Extension of scalars in D & F Abstract Algebra I have troubles in understanding the construction of the extension of scalars as done in Dummit & Foote Abstract Algebra, Chapter 10.4. 
The situation is the following: we have a subring $R$ of a ring $S$ with $1_R=1_S$ and we have a left $R$-module $N$. Then we want to construct an $S$-module in which to embed $N$ as an $R$-submodule, and the best choice seems to be the tensor product $S \otimes_R N$. Now I have some questions:
1) why is it the best choice? Is it because of the universal property? what is the meaning of the universal property?
2) if the tensor product is a quotient by the subgroup of the relations $H$ of the free $\mathbb{Z}$-module on the set $S\times N$, why can we write its elements simply as finite sums of simple tensors, forgetting the fact that there can also be negative multiples of certain simple tensors?
3) I don't understand the left $S$-module structure defined on $S \otimes_R N$ by $s(\sum_{i=1}^p s_i\otimes n_i) = \sum_{i=1}^p (ss_i)\otimes n_i$. In particular I don't understand why this action should be well-defined. What I've understood is that if $\sum s_i\otimes n_i=\sum s'_j\otimes n'_j$ then $\sum (s_i, n_i)-\sum (s'_j, n'_j) \in H$. But now I don't understand why multiplying the first entries of the couples by $s$ corresponds to multiplying by $s$ the first entries of the generators of $H$.
Thanks in advance for your help!
 A: Here is my answer (my memory of these things is somewhat dimmed by time, so I'll certainly appreciate any constructive criticism):
1) "Best" is sort of a loaded term. What this construction has going for it is that it is "canonical"-that is, it doesn't depend on the specific properties of $R,S$ or $N$, it can be done "in the same way" for any situation that satisfies the conditions you state.
The "meaning" of the universal property is essentially a formalization of what I just said, but with one sharpening of it: we can think of the tensor product as a "catch-all" for any $R$-linear map $N \to L$ (where $L$ is an $S$-module, and thus also an $R$-module in the obvious way) in that such a map factors through the tensor product (this is much the same as the way a group homomorphism  (of $G$) that annihilates a certain normal subgroup $K$ factors through the quotient group homomorphism $G \to G/K$). It is also similar to the universal property for free groups, in which we create a "minimal" group from an arbitrary set. In this case, we are creating a kind of "minimal" expansion of an $R$-module into an $S$-module, so that we have an $R$-module homomorphism of $N$ into our new $S$-module, given by $n \mapsto 1_S \otimes n$ (so that $r\cdot n \mapsto 1_S \otimes (r\cdot n) = r\cdot(1_S \otimes n)$).
2) This is pretty obvious, as noted by SpamIAm above, we can write $-(s\otimes n) = (-s)\otimes n$, which is a simple tensor.
3) Let's look at this "one sum at a time":
Suppose $s_1 \otimes n_1 + s_2 \otimes n_2 = s_1' \otimes n_1' + s_2' \otimes n_2'$. By definition, this means:
$(s_1,n_1) + (s_2,n_2) - (s_1',n_1') - (s_2',n_2',) \in H$.
Consider, for any $s \in S$, the following:
$(s(s_1 + s_2),n_1) - (ss_1,n_1) - (ss_2,n_1) = (ss_1 + ss_2,n_1) - (ss_1,n_1) - (ss_2,n_1)$,
$(ss_1,n_1+n_2) - (ss_1,n_1) - (ss_1,n_2)$,
$(s(s_1r),n_1) - (ss_1,rn_1) = ((ss_1)r,n_1) - (ss_1,rn_1)$.
This shows that any generator of $H$, when you (left-) multiply by $s$ in the first coordinate, you get another element (indeed another generator) of $H$. Since elements of $H$ are finite sums of such generators, it follows that the action of $s$ "distributes" over such a sum, that is:
$\sum\limits_i (s_i,n_i) \in H \implies \sum\limits_i (ss_i,n_i) \in H$.
Thus, $(s_1,n_1) + (s_2,n_2) - (s_1',n_1') - (s_2',n_2',) \in H$ implies:
$(ss_1,n_1) + (ss_2,n_2) - (ss_1',n_1') - (ss_2',n_2',) \in H$, and thus:
$ss_1\otimes n_1 + ss_2\otimes n_2 = ss_1'\otimes n_1' + ss_2'\otimes n_2'$, and finally:
$s(s_1\otimes n_1 + s_2\otimes n_2) = s(s_1'\otimes n_1' + s_2'\otimes n_2')$.
Clearly, this can be extended to any finite sum of simple tensors.
