When $A\in M_n(\mathbb{Q})$, we consider the pencil $A-xA^T$. Then $p_A(x)=\det(A-xA^T)$ is a self-reciprocal polynomial. $p_A$ can only be irreducible if $n=2p$ is even.

Question: For every $p$, does there exist a matrix $A\in M_{2p}(\mathbb{Q})$ such that $\mathrm{Gal}(p_A)=C_2\wr S_p$ (which is the maximal Galois group for a self-reciprocal polynomial; it contains a subgroup isomorphic to $S_p$ and has cardinality $2^pp!$) ?

I am sure that the answer is yes; still have to show it !

A candidate family $(A_{2p})$ is as follows.

Let $A_{n}=[a_{i,j}]\in M_n(\mathbb{Q})$, where $a_{i,i}=i,a_{i,i+1}=1$ and otherwise, $a_{i,j}=0$.

$f_{2p}(x)=\det(A_{2p}-x{A_{2p}}^T)$, where $A_{2p}-x{A_{2p}}^T$ is a 1-band matrix.

One has the recurrence formula $f_n(x)=n(1-x)f_{n-1}(x)+xf_{n-2}(x)$.

Experiments seem to "show" that it works. What do you think ? Thanks in advance.

ANSWER to RVD de Bruyn. There are $2$ problems.

$\bullet$ Is $C_2\wr S_p$ realizable as a Galois group of a polynomial in $\mathbb{Q}[x]$ ?

$\bullet$ Can we write such a polynomial in the form $\det(A-xA^T)$ where $A$ is a rational matrix ?

For realization part, there are 2 standard methods

i) We show the required result over $\mathbb{Q}[T]$ (generic polynomial)and we use the Hilbert irred. Theorem to deduce the existence of a suitable specialization (obviously we do not obtain explicitly the polynomial, but no matter).

ii) We explicitly find a suitable family of polynomials. This is what I propose with my $(A_{2p})$'s; but I don't know how to do it.

In general, what is easier? Method i) or ii)? For $S_p$,both methods work. For i), we highlight the presence of a generator of $S_p$ (cf. wiki). For ii) we use Osada's example $x^p-x-1$.

Remark. $C_2\wr S_p$ admits a generator with $3$ elements: a cycle of order $2$, a product of 2 disjoint cycles of order 2, a product of 2 disjoint cycles of order $p$.

I think that $C_2\wr S_p$ is in the list of realizable groups, but, unfortunately, this is not enough to deduce the requested result

  • 2
    $\begingroup$ If you're happy with an inexplicit solution, by the Hilbert irreducibility theorem it's enough to treat the case where $A$ is the 'universal' matrix $(x_{ij})$ for coordinates $x_{ij}$. (To check whether you get the full group for a specific polynomial, I only know some tricks in small cases, e.g. for $S_p$ when $p$ is prime in the non-palindromic case.) $\endgroup$ – R. van Dobben de Bruyn Oct 23 '20 at 22:25
  • $\begingroup$ @R. van Dobben de Bruyn , thanks for your comment. See above my thoughts on this. $\endgroup$ – loup blanc Oct 25 '20 at 17:26
  • $\begingroup$ You probably already realize this, but the comment indicated you can work over the polynomial ring $\mathbb Q[x_{ij}]$, i.e., you can treat the $n^2$ entries of $A$ as independent indeterminates. So in your first bullet point, you wrote $\mathbb Q[x]$, which seems to just be a 1-variable polyniomial ring. I'm not sure if this will help you, but it will give you a lot more freedom to try to create elements of your Galois group. $\endgroup$ – Joe Silverman Oct 25 '20 at 17:47
  • $\begingroup$ Using Magma I checked that for $p \leq 20$, the Galois group of $f_{2p}$ is $C_2 \wr S_p$. $\endgroup$ – François Brunault Oct 25 '20 at 18:42
  • $\begingroup$ @Joe Silverman , yes I agree. $\endgroup$ – loup blanc Oct 26 '20 at 14:25

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