Tensor Products III - Balanced Maps and a Universal Property of the Tensor Product

Balanced maps

Before we state a second universal property of our tensor product construction, we first need to talk about a new class of map between bimodules.

We begin with a slight generalization of the notion of a bimodule, to simply a set $X$ endowed with a left $R$ -action and right $S$ -action for some rings $R$ and $S$ , in a compatible way. In other words, we have set maps $R \times X \to X$ and $X \times S \to X$ such that $r \cdot (x \cdot s) = (r \cdot x) \cdot s$ . Note that there are no "distributive" rules since the set $X$ is not assumed to have any binary operations. Also note that any $(R, S)$ -bimodule $M$ can also be regarded as an $(R, S)$ -set by simply forgetting the addtive operation in $M$ .

A morphism of $(R, S)$ -sets $X$ and $Y$ is simply a set map $f : X \to Y$ that is compatible with the left $R$ - and right $S$ -actions. In other words, $f (r x) = r f (x)$ and $f (x s) = f (x) s$ for every $x \in X$ , $r \in R$ and $s \in S$ .

Now suppose $R$ , $S$ , and $T$ are rings (with unity), $M$ is an $(R, S)$ -bimodule and $N$ is an $(S, T)$ -bimodule. We can form an $(R, T)$ -set denoted $M \times N$ as follows. At the level of sets, it is simply the Cartesian product of the corresponding sets of elements, and so consists of all ordered pairs $(m, n)$ with $m \in M$ and $n \in N$ . We can give this set a left $R$ -action via the $M$ -component and a right $T$ -action via the $N$ -component, i.e., $r \cdot (m, n) = (r m, n)$ and $(m, n) \cdot t = (m, n t)$ .

In this situation, for any $(R, T)$ -module $P$ we say a morphism of $(R, T)$ -sets $f : M \times N \to P$ is:

linear in $M$ if for for every $n \in N$ the map $M \to P$ defined by $m \mapsto f (m, n)$ is a left $R$ -module morphism;
linear in $N$ if for every $m \in M$ the map $N \to P$ defined by $n \mapsto f (m, n)$ is a right $T$ -module morphism;
bilinear if it is both linear in $M$ and in $N$ ; and
$S$ -balanced (or middle balanced) if for every $s \in S$ we have $f (m s, n) = f (m, s n),$ for all $m \in M$ and $n \in N$ .

With this new terminology in place, for any triple of bimodules $M$ , $N$ , $P$ as above, we can talk about the set of bilinear $S$ -balanced $(R, T)$ -maps $f : M \times N \to P$ . These are the maps that satisfy the following conditions:

$f (m_{1} + m_{2}, n) = f (m_{1}, n) + f (m_{2}, n)$ for all $m_{1}, m_{2} \in M$ and $n \in N$
$f (m, n_{1} + n_{2}) = f (m, n_{1}) + f (m, n_{2})$ for all $m \in M$ and $n_{1}, n_{2} \in N$
$f (r m, n) = r f (m, n)$ for all $r \in R$ , $m \in M$ , and $n \in N$
$f (m, n t) = f (m, n) t$ for all $t \in T$ , $m \in M$ , and $n \in N$
$f (m s, n) = f (m, s n)$ for all $s \in S$ , $m \in M$ , and $n \in N$

Finally, we can state the universal property of our tensor product construction.

A universal property of tensor products

Universal property of the tensor product

Suppose $R$ , $S$ , and $T$ are rings (with unity), $M$ is an $(R, S)$ -bimodule, and $N$ is an $(S, T)$ -bimodule. Then there is a natural bijection for every $(R, T)$ -bimodule $P$ between the set of $(R, T)$ -bimodule morphisms $M \otimes_{S} N \to P$ , with the set of $S$ -balanced bilinear $(R, T)$ -maps $M \times N \to P$ .

How does this bijection work? First suppose $f : M \otimes_{S} N \to P$ is an $(R, T)$ -bimodule morphism. Define a set map $g : M \times N \to P$ by $g (m, n) = f (m \otimes n)$ . We claim this is an $S$ -balanced bilinear $(R, T)$ -map. That's a lot to unpack, so let's start verifying properties.

First observe that, using the fact that $f$ is compatible with the additive operation in $M$ , for every $m_{1}, m_{2} \in M$ and $n \in N$ we have

\begin{aligned} g (m_{1} + m_{2}, n) & = f ((m_{1} + m_{2}) \otimes n) \\ = f (m_{1} \otimes n + m_{2} \otimes n) \\ = f (m_{1} \otimes n) + f (m_{2} \otimes n) \\ = g (m_{1}, n) + g (m_{2}, n) . \end{aligned}

The similar argument shows that $g (m, n_{1} + n_{2}) = g (m, n_{1}) + g (m, n_{2})$ for every $m \in M$ and $n_{1}, n_{2} \in N$ . So $g$ is a bilinear map from $M \times N$ .

Next observe that since $f$ is compatible with the left $R$ -action on $M \otimes_{S} N$ , for every $r \in R$ , $m \in M$ , and $n \in N$ we have

g (r m, n) = f (r m \otimes n) = f (r \cdot (m \otimes n)) = r \cdot f (m \otimes n) = r \cdot g (m, n) .

So $g$ is compatible with the left $R$ -action on $M \times N$ . Similarly, $g (m, n t) = g (m, n) \cdot t$ for every $t \in T$ , $m \in M$ , and $n \in N$ , and so $g$ is compatible with the right $T$ -action on $M \times N$ .

Finally, observe that for every $s \in S$ , $m \in M$ , and $n \in N$ we have

g (m s, n) = f (m s \otimes n) = f (m \otimes s n) = g (m, s n) .

Thus, the map $g$ is $S$ -balanced. We've thus verified that our map $g : M \times N \to P$ is indeed an $S$ -balanced $(R, T)$ -map.

Let's try the inverse direction. Suppose $g : M \times N \to P$ is an $S$ -balanced bilinear $(R, T)$ -map. By the universal property of the free $Z$ -module $F (U (M) \times U (N))$ , and since $U (M \times N) ≃ U (M) \times U (N)$ , we immediately have a corresponding $Z$ -module morphism $\tilde{g} : F (U (M) \times U (N)) \to P$ , which is really nothing more than the phrase "extend linearly" applied to $g$ . Moreover, the generators for the subgroup $H$ all lie in the kernel of $\tilde{g}$ , since $g$ is an $S$ -balanced bilinear $(R, T)$ -map. It follows that $\tilde{g}$ factors through the quotient $F (U (M) \times U (N)) / H = M \otimes_{S} N$ , and this is our desired $(R, T)$ -bimodule morphism $f : M \otimes_{S} N \to P$ .

Categorical interpretation

For each $(R, T)$ -bimodule $P$ , let $F (P)$ denote the set of bilinear, $S$ -balanced $(R, T)$ -set maps $g : U_{1} (M) \times U_{2} (N) \to U_{3} (P)$ , where the $U_{i}$ are the appropriate forgetful functors to $Set$ . One can check that $F$ is the object function of a functor $F : (R, T) -Bimod \to Set$ . The universal property above can then be viewed as a natural isomorphism ${Hom}_{(R, T)} (M \otimes_{S} N, -) ≃ F$ . In view of Universal Properties III - Yoneda's Lemma, this characterizes the bimodule $M \otimes_{S} N$ uniquely up to unique isomorphism.

Additional properties of the tensor product

Our universal property of the tensor product can be used to prove many properties, including many similar to basic properties of conventional multiplication, at least when commutative rings are involved. To that end, suppose $R$ , $S$ , and $T$ are commutative rings (with unity).

(Proofs will be added at some point, but for now we will simply state (and allow ourselves to use) each property.)

Identity

Proposition

Suppose $M$ is an $(R, S)$ -bimodule. If we consider the ring $R$ with its usual $(R, R)$ -bimodule structure, then there is a isomorphism of $(R, S)$ -modules

R \otimes_{R} M \tilde{\to} M,

given specifically on simple tensors by $r \otimes m \mapsto r m$ .

Associativity

Proposition

Suppose $M$ , $N$ , and $P$ are $(R, S)$ -, $(S, T)$ -, and $(T, U)$ -bimodules, respectively. Then there is an isomorphism of $(R, U)$ -bimodules

(M \otimes_{S} N) \otimes_{T} P \tilde{\to} M \otimes_{S} (N \otimes_{T} P),

given specifically on simple tensors by $(m \otimes n) \otimes p \mapsto m \otimes (n \otimes p)$ .

Symmetry

Proposition

Suppose $M$ and $N$ are $R$ -modules and we give them the standard $(R, R)$ -bimodule structure. Then there is an $R$ -module isomorphism

M \otimes_{R} N \tilde{\to} N \otimes_{R} M,

given specifically on simple tensors by $m \otimes n \mapsto n \otimes m$ .

Tensor product with a fraction field

The following fact can be occasionally useful:

Proposition

Suppose $D$ is an integral domain with field of fractions $Q$ . The tensor product $Q \otimes_{D} M$ is closely related to the torsion in the module $M$ , in the following ways:

f $j : M \to Q \otimes_{D} M$ is the morphism given by $m \mapsto 1_{Q} \otimes m$ , then $\ker (j) = Tor (M) .$
For any $D$ -module $M$ , the tensor product $Q \otimes_{D} M$ is zero if and only if $M$ is torsion.
For any $D$ -module $M$ , we have an isomorphism of $D$ -modules $Q \otimes_{D} M ≃ Q \otimes_{D} (M / Tor (M))$ .

Tensor products of morphisms

Suppose $ϕ : M_{1} \to M_{2}$ is an $(R, S)$ -bimodule morphism and $ψ : N_{1} \to N_{2}$ is an $(S, T)$ -bimodule morphism. We can define a map $ϕ \otimes ψ : M_{1} \otimes_{S} N_{1} \to M_{2} \otimes_{S} N_{2}$ by

(ϕ \otimes ψ) (m_{1} \otimes n_{1}) = ϕ (m_{1}) \otimes ψ (n_{1}),

and then extending linearly to all tensors. One can check that this is well defined and an $(R, T)$ -bimodule morphism.

Suggested next note

Tensor Products IV - The Adjoint Property