# 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 $H$.

• For (2): if $s \in S$, then $-s \in S$, too, so we can write $-(s \otimes n) = (-s) \otimes n$. Jan 15 '17 at 19:07
• Dear SpamIAm, thanks a lot for your comment, now it's clear! Jan 16 '17 at 13:53

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.

• Dear David, thanks a lot for your answer. I have understood the discussion of points 1 and 2. Instead, I still can't grasp the reason why the multiplication by $s$ can be applied "like a function" on the two members of an equality between elements of the free abelian group on $S\times N$. A priori, multiplying by $s$ the first entries of the sum $h=\sum_i (s_i, n_i) \in H$ could give an entirely different element of the free abelian group. Instead, you say that doing this we obtain exactly the multiplication by $s$ of the first entries of the generators of $H$ of which $h$ is composed. Why? Jan 16 '17 at 13:50
• Inserting $s$ (as a left multiplication) in all the entries of a generator of $H$ gives another generator of $H$-for example in the first line following: "Consider, for any $s\in S$..." we have on the RHS of the equals sign an element of the form: $(u+u',v) - (u,v) - (u'v) \in H$ with $u = ss_1, u' = ss_2$ and $v = n_1$, which (by the definition of $H$) lies in $H$ (since it contains ALL expressions of that form). $H$ is typically "very big", while $FA(S \times N)/H$ is much smaller than $FA(S \times N)$. Jan 16 '17 at 23:33
• Dear David, I understood that multiplication by $s$ of a generator of $H$ gives another generator of $H$. What is still not clear to me is why multiplication by $s$ can be applied as a function to the two members of the equation $(s_1, n_1) + (s_2, n_2) - (s_1', n_1') - (s_2', n_2') = h\in H$. In other words, why can we assume that multiplying by $s$ the first entries of the first member of this equation gives the same thing as multiplying by $s$ the first entries of the generators of which $h$ is composed? A priori they could be totally different elements of the free abelian group... Jan 17 '17 at 10:15
• Our goal is to show that $\sum\limits_i (s_i,n_i) \in H \implies \sum\limits_i (ss_i,n_i) \in H$, from which we can deduce the operation $L_s: S \otimes_R N \to S \otimes_R N$ given by $L_s(\sum\limits_i s_i \otimes n_i) = \sum\limits_i (ss_i \otimes n_i)$ is well-defined, yes? We don't need to show that we get the "same" element of $H$, just that the sum/difference gives another element of $H$ which allows us to conclude the cosets are the same. Jan 18 '17 at 1:22
• Yes, but you have to claim that multiplying by $s$ the first entries in the sum $\sum_i (s_i, n_i)\in H$ gives the same thing as multiplying by $s$ the first entries of the generators which compose the same sum as an element of $H$; otherwise the fact that multiplying by $s$ a generator of $H$ we obtain another generator of $H$ is completely useless. Now the question is: why can we make this claim? Jan 18 '17 at 16:03