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Let ${\mathbb F}_{q}$ be a given finite field. How many couples of $n\times n$ matrices $\left(A,B\right)$ over ${\mathbb F}_{q}$, such that $\gcd\left(\mu_{A}\left(\lambda\right),\mu_{B}\left(\lambda\right)\right)=1$, are there? where $\mu_{A}\left(\lambda\right),\mu_{B}\left(\lambda\right)$ are the minimal polynomials of $A,B$, respectively.

Specifically, how many such couples exist over ${\mathbb F}_{2}$?

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    $\begingroup$ This can probably be estimated accurately using the techniques of subsection 3.2 of arxiv.org/abs/1909.03666 i.e. giving an "arithmetic" formula for the number of matrices with a given characteristic polynomial and then sieving for the pairs of polynomials that are relatively prime (or using contour integral methods). $\endgroup$
    – Will Sawin
    Commented Apr 15, 2023 at 20:28
  • $\begingroup$ I think the asymptotic should be $q^{2n^2} \prod_{\pi \textrm{ irreducible}\in \mathbb F_q[t]} (1 - (1- 1/ \prod_{j=1}^{\infty} (1-q^{-j \deg \pi} ))^2)$ or something like that. $\endgroup$
    – Will Sawin
    Commented Apr 15, 2023 at 20:30
  • $\begingroup$ Dear Will! Thank you for the reference and comments. Note however that you have raised the following question: for a given polynomial, how many matrices are there, that have the given polynomial as their minimal polynomial? Specifically, how can I construct a matrix that has the given polynomial as its minimal polynomial? $\endgroup$
    – Yossi Peretz
    Commented Apr 18, 2023 at 8:40
  • $\begingroup$ That's a nastier question than the characteristic polynomial version but you can use the classification of modules over a PID (since each matrix turns $\mathbb F_q^n$ into a $\mathbb F_q[t]$-module) to attack it. For construction, just make a block-diagonal matrix where one block is the companion matrix and each other block is the companion matrix of a polynomial divisor. $\endgroup$
    – Will Sawin
    Commented Apr 18, 2023 at 10:10
  • $\begingroup$ There doesn't appear to be a reason to not use characteristic polynomials, since $\gcd(\mu_A, \mu_B) = 1$ iff $\gcd(\chi_A, \chi_B) = 1$. $\endgroup$
    – Peter Taylor
    Commented Apr 18, 2023 at 10:33

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