Suppose $M$ is an $R$-module. Considering the "trivial" extension $R\beta \x8a\x86R$, we have constructed an $R$-module $R{\beta \x8a\x97}_{R}M$ extending the $R$-action on $M$ to ... an $R$-action on $M$. It should not be surprising that we have a natural $R$-module isomorphism $R{\beta \x8a\x97}_{R}M\beta \x89\x83M$. We can verify this directly on the level of elements^{[1]}, but we can also observe that it follows directly from our universal property of $R{\beta \x8a\x97}_{R}M$, namely that it is the unique $R$-module (up to unique isomorphism) such that ${\mathrm{Hom}}_{R\mathbf{\text{-Mod}}}(R{\beta \x8a\x97}_{R}M,N)\beta \x89\x83{\mathrm{Hom}}_{R\mathbf{\text{-Mod}}}(M,U(N))$ for every $R$-module $N$. In this instance, the "forgetful" functor $U$ is the identity functor, so that last set is ${\mathrm{Hom}}_{R\mathbf{\text{-Mod}}}(M,N)$. But this exactly says that $M$ itself is an $R$-module with the desired universal property! Thus, $M$ and $R{\beta \x8a\x97}_{R}M$ are isomorphic (through a unique $R$-module isomorphism).

For example, for every abelian group $A$ we have an isomorphism $\mathbf{Z}{\beta \x8a\x97}_{\mathbf{Z}}A\beta \x89\x83A$ of abelian groups. Similarly, if $V$ is any $F$-vector space then $F{\beta \x8a\x97}_{F}V$ is isomorphic to $V$ as an $F$-vector space.

Field extensions

Suppose $F\beta \x8a\x86E$ are fields. We can view $E$ as an $F$-vector space (by remembering the $F$-scaling but forgetting the result of the internal multiplication in $E$), and then we can attempt to "extend scalars" to recover this lost information, by forming the $E$-vector space $E{\beta \x8a\x97}_{F}E$.

If this looks weird, it's because we're hiding a forgetful functor, namely the forgetful functor $U:{\mathbf{\text{Vec}}}_{E}\beta \x86\x92{\mathbf{\text{Vec}}}_{F}$. With this is mind, we're really comparing $E{\beta \x8a\x97}_{F}U(E)$ with $E$. This will, in general, not recover the original field $E$; instead, we will see (once we learn that tensor product commutes with direct sum) that $E{\beta \x8a\x97}_{F}U(E)\beta \x89\x83{E}^{n}$, where $n$ is the dimension of $E$ as an $F$-vector space.

A general property of this tensor construction

Here is a general little property of our construction:

Lemma

Suppose $M$ is an $R$-module and $R$ is a subring of a ring $S$. Then ${1}_{R}\beta \x8a\x97{0}_{m}=0$ in $S{\beta \x8a\x97}_{R}M$, where $0$ is the additive identity of the $S$-module $S{\beta \x8a\x97}_{R}M$.

The result then follows by additive cancellation in the $S$-module $S{\beta \x8a\x97}_{R}M$.

Extending $\mathbf{Z}$-actions to $\mathbf{Q}$-actions for finite abelian groups

Suppose $A$ is a finite abelian group, i.e., a finite $\mathbf{Z}$-module. We claim that the $\mathbf{Q}$-module $\mathbf{Q}{\beta \x8a\x97}_{\mathbf{Z}}A$ is always trivial. To see this, let $n=|A|$ and suppose first we have a simple tensor $q\beta \x8a\x97a$ in $\mathbf{Q}{\beta \x8a\x97}_{\mathbf{Z}}A$. Then observe that

By linearity it follows that every element of $\mathbf{Q}{\beta \x8a\x97}_{\mathbf{Z}}A$ is 0, and hence $\mathbf{Q}{\beta \x8a\x97}_{\mathbf{Z}}A=0$.

Extending scalars for free modules

We will see shortly that the tensor product construction commutes with coproducts, a consequence of which will be that the tensor product of a direct sum is the direct sum of the tensor products. In the case of free modules, this yields a nice result.

Tensor product of free modules

Suppose $R$ is a subring of $S$ and $M$ is a free $R$-module of rank $n$, i.e., $M\beta \x89\x83{R}^{n}=R\beta \x8a\x95R\beta \x8a\x95\beta \x8b\u2015\beta \x8a\x95R$. Then $S{\beta \x8a\x97}_{R}M\beta \x89\x83{S}^{n}$.

For example, we have $\mathbf{Q}{\beta \x8a\x97}_{\mathbf{Z}}{\mathbf{Z}}^{n}\beta \x89\x83{\mathbf{Q}}^{n}$ and $\mathbf{C}{\beta \x8a\x97}_{\mathbf{R}}{\mathbf{R}}^{n}\beta \x89\x83{\mathbf{C}}^{n}$. As another class of special case, if $F\beta \x8a\x86E$ is a field extension and $V$ is an $F$-vector space of dimension $n$, then $V\beta \x89\x83{F}^{n}$ as $F$-vector spaces and $E{\beta \x8a\x97}_{F}V\beta \x89\x83{E}^{n}$ as $E$-vector spaces.