I'm trying to convince myself one part of the proof for a theorem in chapter 10 in DUmmit and Foote.
Let $R$ be a subring of $S$, let $N$ be a left $R$-module and let $\iota:N \to S\otimes_RN$ be the $R$-module homomorphism defined by $\iota(n)=1\otimes n$. Suppose that $L$ is any Left $S$-module and that $\phi:N\to L$ is an $R$-module homomorphism. Then there is a unique $S$-module homomorphism $\Phi: S\otimes_R N \to L$ such that $\phi$ factor through $\Phi$.
In the proof, $\Phi$ is given by universal property on free $\mathbb{Z}$-module $S\otimes_R N$ as $\Phi(s\otimes n)=s\phi(n)$. But the book just mentioned that $\Phi$ is additive. I guess that is because $\phi$ is additive. But how would I go to prove it? I let $s_1 \otimes n_1,s_2\otimes n_2\in S \otimes_R N$. Then show that $$\Phi(s_1\otimes n_1+ s_2\otimes n_2)=\Phi(s_1\otimes n_1)+\Phi(s_2\otimes n_2)$$ But I can't really come up anything logical that can verify $\Phi$ is additive.
Am I not understanding tensor product of modules?
The proof in the book goes like this:
Consider the map $\phi' \colon S \times N \to L$ given by $\phi'(s,n) = s\phi(n)$. Then, if $F$ is the free $\Bbb Z$-module on $S \times N$, there is$^*$ a $\Bbb Z$-module homomorphism (= additive map) $\phi'' \colon F \to L$ such that $$ \phi''(s,n) = \phi'(s,n) $$ for all $(s,n) \in S \times N$. Next, if $H$ is the subgroup generated by all elements of the form (10.3), it is easy to check that $H \subseteq \ker \phi''$. Therefore, there is$^{**}$ a $\Bbb Z$-module homomorphism (= additive map) $\phi''' \colon F/H \to L$ such that $$ \phi'''(x+H) = \phi''(x) $$ for all $x \in F$. Since $S \otimes_R N = F/H$ and $s \otimes n = (s,n)+H$, $\phi'''$ is the desired map $\Phi$. Note that $\Phi$ is additive, trivially.
$^*$This is the universal property of the free module on a set: If $X$ is any set and $F$ is the free module on $X$, then for any module $A$ and any map $f \colon X \to A$, there exists a (unique) module homomorphism $f' \colon F \to A$ such that $f'(x)=f(x)$ for all $x \in X$.
$^{**}$This is the universal property of the quotient module: If $g \colon M \to N$ is a module homomorphism and $S$ is a submodule of $M$ such that $S \subseteq \ker g$, then there exists a (unique) module homomorphism $g' \colon M/S \to N$ such that $g'(x+H)=g(x)$ for all $x \in M$.