tensor product

Motivation

Tensor products are a way to systematically study bilinear maps of modules. They end up having a lot of uses that come from the fact that multilinear functions appear in a lot of places. Tensor products were first introduced to study the following question: Does there exist an $R$-module $T$ such that

$$\left\{\begin{array}{c}R\text{-bilinear maps}\\ M\times N\to P\end{array}\right\}⟷{\text{Hom}}_R(T,P)$$

Thus, tensor products are a way to use what we know of homomorphisms of modules (which are $R$ linear maps) to study multi-linear maps.

Universal Property

A tensor product of $M$ and $N$ is an $R$-module $T$ together with an $R$-bilinear map $M\times N\to T$ which satisfies the following universal property:

For any $R$-bilinear map $f:M\times N\to P$, there exists a unique $R$-linear map $\tilde{f}:T\to P$ which makes the following diagram commute:

Uniqueness

If it exists, $T$ is unique up to unique isomorphism.

Proof

Say $T,\tilde{T}$ are two tensor products. By the universal property of $T$, there exists a unique $R$-module homomorphism $T\to \tilde{T}$ such that the following diagram commutes:

Also,

uniquely. Applying again, the universal property shows that $T\to \tilde{T}$ and $\tilde{T}\to T$ are mutually inverse.

Therefore, we denote the unique tensor product $T$ as $M{\otimes }_{R}N$.

Existence (formal construction)

We can construct an explicit module and bilinear map the satisfies the universal property above.

Let $F$ be the free R-module on the set $M\times N$. Thus, an arbitrary element of $F$ takes the form

$$\sum _{\begin{array}{c}(m,n)\in M\times N\\ \text{finite}\end{array}}{c}_{m,n}(m,n)$$

Let $G\subset F$ be the $R$-submodule generated by the following elements:

1. $(m_1+m_2,n)-(m_1,n)-(m_2,n)$

2. $(m,n_1+n_2)-(m,n_1)-(m,n_2)$

3. $(rm,n)-r(m,n)$ $r\in R$.

4. $(m,rn)-r(m,n)$

The canonical map

$$\begin{array}{rl}i:M\times N& ⟶F/G\\ (m,n)& ⟼[(m,n)]\end{array}$$

is a tensor product

Proof

First, $i$ is bilinear:

$$i(m,n_1+n_2)=[(m,n_1+n_2)]=[(m,n_1)]+[(m,n_2)]$$

by property 2 of $G$. Also,

$$i(m,rn)=[(m,rn)]=r[(m,n)]$$

by property 4. The same idea follows using properties 1 and 3 in the first position.

Now, we just need to show it satisfies the universal property. Let $f:M\times N\to P$ be $R$-bilinear. Define an $R$-linear map

$$\begin{array}{rl}\tilde{f}:F& ⟶P\\ (m,n)& ⟼f(m,n)\end{array}$$

This determines $\tilde{f}$ uniquely based on the universal property of free modules. Note that

$$\begin{array}{rl}& \tilde{f}(m_1+m_2,n)-\tilde{f}(m_1,n)-\tilde{f}(m_2,n)\\ & =f(m_1+m_2,n)-f(m_1,n)-f(m_2,n)\\ & =0\end{array}$$

and

$$\tilde{f}(rm,n)-\tilde{f}(r(m,n))=f(rm,n)-rf(m,n)=0$$

by bilinearity of $f$. This proves that $G$ is in the kernel of $\tilde{f}$. By the universal property of quotients, there exists a unique $R$-linear map $F/G\to P$ such that the following diagram commutes

By the universal properties, $\tilde{f}$ is unique.

Notation

We write $m\otimes n$ for the class of $(m,n)$ in $F/G$. Thus, the canonical map

$$\begin{array}{rl}M\times N& ⟶M{\otimes }_{R}N\\ (m,n)& ⟼m\otimes n.\end{array}$$

The relations 1-4 in $G$ which descend to $M{\otimes }_{R}N$ then read

1. $(m_1+m_2)\otimes n=m_1\otimes n+m_2\otimes n$

2. $m\otimes (n_1+n_2)=m\otimes n_1+m\otimes n_2$

3. $rm\otimes n=r(m\otimes n)=m\otimes rn$

Properties

1. $M{\otimes }_{R}R\simeq M$

2. $M{\otimes }_{R}N\simeq N{\otimes }_{R}M$

3. $(M_1\oplus M_2){\otimes }_{R}N\simeq M_1{\otimes }_{R}N\oplus M_2{\otimes }_{R}N$

4. $R^{\oplus m}{\otimes }_R{R}^{\oplus n}\simeq R^{\oplus mn}$

Tensor products as functors

Let $f:M\to M^{\prime }$ be a morphism of $R$-modules. This induces a morphism of $R$-modules $M{\otimes }_{R}N\to M^{\prime }{\otimes }_{R}N$ as follows

$$M\times N\stackrel{f\times \text{id}}{\to }{M}^{\prime }\times N⟶M^{\prime }{\otimes }_{R}N$$

is bilinear. Hence, by the universal property of the tensor product, there exists a unique $R$-linear map $M{\otimes }_{R}N\to M^{\prime }{\otimes }_{R}N$ such that the following diagram commutes:

Thus, by the universal property, ${\text{id}}_M:M\to M$ induces the identity map

$${\text{id}}_{M{\otimes }_{R}N}:M{\otimes }_{R}N⟶M{\otimes }_{R}N$$

Therefore, for each $R$-module $N$, we get a functor

$$-{\otimes }_{R}N:R\text{-mod}\to R\text{-mod}.$$

We can also define a second functor

$${\text{Hom}}_R(N,-):R\text{-mod}⟶R\text{-mod}.$$

where

Tensor-Hom adjunction

For $R$-modules $M,N,P$, there is a canonical $R$-module isomorphism

$${\text{Hom}}_R(M{\otimes }_{R}N,P)\simeq {\text{Hom}}_R(M,{\text{Hom}}_R(N,P))$$

Proof

Let $f\in {\text{Hom}}_R(M{\otimes }_{R}N,P)$, interpreted as an $R$-bilinear map

$$f:M\times N⟶P.$$

Then $f(m,-)\in {\text{Hom}}_R(N,P)$. This defines a map

$$\begin{array}{rl}\varphi :{\text{Hom}}_R(M{\otimes }_{R}N,P)& ⟶{\text{Hom}}_R(M,{\text{Hom}}_R(N,P))\\ f& ⟼(m\mapsto f(m,-))\end{array}$$

This is $R$-linear.

Similarly, given $g\in {\text{Hom}}_R(M,{\text{Hom}}_R(N,P))$, define an $R$-linear map

$$\begin{array}{rl}M\times N& ⟶P\\ (m,n)& ⟼g(m)n\end{array}$$

This defines a map

$${\text{Hom}}_R(M,{\text{Hom}}_R(N,P))⟶{\text{Hom}}_R(M{\otimes }_{R}N,P)$$

which is inverse to $\varphi$.