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The following is Theorem 8 in Chapter 10.4 (p.362 3rd edition) of Abstract Algebra by Dummit and Foote:

Let $R$ be a subring of $S$, let $N$ be a left $R$-module and let $\iota:N\to S\otimes_R N$ be the $R$-module homomorphism defined by $\iota(n)=1\otimes n$. Suppose that $L$ is any left $S$-module (hence also an $R$-module) and that $\varphi:N\to L$ is an $R$-module homomorphism from $N$ to $L$. Then there is a unique $S$-module homomorphism $\Phi:S\otimes_R N\to L$ such that $\varphi$ factors through $\Phi$, i.e., $\varphi=\Phi\circ\iota.$

The first two sentences in the proof say that

Suppose $\varphi:N\to L$ is an $R$-module homomorphism to the $S$-module $L$. By the universal property of free modules there is a $\mathbb{Z}$-module homomorphism from the free $\mathbb{Z}$-module $F$ on the set $S\times N$ to $L$ that sends each generator $(s,n)$ to $s\varphi(n)$.

Here is my question:

As I understand, $L$ is not necessarily a $\mathbb{Z}$-module. Why is there a $\mathbb{Z}$-module homomorphism from $F(S\times N)$ to $L$?

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    $\begingroup$ Every module is an abelian group, so a $\mathbb{Z}$-module $\endgroup$ – egreg Jan 12 '16 at 23:09
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$\mathbb{Z}$-modules are the same as abelian groups: On every abelian group $A$ there exists a unique $\mathbb{Z}$-module structure via $$ n \cdot a = \begin{cases} \sum_{i=1}^n a & \text{if $n \geq 0$} \\ \sum_{i=1}^n (-a) & \text{if $n \leq 0$}. \end{cases} $$ Given two abelian groups $A$ and $B$ group homomorphisms $A \to B$ are then the same as $\mathbb{Z}$-module homomorphisms.

So using the underlying abelian group of $L$ we can think of $L$ as a $\mathbb{Z}$-module in the above way. A $\mathbb{Z}$-module homorphism $F(S \times N) \to L$ is then the same as a group homomorphism between the abelian group $F(S \times N)$ and the underlying abelian group of $L$. So this is just a fancy way of saying that the map is additive.

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Let $A=S\times N$. By theorem 6, there is a free $\mathbb{Z}$-module $F(A)$ on the set $A$ and $F(A)$ satisfies the universal property: if $M$ is any $\mathbb{Z}$-module and $\varphi: A\to M$ is any map of sets, then there is a unique $\mathbb{Z}$-module homomorphism $\Phi: F(A)\to M$ such that $\Phi(a)=\varphi(a)$, for all $a\in A$. $L$ is an abelian group so a $\mathbb{Z}$-module. Define $\phi: A\to L$ by $(s,n)\mapsto s\varphi(n)$. By the universal property of the $\mathbb{Z}$-module $F(A)$, there is a $\mathbb{Z}$-module homomorphism $\Phi: F(A)\to L$ such that $\Phi(a)=\phi(a)$, for all $a\in A$, that is, $\Phi(s,n)=\phi(s,n)=s\varphi(n)$ for all $(s,n)\in S\times N$.

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  • $\begingroup$ This is an excellent answer to this question because Dummit and Foote leave out the definition of this $\phi$ that you bring up. Bravo! $\endgroup$ – Alex Ortiz Apr 12 '17 at 4:16
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A $\mathbb{Z}$-module is the same thing as an abelian group, because $1\cdot x=x$ and distributivity means there is a unique way to define $n\cdot x$ for any $n\in\mathbb{Z}$ (namely, $n\cdot x$ is the sum of $n$ copies of $x$ if $n\geq 0$, or the negative of the sum of $|n|$ copies of $x$ if $n<0$). Similarly, a homomorphism of $\mathbb{Z}$-modules is the same thing as a homomorphism of abelian groups.

To put it another way, for any ring $S$, there is a unique ring-homomorphism $\mathbb{Z}\to S$ sending $n$ to the sum of $n$ copies of $1$, so any $S$-module is a $\mathbb{Z}$-module as well.

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