# Galois groups associated to matrices

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

• 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.) – R. van Dobben de Bruyn Oct 23 '20 at 22:25
• @R. van Dobben de Bruyn , thanks for your comment. See above my thoughts on this. – loup blanc Oct 25 '20 at 17:26
• 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. – Joe Silverman Oct 25 '20 at 17:47
• Using Magma I checked that for $p \leq 20$, the Galois group of $f_{2p}$ is $C_2 \wr S_p$. – François Brunault Oct 25 '20 at 18:42
• @Joe Silverman , yes I agree. – loup blanc Oct 26 '20 at 14:25