Irreducibility conjecture for the resultant factor of a generic form

The conjecture requires an omitted dimension hypothesis. It is false for binary forms, while in at least three variables the stronger assertion of generic absolute irreducibility holds in every degree.

Write

$$F(T,\mathbf x)=\prod_{j=1}^d(T-r_j),$$

put $y=t_2$, $u=t_1$, and set

$$R=\frac{F(u+iy,\mathbf x)+F(u-iy,\mathbf x)}2, \qquad I=\frac{F(u+iy,\mathbf x)-F(u-iy,\mathbf x)}{2i}.$$

With $C=\binom d2$, the universal resultant identity is

$$\operatorname{Res}_u(R,I) = (-1)^C y^d \prod_{j<k} \left((r_j-r_k)^2+4y^2\right). \tag{1}$$

Indeed, for $A=F(u+iy,\mathbf x)$ and $B=F(u-iy,\mathbf x)$,

$$\operatorname{Res}(A,B)=(-2i)^d\operatorname{Res}(R,I),$$

and evaluating $B$ at the roots $u=r_j-iy$ of $A$ gives (1). For separable $F$, this shows that the extracted exponent is $p=d$, and the residual factor has degree $d(d-1)$.

Suppose first that $n\ge2$ and $d\ge2$. Consider the normalized homogeneous specialization

$$F_*(T,x_1,x_2) = T^d+x_1^{d-1}T+x_1^{d-1}x_2.$$

Writing $T=x_1Z$ and $s=x_2/x_1$ gives

$$q_s(Z)=Z^d+Z+s.$$

This polynomial has Galois group $S_d$ over $\mathbb C(s)$: the cover

$$s=-Z^d-Z$$

is connected, its $d-1$ critical points satisfy

$$dZ^{d-1}+1=0,$$

and their critical values $-(d-1)Z/d$ are pairwise distinct. Thus its finite branch monodromies are transpositions generating a transitive group, which must be $S_d$.

The $d(d-1)$ ordered root differences $r_j-r_k$, $j\ne k$, are distinct. Coincidences sharing a first or last index force equal roots; reversed pairs would give $2(r_j-r_k)=0$; coincidences involving four distinct indices are ruled out by applying a transposition. In the remaining three-index case, a relation

$$2r_j=r_i+r_k$$

and its image under the transposition $(ij)$ imply

$$3(r_i-r_j)=0,$$

again impossible.

Hence $S_d$ acts transitively on all roots

$$y=\frac{i(r_j-r_k)}2$$

of the residual factor. That factor is irreducible over $\mathbb C(x_1,x_2)[y]$. Its nonzero constant leading coefficient in $y$ makes it primitive, so Gauss's lemma gives irreducibility in

$$\mathbb C[x_1,\ldots,x_n,y].$$

For fixed $n,d$, reducible homogeneous forms of a given degree form a Zariski-closed subset of projective coefficient space: they are the finite union of the images of projective multiplication maps. Since $F_*$ lies outside this subset, a generic normalized form has an absolutely irreducible residual factor for every $n\ge2$.

However, the original conjecture does not assume $n\ge2$. For $n=1$ and $d\ge3$, the residual factor is a real homogeneous binary form of degree

$$d(d-1)>2.$$

Every such binary form factors over $\mathbb R$ into linear and quadratic factors, so it is always reducible. Explicitly, take

$$F(T,x)=T(T-x)(T-3x).$$

Formula (1) gives

$$\operatorname{Res}_u(R,I) = -y^3 (x^2+4y^2) (4x^2+4y^2) (9x^2+4y^2),$$

whose residual factor is reducible.

Therefore the unrestricted conjecture is false, whereas the corrected hypothesis $n\ge2$ gives generic absolute irreducibility in every degree $d\ge2$.