Polynomial as a functor

Let $K,L$$K,L$ be two rings and $\varphi :K\to L$$\phi:K\to L$ be a ring homomorphism, this define a $K-$$K-$Algebra sturucture.

Let $X:=(P_i(T){)}_{i\in I}$$X:=(P_i(T))_{i\in I}$ be a family of polynomials over $\varphi (K)$$\phi (K)$.

We claim that we could induce a functor $X_K(-):{\mathrm{Alg}}_K\to \mathrm{Set}$$X_K(-):\mathrm{Alg}_K\to\mathrm{Set}$ from the family of polynomials.

The functor map $X$$X$ to the solutions in $L$$L$.

For example, let $K=\mathbb{R},L=\mathbb{C}$$K=\mathbb R,L=\mathbb C$, $X=T^2+1$$X=T^2+1$. Then $X(\mathbb{C})=\{i,-i\}$$X(\mathbb C)=\{ i,-i\}$.

Another example is consider $K=\mathbb{Z},L=\bar{\mathbb{Q}}$$K=\mathbb Z,L=\overline{\mathbb Q}$, where $\bar{\mathbb{Q}}$$\overline{\mathbb Q}$ is the algebraic closure of $\mathbb{Q}$$\mathbb Q$. The $\varphi$$\phi$ is just inclusion map.

Let $P_i(x)$$P_i(x)$ be the family of all the monic polynomials over $\mathbb{Z}$$\mathbb Z$，then $X_{\mathbb{Z}}(\bar{\mathbb{Q}})$$X_{\mathbb Z}(\overline{\mathbb Q})$ is the algebraic integral ring over $\mathbb{Q}$$\mathbb Q$.

For the morphism, Let $\phi :K\to S$$\varphi:K\to S$ be another $K-$$K-$Algebra.

Let $f:L\to S$$f:L\to S$ be a $K-$$K-$Algebra homomorphism. i.e. $\phi =f\circ \varphi$$\varphi=f\circ\phi$.

If $\mathrm{\forall }i\in I,P_i(\alpha )=\varphi (k_1){\alpha }^n+...+\varphi (k_n)=0$$\forall i\in I,P_i(\alpha)=\phi(k_1)\alpha^n+...+\phi(k_n)=0$, then

$$\begin{array}{}\text{(1)}& f(P_i(\alpha ))=f(\varphi (k_1){\alpha }^n+...+\varphi (k_n))=\phi (k_1){\alpha }^n+...+\phi (k_n)=0\end{array}$$

Hence $X_K(f)$$X_K(f)$ map the solution of $(P_i(T){)}_{i\in I}$$(P_i(T))_{i\in I}$ in $L$$L$ to the solution of $(P_i(T){)}_{i\in I}$$(P_i(T))_{i\in I}$ in $S$$S$.

Well, we could define another functor, that is, ${\mathrm{Hom}}_K(A(X),-)$$\mathrm{Hom}_K(A(X),-)$.

Here $A(X)$$A(X)$ is the coordinate ring of $X$$X$. In other words,$A(X)\cong K[T]/I(X)$$A(X)\cong K[T]/I(X)$, where $I(X)$$I(X)$ is the ideal gennerate by $X$$X$.

For example, Let $K=\mathbb{R}$$K=\mathbb R$, $X:=x^2+y^2-1$$X:=x^2+y^2-1$, $A(X)=\mathbb{R}[x,y]/(x^2+y^2-1)$$A(X)=\mathbb R[x,y]/(x^2+y^2-1)$.

We claim that $X_K(-)$$X_K(-)$ is natural isomorpic to ${\mathrm{Hom}}_K(A(X),-):{\mathrm{Alg}}_K\to \mathrm{Set}$$\mathrm{Hom}_K(A(X),-):\mathrm{Alg}_K\to\mathrm{Set}$.

Firstly, we claim that $X_K(L)\cong {\mathrm{Hom}}_K(A(X),L)$$X_K(L)\cong \mathrm{Hom}_K(A(X),L)$.

Let $t\in X_K(L)$$t\in X_{K}(L)$, then the evaluation map at $t$$t$ define a $h\in {\mathrm{Hom}}_K(A(X),L)$$h\in\mathrm{Hom}_K(A(X),L)$.

Conversely, let $h\in {\mathrm{Hom}}_K(A(X),L)$$h\in \mathrm{Hom}_K(A(X),L)$, it uniquely define a $\iota :K[T]\to A(X)\to L$$\iota:K[T]\to A(X)\to L$.

We also have $\varphi :K\to K[T]\to A(X)\to L$$\phi:K\to K[T]\to A(X)\to L$. Hence $\varphi =\iota \circ i_K$$\phi=\iota\circ i_K$. Where $i_K$$i_K$ is the inclusion map.

Let $\iota :K[T]\to L$$\iota:K[T]\to L$ be the homomorphism, then each $\iota (T)$$\iota(T)$ give you a solution of $X$$X$.

Since let $P_i(T)=\varphi (k_1)T^n+...+\varphi (k_n)$$P_i(T)=\phi(k_1)T^n+...+\phi(k_n)$, $\iota (k_1{T}^n+...+k_n)=\varphi (k_1)\iota (T{)}^n+...+\varphi (k_n)=0$$\iota(k_1T^n+...+k_n)=\phi(k_1)\iota(T)^n+...+\phi(k_n)=0$

Now for proving they are natural isomorphism, we only need to check the diagram commute.

$$\begin{array}{}\text{(2)}& t\to e{v}_t\to e{v}_{f(t)}=t\to f(t)\to e{v}_{f(t)}\end{array}$$

Remark

The notation I use looks like I only consider the one variable case. But indeed, you only need to change the notation then you can get the multivarible case.

Similarly, we could consider the differential ring version.