Introduction to tensor 1: Basic property of tensor product

If you don't know what is $R-$$R-$ module, then just consider the $\mathbb{R}$$\mathbb R$ or $\mathbb{C}-$$\mathbb C-$ vector space.

We will introduce two versions of the tensor product, and we will have more interest in the bilinear tensor.

The blog does not involve any commutative diagrams, readers should draw them by their hands.

 

Balanced product and Bilinear map

Balanced product

Definition 1.1

Let $R$$R$ be a ring, consider the module $M_R{,}_{R}N$$M_R,_RN$ and abelian group $(A,+)$$(A,+)$, if a map

$$\begin{array}{}\text{(1)}& B:M\times N\to A\end{array}$$

satisfies the following conditions, we will call $B$$B$ balanced product.

$†.$$\dagger.$ $B(x+x^{\prime },y)=B(x,y)+B(x^{\prime },y),B(x,y+y^{\prime })=B(x,y)+B(y,y^{\prime })$$B(x+x',y)=B(x,y)+B(x',y),B(x,y+y')=B(x,y)+B(y,y')$

$†.B(xr,y)=B(x,ry)$$\dagger. B(xr,y)=B(x,ry)$

Where $x,x^{\prime }\in M,y,y^{\prime }\in N,r\in R$$x,x'\in M,y,y'\in N, r\in R$.

Proposition 1.2

All the balanced product $B:M\times N\to A$$B:M\times N\to A$ form an abelian group $\mathfrak{A}$$\mathfrak A$.

Proof.

Let $B_{,}{B}^{\prime }\in \mathfrak{A}$$B_,B'\in\mathfrak A$, $(B+B^{\prime })(x+x^{\prime },y)=B(x+x^{\prime },y)+B(x+x^{\prime },y)=B(x,y)+B(x^{\prime },y)+B^{\prime }(x,y)+B^{\prime }(x^{\prime },y)$$(B+B')(x+x',y)=B(x+x',y)+B(x+x',y)=B(x,y)+B(x',y)+B'(x,y)+B'(x',y)$

Using the fact that $A$$A$ is an abelian group we get

$$\begin{array}{}\text{(2)}& B(x,y)+B(x^{\prime },y)+B^{\prime }(x,y)+B^{\prime }(x^{\prime },y)=(B+B^{\prime })(x,y)+(B+B^{\prime })(x^{\prime },y)\end{array}$$

Similarly for the $y$$y$ side.

To see $(B+B{)}^{\prime }(xr,y)=(B+B^{\prime })(x,ry)$$(B+B)'(xr,y)=(B+B')(x,ry)$

$$\begin{array}{}\text{(3)}& (B+B{)}^{\prime }(xr,y)=B(xr,y)+B^{\prime }(rx,y)=B(x,ry)+B^{\prime }(x,ry)=(B+B^{\prime })(x,ry)\end{array}$$

It is easy to see that $nB\in \mathfrak{A}$$n B\in\mathfrak A$. $◻$$\square$

The balanced product gives us a functor $\mathrm{Bal}(M\times N:-):\mathrm{Ab}\to \mathrm{Ab}$$\mathrm{Bal}(M\times N:-):\mathrm{Ab}\to\mathrm{Ab}$

For the morphism, Let $f:A\to A^{\prime }$$f:A\to A'$ be a group homomorphism, then

$$\begin{array}{}\text{(4)}& \mathrm{Bal}(M\times N:f)=f_{\ast },f_{\ast }(B)=f\circ B=B^{\prime }\end{array}$$

Here $\mathrm{Bal}(M\times N:\ast )$$\mathrm{Bal}(M\times N:*)$ define a new category.

The object is balanced map $M\times N\to A$$M\times N\to A$. Morphism is $f_{\ast }$$f_*$.

Example of balanced map.

Let $T$$T$ be a self-adjoint linear map over a $\mathbb{R}$$\mathbb{R}$ vector space $V$$V$. Let $\mathbb{R}[T]$$\mathbb R[T]$ act on $V$$V$, we get a $\mathbb{R}[T]-$$\mathbb R[T]-$ module.

Let $p(T)\in \mathbb{R}[T]$$p(T)\in\mathbb R[T]$, $p(T{)}^{\ast }=p(T^{\ast })=p(T)$$p(T)^*=p(T^*)=p(T)$. Then the inner product $⟨-,-⟩$$\langle -,-\rangle$ is a balanced product over the $\mathbb{R}[T]-$$\mathbb{R}[T]-$ module.

Since

$$\begin{array}{}\text{(5)}& ⟨p(T)v,w⟩=⟨v,p(T)w⟩\end{array}$$

Bilinear map

Definition 1.3

Let $R$$R$ be a commutative ring, consider three $R-$$R-$ modules $M,N,A$$M,N,A$.

If a map $\mathfrak{B}:M\times N\to A$$\mathfrak B:M\times N\to A$ satisefies the following conditions:

$†.$$\dagger.$ $\mathfrak{B}(x+x^{\prime },y)=\mathfrak{B}(x,y)+\mathfrak{B}(x^{\prime },y),\mathfrak{B}(x,y+y^{\prime })=\mathfrak{B}(x,y)+\mathfrak{B}(x,y^{\prime })$$\mathfrak B(x+x',y)=\mathfrak B(x,y)+\mathfrak B(x',y),\mathfrak B(x,y+y')=\mathfrak B(x,y)+\mathfrak B(x,y')$

$†.$$\dagger.$ $\mathfrak{B}(rx,y)=\mathfrak{B}(x,ry)=r\mathfrak{B}(x,y)$$\mathfrak B(rx,y)=\mathfrak B(x,ry)=r\mathfrak B(x,y)$

Then we call $\mathfrak{B}$$\mathfrak B$ a bilinear map.

Proposition 1.4. All the bilinear map from $M\times N\to A$$M\times N\to A$ , $\mathrm{Bil}(M\times N,A)$$\mathrm{Bil}(M\times N,A)$ is a $R-$$R-$module.

Proof. Same things as balanced map.

Proposition 1.5. $\mathrm{Bil}(M\times N,-):R-$$\mathrm{Bil}(M\times N,-):R-$ module to $R-$$R-$ module is a functor.

Proof. same.

Tensor product for balanced map

Definition 2.1.

The tensor product $M\times N⟶M{\otimes }_{R}N$$M\times N\longrightarrow M\otimes_{R} N$ is the initial object of this category.

Sometimes we will omit the arrow $⟶$$\longrightarrow$ and $R$$R$, just say that $M\otimes N$$M\otimes N$ is the tensor product.

Remark.

The sturcture of $M\otimes N$$M\otimes N$ is not funny. We always need to consider the arrow when we deal with the property of tensor product.

Proposition 2.2 For any $M_R{,}_{R}N$$M_R,_RN$, $M\otimes N$$M\otimes N$ exsist.

Let us consider the free $\mathbb{Z}-$$\mathbb Z-$ module over $M\times N$$M\times N$, i.e. $F$$F$

The elements

$$\begin{array}{}\text{(6)}& \begin{array}{c}(x+x^{\prime },y)-(x,y)-(x^{\prime },y)\\ (x,y+y^{\prime })-(x,y)-(x,y^{\prime })\\ (xr,y)-(x,ry)\end{array}\end{array}$$

will generate a sub $\mathbb{Z}-$$\mathbb Z-$ module $I$$I$.

The map

$$\begin{array}{}\text{(7)}& M\times N⟶\frac{F}{I}\end{array}$$

satisfies the universal property of $M\times N⟶M\otimes N$$M\times N\longrightarrow M\otimes N$.

Since $B$$B$ is a balanced map if and only if $B$$B$ vanish at $I$$I$.

Tensor product for bilinear map

Proposition 2.3. $\mathrm{Bil}(M\times N,\ast )$$\mathrm{Bil}(M\times N,*)$ define a category. Here the object is $M\times N\to A$$M\times N\to A$, morphism is $f_{\ast }$$f_*$

Definition 2.4. Tensor product of $M,N$$M,N$ is the initial object in $\mathrm{Bil}(M\times N,\ast )$$\mathrm{Bil}(M\times N,*)$.

Proposition 2.5. Tensor product of $M,N$$M,N$ exists for any $R-$$R-$ module.

Proof. Consider the free $R$$R$ module of $M\times N$$M\times N$ , $F$$F$ and consider the submodule generated by

$$\begin{array}{}\text{(8)}& \begin{array}{c}(x+x^{\prime },y)-(x,y)-(x^{\prime },y)\\ (x,y+y^{\prime })-(x,y)-(x,y^{\prime })\\ (rx,y)-r(x,y),(x,ry)-r(x,y)\end{array}\end{array}$$

That is, $I$$I$.

Then $F/I$$F/I$ satisfies the universal property of the tensor product.

Since $\mathfrak{B}$$\mathfrak B$ is bilinear if and only if $\mathfrak{B}$$\mathfrak B$ vanishing at $I$$I$.

Bifunctor

Let $\mathrm{C},D$$\mathrm C,D$ be two categories , bifunctor is a functor $(-)\otimes (-):\mathrm{C}\times \mathrm{C}\to D$$(-)\otimes(-) :\mathrm{C}\times\mathrm{C}\to D$.

In particular, we are interested in the case $D=C$$D=C$.

Example. 3.1

Let $\mathrm{C}$$\mathrm{C}$ be a category that finite product/coproduct exists. Then $(-)\times (-):\mathrm{C}\times \mathrm{C}\to \mathrm{C}$$(-)\times(-):\mathrm{C}\times\mathrm{C}\to\mathrm{C}$ is a bifunctor.

For example, let $\mathrm{Prop}$$\mathrm{Prop}$ be the category of propositional logic,

The object is the proposition and the morphism is $⟹$$\implies$.

Then $\wedge ,\vee$$\wedge,\vee$ are bifunctors. $\wedge :(p,q)\to p\wedge q,\vee :(p,q)\to p\vee q$$\wedge:(p,q)\to p\wedge q,\vee:(p,q)\to p\vee q$

For the morphism, $\wedge (p⟹p^{\prime },q⟹q^{\prime }):=p\wedge q⟹p^{\prime }\wedge q^{\prime }$$\wedge(p\implies p',q\implies q' ):= p\wedge q\implies p'\wedge q'$

Similarly, $\vee (p⟹p^{\prime },q⟹q^{\prime }):=p\vee q⟹p^{\prime }\vee q^{\prime }$$\vee (p\implies p',q\implies q'):=p\vee q\implies p'\vee q'$.

Consider the groupoid of $\mathrm{Prop}$$\mathrm{Prop}$, the De Morgan Law is a Natural isomorphism between $\wedge$$\wedge$ and $\vee$$\vee$.

Readers could generalise it to lattice. For example, the divisor of $n$$n$, $n\downarrow$$n\downarrow$, and consider $gcd,\mathrm{lcm}$$\gcd,\mathrm{lcm}$.

Now let $\mathrm{C}$$\mathrm{C}$ be the category of $R$$R$-module, where $R\in \mathrm{CRing}$$R\in\mathrm{CRing}$.

Proposition 3.2

The product and tensor product define two bifunctors in $R-\mathrm{Mod}$$R-\mathrm{Mod}$.

$\times :(M,N)\to M\times N,(\phi ,\psi )\to \phi \times \psi$$\times:(M,N)\to M\times N,(\varphi,\psi)\to\varphi\times\psi$.

$$\begin{array}{}\text{(9)}& \otimes :(M,N)\to M\otimes N,(\phi ,\psi )\to (\phi \otimes \psi )\end{array}$$

Proposition 3.3

Then the balanced/bilinear map

$$\begin{array}{}\text{(10)}& M\times N⟶M\otimes N\end{array}$$

Define a natural transformation $\eta :(-)\times (-)⟶(-)\otimes (-)$$\eta:(-)\times(-)\longrightarrow (-)\otimes(-)$.

Proof.

Consider $\phi :M⟶M^{\prime },\psi :N⟶N^{\prime }$$\varphi:M\longrightarrow M',\psi:N\longrightarrow N'$.

Then $B:M\times N⟶M^{\prime }\times N^{\prime }⟶M^{\prime }\otimes N^{\prime }$$B:M\times N\longrightarrow M'\times N'\longrightarrow M'\otimes N'$, $B(m,n)=\phi (m)\otimes \psi (n)$$B(m,n)=\varphi(m)\otimes\psi(n)$ is a balanced/bilinear map.

By universal property of tensor product, $\mathrm{\exists }!\phi \otimes \psi :(M\times N\to M\otimes N)⟶(M\times N\to M^{\prime }\otimes N^{\prime })$$\exists !\varphi\otimes \psi:(M\times N\to M\otimes N)\longrightarrow (M\times N\to M'\otimes N')$ makes the diagram commute. $◻$$\square$

The diagram commute tell us that $\phi \otimes \psi (m\otimes n)=\phi (m)\otimes \psi (n)$$\varphi\otimes\psi(m\otimes n)=\varphi(m)\otimes\psi(n)$.

Hence $(\phi \otimes \psi )\circ ({\phi }^{\prime }\otimes {\psi }^{\prime })=(\phi \circ {\phi }^{\prime }\otimes \psi \circ {\psi }^{\prime })$$(\varphi\otimes\psi)\circ(\varphi'\otimes\psi')=(\varphi\circ \varphi'\otimes\psi\circ\psi')$.

Tensor-Hom adjoint for bilinear map

Lemma 4.1 Let $N$$N$ be a R-module, then $(-)\otimes N:R-\mathrm{Mod}\to R-\mathrm{Mod}$$(-)\otimes N:R-\mathrm{Mod}\to R-\mathrm{Mod}$ is a functor.

Proof. It follows from $\otimes$$\otimes$ is a bifunctor directly.

Proposition. 4.2

$$\begin{array}{}\text{(11)}& {\mathrm{Hom}}_{R-\mathrm{Mod}}(M\otimes N,K)\cong {\mathrm{Hom}}_{R-\mathrm{Mod}}(M,{\mathrm{Hom}}_{R-\mathrm{Mod}}(N,K))\end{array}$$

Proof.

Define $\varphi :{\mathrm{Hom}}_{R-\mathrm{Mod}}(M\otimes N,K)⟶{\mathrm{Hom}}_{R-\mathrm{Mod}}(M,{\mathrm{Hom}}_{R-\mathrm{Mod}}(N,K))$$\phi: \mathrm{Hom}_{R-\mathrm{Mod}}(M\otimes N,K)\longrightarrow\mathrm{Hom}_{R-\mathrm{Mod}}(M,\mathrm{Hom}_{R-\mathrm{Mod}}(N,K))$

For any $f:M\otimes N\to K$$f:M\otimes N\to K$, and $x\in M$$x\in M$, $\varphi (f)(x):=f(x\otimes n)$$\phi(f)(x):=f(x\otimes n)$. Here $f(x\otimes n):N\to K$$f(x\otimes n):N\to K$ and $\varphi (f)(x):M\to (N\to K)$$\phi(f)(x):M\to(N\to K)$.

Firstly we need to prove that $\varphi (f)(x)$$\phi(f)(x)$ is a module homomorphism $N\to K$$N\to K$.

$$\begin{array}{}\text{(12)}& f(x\otimes (n+n^{\prime }))=f(x\otimes n+x\otimes n^{\prime })=f(x\otimes n)+f(x\otimes n^{\prime })\end{array}$$

$$\begin{array}{}\text{(13)}& f(x\otimes (rn))=f(r(x\otimes n))=rf(x\otimes n)\end{array}$$

Now we need to check that $\varphi :\varphi :{\mathrm{Hom}}_{R-\mathrm{Mod}}(M\otimes N,K)⟶{\mathrm{Hom}}_{R-\mathrm{Mod}}(M,{\mathrm{Hom}}_{R-\mathrm{Mod}}(N,K))$$\phi: \phi: \mathrm{Hom}_{R-\mathrm{Mod}}(M\otimes N,K)\longrightarrow\mathrm{Hom}_{R-\mathrm{Mod}}(M,\mathrm{Hom}_{R-\mathrm{Mod}}(N,K))$ is a $R-$$R-$ module homomorphism.

$$\begin{array}{}\text{(14)}& \varphi (f+g)(x)(n(=(f+g)(x\otimes n)=f(x\otimes n)+g(x\otimes n)=\varphi (f)(x)(n)+\varphi (g)(x)(n)\end{array}$$

$$\begin{array}{}\text{(15)}& \varphi (rf)(x)(n)=(rf)(x\otimes n)=rf(x\otimes n)=r\varphi (f)(x)(n)\end{array}$$

Then we need to check that $\varphi$$\phi$ is bijective.

Let $\varphi (f)\ne \varphi (g)$$\phi(f)\ne\phi(g)$, then $\mathrm{\exists }x,f(x\otimes n)\ne g(x\otimes n)$$\exists x, f (x\otimes n)\ne g(x\otimes n)$, hence $f\ne g$$f\ne g$.

To see it is surjective, let $h\in {\mathrm{Hom}}_{R-\mathrm{Mod}}(M,{\mathrm{Hom}}_{\mathrm{Ab}}(N,K))$$h\in \mathrm{Hom}_{R-\mathrm{Mod}}(M,\mathrm{Hom}_{\mathrm{Ab}}(N,K))$,

Observe that $h$$h$ is bilinear map since $h$$h$ is a $R-$$R-$ mod homomorphism and $h(x)$$h(x)$ is a $R-$$R-$ module homomorphism.

$$\begin{array}{}\text{(16)}& h(x+x^{\prime })(n)=h(x)(n)+h(x^{\prime })(n),h(x)(n+n^{\prime })=h(x)(n)+h(x)(n^{\prime })\end{array}$$

$$\begin{array}{}\text{(17)}& h(rx)(n)=rh(x)(n)=h(x)(rn)\end{array}$$

By the universal property of tensor product,

$$\begin{array}{}\text{(18)}& \mathrm{\exists }!f,f(x\otimes n)=h(x)(n)\end{array}$$

$$\begin{array}{}\text{(19)}& \varphi (f)=f(x\otimes -)=h(x)(-)◻\end{array}$$

Tensor product as operation

We would like to prove the following result:

$$\begin{array}{}\text{(20)}& (M\otimes N)\otimes P\cong M\otimes (N\otimes P),M\otimes N\cong N\otimes M,\left(\underset{i\in I}{⨁}{M}_i\right)\otimes N=\underset{i\in I}{⨁}{M}_i\otimes N,R\otimes M\cong M\end{array}$$

The assotiative law and commutative law leaves to reader.

For the third one, the left adjoint preserves colimit.

For the last one,

$$\begin{array}{}\text{(22)}& R\otimes M\to M:r\otimes m,M\to R\otimes M:m\to 1\otimes m\end{array}$$

Zero divisor

Proposition.5.2View $\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/n\mathbb{Z}$$\mathbb Z/m\mathbb Z,\mathbb Z/n\mathbb Z$ as two $\mathbb{Z}-$$\mathbb Z-$ module, if $gcd(m,n)=1$$\gcd(m,n)=1$, then $\mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z}=0$$\mathbb Z/m\mathbb Z\otimes\mathbb Z/n\mathbb Z=0$

Since $m\mathbb{Z}+n\mathbb{Z}=\mathbb{Z}$$m\mathbb Z+n\mathbb Z=\mathbb Z$,(by the lattice anti isomorphism is PID ) there exists $t,s\in \mathbb{Z},tm+sn=1$$t,s\in\mathbb Z,tm+sn=1$

Hence $\mathrm{\forall }a\otimes b\in \mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z}$$\forall a\otimes b\in \mathbb Z/m\mathbb Z\otimes\mathbb Z/n\mathbb Z$

$a\otimes b=(tm+sn)a\otimes b=t(ma)\otimes b+s(a\otimes nb)=0\otimes b+a\otimes 0=(0\times 0)\otimes b+a\otimes (0\times 0)=0+0=0$$a\otimes b=(tm+sn)a\otimes b=t(ma)\otimes b+s(a\otimes nb)=0\otimes b+a\otimes 0=(0\times 0)\otimes b+a\otimes(0\times 0)=0+0=0$