# Commuting ODE's implies existence of nonzero vanishing two variable polynomial?

Write $\partial := d/dt$, fix $m, n > 0$, and let$$F = \partial^n + f_1(t)\partial^{n-1} + \dots + f_{n-1}\partial + f_0,\text{ }G:= \partial^m + g_1(t)\partial^{m-1} + \dots + g_{m-1}\partial + g_0$$be a pair of ordinary differential operators with rational coefficients (i.e. with coefficients in $\mathbb{C}(t)$). Assume that $F$ and $G$ commute (as operators). Then, for any polynomial$$p = \sum_{i, j} c_{i, j} x^i y^j \in \mathbb{C}[x, y],$$there is a well-defined differential operator$$p(F, G) = \sum_{i, j} c_{i, j} F^i \circ G^j,$$where$$F^i := F \circ \dots \circ F\text{ }(i \text{ times}),\text{ }G^j := G \circ \dots \circ G\text{ }(j \text{ times}).$$My question is, if $F$ and $G$ commute, does there exist a nonzero polynomial $p \in \mathbb{C}[x, y]$ such that one has $p(F, G) = 0$?

EDIT: $n = 2$ is not hard. For each $\lambda \in \mathbb{C}$, let $V_\lambda$ be the space of solutions $\psi \in C^\infty(a, b)$ of the differential equation $F(\psi) = \lambda \cdot \psi$, on a small enough segment $[a, b] \subset \mathbb{R}$. The operator $G$ acts on $V_\lambda$. We use that there exists a polynomial $p_\lambda \in \mathbb{C}[y]$ such that $p_\lambda(G|_{V_\lambda}) = 0$. Then, we analyze the dependence of solutions $\psi$ on the initial condition at the point $a$ to show that we can find $p_\lambda$ in a way that the function $(\lambda, \mu) \mapsto p_\lambda(\mu)$ gives a polynomial $p \in \mathbb{C}[x, y]$ such that $p(F, G) = 0$.

• Do you mean to have $c_{i,j}$ twice in the definition of $p(F,G)$, or is that a typo? Aug 31, 2015 at 13:34
• Typo, sorry. Fixed now.
– user61522
Aug 31, 2015 at 15:11
• Just a remark. If there exist univariate polynomials $P(x)$ and $Q(y)$ such that $P(f) = Q(g)$, then obviously $p(x,y) = P(x) - Q(y)$ has the property that you want. Aug 31, 2015 at 15:44