LEC ALG4 12

Tensor products

See also Bourbaki (1974) ch 3, 4

Definition 418.1(Bilinear map).

Let $M,N,P$ be $R$-modules. A map $\phi :M\times N\to P$ is called $R$-bilinear if $\phi (x,-):N\to P$ and $\phi (-,y):M\to P$ are $R$-linear for all $x\in M$ and $y\in N$.

Note that $\phi$ itself is not linear.

Definition 418.2(Universal property of tensor products).

Let $M,N$ be $R$-modules. We call a pair $(T,\tau )$ the tensor product of $M$ and $N$, where $T$ is an $R$-module and $\tau :M\times N\to T$ is $R$-bilinear, if for all bilinear $\phi :M\times N\to P$, there exists unique $R$-linear $\overline{\phi }:T\to P$ such that the following diagram commutes.

We denote $\tau (x,y)$ by $x\otimes y$.

2dca86

I do not know where you got the idea from, but $\tau$ is not an inclusion!

Definition 418.3(Definition).

Let $M$ and $N$ be $R$-modules. Denote by $M{\otimes }_{R}N$ the quotient module of the free $R$-module $F^R(M\times N)=R^{\oplus (M\times N)}$ with the set $M\times N$ as a basis modulo the submodule $K$ generated by

$$\begin{array}{rl}& \{e_{(x+x^{\prime },y)}-e_{(x,y)}-e_{(x^{\prime },y)}:x,x^{\prime }\in M,y\in N\}\\ \cup & \{e_{(x,y+y^{\prime })}-e_{(x,y)}-e_{(x,y^{\prime })}:x\in M,y,y^{\prime }\in N\}\\ \cup & \{e_{(rx,y)}-r{e}_{(x,y)}:r\in R,x\in M,y\in N\}\\ \cup & \{e_{(x,ry)}-r{e}_{(x,y)}:r\in R,x\in M,y\in N\}.\end{array}$$

Endow it with the map $\otimes :M\times N\to M{\otimes }_{R}N$ obtained by composing the set map $j:M\times N\to F^R(M\times N)$ with the natural projection $\pi$ onto $M{\otimes }_{R}N$. The element $\otimes (m,n)$ (that is, the class of $j(m,n)$ modulo $K$) is denoted $m\otimes n$.

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Proposition 418.4(Proposition).

Let $M$ and $N$ be $R$-modules. $M{\otimes }_{R}N$ from Def 3 satisfies the universal property in Def 2.

Proof.

It is evident that $(m,n)\to m\otimes n$ defined an $R$-bilinear map.

If $\phi :M\times N\to P$ is any $R$-linear map, we have a unique induced $R$-linear map $\tilde{\phi }$ from the free $R$-module, by the universal property of the latter.

It is easily seen that $K$ is contained in $\ker \tilde{\phi }$. For instance,

$$\begin{array}{rl}\tilde{\phi }(e_{(x+x^{\prime },y)}-e_{(x,y)}-e_{(x^{\prime },y)})& =\tilde{\phi }(e_{(x+x^{\prime },y)})-\tilde{\phi }(e_{(x,y)})-\tilde{\phi }(e_{(x^{\prime },y)})\\ & =\phi (x+x^{\prime },y)-\phi (x,y)-\phi (x^{\prime },y)\\ & =0.\end{array}$$

By the universal property of quotients, there exists a unique $R$-module homomorphism $\overline{\phi }:M{\otimes }_{R}N\to P$ such that $\overline{\phi }\circ \pi =\tilde{\phi }$.

How do we know that $\overline{\phi }$ is unique? Well, suppose $\psi$ is another map such that $\psi \circ \otimes =\phi$. Since $\otimes =\pi \circ j$, we have $\phi =\psi \circ (\pi \circ j)=(\psi \circ \pi )\circ j$ . This forces $\psi \circ \pi =\tilde{\phi }$, which then forces $\psi =\overline{\phi }$!□

Warning

Elements of the form $m\otimes n$ are called pure tensors. Not every element of the tensor product is a pure tensor! Pure tensors are nevertheless very useful, as a set of generators for the tensor product. For example, if two homomorphisms $\alpha ,\beta :M{\otimes }_{R}N\to P$ coincide on pure tensors, then $\alpha =\beta$.

Basic properties

Proposition 418.5(Proposition).

Let $M$ and $N$ be $R$-modules. If $\{x_{\lambda }\in M:\lambda \in \Lambda \}$ generates $M$ and $\{y_{\gamma }:\gamma \in \Gamma \}$ generates $N$ then $\{x_{\lambda }\otimes y_{\gamma }:\lambda \in \Lambda ,\gamma \in \Gamma \}$ generates $M{\otimes }_{R}N$.

Proposition 418.6(Proposition).

Let $M_1,M_2,N_1,N_2$ be $R$-modules. Let $\phi :M_1\to M_2$ be $R$-linear. Then there exists an $R$-linear map

$$\begin{array}{cccc}\phi \otimes {\text{id}}_N:& M_1{\otimes }_{R}N& \to & M_2{\otimes }_{R}N,\\ & x\otimes y& \mapsto & \phi (x)\otimes y.\end{array}$$

More generally, if $\psi :N_1\to N_2$ is $R$-linear then there exists $R$-linear map

$$\begin{array}{cccc}\phi \otimes \psi :& M_1{\otimes }_R{N}_1& \to & M_2{\otimes }_R{N}_2,\\ & x\otimes y& \mapsto & \phi (x)\otimes \psi (y).\end{array}$$

Proof.

It is easy to verify that the map $\Phi :M_1\times N_1\to M_2{\otimes }_R{N}_2$ defined by $(x,y)\mapsto \phi (x)\otimes \psi (y)$ is bilinear. The universal property then yields the required $R$-linear map.

□

Since the diagram

clearly commutes, it is evident that

Proposition 418.7(Proposition).

$\_{\otimes }_{R}N:R\text{-Mod}\to R\text{-Mod}$ is a functor; it is defined by $M\mapsto M{\otimes }_{R}N$ and $\phi \mapsto \phi {\otimes }_R{\mathrm{id}}_N$.

Proposition 418.8(The Hom-Tensor adjunction).

Consider the three functors $(R\text{-Mod}{)}^{op}\times (R\text{-Mod}{)}^{op}\times R\text{-Mod}\to R\text{-Mod}$ defined by

$$\begin{array}{r}(M,N,P)\mapsto \{\begin{array}{l}{\text{Hom}}_{R\text{-Mod}}(M,{\text{Hom}}_{R\text{-Mod}}(N,P))\\ {\text{Bi-linear}}_R(M\times N,P)\\ {\text{Hom}}_{R\text{-Mod}}(M{\otimes }_{R}N,P).\end{array}\end{array}$$

These are naturally isomorphic. In particular, if we consider the covariant functors $R\text{-Mod}\to R\text{-Mod}$ given by $\mathcal{F}={\text{Hom}}_R(N,\_)$ and $\mathcal{G}=\_{\otimes }_{R}N$, we have

$${\text{Hom}}_R(M,\mathcal{F}(P))\cong {\text{Hom}}_R(\mathcal{G}(M),P),$$

i.e, $\mathcal{F}$ and $\mathcal{G}$ are adjoint. In words, the tensor product is left adjoint to $\text{Hom}$.

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Proposition 418.9(Proposition).

The map $M{\otimes }_{R}N\to N{\otimes }_{R}M$ defined on pure tensors by $x\otimes y\mapsto y\otimes x$ is an $R$-module isomorphism.

Proof.

The map $M\times N\to N{\otimes }_{R}M$ defined by $(x,y)\mapsto y\otimes x$ is $R$-bilinear. It is easily seen that the induced map is bijective.□

Proposition 418.10(Proposition).

The tensor product is associative: $(M_1{\otimes }_R{M}_2){\otimes }_R{M}_3\cong M_1{\otimes }_R(M_2{\otimes }_R{M}_3)$.

Proof.

For fixed $x\in X$, define a bilinear map $Y\times Z\to (X\otimes Y)\otimes Z$ by

$$(y,z)\mapsto (x\otimes y)\otimes z.$$

The universal property induces a linear map $A_x:Y\otimes Z\to (X\otimes Y)\otimes Z$ such that $A_x(y\otimes z)=(x\otimes y)\otimes z$ for every $y\in Y$, $z\in Z$. Next define the bilinear map $X\times (Y\otimes Z)\to (X\otimes Y)\otimes Z$ by

$$(x,\sum _{i=1}^n{y}_i\otimes z_i)\mapsto A_x(\sum _{i=1}^n{y}_i\otimes z_i).$$

Pass this map through the universal property to obtain a $R$-linear map $X\otimes (Y\otimes Z)\to (X\otimes Y)\otimes Z$ which maps pure tensors as $x\otimes (y\otimes z)\mapsto (x\otimes y)\otimes z$. We can repeat the same construction to obtain an inverse $(X\otimes Y)\otimes Z\to X\otimes (Y\otimes Z)$.□

Remark 418.11(Universal property of$n$-fold tensors).

Thus, we can talk of $M_1\otimes \cdots \otimes M_n$ for any $n$. The universal property of tensor products extends to $n$-linear maps in the obvious way. We’ll work out the details for threefold products; higher products are handled by induction. Suppose we are given a trilinear map $\phi :M_1\times M_2\times M_3\to P$. For each $z\in M_3$, $\phi (-,-,z):M_1\times M_2\to P$ is a bilinear map. The universal property gives unique linear maps ${\phi }_z:M_1\otimes M_2\to P$ for each $z\in M_3$. Using these, we can define a bilinear map $\psi :(M_1\otimes M_2)\times M_3\to P$ by

$$\left(\sum _{i=1}^n{x}_i\otimes y_i,z\right)\mapsto {\phi }_z\left(\sum _{i=1}^n{x}_i\otimes y_i\right)=\sum _{i=1}^n\phi (x_i,y_i,z).$$

The universal property gives a unique linear map $M_1\otimes M_2\otimes M_3\to P$.

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Proposition 418.12(Proposition).

The tensor product distributes over direct sums, i.e,
$M{\otimes }_R(N_1\oplus N_2)=M{\otimes }_R{N}_1\oplus M{\otimes }_R{N}_2$.

Proof.

Consider the exact sequence

$$0\to N_1\stackrel{i_1}{\hookrightarrow }{N}_1\oplus N_2\stackrel{{\pi }_2}{\twoheadrightarrow }{N}_2\to 0.$$

Since tensor is right exact, we get that

$$M\otimes N_1\to M\otimes (N_1\oplus N_2)\to M\otimes N_2\to 0$$

is exact; we need to show that ${\mathrm{id}}_M\otimes i_1$ is injective, since tensor products usually do not preserve that. This follows from functoriality: ${\pi }_1\circ i_1={\mathrm{id}}_{N_1}$, so $({\mathrm{id}}_M\otimes {\pi }_1)\circ ({\mathrm{id}}_M\otimes i_1)={\mathrm{id}}_{M\otimes N_1}$, so ${\mathrm{id}}_M\otimes i_1$ must be injective. Thus,

$$0\to M\otimes N_1\to M\otimes (N_1\oplus N_2)\to M\otimes N_2\to 0$$

is exact; furthermore, while proving injectivity, we have also shown than ${\mathrm{id}}_M\otimes i_1$ has a left inverse. Thus, the above sequence is a split exact sequence, and we are done.□

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Remark 418.13(Remark).

Arguing as in the proof of Prp 12, we can show that ==tensor preserves split exact sequences==.

Proposition 418.14(Tensor is right exact).

Let $\phi :M_1\to M_2$ be $R$-linear. Then $\text{coker}\phi \otimes N\cong \text{coker}(\phi \otimes {\text{id}}_N)$.

In other words, given an exact sequence

$$M_1\stackrel{\phi }{\hookrightarrow }{M}_2\stackrel{ϵ}{\twoheadrightarrow }{M}_3\to 0,$$

the following sequence is also exact:

$$M_1\otimes N\stackrel{\phi \otimes {\mathrm{id}}_N}{\to }{M}_2\otimes N\stackrel{ϵ\otimes {\mathrm{id}}_N}{\to }{M}_3\otimes N\to 0.$$

Equivalently, the functor $\_{\otimes }_{R}N$ is right exact.

Proof.

1. Since $ϵ$ is surjective, it is easy to see from the definition of $ϵ\otimes {\mathrm{id}}_N$ that it is surjective too.
2. Since $\phi ϵ=0$, functoriality yields $\phi \otimes {\mathrm{id}}_Nϵ\otimes {\mathrm{id}}_N=0$, so $\mathrm{im}(\phi \otimes {\mathrm{id}}_N)\subseteq \ker (ϵ\otimes {\mathrm{id}}_N)$.

We need to show that the containment above is in fact an equality.

The mapping property of quotient modules gives us $\theta$. Since $ϵ\otimes {\mathrm{id}}_N$ is surjective, $\lambda$ is an isomorphism. We contend that $\theta \lambda$ is an isomorphism; we will show this by constructing an inverse.

First, suppose that $p_1,p_2\in M_2$ are such that $ϵ(p_1)=ϵ(p_2)$. Then,

$$\begin{array}{rlr}& ϵ(p_1-p_2)=0& \\ ⟹& p_1-p_2\in \mathrm{im}\phi & \\ ⟹& (p_1-p_2)\otimes n\in \mathrm{im}(\phi {\otimes }_R{\mathrm{id}}_N)& \forall n\in N\\ ⟹& \pi ((p_1-p_2)\otimes n)=0& \forall n\in N\\ ⟹& \pi (p_1\otimes n)=\pi (p_2\otimes n)& \forall n\in N\end{array}$$

Thus, $\psi$ maps the entire set $(ϵ\times {\mathrm{id}}_N{)}^{-1}(p,n)$ to a single element; we can thus define $\nu$ to be the composite $(ϵ\times {\mathrm{id}}_N{)}^{-1}\psi$. It is clear that $\nu$ is linear in the second slot. It is also linear in the first slot: if $(r{p}_1+p_2,n)\in M_3\times N$ and $q_1,q_2\in M_2$ are such that $ϵ(q_1)=p_1$ and $ϵ(q_2)=p_2$, we have

$$\begin{array}{r}\nu (r{p}_1+p_2,n)=\pi ((r{q}_1+q_2)\otimes n)=r\pi (q_1\otimes n)+\pi (q_2\otimes n)=\nu (p_1,n)+\nu (p_2,n).\end{array}$$

Thus, $\nu$ is bilinear; let $\overline{\nu }$ be the map induced by the universal property.

We contend that $\overline{\nu }=(\theta \lambda {)}^{-1}$. Indeed, let $p\otimes n\in M_3\otimes N$. Let $q\in M_2$ be such that $ϵ(q)=p$. Then, $\overline{\nu }(p\otimes n)=\pi (q\otimes n)$. Now,

$$\begin{array}{rl}\lambda \circ \theta (\pi (q\otimes n))& =\lambda \circ \theta (q\otimes n+\mathrm{im}(\phi \otimes {\mathrm{id}}_N))\\ & =\lambda (q\otimes n+\ker (ϵ\otimes {\mathrm{id}}_N))\\ & =ϵ(q)\otimes n\\ & =p\otimes n.\end{array}$$

Similarly, it is evident that $\overline{\nu }(ϵ(q)\otimes n)=\pi (q\otimes n)$. We are done.□

Proof.

We know that tensor is left adjoint; it thus preserves colimits. In particular, it preserves cokernels.□

Proposition 418.15(Upgrading structure, I).

Let $S$ be an $R$-algebra ($\phi :R\to S$) and $M$ be an $R$-module. The $R$-module structure of $S{\otimes }_{R}M$ can be extended to that of an $S$-module, with the action of $S$ defined by

$$s\left(\sum t_i\otimes m_i\right):=\sum (s{t}_i\otimes m_i).$$

$S{\otimes }_{R}M$ also satisfies the following upgraded universal property: for all $S$-modules $N$ and maps $f:S\times M\to N$ which are $R$-bilinear and $S$-linear in the first slot, there exists a unique $S$-linear map $\tilde{f}:S{\otimes }_{R}M\to N$ which makes the following diagram commute.

Proof.

We know that $S{\otimes }_{R}M$ is an $R$-module; we are attempting to upgrade its structure to that of an $S$-module. To so this, we need to define an $S$-action on $S{\otimes }_{R}M$. Say we define

$$s\left(\sum t_i\otimes m_i\right):=\sum (s{t}_i\otimes m_i).$$

Note that we need to first check that this is well defined: if $\sum t_i\otimes m_i=\sum t_i^{\prime }\otimes m_i$, then we must have $\sum s{t}_i\otimes m_i=\sum s{t}_i^{\prime }\otimes m_i$. It suffices to show that if $\sum t_i\otimes m_i=0$, then $\sum s{t}_i\otimes m_i=0$. Here’s what we will do: we will show that the map $S{\otimes }_{R}M\to S{\otimes }_{R}M$ defined by its values on pure tensors as $t\otimes m\mapsto st\otimes m$ is well-defined (do you see why this accomplishes our goal?). Start with defining a map $S\times M\to S{\otimes }_{R}M$ by $(t,m)\mapsto st\otimes m$; this is clearly $R$-bilinear. Using the universal property of tensor products in $R\text{-Mod}$, we get a map $S{\otimes }_{R}M\to S{\otimes }_{R}M$ which maps pure tensors as $t\otimes m\mapsto st\otimes m$ - exactly what we want¹!

Okay, so we have a well defined action of $S$ on $S{\otimes }_{R}M$. We are not done yet - we still need to show that this action is an

1. abelian group homomorphism for each $s\in S$, and
2. the map $S\to {\text{End}}_{\text{Ab}}(S{\otimes }_{R}M)$ is a ring homomorphism.

In this case, it is much easier to check that $s(\alpha +\beta )=s\alpha +s\beta$, $(s_1+s_2)\alpha =s_1\alpha +s_2\alpha$, $(s_1{s}_2)\alpha =s_1(s_2\alpha )$. That this $S$-module structure extends the preexisting $R$-module structure on $S{\otimes }_{R}M$ is evident.

As for the upgraded universal property, we first use the universal property of tensor products in $R\text{-Mod}$ (recall: $S$-modules can be seen as $R$-modules, $S$-linear maps as $R$-linear maps) to get an $R$-linear map $\tilde{f}$. We then show that $\tilde{f}$ is actually $S$-linear; this easily follows from the hypothesis that $f$ is $S$-linear in the first slot.□

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Corollary 418.16(Corollary).

Suppose that

$$\begin{array}{r}{R}^m\stackrel{[a_{ij}]}{\to }{R}^n\to M\to 0\end{array}$$

is exact; then

$$S^m\stackrel{[\phi (a_{ij})]}{\to }{S}^n\to S{\otimes }_{R}M\to 0$$

is exact.

Proposition 418.17(Upgrading structure, II).

Let $A,B$ be $R$-algebras: there exist ring maps $\alpha :R\to A$, $\beta :R\to B$, and $\phi \in {\text{Hom}}_{R\text{-Alg}}(A,B)$ means $\phi :A\to B$ is a ring homomorphism such that

commutes. We’ve seen that $A{\otimes }_{R}B\in R\text{-Mod},A\text{-Mod},B\text{-Mod}$. We can upgrade it to an $R$-algebra: the multiplication $(a_1\otimes b_1)\cdot (a_2\otimes b_2):=a_1{a}_2\otimes b_1{b}_2$ makes it into a ring, and the map $R\to A{\otimes }_{R}B$ given by

$$r\mapsto \alpha (r)\otimes 1_B=(r\cdot 1_A)\otimes 1_B=1_A\otimes (r\cdot 1_B)=1_A\otimes \beta (r).$$

Proof.

To show that the multiplication $(A{\otimes }_{R}B)\times (A{\otimes }_{R}B)\to A{\otimes }_{R}B$ defined above is indeed well defined, by use the universal property from Rmk 11:

It remains to check that $m$ and $+$ make $A{\otimes }_{R}B$ into a ring□

Proposition 418.18(Proposition).

$A{\otimes }_{R}B$, along with the $R$-algebra maps² $A\to A{\otimes }_{R}B$, $a\mapsto a\otimes 1$ and $B\to A{\otimes }_{R}B$, $b\mapsto 1\otimes b$, is the coproduct of $A$ and $B$ in $R\text{-Alg}$.

Proof.

It is clear that the coproduct maps specified are indeed $R$-algebra maps.

We have to show that $A{\otimes }_{R}B$ satisfies the universal property of coproducts in $R\text{-Alg}$. Suppose we have $R$-algebra maps $\phi :A\to C$ and $\psi :B\to C$. Suppose $\theta :A{\otimes }_{R}B\to C$ is a map which makes the diagram commute. Then, commutativity of the diagram forces $\theta$ to map $a\otimes 1\mapsto \phi (a)$ and $1\otimes b\mapsto \psi (b)$, and thus $\theta$ is determined on all pure tensors (since $a\otimes b=(a\otimes 1)(1\otimes b)$), and hence on $A{\otimes }_{R}B$. Of course, none of this means anything till we show that $\theta$ is well-defined; we proceed as before by defining a suitable bilinear map and using the universal property of tensors in $R\text{-Mod}$.

Thus, $\theta$ does indeed exist, and is unique. However, the universal property only tells us that $\theta$ is an $R$-module homomorphism, so we have to show that $\theta$ is actually an $R$-algebra homomorphism. This is an easy check.□

Footnotes

1. This is a common theme: to construct maps out of a tensor product, you first define a map from the cartesian product which when passed though the universal property would give you your desired map; then, you show that this map is bilinear, which makes the universal property applicable. ↩

2. Note that the universal property of coproducts does NOT require the maps into the coproduct to be injective - indeed, this is not the case in $R\text{-Alg}$. For example, consider $\mathbb{Z}/4\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/6\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}$ - there’s no way to have an injection from $\mathbb{Z}/4\mathbb{Z}$ to $\mathbb{Z}/2\mathbb{Z}$! ↩

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References

Bourbaki, N. (1974). Algebra I. Hermann.