Theorem 7, Section 4.5 of Hungerford’s Algebra

Question

If $R$ is a ring with identity and $A_R$, ${}_{R}B$ are unitary $R$-modules, then there are $R$-module isomorphisms $$A\otimes_R R\cong R \text{ and } R\otimes_R B\cong B$$

Sketch of proof: Since $R$ is an $R\text{-}R$ bimodule $R \otimes_R B$ is a left $R$-module by Theorem 5.5. The assignment $(r,b) \mapsto rb$ defines a middle linear map $R\times B \to B$. By Theorem 5.2 there is a group homomorphism $\alpha : R\otimes_R B \to B$ such that $\alpha (r\otimes b)=rb$. Verify that $\alpha$ is in fact a homomorphism of left $R$-modules. Then verify that the map $\beta :B\to R\otimes_R B$ given by $b\mapsto 1_R\otimes b$ is an $R$-module homomorphism such that $\alpha \beta =1_B$ and $\beta \alpha = 1_{R\otimes_R B}$. Hence $\alpha : R \otimes_R B\cong B$. The isomorphism $A\otimes_R R\cong A$ is constructed similarly.

Can we prove this theorem without defining map $\alpha :R\otimes_R B\to B$? To be specific, we will show $\beta:B\to R\otimes_R B$ is left $R$-module homomorphism, surjective and injective. One of the benefit of considering only $\beta: B\to R\otimes_R B$ is that we don’t have to show $\beta$ is well defined.

Let $r\in R$ and $b\in B$. Then $$\beta (rb)=1_R\otimes rb=1_Rr\otimes b=r1_R\otimes b=r(1_R\otimes b)=r\beta (b).$$ Thus $\beta$ is left $R$-module homomorphism. Let $r\otimes b\in R\otimes_R B$. Then $\beta (rb)=r\otimes b\in \text{Im }\beta$. Thus $\text{Im } \beta$ contain all generators of $R\otimes_R B$. So $\text{Im } \beta=R\otimes_R B$ and $\beta$ is surjective.

Let $b\in \text{Ker }\beta$. Then $\beta(b)=1_R\otimes b=0$, i.e. $(1_R,b)\in K$. How to show $b=0$? Why is it so difficult to work with element of $A\otimes_R B$?

Answer 1

As you asked, I will convert my comments in an answer.

I guess you can prove the theorem showing directly that $\beta$ is bijective but what's the point of that? I mean, you have defined tensor product using an extremely powerful universal property, why you don't want to use it? If you think about the contruction of tensor product, it should be clear why it is difficult to work with elements of it. The contruction I know of the tensor product (but there are others) define $A\otimes_R B$ as a quotient of an enormous free $\mathbb{Z}$-module by some submodule which reflects the property you want tensor product to have, i.e. transforming bilinear maps into linear ones. So what you care about when working with tensor product it's usually only its universal property.

What I'm trying to say is that tensor product is defined by its universal property, so most of the constructions you do with tensor product just rely on this universal property. Thus there really is no point in working with elements explicitly, it only complicates things that otherwise are natural.

I think it's natural to feel uncomfortable at first about universal property, but all the effort you do now to understand it correctly will be of much help in the future approaching subjects like category theory. The universal property uniquely determines (up to isomorphism) the tensor product, so it tells you all you need to know about it as a module (or abelian group, depending on the setting).

If you want to feel more more comfortable about universal properties, try to look for the definitions using universal property of things you already know about such as free modules, direct sum, direct product,...