# Theorem 5, Section 4.5 of Hungerford’s Algebra

Let $$R$$ and $$S$$ be rings and $${}_{S}A_{R}$$, $${}_{R}B$$, $$C_{R}$$, $${}_{R}D_{S}$$ (bi)modules as indicated.

(i) $$A\otimes_R B$$ is a left $$S$$-module such thats $$s(a\otimes b)= sa\otimes b$$ for all $$s\in S$$, $$a\in A$$, $$b\in B$$.

Sketch of proof: (i) For each $$s\in S$$ the map $$A\times B\to A\otimes_R B$$ given by $$(a,b)\mapsto sa\otimes b$$ is $$R$$-middle linear, and therefore induces a unique group homomorphisma. $$\alpha_s : A\otimes_R B\to A\otimes_R B$$ such that $$\alpha_s(a\otimes b)= sa\otimes b$$. For each element $$u=\sum_{i=1}^na_i\otimes b_i\in A\otimes_R B$$ define $$su$$ to be the element $$\alpha_s(u)=\sum_{i=1}^n \alpha_s (a_i\otimes b_i) = \sum_{i=1}^n s a_i \otimes b_i$$. Since $$\alpha_s$$ is a homomorphism, this action of $$S$$ is well defined (that is, independent of how $$u$$ is written as a sum of generators). It is now easy to verify that $$A\otimes_R B$$ is a left $$S$$-module.

(1) Is statement of theorem precise? I mean scalar multiplication is only defined for generators of $$A\otimes_R B$$. By axiom of left $$S$$-module, we have $$s(\sum_{i=1}^na_i\otimes b_i)= \sum_{i=1}^n s(a_i\otimes b_i)= \sum_{i=1}^n s a_i\otimes b_i.$$ But well-defined of scalar multiplication on generator don’t extend over to arbitrary element of $$A\otimes_R B$$.

(2) Author’s proof of theorem is really cleaver; scalar multiplication is well-defined follows from $$\alpha_s$$ is well-defined. I am trying to prove scalar multiplication is well-defined on generators of $$A\otimes_R B=F/K$$ using direct approach. If $$a\otimes b=c\otimes d$$, then $$sa\otimes b = sc\otimes d$$. Suppose $$a\otimes b=c\otimes d\iff(a,b)-(c,d)\in K$$. We need to show $$(sa,b)-(sc,d)\in K$$. How to progress from here?

(1) The statement is precise. To define any homomorphism on the tensor product, it is sufficient to define it on generators $$a\otimes b$$. The general action of the homomorphism is then given by linearity, i.e. given the action on generators, there is only one way the homomorphism can act on an arbitrary element of the tensor product if it wants to be a homomorphism. To be precise, this way is to simply distribute over the sum of simple tensors, exactly as is shown in the proof. However, the proof does define it on all elements, when defining $$\alpha_s(u)$$.
I do not understand what you mean by "well-defined of scalar multiplication on generator don't extend over to arbitrary element..." That is exactly what it does. Remember that a general element of the tensor product is always a linear combination of the generators, which are simple tensors $$a\otimes b$$, and that a linear combination is always finite, i.e. a finite sum.
(2) If you know that $$a\otimes b = c\otimes d$$, then $$(a,b)-(c,d)\in K$$. Now you simply use the definition of $$K$$, in particular that $$s(v,w)-(sv,w)\in K$$. Indeed, we see that $$(sa,b)-(sc,d)=s(a,b)-s(c,d) = s((a,b)-(c,d)).$$ Since $$K$$ is a submodule of $$F$$ (again, by definition), it is in particular closed under scaling, so $$s((a,b)-(c,d))\in K$$, completing the proof.
• Thank you so much for the answer. (1) left scalar multiplication is not homomorphism. (2) what is $s(v,w)$? Left scalar multiplication is not defined on free abelian group $F$. For clarification, here is definition of tensor product given in Hungerford’s algebra. Feb 15 at 15:25
• Why is (left) scalar multiplication not a homomorphism? By definition, it distributes. And in the case of the tensor product, in the proof it is literally defined as $\alpha_s$, which by construction is a homomorphism. Feb 15 at 17:04