Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / A / A2 / b

Over ${\mathbb{F}}_2$, the cubic takes value one at both zero and one, so it has no root. A reducible cubic over a field has a linear factor and therefore a root; hence this cubic is irreducible. Adjoin a root:This extension has basis $1,\alpha ,{\alpha }^2$, degree three and eight elements. Squaring preserves polynomial equations with coefficients in ${\mathbb{F}}_2$, so ${\alpha }^2$ and ${\alpha }^4$ are roots too. The nonzero element $\alpha$ is not one; the multiplicative group has prime order seven, so $\alpha$ has order seven. The exponents $1,2,4$ are distinct modulo seven, making these three roots distinct. Thus they are all the roots and already lie in $K$. Therefore