2026.08.19

When a polynomial differential equation forces a straight line

One short identity says that a polynomial whose derivative is balanced against a second polynomial in a particular way can only be linear, and that rigidity is what closes a block in the strip reduction.

Take two polynomials in one variable, $A$ and $C$ , and form the combination

A\,C' - \nu\,A'\,C,

where $\nu$ is a positive integer. Suppose the result is a nonzero constant. What does that force?

For $\nu=1$ the combination is the Wronskian $AC' - A'C$ , the classical test for whether two functions are linearly independent. Asking for it to be a nonzero constant is a strong demand. The answer, for every $\nu$ , is surprisingly rigid: $A$ has to be linear or constant. No polynomial of degree two or more can appear.

A first look at the top terms

Write $a = \deg A$ and $m = \deg C$ , with leading coefficients $\alpha$ and $\gamma$ . The two products have the same degree:

\deg(A\,C') = a + m - 1, \qquad \deg(\nu A'C) = a + m - 1 .

Their leading coefficients are $m\alpha\gamma$ and $\nu a\,\alpha\gamma$ . So the coefficient of $y^{a+m-1}$ in the combination is

(m - \nu a)\,\alpha\gamma .

If $m \neq \nu a$ , this is nonzero, so the combination has degree exactly $a+m-1$ . A nonzero constant has degree zero, which forces $a + m = 1$ . The only options are $a=0$ with $m=1$ , or $a=1$ with $m=0$ . Either way $\deg A \le 1$ , which is the conclusion.

So the entire difficulty sits in one case: $m = \nu a$ , where the top terms cancel and the degree count says nothing.

The identity that removes the hard case

Write $E(A,C) = A C' - \nu A' C$ . It is linear in $C$ . The useful fact is what it does to one particular input:

E(A, A^{\nu}) = A\cdot \nu A^{\nu-1}A' - \nu A' \cdot A^{\nu} = 0 .

The combination annihilates $A^{\nu}$ . By linearity, we may subtract any multiple of $A^{\nu}$ from $C$ without changing the value at all:

E(A,\; C - \lambda A^{\nu}) = E(A,C).

Now the hard case takes care of itself. Suppose $a \ge 2$ . The degree count already showed $m = \nu a$ . Since $\deg A^{\nu} = \nu a = m$ , choosing $\lambda = \gamma/\alpha^{\nu}$ cancels the leading term of $C$ , producing a $C_1$ with $\deg C_1 < \nu a$ and the same value $E(A,C_1) = c$ . If $C_1 = 0$ then $c = 0$ , which is excluded. Otherwise $\deg C_1 \neq \nu a$ , so the first argument applies and gives $a + \deg C_1 = 1$ . That contradicts $a \ge 2$ .

Figure 1

Every solution has to sit somewhere on this lattice. Off the dashed line the leading terms survive, so the degree must fall to zero and only $\deg A + \deg C = 1$ is left. On the dashed line, subtracting a multiple of $A^{\nu}$ moves straight down off it without changing anything, so that case collapses into the first one. The two ringed points are all that remain.

That is the whole proof. It uses nothing beyond the degree of a product and one line of differentiation.

The statement

Theorem A. Let $F$ be an integral domain of characteristic zero, let $\nu \ge 1$ be an integer, and let $A, C \in F[y]$ satisfy

A\,C' - \nu\,A'\,C = c, \qquad c \in F,\ c \neq 0 .

Then $\deg A \le 1$ .

This is formalized in Lean 4 and kernel-checked, and registered as PALOMAR-2026-08-19-000005.

Where the characteristic matters

The proof leans on $(m - \nu a)\alpha\gamma$ being nonzero. Over a field of characteristic $p$ that step can fail, because $m - \nu a$ can be a nonzero integer that is still zero in the field. Bounding both quantities below $p$ repairs it, and that is exactly the hypothesis of the positive-characteristic version: the same conclusion holds in characteristic $p$ provided $\nu \deg A < p$ and $\deg C < p$ .

Some bound is genuinely necessary. Take $\nu = 1$ , $C = 1$ , and

A = y^{p} + y .

In characteristic $p$ the derivative is $A' = p\,y^{p-1} + 1 = 1$ , so

A\,C' - A'\,C = 0 - 1 = -1 ,

a nonzero constant, while $\deg A = p$ . The theorem fails without a bound, and this example shows the bound $\nu \deg A < p$ is doing real work rather than being an artifact of the proof.

What it is used for

The identity is the rigidity engine in the depth-two block of the strip reduction at Newton-degree pair $(72,108)$ . Through the bridge from that block to the differential equation above, Theorem A forces the edge polynomial of a hypothetical Keller counterexample to be a binomial. A general polynomial edge has many coefficients to play with; a binomial has almost none. That collapse is what makes the surrounding argument finite.

Scope, and what is not claimed

The characteristic-zero statement is elementary, and for $\nu = 1$ it is the classical fact that two polynomials with constant nonzero Wronskian cannot both be nonlinear. We do not claim the characteristic-zero case as new. What the campaign contributes here is an independent proof, the machine-checked formalization linked above, and the positive-characteristic extension with its bound. If you know a prior reference for the general $\nu$ statement, please send it and we will credit it here.

The block-variety theorem that consumes this result is a separate statement and is not formalized. The Lean development covers Theorem A only.

A question to take away

The positive-characteristic bound $\nu \deg A < p$ is sufficient, and the example $A = y^{p}+y$ shows some bound is needed. Is $\nu \deg A < p$ sharp? For $\nu = 1$ the example sits exactly at $\deg A = p$ , so nothing between is ruled out by it. A counterexample with $1 < \deg A < p$ , or a proof that none exists, would pin the statement down.