Here is the theorem giving me trouble:
Here is theorem 6, which is invoked in the proof of theorem 6:
Here is the proof of theorem 8:
Here is the sentence giving me trouble so far:
"Since $\varphi$ is an $R$-module homomorphism, the generators of the subgroup $H$ in equation (3) all map to zero in $L$."
Note that $H$ is generated by elements of the following three forms: $(s_1 + s_2, n) - (s_1,n) - (s_2,n)$ and $(s,n_1+n_2) - (s,n_1) - (s,n_2)$ and $(sr,n)-(s,rn)$ for $s,s_1,s_2 \in S$, $r \in R$, and $n,n_1,n_2 \in N$.
Here is my understanding of the proof up to that sentence. $F(S \times N)$ is taken to be a $\Bbb{Z}$-module, as described in theorem 6, and note that $L$, being an abelian group, has a natural $\Bbb{Z}$-module structure. Now consider the map $f : S \times N \to L$ defined by $f(s,n) = s\varphi(n)$. By theorem 6, there exists a unique $\Bbb{Z}$-module $g : F(S \times N) \to L$ such that $g(s,n) = f(s,n) = s\varphi(n)$ for all $(s,n) \in S \times N$. Now, I interpret the above sentence as saying that $g$ is zero on the $H$, a subgroup of $F(S \times N)$. To show this, it suffices to show $g$ annihilates the $H$'s generators. For simplicity, let's show it annihilates $(sr,n)-(s,rn)$:
\begin{align} g((sr,n)-(s,rn)) &= g((sr-s,n-rn)) \\ &= (sr-s)\varphi(n-rn) \\ &= sr \varphi(n-rn) -s \varphi(n-rn) \\ &= sr \varphi(n) - sr\varphi(rn) - s \varphi (n) - s \varphi(rn) \\ &= sr \varphi (n) - sr^2 \varphi(n) - s \varphi(n) - sr \varphi(n)\\ &= - sr^2 \varphi(n) - s \varphi(n) ... \end{align} ...This doesn't look like zero to me...What did I do wrong? What am I misunderstanding?
It is not true that $(sr,n)-(s,rn)=(sr-s,n-rn)$! Remember that these are elements of $F(S\times N)$, so $(sr,n)$ and $(s,rn)$ are just two of our free generators. Our addition operation is the addition of the free $\mathbb{Z}$-module $F(S\times N)$, not the usual product group operation on $S\times N$.
Instead, you can use linearity of $g$ to say that $$g((sr,n)-(s,rn))=g((sr,n))-g((s,rn))=sr\varphi(n)-s\varphi(rn)=sr\varphi(n)-s(r\varphi(n))=0.$$ Here we used the $R$-linearity of $\varphi$ to conclude that $\varphi(rn)=r\varphi(n)$.