Given a finite field $\mathbb{F}$ of order $2^n$, how to construct a field of order $2^{2n}$?

Question

Specifically, I would like to construct a field of order $2^{2n}$ with elements being $2\times2$ matrices whose entries are elements of $\mathbb{F}$.

I know the complex numbers can be represented as $2\times2$ real matrices, and I was trying to do something similar. I think the key is to find an expression $f(a, b)$ with $a, b\in\mathbb{F}$ such that $f(a, b)=0$ if and only if $a=b=0$. Then make sure the expression of the determinant is exactly $f(a, b)$ to get $p^{2n}-1$ invertible $2\times2$ matrices. I haven't had any luck with this idea though.

Any suggestions would be appreciated. I don't really want a specific answer since I want to do the thinking, but I'm hoping someone could guide me towards the right direction.

Answer 1

In general this takes a bit of tinkering, and more often than not those finite fields are constructed a bit differently. Offering hints/ideas for the following two special cases (both have plenty of details remaining for you!):

• If $n$ is odd, then $f(a,b)=a^2+ab+b^2$ works. This is because if $f(a,b)=0, a\neq0,$ then it follows that $(b/a)^3=1$. But in a field of order $2^n, 2\nmid n,$ this means that $(b/a)=1$, and that helps prove the claim.
• There is a recursive construction that covers the cases when $n$ is a power of two. It is outlined in this answer by yours truly. In that latter case you can use the polynomial $f(a,b)=a^2+ab+\alpha_kb^2$ instead. See that question for the description of $\alpha_k\in\Bbb{F}_{2^{2^k}}$.

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Edit: Sorry about not suggesting the following right away. Consider the polynomial (aka the trace) $$ tr(x)=x+x^2+x^4+x^8+\cdots+x^{2^{n-1}}. $$ It is of degree $2^{n-1}$, so it cannot vanish identically on $\Bbb{F}_{2^n}$. Let $\beta\in\Bbb{F}_{2^n}$ be such that $tr(\beta)\neq0$. Then it follows that $tr(\beta)=1$, and also that $x^2+x+\beta$ has no zeros in $\Bbb{F}_{2^n}$. Therefore $$ f(a,b)=a^2+ab+\beta b^2 $$ works the same way as in the other cases.