Möbius transformation that permutes roots of a cubic polynomial

Question

The roots of the polynomial $x^3-3x-1$ can be permuted by the function $z\mapsto \dfrac{-1}{1+z}$ which is easily checked by a direct calculation.

Is there a simple formula for a Möbius transformation that permutes the roots of $x^3+px+q$?

Answer 1

Actually, yes.

This is from section number 511 in J.-A. Serret's 1877 work, "Cours D'Algèbre Supérieure". As far as I can tell, it's never been translated into English (which is surprising, given how important it was in the 19th century), but it is available online from various libraries. Anyways, here's his result.

For the polynomial $x^3 + Px +Q$, let's recall that its discriminant $\Delta$ can be written as $\Delta = -(4P^3 + 27Q^2)$.

With this in mind, and so long as $\Delta \not= 0$ (which is the same as saying that the polynomial has no repeated roots), let's define \begin{align*} a &= \frac{\sqrt{\Delta} - 9Q}{2\sqrt{\Delta}} \\[1.2ex] b &= \frac{2P^2}{2\sqrt{\Delta}} \\[1.2ex] c &= \frac{6P}{2\sqrt{\Delta}} \\[1.2ex] d &= \frac{\sqrt{\Delta} + 9Q}{2\sqrt{\Delta}} \end{align*} Then, Serret showed that $m(z) = \displaystyle\frac{a z + b}{c z + d}$ is of order three under composition, and it permutes the roots of the cubic $x^3 + Px + Q$. Interestingly, this is true for all cubics without repeated roots. It works for irreducible polynomials and for reducible polynomials. It works for all possible coefficients. It even works for polynomials with complex roots; in that case, $\Delta<0$ and so $\sqrt{\Delta}$ is imaginary and thus $m(z)$ can take a real root to a complex root.

(Actually, Serret did this in full generality for the non-depressed cubic $x^3 + Px^2 + Qx + R$, but I've reduced the formula down to the depressed cubic you were interested in.)

Let's check it for your polynomial $x^3 -3x -1$. You have $P=-3$ and $Q=-1$, so you get $\Delta = 81$. Thus, your values for $a,b,c,d$ are: \begin{align*} a &= \frac{\sqrt{\Delta} - 9Q}{2\sqrt{\Delta}} &= 1 \\[1.2ex] b &= \frac{2P^2}{2\sqrt{\Delta}} &= 1 \\[1.2ex] c &= \frac{6P}{2\sqrt{\Delta}} &=-1\\[1.2ex] d &= \frac{\sqrt{\Delta} + 9Q}{2\sqrt{\Delta}} &=0 \end{align*} and so you have $m(z) = \displaystyle\frac{ z + 1}{- z} = -1 - \frac{1}{z}$. Of course, this is just the inverse of the map you gave in your question, $z \mapsto \displaystyle \frac{-1}{z+1}$.