Examples of extending scalars

Trivial extensions

Suppose $M$ is an $R$ -module. Considering the "trivial" extension $R \subseteq R$ , we have constructed an $R$ -module $R \otimes_{R} M$ 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 $R \otimes_{R} M ≃ M$ . 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 $R \otimes_{R} M$ , namely that it is the unique $R$ -module (up to unique isomorphism) such that ${Hom}_{R -Mod} (R \otimes_{R} M, N) ≃ {Hom}_{R -Mod} (M, U (N))$ for every $R$ -module $N$ . In this instance, the "forgetful" functor $U$ is the identity functor, so that last set is ${Hom}_{R -Mod} (M, N)$ . But this exactly says that $M$ itself is an $R$ -module with the desired universal property! Thus, $M$ and $R \otimes_{R} M$ are isomorphic (through a unique $R$ -module isomorphism).

For example, for every abelian group $A$ we have an isomorphism $Z \otimes_{Z} A ≃ A$ of abelian groups. Similarly, if $V$ is any $F$ -vector space then $F \otimes_{F} V$ is isomorphic to $V$ as an $F$ -vector space.

Field extensions

Suppose $F \subseteq E$ 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 $E \otimes_{F} E$ .

If this looks weird, it's because we're hiding a forgetful functor, namely the forgetful functor $U : {Vec}_{E} \to {Vec}_{F}$ . With this is mind, we're really comparing $E \otimes_{F} U (E)$ 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 $E \otimes_{F} U (E) ≃ E^{n}$ , 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 $1_{R} \otimes 0_{m} = 0$ in $S \otimes_{R} M$ , where $0$ is the additive identity of the $S$ -module $S \otimes_{R} M$ .

The proof is very short. Simply observe that

1_{R} \otimes 0_{m} = 1_{R} \otimes (0_{m} + 0_{m}) = 1_{R} \otimes 0_{m} + 1_{R} \otimes 0_{m} .

The result then follows by additive cancellation in the $S$ -module $S \otimes_{R} M$ .

Extending $Z$ -actions to $Q$ -actions for finite abelian groups

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

q \otimes a = \frac{q n}{n} \otimes a = (\frac{q}{n} \cdot n) \otimes a = \frac{q}{n} \otimes n a = \frac{q}{n} \otimes 0_{A} = 0.

By linearity it follows that every element of $Q \otimes_{Z} A$ is 0, and hence $Q \otimes_{Z} A = 0$ .

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., $M ≃ R^{n} = R \oplus R \oplus \dots \oplus R$ . Then $S \otimes_{R} M ≃ S^{n}$ .

For example, we have $Q \otimes_{Z} Z^{n} ≃ Q^{n}$ and $C \otimes_{R} R^{n} ≃ C^{n}$ . As another class of special case, if $F \subseteq E$ is a field extension and $V$ is an $F$ -vector space of dimension $n$ , then $V ≃ F^{n}$ as $F$ -vector spaces and $E \otimes_{F} V ≃ E^{n}$ as $E$ -vector spaces.

Suggested next notes

Bimodules
Tensor Products II - Tensor products of bimodules

First show that every element in $R \otimes_{R} M$ can be expressed as simple tensor of the form $1_{R} \otimes m$ for some $m \in M$ , then define a map by $1_{R} \otimes m \mapsto m$ . You then need to verify this map is well defined (since tensors are really cosets!) and is an $R$ -module isomorphism, a fair number of details to check. ↩︎