \to R$ such that for all $x \in R_1, y \in R_2$ we have $f(x, y) = g(x)$;
(b) there exists a ring homomorphism $h : R_2 \to R$ such that for all $x \in R_1, y \in R_2$ we have $f(x, y) = h(y)$.
My thinking: from one concrete example $\mathbb Z/3 \times \mathbb Z/4 \to \mathbb Z/2$, I guess that one of $R_1$ or $R_2$ should be the subset of $\operatorname{Ker}(f)$, but have no idea how to prove it.
For each $a\in R_1$ and $b\in R_2$, we have $(0,0)=(a,0)(0,b)$. So: $$0=f(0,0)=f\bigl( (a,0)(0,b)\bigr) = f(a,0)f(0,b).$$ Since $R$ is a domain, one of the factors is $0$. If there is an $a$ for which $f(a,0)\neq 0$, then we must have $f(0,b)=0$ for all $b\in R_2$. Thus we conclude that either $f(a,0)=0$ for all $a\in R_1$, or $f(0,b)=0$ for all $b\in R_2$.
In the latter case, note that for all $x\in R_1$, $y\in R_2$, $$f(x,y) = f\bigl( (x,0)+(0,y)\bigr) = f(x,0)+f(0,y) = f(x,0).$$ So define $g(x)=f(x,0)$.
The former case is symmetrical.
Alternatively, once you know tht at, say, $\{0\}\times R_2\leq \ker(f)$, you know that $f$ factors through $(R_1\times R_2)/(\{0\}\times R_2)\cong R_1$, so $g$ is the induced map.