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Sorry if this question is too elementary for MO.

Let $p$ be a prime and $F_p$ a field with $p$ elements and $F_{p^n}$ the field with $p^n$ elements Then we can choose an irreducible factor $f$ of degree n of a suitable cyclotomic polynomial in $F_p [x]$, take the companion matrix $C_f$ of $f$ and then $F_{p^n}$ can be representated as $\{0,C_f,C_f^2,...,C_f^{p^n-1} \}$.

In practise this gives a pretty nice presentation of $F_{p^n}$ via matrices over $F_p$ in my opinion.

For example, look at $F_3$ and $f:=x^2+x+2$. Then $C_f= \begin{pmatrix} 0 & 1 \\ 1 & 2 \end{pmatrix}$ and $F_9= \{ 0, C_f^i | i=1,...,8 \}.$

Question: Is there a nice explicit presentation of the algebraic closure of $F_p$ via matrices over $F_p$ that is discussed somewhere in the literature?

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    $\begingroup$ If by "finitely supported matrices" you mean matrices with all but finitely many entries nonzero, it is not possible for such matrices to represent a unital ring, since (if the matrices become arbitrarily large) the unit must be the identity matrix which is not finitely supported. You can in principle write down a matrix representation of $\overline{\mathbb{F}_p}$ by choosing a basis for it over $\mathbb{F}_p$. On that theme see: mathoverflow.net/questions/483353/… $\endgroup$ Commented Jul 10 at 19:25
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    $\begingroup$ I'd be interested in any natural description of the algebraic closure of $\mathbb{F}_p$. You can identify its elements with periodic sequences of elements of $\mathbb{F}_p$; then addition is entrywise addition, and Frobenius (= taking the $p$-th power) is cyclic shift by $1$ step, but I have no idea how to describe multiplication (and there is probably not only one way). But I suspect that no one else knows either, or else there would be no need for such an awkward kludge as Conway polynomials. $\endgroup$ Commented Jul 10 at 19:27
  • $\begingroup$ @QiaochuYuan Right, thanks. I deleted "finitely supported". $\endgroup$ Commented Jul 10 at 20:39
  • $\begingroup$ @darijgrinberg isn't the union $\bigcup_{n}\mathbb{F}_{q^{n!}}$ natural as an algebraic closure of $\mathbb{F}_{q}$? $\endgroup$ Commented Jul 10 at 20:44
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    $\begingroup$ @Ben: But you need to construct that union in the first place. $\endgroup$ Commented Jul 10 at 20:53

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I don't know if this counts as nice or explicit, but I'll take a shot. For a matrix $A\in M_n(\mathbb F_p)$ and $n|m$ or $m=\infty$ denote by $d_m(A)$ the matrix of the form $$\left[\begin{array}{ccc}A&&\\&A&\\&&\ddots\end{array}\right]\in M_m(\mathbb F_p).$$

First choose $f_2\in\mathbb F_p[x]$ monic irreducible of degree 2 and let $C_2=C_{f_2}$ be its companion matrix. Let $\mathbb F_{p^2}\subset M_2(\mathbb F_p)$ be the $\mathbb F_p$-subspace spanned by $I_2,C_2$. Now choose $f_3\in\mathbb F_{p^2}[x]$ monic irreducible of degree 3 and let $C_3=C_{f_2}\in M_3(\mathbb F_{p^2})=M_6(\mathbb F_p)$ be its companion matrix. We take $$\mathbb F_{p^6}=\langle d_6(C_2)^iC_3^j\,:\,0\le i\le 1,\,0\le j\le 2\rangle_{\mathbb F_p}\subset M_\infty(\mathbb F_p)$$ and continue recursively: at each step we have $\mathbb F_{p^{n!}}$ already constructed, choose a monic irreducible $f_{n+1}\in\mathbb F_{p^{n!}}[x]$ of degree $n+1$, let $C_{n+1}=C_{f_{n+1}}\in M_{n+1}(\mathbb F_{p^{n!}})= M_{(n+1)!}(\mathbb F_p)$ be its companion matrix and define $$\mathbb F_{p^{(n+1)!}}=\langle d_{(n+1)!}(C_2)^{i_2}\cdots d_{(n+1)!}(C_{n+1})^{i_{n+1}}\,:\,0\le i_2\le 1,\ldots,0\le i_{n+1}\le n\rangle_{\mathbb F_p}\subset M_{(n+1)!}(\mathbb F_p).$$ Finally take $\overline{\mathbb F_p}=\bigcup_{n=1}^\infty\{d_{\infty}(A):\,A\in\mathbb F_{p^{n!}}\}\subset M_\infty(\mathbb F_p)$

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