Let $R$ be a commutative ring with 1. Let $f:R\times R\to R$ be a map.
The claim: $f$ is an homomorphism of $R$-modules iff $\exists\,\alpha,\beta\in R: f((x,y))=\alpha x+\beta y\,\forall\,x,y\in R$.
I'm working on the if direction.
What I got so far: One of the things we want to show is $f((a,b)+(c,d))=f(a,b)+f(c,d)$.
We know that $f((a,b)+(c,d))=f((a+c,b+d))$. Then by the assumption, $\exists\,\alpha_{0},\beta_{0}\in R:f((a+c,b+d))=\alpha_{0}(a+c)+\beta_{0}(b+d)=\alpha_{0}a+\alpha_{0}c+\beta_{0}b+\beta_{0}d$.
Then we want this to be equal to $f(a,b)+f(c,d)$. By assumption, $\exists\,\alpha_{1},\beta_{1},\alpha_{2},\beta_{2}\in R:f(a,b)+f(c,d)=\alpha_{1}a+\beta_{1}b+\alpha_{2}c+\beta_{2}d$
If we want $f((a,b)+(c,d))=f(a,b)+f(c,d)$, then we'd need to have $\alpha_{0}=\alpha_{1}=\alpha_{2},\beta_{0}=\beta_{1}=\beta_{2}$, but I don't understand why that would have to be true.
Sorry if this is a dumb question . . . the book I'm reading says this direction is clear, but I don't see it.