Find root of polynomial over finite field

Question

Let $\mathbb{k} = \mathbb{F}_2[\alpha]$, where $\alpha$ is a root of $x^4+x+1$. I'm stuck with finding roots of $x^2 + x + 1$ in $\mathbb{k}$.

I'd be greateful for any advice.

Answer 1

It is easy to see that there $x^2+x+1$ splits in $\Bbb{k}$, since $\Bbb{F}_4$ sits naturally inside $\Bbb{k}$.

Finding the roots explicitly can be done in this case by inspection. We have $\alpha^4=\alpha+1$, so $(\alpha^2+\alpha)^2=\alpha^4+\alpha^2=(\alpha^2+\alpha)+1$, i.e. $\alpha^2+\alpha$ is a root. The other root is $(\alpha^2+\alpha)^2$.

Answer 2

Another approach, slightly more systematic than Sarastro’s: Granting that $X^4+X+1$ is irreducible over $\Bbb F_2$, you look at your element $\alpha$ such that $\alpha^4+\alpha+1=0$, and look at its powers. You know that $\alpha^{15}=1$, since $\Bbb F_{16}^*$ is cyclic of order $15$, and check that neither $\alpha^3$ nor $\alpha^5$ is equal to $1$, so that $\alpha$ itself is of order $15$ multiplicatively. Then its fifth power must be a primitive cube root of unity, and the two such are the roots of $X^2+X+1$. As it happens $\alpha^5=\alpha^2+\alpha$, and your other root of $X^2+X+1$ is $\alpha^2+\alpha+1$.

Answer 3

You can check that the polynomial for which $\alpha$ is a solution is irreducible, so $\alpha$ is of degree 4. This means that $\mathbb{F}[\alpha]$ is generated by $\langle \alpha^0, \alpha, \alpha^2 \alpha^3\rangle$ since $\alpha$ satisfies $\alpha^4 +\alpha+1=0$, what would we get by plugging in $\alpha^2$ into our equation of interest?