Tensor Products I - Extending scalars

Restriction 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.^[1] You can verify this gives $N$ the structure of an $R$ -module and in fact is the object function of a functor $S -Mod \to R -Mod$ . We call this functor the restriction of scalars from $S$ to $R$ .

Extension of scalars

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 $M$ ?

Sadly, not in general, as the example below illustrates:

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^[2] $Z$ -action, i.e., $2 z = z + z$ . Since there is no integer $z$ such that $z + z = 1$ , we reach a contradiction.

Can we try to do the next best thing, which is to 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 no (at least in general) as the next example illustrates.

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 $N$ 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 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? More formally, and still under the working assumption that $R$ is a subring of $S$ , let $U : S -Mod \to R -Mod$ be the restriction of scalars functor described above. ^[3] We are ideally looking for a functor $T : R -Mod \to S -Mod$ that is left adjoint to $U$ , in that 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 the universal property of the free module, and so it is perhaps no surprise that the construction of the mystery module $T (M)$ is very similar to that of the free module. In fact, the free module construction will make a return shortly.

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 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

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 $S \times M$ .^[4] 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:

$(s, m_{1} + m_{2}) - (s, m_{1}) - (s, m_{2})$
$(s_{1} + s_{2}, m) - (s_{1}, m) - (s_{2}, m)$
$(s r, m) - (s, r m)$

for $s_{i} \in S$ and $m, m_{i} \in M$ . 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 $R$ . Note that by construction it "remembers" the left $R$ -module structure of $M$ and right $R$ -module structure of $S$ .

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

$s \otimes (m_{1} + m_{2}) = s \otimes m_{1} + s \otimes m_{2}$
$(s_{1} + s_{2}) \otimes m = s_{1} \otimes m + s_{2} \otimes m$
$s r \otimes m = s \otimes r m$

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:$$\displaystyle s\left(\sum s_i\otimes m_i\right) := \sum ss_i\otimes m$$
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 $$\displaystyle f_s\left(\sum (s_i, m_i)\right) := \sum (ss_i,m_i)$$
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 $j : M \to U (S \otimes_{R} M)$ defined by $m \mapsto 1_{S} \otimes m$ , where $U$ is the forgetful functor from $S -Mod$ to $R -Mod$ . 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 is a natural bijection

t a u_{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 $j : M \to U (S \otimes_{R} M)$ be the $R$ -module morphism defined above. Then $M / \ker (j)$ 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 $M / \ker (j)$ is isomorphic to the image of $j$ , 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 $j$ , i.e., we have a commutative diagram

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

Suggested next note

Examples of extending scalars

More generally, suppose $f : R \to S$ is any ring morphism. We can then view $N$ as an $R$ -module using the ring morphism $f$ and the action of $S$ on $N$ , by defining $r \cdot n = f (r) n$ . ↩︎
In fact, the only possible $Z$ -action! ↩︎
This is really an example of a forgetful functor, since we are simply forgetting part of the $S$ -action, hence our choice for the letter $U$ . ↩︎
If we're being pedantic, we mean the set $U (S) \times U (M)$ , where $U$ is the forgetful functor from $R -Mod \to Set$ . ↩︎