In section 10.4 of D+F, in Theorem 8, there is the following use of the universal property of the free module.
We have that $R$ is a subring of $S$, and $N$ is a left R-module, and L is a left S-module.
(slightly modified wording)
"Let $\varphi:N \to L$ be a map of R-Modules. By the universal property of the free module, there is a $\mathbb{Z}$-module homomorphism from the free module on the set $S \times N$ to $L$ such that $(s,n) \mapsto s\varphi(n)$..."
First off, I think without mentioning it, they are thinking of extending the map $\varphi:N \to L$ to a map (abusing notation) $\varphi:S\times N \to L$ where $(s,n) \mapsto s\varphi(n)$. This I am fine with. But when they invoke the universal property of the free module, they say that they get a $\mathbb{Z}$-module homomorphism. But doesn't this assume that $L$ is also a $\mathbb{Z}$-module (which we didn't assume)?
Could somebody please point out what I am missing here?
A $\mathbb{Z}$-module is just an abelian group. So $L$, being an $S$-module, can also be considered as a $\mathbb{Z}$-module, by just forgetting the scalar multiplication operation.