Differential operator polynomial as a functor

Preliminary

Math Essays: Polynomial as a functor (wuyulanliulongblog.blogspot.com)

Math Essays: Differential Ring (1): Some general result and interesting application (wuyulanliulongblog.blogspot.com)

Differential operator polynomial as a functor

Let $(K,d_K)$$(K,d_K)$ be a differential ring.

Define the $K$$K$ differential algebra be ${\varphi }_A:(K,d_K)\to (A,d_A)$$\phi_A:(K,d_K)\to (A,d_A)$, where $\varphi (K)\subseteq \mathrm{Ker}(d_A)$$\phi(K)\subseteq\mathrm{Ker}(d_A)$.

Let $p(x)$$p(x)$ be a differential operator polynomial, then $p(d_K)f=0$$p(d_K)f=0$ define an ODE.

Then the polynomial $p(x)$$p(x)$ induce a functor $X_p(-):K-\mathrm{DiffAlg}\to K-\mathrm{Mod}$$X_p(-):K-\mathrm{DiffAlg}\to K-\mathrm {Mod}$.

Let $p(d_K)=(k_n{d}_K^n+...+k_0)$$p(d_K)=(k_nd_K^n+...+k_0)$.

For a $K$$K$-differential algebra $(R,d_R)$$(R,d_R)$, we can induce $p(d_R)={\varphi }_R(k_n)d_R^n+...+{\varphi }_R(k_0)$$p(d_R)=\phi_R(k_n)d^n_R+...+\phi_R(k_0)$ , which gives you a $R$$R$ differential operator polynomial.

For the object, $X_p(R)$$X_p(R)$ gives you the solution module of $p(d_k)$$p(d_k)$ in $R$$R$.

Let $f:(R,d_R)\to (S,d_S)$$f:(R,d_R)\to(S,d_S)$ be a $K$$K$ differential algebra homomorphism, i.e. $f\circ d_R=d_S\circ f$$f\circ d_R=d_S\circ f$, $f\circ {\varphi }_R={\varphi }_S$$f\circ\phi_R=\phi_S$.

For the morphism, $X_p(f)$$X_p(f)$ is a $K$$K$-module homomorphism, map solutions in $R$$R$ to solution in $S$$S$.

If $\alpha$$\alpha$ is one solution of $p$$p$ in $R$$R$, then $f({\varphi }_R(k_n)d_R^n\alpha +...+{\varphi }_R(k_0)\alpha )=f(0)=0={\varphi }_S(k_n)d_S^{n}f(\alpha )+...+{\varphi }_S(k_0)f(\alpha )=0$$f(\phi_R(k_n)d^n_R\alpha+...+\phi_R(k_0)\alpha)=f(0)=0=\phi_S(k_n)d_S^nf(\alpha)+...+\phi_S(k_0)f(\alpha)=0$