How to prove that the highest root of a polynomial does not have a formula in radicals

Question

Let $P(x) = 720 x^7-1800 x^6+1560 x^5-540 x^4+62x^3-x^2$.

I have already shown that

1. all roots of $P(x)$ are in $[0,1)$ and all non-zero roots distinct,
2. the highest root is $x^*=\frac{1}{30}\left(15 + \sqrt{15 (10 + \sqrt{15})}\right) \approx 0.98085$, and
3. $P(1)=1$.

Let $P^{-1}(y)$ be the highest root of $P(x)-y=0$ for $y \in [0,1]$. By the properties above, it is a well-defined strictly increasing and continuous function, with $P^{-1}(0)=x^*$, $P^{-1}(1)=1$.

I would like to prove that $P^{-1}(y)$ does not have a formula in radicals.

I presume that this should be possible to show using Galois theory, but unfortunately, my background in abstract algebra is not strong enough, so I could use some pointers. Just for background information: I actually have a class of polynomials and I want to show that at least for some of them there is no formula for the inverse function. This particular one seems to be the shortest one where things get complicated. If there is an argument that works generally, it would be even better.

Answer 1

We'll show for a particular choice of $y$ that the Galois group $\textrm{Gal}(Q)$ of $Q(x) := P(x) - y$ contains both a transposition and a $7$-cycle. Those elements together generate $S_7$, which hence is the Galois group of $Q(x)$. In particular, this group is not solvable, so for that $y$ no root of $Q(x)$ is expressible in radicals, and hence there is no general formula.

The method is elementary, but it's a great help to have a CAS to find a suitable $y$ (and to carry out some evaluations $Q(x_0)$ later).

Transposition By Descartes' Rule of Signs, $Q(x)$ has at most $5$ positive roots and $Q(-x)$ no positive positive roots. For $y \neq 0$, $0$ is not a root of $Q$, so because $Q$ has degree $7$, it has at least $2$ nonreal roots.

Now, take $y = \frac{1}{73}$. Checking signs of $Q(x_0)$ at suitable points $x_0$ (e.g., $0$, $\frac{1}{8}, \frac{1}{3}, \frac{2}{3}, \frac{5}{6}, 1$) shows that $Q(x)$ has at least $5$ real roots, hence by the above it has exactly $5$ roots. Thus, the complex conjugation map is a transposition in $\textrm{Gal}(Q)$.

$7$-cycle Reducing $73 Q(x)$ modulo $7$ gives $$3\bar{Q}(x) = -(x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + 5) \in \Bbb F_7[x] .$$ Checking possible factorizations shows $3\bar Q$ has no factors of degree $\leq 3$ modulo $7$, hence it is irreducible, hence $Q$ is irreducible. (One can see quickly that $3\bar Q$ only takes on the values $4, 5$ using Fermat's Little Theorem, quickly eliminating the possibility that it has a linear factor.) In particular, $\textrm{Gal}(Q)$ contains a $7$-cycle.

The value $73$ occurring in the denominator of $y$ is, by the way, not totally arbitrary: It is the smallest value such that $Q$ has five real roots and is irreducible modulo $7$.

Answer 2

$P$ can be interpreted geometrically as a branched covering between Riemann spheres $\{(y,x) \in P^1(\Bbb C)^2 \mid y = P(x) \} \xrightarrow{\pi_1} P^1(\Bbb C)$.

$P'$ has six distinct distinct roots in $[0;1]$, and looking at its graph we see that their images by $P$ have $6$ distinct values (the critical values of $P$). Therefore this covering has $6$ simple branch points at those values and a $7$-cycling branch point of order $6$ at the point at infinity.

This is the generic covering for a degree $7$ polynomial, which implies that its Galois group, which is also the Galois group of the field extension $\Bbb C(f(x)) \subset \Bbb C(x)$ is $S_7$.

With analytic continuation, any radical expression giving you the largest root for $y \in [0;1]$ in terms of $y$ would extend to a multivalued radical expression giving all the roots for every complex number, and this would imply that the field extension is solvable by radicals.

But since $S_7$ is not a solvable group, this is impossible.

In general you can read off the Galois group of the extension from the geometry of the covering, and over $\Bbb C$ this is done pretty easily.