# Splitting the Resultant, as when the Determinant becomes the square of the Pfaffian

The Determinant of an $n\times n$ matrix, viewed as a polynomial in the entries, is irreducible. But when it is restricted to the subspace of alternate matrices, it becomes reducible, actually the square of a polynomial known as the Pfaffian.

Likewise, the Resultant of a pair $(P,Q)$ of polynomials of respective degrees $\le n,m$, when viewed as a polynomial in the coefficients of $P$ and $Q$, is an irreducible polynomial ($n+m+2$ indeterminates, degree $n+m$).

Is there a natural subspace $E$ of $k_n[X]\times k_m[X]$, such that the restriction of the Resultant to $E$, viewed as a polynomial in the coordinates, splits in a non-trivial way ?

Low temperature example: Let $n,m$ be even, and $E$ be the space of pairs of even polynomials. Then the restriction of the Resultant over $E$ is a square. If $P(X)=p(X^2)$ and $Q(X)=q(X^2)$, then ${\rm Res}(P,Q)=({\rm Res}(p,q))^2$. This follows from the formula ${\rm Res}(P,Q)=\prod(x_i-y_j)$ in terms of the roots $x_i$ and $y_j$ (to be adapted if $P,Q$ are not monic).

Less cold example: Let $n,m$ be even, and $F$ be the space of pairs of reciprocal polynomials. Then the restriction of the Resultant over $F$ is a square. If $P(X)=X^np(X+\frac1X)$ and $Q(X)=X^mq(X+\frac1X)$, then ${\rm Res}(P,Q)=({\rm Res}(p,q))^2$.

• Polynomials in $X^2$. – Felipe Voloch Sep 26 '17 at 21:32
• @Denis: where does the hot versus cold terminology come from? – Abdelmalek Abdesselam Sep 27 '17 at 22:04
• @AbdelmalekAbdesselam. I just coined it. – Denis Serre Sep 28 '17 at 5:53

I think it is better to homogenize and view $P$, $Q$ as binary forms in $\mathbb{C}[X_1,X_2]$. An invertible $2\times 2$ matrix $g$ acts on the variables $x_i$ by $(gx)_i=\sum_{j=1,2}g_{ij}x_j$ and on forms by $(gP)(x)=P(g^{-1}x)$. The resultant is invariant: $${\rm Res}_{p,q}(gP,gQ)=({\rm det}(g))^{-pq}\times {\rm Res}_{p,q}(P,Q)$$ Both examples you gave are given by eigenspaces of a fixed matrix $g$ such as $$\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)$$ or $$\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$$ Note one can also consider mutlidimensional resultants ${\rm Res}_{p_1,\ldots,p_n}(P_1,\ldots,P_n)$ and they satisfy tons of base change formulas (see Jouanolou's article). Hope this helps for the generalization you are looking for.
What I said above shows that the two examples are the same. For the first one this follows from the base change formula in Section 5.12 of the article by Jouanolou, which is a vast nonlinear generalization of the transformation formula I wrote above. One has: $${\rm Res}_{dp_1,\ldots,dp_n}(P_1\circ G,\ldots,P_n\circ G)=$$ $${\rm Res}_{p_1,\ldots,p_n}(P_1,\ldots,P_n)^{d^{n-1}}\ \times \ {\rm Res}_{d,\ldots,d}(G_1,\ldots,G_n)^{p_1\cdots p_n}$$ where each $P_i$ is a homogenous form of degree $p_i$ in $n$ variables and $G$ is a polynomial map $\mathbb{C}^n\rightarrow\mathbb{C}^n$ with components $G_i$ that are homogeneous polynomials of the same degree $d$. The first example is given by $$\left\{ \begin{array}{ccc} G_1(x_1,x_2) & = & x_1^2\\ G_2(x_1,x_2) & = & x_2^2 \end{array}\right.$$ Spaces of forms of the form (sounds horrible I know) $P\circ G$ should give you tons of examples of this factorization phenomenon.
Another thought: this is vaguely reminiscent of the Frobenius determinant factorization. The two matrices $g$ above are involutions and correspond to the discrete group $\mathbb{Z}_2$. I am not familiar with the Galois theory of covers like the one given by $G$ but I think it may help.