Tensor Products I - Extending scalars

Extension of scalars

Suppose $N$ is an $S$ -module and $R \subseteq S$ is a subring. By restricting the action of $S$ on $N$ to an action of $R$ on $N$ , we can view $N$ as an $R$ -module. You can verify this gives $N$ the structure of an $R$ -module and in fact is the object function of a forgetful functor $S -Mod \to R -Mod$ . Historically, this process was called the restriction of scalars from $S$ to $R$ .

In light of the above process, one might ask the following:

Question

Is it possible to go the other way? In other words, if $M$ is an $R$ -module and $R$ is a subring of $S$ , is it possible to extend the action of $R$ on $M$ to an action of $S$ on $M$ ?

Categorically speaking, we're asking for a functor $R -Mod \to S -Mod$ that "nicely complements" the forgetful functor above, where "nicely complements" is momentarily ambiguous and ill defined.

Taking our question at its most naive hope (to put an $S$ -action directly on $M$ ), the answer is sadly a negative.

Frustrating example

Consider the ring $Z$ when viewed as a $Z$ -module (i.e., an abelian group). Even though $Z$ is a subring of $Q$ , we cannot extend the $Z$ -module structure on $Z$ to a $Q$ -module structure. Why not? Suppose we would, and consider the action of $\frac{1}{2}$ on the integer $1$ . There would need to be an integer $z$ such that $\frac{1}{2} ⋆ 1 = z$ . Since our $Q$ -action is supposed to extend our $Z$ -action, it would follow that $2 z = 1$ in $Z$ , where here $2 z$ denotes the usual^[1] $Z$ -action, i.e., $2 z = z + z$ . Since there is no integer $z$ such that $z + z = 1$ , we reach a contradiction.

If we can't expect to extend the $R$ -action on $M$ to an $S$ -action on $M$ , what's the next best thing we could hope? Perhaps we can embed the original $R$ -module $M$ as an $R$ -submodule of a larger $R$ -module $N$ that also has the structure of an $S$ -module (extending the action of $R$ ).

Sadly this is also not always possible.

Second frustrating example

Suppose $M$ is the $Z$ -module $Z_{2}$ , $N$ is a $Q$ -module, and $f : M \to U (N)$ is a $Z$ -module morphism, where $U : Q-Mod \to Z-Mod$ is the forgetful functor. Then $N$ is a $Q$ -vector space, so every nonzero element in $N$ has infinite additive order. Since both elements in $M$ have finite order, this implies their images in $N$ must be zero. In other words, every $Z$ -module morphism from $M$ to $U (N)$ must be the zero map, and so there cannot be any embeddings of $M$ into a $Q$ -module.

Category theory insight

In light of the second example, let's widen our scope further and instead consider all $R$ -module morphisms $f : M \to U (N)$ , where $N$ is an $S$ -module. Can we find a "best possible" $S$ -module through which all $R$ -module morphisms from $M$ factor? Based on our work with free modules, it sounds reasonable to consider the existence of a functor $T : R -Mod \to S -Mod$ that is left adjoint to $U$ ; i.e., for which there are natural bijections

τ_{M, N} : {Hom}_{S -Mod} (T (M), N) \tilde{\to} {Hom}_{R -Mod} (M, U (N)) .

This looks very similar to our universal property of the free module, so it is perhaps no surprise that the construction of the mystery module $T (M)$ is very similar to that of the free module.

Before we move on to the actual construction, it's worthwhile to consider the choice we've made above, which is that we're specifically looking for a functor $T : R -Mod \to S -Mod$ that is left adjoint to the forgetful functor $U$ (the so-called "restriction of scalars" functor). What about a functor that is right adjoint to $U$ ? It turns out such a functor also exists, and it is sometimes called the co-extension of scalars.

The construction of $T (M)$

Given our $R$ -module $M$ and the fact that we have an inclusion of rings $R \subseteq S$ , our first step towards extending the $R$ -action on $M$ is to look at the free $Z$ -module on the set $S \times M$ .^[2] In this abelian group $F (S \times M)$ we then want to impose relations so that the quotient of this abelian group by the subgroup generated by those relations has the structure of an $S$ -module in which the $S$ -action extends the $R$ -action on $M$ . To that end, let $H$ be the subgroup of $F (S \times M)$ generated by all elements of the following form:

$(s, m_{1} + m_{2}) - (s, m_{1}) - (s, m_{2})$ , for all $s \in S$ , $m_{1}, m_{2} \in M$
$(s_{1} + s_{2}, m) - (s_{1}, m) - (s_{2}, m)$ , for all $s_{1}, s_{2} \in S$ , $m \in M$
$(s r, m) - (s, r m)$ , for all $r \in R, s \in S, m \in M$

Note that we are using the natural identification of the set $S \times M$ with a subset of $F (S \times M)$ .

A small but important note

Although the free $Z$ -module $F (S \times M)$ is constructed from the sets of elements of $S$ and $M$ without knowledge of the algebraic structure (i.e., their internal operations), the subgroup $H$ defined above does use the information about their algebraic structures. The first bullet point uses the additive structure in $M$ , the second uses the additive structure in $S$ , and the third uses both the right $R$ -module structure of S and the left $R$ -module structure of $M$ .

We denote the resulting quotient group by $S \otimes_{R} M$ and call it the tensor product of $S$ and $M$ over $R$ .

It is common to let $s \otimes m$ denote the coset containing $(s, m)$ in the quotient group $S \otimes_{R} M$ . Using this notation, every element in $S \otimes_{R} M$ can be written (non-uniquely!) as a finite sum of the form $\sum_{i} s_{i} \otimes m_{i}$ . Elements that can be written simply as $s \otimes m$ are called simple tensors (or sometimes pure tensors).

In this new notation, our construction of $S \otimes_{R} M$ has forced the relations

$s \otimes (m_{1} + m_{2}) = s \otimes m_{1} + s \otimes m_{2}$ , for all $s \in S$ , $m_{1}, m_{2} \in M$
$(s_{1} + s_{2}) \otimes m = s_{1} \otimes m + s_{2} \otimes m$ , for all $s_{1}, s_{2} \in S$ , $m \in M$
$s r \otimes m = s \otimes r m$ , for all $r \in R, s \in S, m \in M$

The $S$ -module structure on $S \otimes_{R} M$

We have constructed the abelian group $S \otimes_{R} M$ to have an obvious $S$ -action:

s (\sum s_{i} \otimes m_{i}) := \sum s s_{i} \otimes m

As with any action (or function) defined on a quotient group by a formula using a choice of coset representative, we should check that this action is well defined. In other words, we should really verify that for each $s \in S$ the map $f_{s} : F (S \times M) \to S \otimes_{R} F$ defined by

f_{s} (\sum (s_{i}, m_{i})) := \sum (s s_{i}, m_{i})

is a morphism of abelian groups with $H \leq \ker (f)$ . We'll leave the details for now and instead move on to various properties of the $S$ -module $S \otimes_{R} M$ we have created.

Properties of our construction

There is a (natural) $R$ -module morphism $η_{M} : M \to U (S \otimes_{R} M)$ defined by $m \mapsto 1_{S} \otimes m$ , where $U$ is the forgetful functor from $S -Mod$ to $R -Mod$ . (These morphisms are exactly the components of the unit map $η : I_{R -Mod} \to U T$ , where $T = S \otimes_{R} -$ is our tensor product functor.) Since our tensor product involved taking a quotient group, it should not be expected that this morphism is injective. However, the module we created does possess the desired universal property, in that there are natural bijections

τ_{M, N} : {Hom}_{S -Mod} (S \otimes_{R} M, N) \tilde{\to} {Hom}_{R -Mod} (M, U (N))

We'll add the details of this argument later, but for now we'll simply note that the construction follows from a universal property of the free $Z$ -module on $S \times M$ , a universal property of quotient groups, and the fact that our subgroup $H$ of desired relations automatically is in the kernel of any module morphism to an $S$ -module.

Here is a nice corollary of our universal property:

Corollary

Let $η_{M} : M \to U (S \otimes_{R} M)$ be the $R$ -module morphism defined above. Then $M / \ker (η_{M})$ is the maximal quotient of $M$ that can be embedded into an $S$ -module.

Why is this true? First observe that by the First Isomorphism Theorem the quotient $M / \ker (η_{M})$ is isomorphic to the image of $η_{M}$ , which is an $R$ -submodule of $U (S \otimes_{R} M)$ . Now suppose $N$ is an $S$ -module and $f : M \to U (N)$ is an $R$ -module morphism (and so $M / \ker (f)$ is isomorphic to a submodule of $N$ ). By our universal property of the tensor product, we have a corresponding $S$ -module morphism $g : S \otimes_{R} M \to N$ ; the natural bijection on the hom-sets also tells us that $f$ factors through $U (g)$ via $η_{M}$ , i.e., we have a commutative diagram

This implies that $\ker (η_{M}) \subseteq \ker (f)$ as submodules of $M$ , so by the Isomorphism Theorems for Modules we have that $N / \ker (f)$ is a quotient of $N / \ker (η_{M})$ (by the submodule $\ker (f) / \ker (η_{M})$ , specifically).

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 $S \otimes_{R} M$ .

Examples

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^[3], 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 a (unique) 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.

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 (isomorphic to) 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

In fact, the only possible $Z$ -action! ↩︎
If we're being pedantic, we mean the set $U_{1} (S) \times U_{2} (M)$ , where $U_{1}, U_{2}$ are the appropriate forgetful functors from to $Set$ . ↩︎
Show the unit morphism $η_{M} : M \to R \otimes_{R} M$ is an isomorphism. ↩︎