Introduction to tensor (5): Tensor product of R-alg(coproduct), polynomial functor and its adjoint

Polynomial functor and its adjointTensor product of R-AlgThe adjoint between extension of scalars via tensor product and restriction of scalars

Polynomial functor and its adjoint

Let $R$$R$ be a commutative ring and $X$$X$ be a set,

$$\begin{array}{}\text{(1)}& R[]:S\to R[S]\end{array}$$

gives us a functor from category of set to category of $R-$$R-$​algebra.

The right adjoint of $R[]$$R[\quad]$ exists, indeed, it is the forgetful functor $F$$F$ from $R-\mathrm{Alg}\to \mathrm{Set}$$R-\mathrm{Alg}\to\mathrm{Set}$.

$$\begin{array}{}\text{(2)}& {\mathrm{Hom}}_{R-\mathrm{Alg}}(R[S],L)\cong {\mathrm{Hom}}_{\mathrm{Set}}(S,F(L))\end{array}$$

Proof.

Let $f:R[S]\to L$$f:R[S]\to L$ be a $R-\mathrm{Alg}$$R-\mathrm{Alg}$ homomorphism, then $f$$f$ is totally determined by the value at $f(s),\mathrm{\forall }s\in S$$f(s),\forall s\in S$​.

Hence we get an injective from ${\mathrm{Hom}}_{R-\mathrm{Alg}}(R[S],L)\to {\mathrm{Hom}}_{\mathrm{Set}}(S,F(L))$$\mathrm{Hom}_{R-\mathrm{Alg}}(R[S],L)\to\mathrm{Hom}_{\mathrm{Set}}(S,F(L))$. To see it is a surjectiion, notice that every function $g:S\to F(L)$$g:S\to F(L)$ gives us a unique $R-$$R-$algebra homomorphism, i.e. evaluation map. The naturalness leaves to readers. $◻$$\square$

Tensor product of R-Alg

Let $L,K$$L,K$ be two $R-$$R-$algebra, the tensor product of $L,K$$L,K$ is defined as $L{\otimes }_{R}K$$L\otimes_R K$.

The multiplication in $L{\otimes }_{R}K$$L\otimes_R K$ is defined as:

$$\begin{array}{}\text{(3)}& (l\otimes k)(l^{\prime }\otimes k^{\prime })=l{l}^{\prime }\otimes k{k}^{\prime }\end{array}$$

The identity is $1_L\otimes 1_K$$1_L\otimes 1_K$.

Proposition. Tensor product of two $R$$R$-algebras is coproduct in category of $R-$$R-$algebra.

Proof.

Let $i_1:L\to L{\otimes }_{R}K,l⟼l\otimes 1_K$$i_1:L\to L\otimes_R K,l\longmapsto l\otimes 1_K$, similarly for $i_2$$i_2$. For any $f_j:X_j\to Y$$f_j:X_j\to Y$, we have $f_j=f\circ i_j$$f_j=f\circ i_j$, i.e.

$$\begin{array}{}\text{(4)}& f(l\otimes 1_K)=f_1(l),f(1_L\otimes k)=f_2(k)\end{array}$$

Extend it to a ring homomorphism, it uniquely define the $f$$f$​

$$\begin{array}{}\text{(5)}& f:L{\otimes }_{R}K\to Y,f(l\otimes k)=f((l\otimes 1_K)(1_L\otimes k))=f_1(l)f_2(k)\end{array}$$

$$\begin{array}{}\text{(6)}& f(l\otimes k+l^{\prime }\otimes k^{\prime })=f(l\otimes k)+f(l^{\prime }\otimes k^{\prime })\end{array}$$

Corollary

Since $R[]$$R[\quad]$ is a left adjoint functor, hence it preserve colimit:

$$\begin{array}{}\text{(7)}& R[X\coprod Y]\cong R[X]{\otimes }_{R}R[Y]\end{array}$$

The isomorphism is given by

$$\begin{array}{}\text{(8)}& h(x^i\otimes y^j)=x^i{y}^j\end{array}$$

Remark. Let $f:L\to Y,g:K\to Y$$f:L\to Y,g:K\to Y$ be two $R$$R$-algebra homomorphism, then the image of the universal map

$$\begin{array}{}\text{(9)}& L{\otimes }_{R}K\to Y,l\otimes k\to f(l)g(k)\end{array}$$

is generated by $f(L),g(K)$$f(L),g(K)$​.

Here $f(L),g(K)$$f(L),g(K)$ is two subalgebra of $Y$$Y$, and the subalgebra $f(L)g(K)$$f(L)g(K)$ could be viewed as a quotient algebra of $L{\otimes }_{R}K$$L\otimes_R K$.

Proposition. $(A/I){\otimes }_R(B/J)\cong A{\otimes }_{R}B/(A{\otimes }_{R}J+I{\otimes }_{R}B)$$(A/I)\otimes_R(B/J)\cong A\otimes_R B/(A\otimes_R J+I\otimes_R B)$​.

Proof. Let us use the universal property of coproduct.

Let $i_1:a+I⟼a\otimes 1_B+(A{\otimes }_{R}J+I{\otimes }_{R}B),i_2:b+J⟼1_A\otimes b+(A{\otimes }_{R}J+I{\otimes }_{R}B)$$i_1:a+I\longmapsto a\otimes 1_B+(A\otimes_R J+I\otimes_R B),i_2:b+J\longmapsto1_A\otimes b+(A\otimes_R J+I\otimes_R B)$

For any $f_1:A/I\to Y,f_2:B/J\to Y$$f_1:A/I\to Y,f_2:B/J\to Y$, and define $f\circ i_1=f_1,f\circ i_2=f_2$$f\circ i_1=f_1,f\circ i_2=f_2$, i.e. $f\circ i_1(a+I)=f_1(a\otimes 1_B+(A{\otimes }_{R}J+I{\otimes }_{R}B)),f\circ i_2(1_A\otimes b+(A{\otimes }_{R}J+I{\otimes }_{R}B))=f_2(b+J)$$f\circ i_1(a+I)=f_1(a\otimes 1_B+(A\otimes_R J+I\otimes_R B)),f\circ i_2(1_A\otimes b+(A\otimes_R J+I\otimes_R B))=f_2(b+J)$.

Hence $f$$f$ is defined as

$$\begin{array}{}\text{(10)}& f(a\otimes b+(A{\otimes }_{R}J+I{\otimes }_{R}B))=f_1(a\otimes 1_B+(A{\otimes }_{R}J+I{\otimes }_{R}B))f_2(1_A\otimes b+(A{\otimes }_{R}J+I{\otimes }_{R}B))\end{array}$$

It is well defined and unique, hence

$$\begin{array}{}\text{(11)}& A{\otimes }_{R}B/(A{\otimes }_{R}J+I{\otimes }_{R}B)\end{array}$$

is coproduct of $(A/I),(B/J)$$(A/I),(B/J)$, hence it is isomorphic to $(A/I){\otimes }_R(B/J)$$(A/I)\otimes_R(B/J)$. $◻$$\square$

Corollary.

Let $p_1,...,p_n\in R[X_1,...,X_n],q_1,....,q_m\in R[Y_1,...,Y_n]$$p_1,...,p_n\in R[X_1,...,X_n],q_1,....,q_m\in R[Y_1,...,Y_n]$, then

$$\begin{array}{}\text{(12)}& R[X_1...X_n]/(p_1,...,p_n){\otimes }_{R}R[Y_1,...,Y_m]/(q_1,...,q_m)\cong R[X_1,...,X_n,Y_1,...,Y_m]/(p_1,...,p_n,q_1,...,q_m)\end{array}$$

The adjoint between extension of scalars via tensor product and restriction of scalars

Let $f:A\to B$$f:A\to B$ be an $R-$$R-$alg, then we can view $B$$B$ as a $A-$$A-$module via $\phi (a)b$$\varphi(a)b$.

For an $A-$$A-$module $M$$M$, we could give an $B-$$B-$module for $M$$M$ by considering $B{\otimes }_{A}M$$B\otimes_{A} M$ via $b\cdot (b^{\prime }\otimes m)=b{b}^{\prime }\otimes m$$b\cdot(b'\otimes m)=bb'\otimes m$

It gives us a functor

$$\begin{array}{}\text{(13)}& f_{!}:\mathrm{Mod}(A)\to \mathrm{Mod}(B)\end{array}$$

$$\begin{array}{}\text{(14)}& f_{!}M=B{\otimes }_{A}M,\text{for an }A-\text{linear map}\phi :M\to N,f_{!}(\phi )(b\otimes m)=b\otimes \phi (m)\end{array}$$

Also, for an $B-$$B-$module $M$$M$, we can define an $A-$$A-$module structure on $M$$M$ via $f(a)m$$f(a)m$.

This gives us a functor

$$\begin{array}{}\text{(15)}& f^{\ast }:\mathrm{Mod}(B)\to \mathrm{Mod}(A)\end{array}$$

Propostion. $f_{!}$$f_!$ is the left adjoint of $f^{\ast }$$f^*$.

Proof.

We need to prove:

$$\begin{array}{}\text{(16)}& {\mathrm{Hom}}_{\mathrm{Mod}(B)}(f_{!}M,N)\cong {\mathrm{Hom}}_{\mathrm{Mod}(A)}(M,f^{\ast }N)\end{array}$$

For convenience, we use $A,B$$A,B$ for $\mathrm{Mod}(A),\mathrm{Mod}(B)$$\mathrm{Mod}(A),\mathrm{Mod}(B)$.

$$\begin{array}{}\text{(17)}& \mathrm{\Phi }:{\mathrm{Hom}}_B(B{\otimes }_{A}M,N)\to {\mathrm{Hom}}_A(M,f^{\ast }N),\end{array}$$

By

$$\begin{array}{}\text{(18)}& \mathrm{\Phi }(\varphi )(m)=\varphi (1_B\otimes m)\end{array}$$

This is an $A-$$A-$linear map since

$$\begin{array}{}\text{(19)}& \mathrm{\Phi }(\varphi )(am)=\varphi (1_B\otimes am)=\varphi (f(a)1_B\otimes m)=f(a)\varphi (1_B\otimes m)=a\mathrm{\Phi }(\varphi )(m)\end{array}$$

$$\begin{array}{}\text{(20)}& \mathrm{\Psi }:{\mathrm{Hom}}_A(M,f^{\ast }N)\to {\mathrm{Hom}}_B(B{\otimes }_{A}M,N)\end{array}$$

By

$$\begin{array}{}\text{(21)}& \mathrm{\Psi }(\phi )(b\otimes m)=b\phi (m)\end{array}$$

To see it is well-defined, we need to check that

$$\begin{array}{}\text{(22)}& \mathrm{\Psi }(\phi )(ba\otimes m)=\mathrm{\Psi }(\phi )(b\otimes am)\end{array}$$

Since $\mathrm{\Psi }(\phi )(ba\otimes m)=ba\phi (m)=b\phi (am)=\mathrm{\Psi }(\phi )(b\otimes am)$$\Psi(\varphi)(ba\otimes m)=ba\varphi(m)=b\varphi(am)=\Psi(\varphi)(b\otimes am)$, hence it is well defined.

To see it is $B-$$B-$linear,

$$\begin{array}{}\text{(23)}& \mathrm{\Psi }(\phi )(b^{\prime }(b\otimes m))=\mathrm{\Psi }(\phi )(b^{\prime }b\otimes m)=b^{\prime }b\phi (m)=b^{\prime }\mathrm{\Psi }(\phi )(b\otimes m)\end{array}$$

and they are pair of inverse since

$$\begin{array}{}\text{(24)}& \mathrm{\Psi }\circ \mathrm{\Phi }(\varphi )(m)=\mathrm{\Psi }(\varphi (1_B\otimes m))=\varphi (m)\end{array}$$

and

$$\begin{array}{}\text{(25)}& \mathrm{\Phi }\circ \mathrm{\Psi }(\phi )(b\otimes m)=\mathrm{\Phi }(b\phi (m))=b\mathrm{\Phi }(\phi (m))=b\phi (1_B\otimes m)=\phi (b\otimes m)\end{array}$$

The naturalness leave to readers $◻$$\square$