I need an example for a field which have characteristic two and seven different elements which satisfies the equation $a^7 = 1$. I've been trying to find an answer, by brute force, searching over the web with no luck.
I think it should be some isomorphic field to $GF(2^3)$, am I right?
Thanks.
How about the field with $8$ elements, $GF(2^3)$? The multiplicative group has order $7$, so they are the $7^{th}$ roots of $1$ you require, i.e. all non-zero elements satisfy $x^7=1$--recall $GF(p^n)$ is exactly the splitting field of $x^{p^n}-x$, so all elements of $GF(2^3)$ are roots of $x(x^7-1)$. And any irreducible cubic will generate it, so since $x^3+x+1$ has no roots in $\Bbb F_2$ we see that $\Bbb F_2[x]/(x^3+x+1)$ is such a field.