2

For $m\geq 2$, let $P_m=4X^{4m}+X^{4m-2}+X^2+4$. I have checked that for $1\leq m \leq 10$,

1) $P_m$ is irreducible over $\mathbb Q$

2) All its roots have modulus $1$

3) Its Galois group has order $4^m(m!)$ (computation made with online MAGMA calculator )

My questions : does anyone know a proof or counterexample for any of those claims ?

What makes this problem interesting in my view is that (according to MAGMA) the Galois group is non-abelian for $m\geq 2$. So we're not in an entirely cyclotomic context here, contrary to what might be expected from the fact that the roots have modulus $1$.

Update : There is a unique polynomial $Q_m$ of degree $m$ such that $P_m(X)=X^{2m}Q_m(X^2+\frac{1}{X^2})$. Some small values of $Q_m$ follow below. As guessed by in user760780's comment below, the Galois group of $Q_m$ is exactly $S_m$.

$$ \begin{array}{lcl} Q_1(y) & = & 4y + 2 \\ Q_2(y) & = & 4y^2 + y - 8 \\ Q_3(y) & = & 4y^3 + y^2 - 12y - 2 \\ Q_4(y) & = & 4y^4 + y^3 - 16y^2 - 3y + 8 \\ Q_5(y) & = & 4y^5 + y^4 - 20y^3 - 4y^2 + 20y + 2 \\ Q_6(y) & = & 4y^6 + y^5 - 24y^4 - 5y^3 + 36y^2 + 5y - 8 \\ Q_7(y) & = & 4y^7 + y^6 - 28y^5 - 6y^4 + 56y^3 + 9y^2 - 28y - 2 \\ Q_8(y) & = & 4y^8 + y^7 - 32y^6 - 7y^5 + 80y^4 + 14y^3 - 64y^2 - 7y + 8 \\ Q_{9}(y) & = & 4y^9 + y^8 - 36y^7 - 8y^6 + 108y^5 + 20y^4 - 120y^3 - 16y^2 + 36y + 2 \\ Q_{10}(y) & = & 4y^{10} + y^9 - 40y^8 - 9y^7 + 140y^6 + 27y^5 - 200y^4 - 30y^3 + 100y^2 + 9y - 8 \\ \end{array} $$

Ewan Delanoy
  • 61,600
  • Interesting! $P_m(X)$ is even and palindromic. Both facts place constraints on the Galois group. See this old thread. The idea being that the roots form quadruplets $\pm\alpha^{\pm1}$. The Galois group will then permute quadruplets, but if you know the automorphic image of one member of a quad, the others are forced. – Jyrki Lahtonen May 20 '20 at 14:12
  • Wonder whether this comes in handy? – Jyrki Lahtonen May 20 '20 at 14:23
  • 2
    Item 2 is easy. Substitute $x=e^{it}$. The equation $P_m(e^{it})=0$ is equivalent to (divide by $X^{2m}$) $$4\cos(2mt)+\cos([2m-2]t)=0.$$ The first term takes alternating values $+4,-4,\ldots$ at the points $0,\pi/(2m), 2\pi/(2m), \ldots,2\pi$. The second term cannot change the sign at those points because its absolute value is $\le 1$. Therefore $P_m(e^{it})$ has a zero $t$ in each open interval $(k\pi/(2m),[k+1]\pi/(2m))$, $k=0,1,\ldots,4m-1$. That's $4m$ zeros of $P_m(x)$ right there, so that's all. – Jyrki Lahtonen May 20 '20 at 14:55
  • 1
    There's no reason to expect any relation to cyclotomic polynomials just because the roots all have absolute value one because the polynomial is not monic. Take $\alpha$ to be any element of a totally real field $K$. Then all the roots of $$\prod_{\sigma:K \rightarrow \mathbf{R}} (x^2 - (\alpha/N) x + 1) \in \mathbf{Q}[x]$$ have absolute value $1$ for $N$ large enough, and the splitting field contains the field $K$. The Galois group claim is very close just to saying that $\mathbf{Q}(x^2 + x^{-2})$ has Galois group $S_m$, which is pretty much one would expect at random. – user760870 May 20 '20 at 15:27
  • @user760870 Thanks for the feedback. You've pretty much shredded the glamour and mystery of this problem, now all that's left is the computational details ... – Ewan Delanoy May 20 '20 at 15:48
  • Yeah. I'm willing to wager that the Galois group ends up being $V_4\wr S_m$. – Jyrki Lahtonen May 20 '20 at 16:20
  • @JyrkiLahtonen I've asked a separate question about your wager : https://math.stackexchange.com/questions/3683952/relating-galois-group-of-evenized-reciprocalized-version-to-original-galois-grou – Ewan Delanoy May 20 '20 at 16:36

0 Answers0