Tensor Products IV - The Adjoint Property

#module_theory

You might be wondering if there is a way to avoid inventing these new objects (the $(R, T)$ -sets) together with the extremely specific notion of balanced bilinear maps between such objects. The answer is yes, through the following result:

Alternate (better?) universal property of the tensor product

Suppose $M$ is an $(R, S)$ -bimodule and $N$ is an $(S, T)$ -bimodule. Then there is a natural bijection for every $(R, T)$ -bimodule $P$

τ_{P} : {Hom}_{(R, T)} (M \otimes_{S} N, P) \tilde{\to} {Hom}_{(R, S)} (M, {Hom}_{T} (N, P)) .

Let's investigate this bijection. First suppose $f : M \otimes_{S} N \to P$ is an $(R, T)$ -bimodule morphism. By the original universal property of the tensor product, this corresponds uniquely to an $S$ -balanced bilinear $(R, T)$ -map $g : M \times N \to P$ with $g (m, n) = f (m \otimes n)$ . We will now produce an $(R, S)$ -bimodule morphism $h : M \to {Hom}_{T} (N, P)$ . For each $m \in M,$ consider the map $h_{m} : N \to P$ defined by $h_{m} (n) = g (m, n) .$ The linearity of $g$ in the second factor implies $h_{m}$ is additive; the compatibility of $g$ with the right $T$ -actions on $N$ and $P$ implies $h_{m}$ is a right $T$ -module morphism. We can therefore define $h (m) = h_{m}$ . In terms of the original morphism $f$ , we have $h_{m} = f (m \otimes -)$ .

Why is $h$ an $(R, S)$ -bimodule morphism? The additivity of $h$ follows from the linearity of $g$ in the first factor:

\begin{aligned} h (m_{1} + m_{2}) & = h_{m_{1} + m_{2}} \\ = g (m_{1} + m_{2}, -) \\ = g (m_{1}, -) + g (m_{2}, -) \\ = h_{m_{1}} + h_{m_{2}} \\ = h (m_{1}) + h (m_{2}) . \end{aligned}

Here we've used a dash ( $-$ ) to indicate where the input (from $N$ ) would go. This is to minimize the number of parentheses involved, so that we don't have to write awkward things like $h (m) (n)$ .

Before we check the compatibilities with the ring actions, recall how the abelian group ${Hom}_{T} (N, P)$ was given the structure of an $(R, S$ )-bimodule: the left $R$ -action arose from the left $R$ -action on $P$ , while the right $S$ -action arose from the left $S$ -action on $N$ . In terms of elements, if $f : N \to P$ is morphism of right $T$ -modules, then for each $r \in R$ we defined the map $r \cdot f$ by

(r \cdot f) (n) = r \cdot f (n),

while for each $s \in S$ we defined $f \cdot s$ by

(f \cdot s) (n) = f (s n) .

We now return to our situation. To see that $h$ is compatible with the left $R$ -action, observe that

\begin{aligned} r \cdot h (m) & = r \cdot h_{m} \\ = r \cdot f (m \otimes -) \\ = f (r \cdot (m \otimes -)) \\ = f (r m \otimes -) \\ = h_{r m} \\ = h (r m) . \end{aligned}

To see that $h$ is compatible with the right $S$ -action, observe that

\begin{aligned} h (m) \cdot s & = h_{m} \cdot s \\ = h_{m} (s \cdot -) \\ = f (m \otimes (s \cdot -)) \\ = f (m s \otimes -) \\ = h_{m s} \\ = h (m s) \end{aligned}

We've finally verified that our map $h : M \to {Hom}_{T} (N, P)$ really is an $(R, S)$ -bimodule morphism. This association $f \mapsto h$ is our set map

τ_{P} : {Hom}_{(R, T)} (M \otimes_{S} N, P) \to {Hom}_{(R, S)} (M, {Hom}_{T} (N, P)) .

We have not yet shown that $τ_{P}$ is natural (in $P$ ), nor that it is bijective. We'll leave verifying naturality as an exercise, but we will verify $τ_{P}$ is a bijection by constructing the inverse map.

To that end, suppose $h : M \to {Hom}_{T} (N, P)$ is an $(R, S)$ -bimodule morphism. As above, for each $m \in M$ let's write $h_{m}$ for $h (m)$ , to cut down on parentheses. Define a set map $g : M \times N \to P$ by $g (m, n) = h_{m} (n) .$ This set map is linear in $M$ because $h$ is additive:

g (m_{1} + m_{2}, -) = h (m_{1} + m_{2}) = h (m_{1}) + h (m_{2}) = g (m_{1}, -) + g (m_{2}, -) .

It is linear in $N$ because each $h_{m}$ is additive:

g (m, n_{1} + n_{2}) = h_{m} (n_{1} + n_{2}) = h_{m} (n_{1}) + h_{m} (n_{2}) = g (m, n_{1}) + g (m, n_{2}) .

To see $g$ is $S$ -balanced, first note that $g (m s, -) = h (m s) = h (m) \cdot s = h_{m} \cdot s$ , and so

g (m s, n) = (h_{m} \cdot s) (n) = h_{m} (s n) = g (m, s n) .

It is compatible with the left $R$ -actions:

g (r m, n) = h_{r m} (n) = (r \cdot h_{m}) (n) = r \cdot h_{m} (n) = r \cdot g (m, n) .

And it is compatible with the right $T$ -actions:

g (m, n t) = h_{m} (n t) = h_{m} (n) \cdot t = g (m, n) \cdot t .

It now follows from the original universal property of the tensor product that the corresponding map $f : M \otimes_{S} N \to P$ defined on simple tensors by $f (m \otimes n) = g (m, n)$ is an $(R, T)$ -bimodule morphism. This association $h \mapsto f$ is our set map

η_{P} : {Hom}_{(R, S)} (M, {Hom}_{T} (N, P)) \to {Hom}_{(R, T)} (M \otimes_{S} N, P) .

From our explicit constructions one can verify that the set maps $τ_{P}$ and $η_{P}$ are mutual inverses, and hence both are bijections.

An immediate consequence

Now that we know the tensor product functor $M \otimes_{S} -$ is a left adjoint, we also know it commutes with all colimits, such as direct sums.

The tensor product commutes with direct sums

Suppose $M$ is an $(R, S)$ -bimodule and ${N_{i} ∣ i \in I}$ is a family of $(S, T)$ -bimodules. Then there is a unique isomorphism of $(R, T)$ -bimodules

M \otimes_{S} (⨁_{i \in I} N_{i}) ≃ ⨁_{i \in I} (M \otimes_{S} N_{i})

The analogous result is true with the positions of the tensor product and direct sum exchanged, i.e., right-tensoring distributes across direct sums. Because of this, we say that tensor product commutes with direct sums. As finite direct products are isomorphic to the corresponding direct sums, this also implies that tensor product commutes with finite direct products.

Extending scalars on free modules

A special case of the above property is that "extending scalars" commutes with the free module construction. More precisely:

Corollary

Suppose $R \subseteq S$ are rings and $R^{n} = R \oplus \dots \oplus R$ is the free module on a set of size $n$ .
Then there is an $S$ -module isomorphism

S \otimes_{R} R^{n} ≃ S^{n} .

Indeed, observe that

\begin{aligned} S \otimes_{R} R^{n} & ≃ S \otimes_{R} (R \oplus \dots \oplus R) \\ ≃ (S \otimes_{R} R) \oplus \dots (S \otimes_{R} R) \\ ≃ S \oplus \dots S \\ ≃ S^{n} \end{aligned}

Tensor products of free modules

Another consequence of the above property is that the tensor product of two free $R$ -modules is again a free $R$ -module. More precisely:

Corollary

Suppose $M ≃ R^{a}$ and $N ≃ R^{b}$ are free $R$ -modules of ranks $a$ and $b$ , respectively. We then have an $R$ -module isomorphism

M \otimes_{R} N ≃ R^{a b} .

Furthermore, if ${m_{1}, \dots, m_{a}}$ and ${n_{1}, \dots, n_{b}}$ are bases for $M$ and $N$ as free $R$ -modules, respectively, then a basis for $M \otimes_{R} N$ as a free $R$ -module is ${m_{i} \otimes n_{j} ∣ 1 \leq i \leq a, 1 \leq j \leq b}$ .

The statement about bases follows from our explicit isomorphism (involving tensor products of direct sums) above.

Suggested next notes

Universal Properties III - Yoneda's Lemma
The tensor product functor and flat modules