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$\DeclareMathOperator{\split}{split}$ I am interested in a list consisting of different points of view that one could take to look at finite fields. So far I found four different aspects (see the list below). I would appreciate it if this list could be checked for correctness and expanded by other points of view.

Let $\mathbb{F}_{p^n}$ be a finite field, $\eta \in \mathbb{F}_{p^n}$ a primitive element, $f \in \mathbb{F}_p[x]$ an irreducible polynomial of degree $n$ dividing $x^{p^n}-x$ and $\alpha_i,\dots,\alpha_n$ all the roots of $f$. Denote the splitting field of a polynomial $g$ over some field $\mathbb{F}$ by $\split(g,\mathbb{F})$. Then $\mathbb{F}_{p^n}$ is isomorphic to

  • $\split(x^{p^n}-x,\mathbb{F}_p)$
  • $\mathbb{F}_p(\alpha_1,\dots,\alpha_n)$
  • $\mathbb{F}_p(\eta)$
  • $\mathbb{F}_p[x]/(f)$
Schief
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  • $\mathbb F_p[\alpha_i]$ is all of the field, for any one $\alpha_i,$ whether $\alpha_i$ is primitive or not. – Thomas Andrews Jan 23 '22 at 15:04
  • Everywhere you’ve written $\mathbb F_p(\cdot)$ you could have written $\mathbb F_p[\cdot].$ – Thomas Andrews Jan 23 '22 at 15:06
  • More generally, if $d\mid n$ then there is is an irreducible polynomial $f_d\in\mathbb F_{p^d}[x]$ of degree $n/d$ with $$\mathbb F_{p^n}\cong \mathbb F_{p^d}[x]/(f_d).$$ – Thomas Andrews Jan 23 '22 at 15:10
  • In fact, every where you’ve used $\mathbb F_p,$ there is an equivalent for $\mathbb F_{p^d}.$ – Thomas Andrews Jan 23 '22 at 15:15
  • @ThomasAndrews Ouch, I misread what $\alpha_i$ were. Sorry. –  Jan 23 '22 at 15:19
  • You can always realize a finite extension $L/K$ as matrices, which gives you yet another way to look at finite fields. – schiepy May 28 '25 at 19:39
  • I'm not sure if this qualifies, but a normal basis tower binary field for $GF(2^n)$, where $n$ is even, has elements of the form like $( a_0 X^2 + a_1 X)$, where $a_0$ and $a_1$ are $n / 2$ bit sub-fields, while $X$ is an element of some other field. Link to a question showing an actual example. – rcgldr May 28 '25 at 22:02
  • The list is missing the possibility $\Bbb{F}\simeq\mathcal{O}/\mathfrak{p}$, where $\mathcal{O}$ is the ring of algebraic integers of a number field $K$, and $\mathfrak{p}$ is a prime ideal of $\mathcal{O}$ containing the (rational) integer $p$. The catch here is that controlling $n$ is not as straight forward as one might wish. For example, with $\mathcal{O}=\Bbb{Z}[i]$ (=the ring of integers of the quadratic extension field $\Bbb{Q}(i)$) yields (by this process) either $\Bbb{F}p$ or $\Bbb{F}{p^2}$ according to the residue class of $p$ modulo $4$. – Jyrki Lahtonen May 29 '25 at 02:40

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