TITLE: How to find a root swapper polynomial?
QUESTION [5 upvotes]: IMC 2020 problem 6 was based on the following idea: if $P(x)=x^3-3x+1$ then $Q(x)=x^2-2$ has the property that it cyclically rotates the roots of $P$. That is, if we call them $x_1$, $x_2$, $x_3$ then $Q(x_1)=x_2$, $Q(x_2)=x_3$ and $Q(x_3)=x_1$. I want to find polynomials like this in general.
If $P(x)=(x-x_1)(x-x_2)(x-x_3)=x^3-ax^2+bx-c$ and $Q(x)=kx^2+lx+m$ and $Q(x_1)=x_2$, $Q(x_2)=x_3$ and $Q(x_3)=x_1$ then
$$\begin{bmatrix}
x_1^2 & x_1 & 1 \\
x_2^2 & x_2 & 1 \\
x_3^2 & x_3 & 1
\end{bmatrix}
\begin{bmatrix} 
k \\
l \\
m
\end{bmatrix}=
\begin{bmatrix} 
x_2 \\
x_3 \\
x_1
\end{bmatrix}.$$
The determinant of the Vandermonde matrix is the product of the pairwise differences which is the square root of the discriminant, so it's expressible as a function of $a,b,c$. $k$ is expressible as a symmetric polynomial too, so we can write $k=\frac{a^2-3b}{\sqrt{|\Delta|}}$, however $l$ and $m$ have expressions as a function of the roots which are not symmetric, so we can't write them as a function of the coefficients:
$$l=\frac{-x_1^2-x_2^3-x_3^3+x_1^2x_3+x_2^2x_1+x_3^2x_2}{\sqrt{|\Delta|}}$$
$$m=\frac{x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2-x_1^3x_2-x_2^3x_3-x_3^3x_1}{\sqrt{|\Delta|}}.$$
The problem with $l$ is the $x_1^2x_3+x_2^2x_1+x_3^2x_2$ bit, because it's missing the other half $x_1^2x_2+x_2^2x_3+x_3^2x_1$ which we would need to add to make it symmetric. $m$ is similar.
So my questions are:

When are the expressions for $l$ and $m$ "nice"? For example if the coefficients are rational when are $l,m$ rational?

Can we require something of the roots which makes $l,m$ expressible as a function of the coefficients? It seems like they're "nearly" symmetric, so it feels like it shouldn't take much to get there.

What happens in higher degrees?

So far I've been thinking about $Q$ having degree one less than $P$. Does this always allow for the "cleanest" solution? In particular the case where $P(x)=(x-x_1)(x-x_2)=x^2-ax+b$ is easy: $Q(x)=a-x$.

REPLY [2 votes]: Both $l$ and $m$ are rational when the discriminant $d$ of the cubic $P$ is a rational square. For example, let $$\lambda = x_1^2x_2 + x_2^2x_3 + x_3^2x_1$$ and $\overline \lambda$ its “symmetric conjugate” by swapping any two roots.  Then $ (x-\lambda)(x-\overline \lambda)$ is a rational polynomial with discriminant $d$, a rational square, so both $\lambda$ and $\overline \lambda$ are rational.
Similarly for $$\mu = x_1^3x_2 + x_2^3x_3 + x_3^3x_1$$ the discriminant of $(x - \mu)(x - \overline \mu)$ is $ d \cdot (x_1 + x_2 + x_3)^2$, a rational square. Conclusion: if $d$ is a rational square then the coefficients of the permutation polynomials can indeed be expressed in the coefficients of $P$.  The choice $\pm \sqrt d$ leads to two non~trivial permutations.
The condition that $d$ is a rational square is also required to ensure that the splitting field of $P$ has degree either $1$ or $3$ over $\mathbb Q$.