# Weyl Groups as Galois groups

I am looking for explicit examples (for all positive integers $n \ge 5$) of degree $2n$ even polynomials $f(x)=h(x^2)$ over the field $\mathbb{Q}$ of rational numbers such that the Galois groups of $f(x)$ over $\mathbb{Q}$ are the Weyl groups $W(B_n), W(D_n)$ or their normal subgroups of small index 2 or 4. (In this case the Galois group of $h(x)$ should be the full symmetric group $S_n$ or the alternating group $A_n$.)

• You didn't ask for this, but Shioda gave a beautiful construction of even polynomials with Galois groups $W(E_n)$ with $n=6,7,8$. Mar 22, 2016 at 0:17
• Jouve, Kowalski and Zywina constructed a degree 240 reciprocal polynomial with Galois group $W(E_8)$. Mar 22, 2016 at 16:13

It's possible to produce examples using congruence conditions. If there is a prime modulo which $h(x)$ is irreducible, a prime modulo which it splits into the product of a linear polynomial and an irreducible polynomial, and a prime modulo which it splits into the product of a linear polynomial and $n-2$ irreducible polynomials, then the Galois group of $h(x)$ is $S_n$ (because every subgroup of $S_n$ containing an $n$-cycle, an $n-1$-cycle, and a transposition is all of $S_n$).
Under these conditions, the kernel of the map from the Galois group to $S_n$ is an $S_n$-invariant subgroup of $(\mathbb Z/2)^n$. As long as there is a prime modulo which $h(x)$ splits completely and some of the roots, but not all, are squares, this kernel will contain the index $2$ normal subgroup of elements that sum to $0$, and if the number of non-squares is odd, it will contain the whole group. So if you want $W(B_n)$ you want the number of non-squares to be odd, and if you want $W(D_n)$ you want it to be even.
To construct explicitly a polynomial with $W(B_n)$ Galois group, take 4 primes and modulo each prime find a polynomial satisfying one of the four conditions, then use the Chinese remainder theorem to find an integer polynomial that reduces mod $p$ to the desired polynomials.
This applies immediately for $W(B_n)$, because we have shown the kernel contains all of $(\mathbb Z/2)^n$.
For $W(D_n)$, the congruence conditions force the kernel to contain an index $2$ subgroup, but to get exactly $W(D_n)$ you also need the constant term to be a perfect square. This will still work with the Chinese remainder theorems as long as the constant terms of your polynomials mod $p$ are squares mod $p$. It should be possible to find polynomials satisfying the first three conditions with the desired reduction mod $p$ (it is trivial if $n$ is odd as you can just scale the polynomial appropriately) and a polynomial that splits into linear factors will ave a perfect square constant term as long as the number of non-squares is even.
• Thanks, Will! Actually, I've came up with similar examples for $W(B_{2g+1})$, using Mori trinomials $h(x)=x^{2g+1}-bx -pc/4$ when $p$ does not divide $c$. (For Mori polynomials, see European J. Math. 2 (2016), no. 1, 360-381). Dec 12, 2016 at 16:10