Factorization conjecture for the determinant of the Painlevé coefficient matrix

In fact, there is an explicit complete irreducible factorization for every $n\ge3$.

Let

$$g(X)=\sum_{r=0}^{n-1}f_rX^r,$$

and define the source's coefficient matrix $F_n=(F_{k,r})$ by

$$g(X)^k\equiv\sum_{r=0}^{n-1}F_{k,r}X^r \pmod{X^n-z}, \qquad 1\le k\le n.$$

Choose a primitive $n$th root of unity $\zeta$, let $t^n=z$, and put $t_j=\zeta^jt$. Define

$$H(X,Y)=\frac{g(X)-g(Y)}{X-Y}.$$

For every divisor $d>1$ of $n$, put

$$D_d= \prod_{\substack{0\le i<j<n\\ \operatorname{ord}(t_j/t_i)=d}} H(t_i,t_j),$$

and put

$$D_1=\prod_{j=0}^{n-1}g(t_j) =\operatorname{Res}_X(X^n-z,g(X)).$$

Then

$$\boxed{\det F_n=\prod_{d\mid n}D_d,}$$

where every $D_d$ belongs to $\mathbb Z[z,f_0,\ldots,f_{n-1}]$, is irreducible there, and the factors are pairwise nonassociate. Moreover,

$$\deg_z D_1=n-1,\qquad \deg_zD_d=\frac{(n-2)\varphi(d)}2\quad(d>1).$$

To prove the factorization, work over

$$L=\mathbb Q(z)(t,\zeta)$$

and introduce

$$V_{r,j}=t_j^r,\qquad M_{k,j}=g(t_j)^k.$$

Evaluation at the roots of $X^n-z$ gives $F_nV=M$. The two Vandermonde determinants yield

$$\det F_n =\left(\prod_jg(t_j)\right) \prod_{i<j}\frac{g(t_j)-g(t_i)}{t_j-t_i},$$

which is the asserted product after grouping unordered pairs by the order of $t_j/t_i$.

The splitting-field Galois group is

$$\operatorname{Gal}(L/\mathbb Q(z)) =C_n\rtimes(\mathbb Z/n\mathbb Z)^\times,$$

acting by

$$t_j\longmapsto t_{uj+a}, \qquad a\in\mathbb Z/n\mathbb Z,\quad u\in(\mathbb Z/n\mathbb Z)^\times.$$

It acts transitively on the roots $t_j$, and separately transitively on each set of unordered pairs with fixed ratio order $d$. Such a set has

$$\frac{n\varphi(d)}2$$

pairs. The orbit products $D_d$ are therefore Galois invariant. Their coefficients belong initially to $\mathbb Z[\zeta,t]$; invariance under $t\mapsto\zeta t$ leaves only powers of $t^n=z$, and invariance under $\zeta\mapsto\zeta^u$ then gives coefficients in $\mathbb Z[z]$.

The linear factors $g(t_j)$ are pairwise nonassociate because their $f_0$-coefficients equal $1$, while their $f_1$-coefficients are the distinct $t_j$. Transitivity therefore makes $D_1$ irreducible over $\mathbb Q(z)$.

Likewise, each $H(t_i,t_j)$ is linear in $f_1,\ldots,f_{n-1}$ and has $f_1$-coefficient $1$. For $n\ge4$, its next two coefficients are

$$[f_2]H(t_i,t_j)=t_i+t_j, \qquad [f_3]H(t_i,t_j) =(t_i+t_j)^2-t_it_j.$$

These recover the unordered pair $\{t_i,t_j\}$, so all the linear factors are distinct. For $n=3$, the sum $t_i+t_j$ already determines the omitted third root. Thus transitivity of each pair orbit makes every $D_d$ irreducible over $\mathbb Q(z)$.

Finally, $D_1$ contains $f_0^n$ with coefficient $1$, and $D_d$ contains

$$f_1^{n\varphi(d)/2}$$

with coefficient $1$. Hence all the factors are primitive over the UFD $\mathbb Z[z]$, and Gauss's lemma upgrades their irreducibility to

$$\mathbb Z[z,f_0,\ldots,f_{n-1}].$$

Distinct divisor orbits give distinct factors, while $D_1$ is distinguished by its dependence on $f_0$.

The $z$-degree of $D_1$ is $n-1$, with leading term

$$(-1)^{n-1}f_{n-1}^nz^{n-1}.$$

For $d>1$, each divided-difference factor has nonzero highest $t$-degree $n-2$, and there are $n\varphi(d)/2$ such factors. Since $t^n=z$,

$$\deg_zD_d=\frac{(n-2)\varphi(d)}2.$$

Therefore $\det F_n$ has exactly $\tau(n)$ irreducible factors for every $n\ge3$, including the required factor of $z$-degree $n-1$.