Every finite field of characteristic $2$ ist given by $\mathbb{F}\_2[x]/P(x)$ for some irreducible polynomial $P\in \mathbb{F}\_2[x]$.


For small degree, a simple algorithm gives a way to find $P$. Is there a way to give explicitely the polynomial $P$ for large degrees? I would be interested in a formula like
$$x^n+a\_{n-1} x^{n-1}+\cdots +a\_1 x+a\_0$$
where $a\_0,\ldots,a\_{n-1}\in \mathbb{F}\_2[n]$. This might be too much to ask for all degrees, so an infinite sequence of irreducible polynomials would be good.


EDIT: Some sequences have already be given. If you have more to share (with the proof of the irreducibility, either by giving the reference or explaining it), please go ahead.



Explicit large finite fields in characteristic $2$
For $n=3^k$, the polynomial $p=x^{2n}+x^n+1$ is irreducible over $\mathbb{F}\_2$.


**Proof:** By Rabin's irreducibilty test, it suffices to check that $p|x^{2^{2n}}-x$ and $\gcd(p,x^{2^{2n/3}}-x)=1$.


Note that the order of $x$ mod $p$ is $3n=3^{k+1}$. Hence, since $3^{k+1}|4^{n}-1$ by lifting-the-exponent lemma, we have $p|x^{2^{2n}-1}-1$.


Again by lifting-the-exponent lemma, we have $4^{3^{k-1}}-1=3^km$ for $m$ not divisible by $3$. Hence $x^{4^{3^{k-1}}-1}=x^{nm}\equiv x^n\ \text{or}\ x^{2n} \pmod{p}$ since $x^n$ has order $3$. As $(x^n-1)x^n=(x^{2n}-1)x^{2n}=1$, this mean that $x^{2^{2n/3}-1}-1$ is invertible in $\mathbb{F}\_2[x]/(p)$. As $x$ is also invertible, we have shown that $\gcd(p,x^{2^{2n/3}}-x)=1$, as wanted.


