Number of elements of an extension

Question

Given an irreducible polynomial, for example $h=x^4+x+1$ over $F=\{0,1\}$ with $E$ being the extension of $F$, what is the method of calculating the number of elements of $E$?

I'm really confused on how to actually calculate how many elements there are in the extension with just knowing the irreducible polynomial as well as the field. Any help would be appreciated

Answer 1

Let $E=\mathbf F_2[x]/(x^4+x+1)$. $E/\mathbf F_2$ is Galois. Indeed it is separable since $h'(x)=1$ and it is the splitting field of $h$.

Denote $\omega$ the congruence class of $x$ in the quotient. One checks the other roots are $\omega^2$, $1+\omega$ and $1+\omega^2$. To see this, note the equation can be rewritten as $x^4=x+1$ or $x^4+1=x$.

• $(\omega^2)^4=(\omega^4)^2=(\omega+1)^2=\omega^2+1$;
• $(1+\omega)^4=1+\omega^4=1+\omega+1=\omega$;
• $(1+\omega^2)^4=1+(\omega^4)^2=1+(\omega+1)^2=(1+\omega^2)+1.$