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The finite field of order $p^n$ is isomorphic to $(\mathbb Z/p \mathbb Z)[X]/(P)$, where $P$ is an irreducible polynomial in $(\mathbb Z/p \mathbb Z)[X]$ of degree $n$. This describes every finite field up to isomorphism.

My question is, what are alternate ways of describing these finite fields?

For example, Conway gave an alternate way of describing the fields $GF(2^{2^n})$ (see this example answer).

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    $\begingroup$ Try $A/\mathfrak m$ for $A$ any finitely generated $\mathbf Z$-algebra and $\mathfrak m$ a maximal ideal in $A$. $\endgroup$ – KConrad Jun 8 '18 at 21:15
  • $\begingroup$ @KConrad, as suggested perhaps by the username, it seems that @‍PyRulez might be searching for a computationally effective presentation. $\endgroup$ – LSpice Jun 8 '18 at 23:04
  • $\begingroup$ Maybe you can define a multiplication on $\Bbb F_p^2$ via $ (x,y)(x',y') = (xx'-yy', xy' + x'y)$ and this might give you a field isomorphic to $\Bbb F_{p^2}$. $\endgroup$ – Watson Jun 12 '18 at 19:54
  • $\begingroup$ @Watson, of course "maybe" here means "if and only if $p \equiv -1 \pmod4$" (i.e., if and only if $-1$ is not a square already in $\mathbb F_p$). $\endgroup$ – LSpice Jun 13 '18 at 8:53
  • $\begingroup$ @LSpice : yes, obviously. I just wanted the OP to have a thought about it. $\endgroup$ – Watson Jun 13 '18 at 9:28
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In ONAG chapter 6, Conway gives a way of defining the finite field $2^{2^n}$ for any natural $n$. The domain is $\{0,1,\dots, 2^{2^n}-1\}$. Addition is defined by the two rules:

  1. The sum of $2^x$ and $2^y$ with $x \ne y$ is just the regular $2^n+2^m$. (So $13 = 8 + 4 + 1$.)
  2. The sum of two equal numbers is $0$.

Multiplication is likewise defined by two rules.

  1. The product of $2^{2^x}$ and $2^{2^y}$ with $x \ne y$ is just their ordinary product.
  2. The square of $2^{2^x}$ is the ordinary product of $\frac 32$ and $2^{2^x}$.

This uniquely defines the field. For example, using only the field axioms and above rules, we conclude $$5 \times 9 = (4+1)(4\times2+1) =\\ 4^2 \times 2 + 4 \times 2 + 4 + 1 = 6 \times 2 + 8 + 4 + 1 =\\ (4 + 2) \times 2 + 13 = 4 \times 2 + 2^2 + 13 =\\8 + 3 + 13 = 6$$

Also see this addition and multiplication table.

Conway goes on to extend this to the set $\{\alpha:\alpha < \beta\}$ for certain ordinals $\beta$, and to the class of all ordinals to create the curious Field $\text{On}_2$.

One nice thing about this construction is it shows show $GF(2^{2^n})$ is isomorphic to a subfield of $GF(2^{2^m})$ for any $n \le m$.

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    $\begingroup$ It seems a bit strange to post a question and an answer simultaneously. Rather, it probably makes more sense to make this part of the question itself, so that people can see what kind of answer you want (which wasn't clear to me from the question, at least). $\endgroup$ – LSpice Jun 8 '18 at 20:34
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    $\begingroup$ @LSpice I edited the question to link to this answer. I thought the answer was a little big to include in the question. $\endgroup$ – PyRulez Jun 8 '18 at 20:36

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