I'm reading Dummit and Foote chapter 10 section 4 on tensors. They write
It is ... natural to consider the free $\mathbb Z$-module ... on the set $S\times N$, i.e. the collection of all finite commuting sums of elements of the form $(s_i,n_i)$ where $s_i\in S$ and $n_i\in N$. This is an abelian group where there are no relations between any distinct pairs $(s,n)$ and $(s',n')$, i.e. no relationship between the ``formal products'' $sn$, and in this abelian group the original module $N$ has been thoroughly distinguished from the new "coefficients" from $S$. To satisfy the relations necessary for an $S$-module structure imposed [above] and the compatibility relation with the action of $R$ on $N$ [above], we must take the quotient of this abelian group by the subgroup $H$ generated by all elements of the form
$$ (s_1+s_2,n) - (s_1,n)-(s_2,n) $$ $$ (s,n_1+n_2) - (s,n_1)-(s,n_2) $$ $$ (sr,n) - (s,rn) $$
Now my question is basically: isn't $(s_1+s_2,n) - (s_1,n) - (s_2,n)$ the same as $(0,-n)$? That just doesn't seem possible though, since if it were true we would then just be taking all elements of the form $(0,n)$ and $(s,0)$ as well as elements $(sr-s,n-rn)$. This wouldn't generally even be closed under addition.
I'm sure I must be misunderstanding the nature of the + operation on two pairs $(s_1,n_1)+(s_2,n_2)$. But in this section they do say that we take the product and that it's an abelian group. Well ... don't we define the plus operation on the product, coordinatewise? What has indicated so far that we don't use coordinate-wise addition, and what is the alternate notion of addition that we are instead using? The semi-direct plus from group theory?