Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / A (Introduction to Fields)

A splitting field of $f$ over $F$ is an extension field in which $f$ factors into linear factors and which is generated over $F$ by those roots. The generation requirement excludes unnecessary extra elements.

For existence, induct on the degree. An irreducible factor $p$ of $f$ gives a field $F_1=F[X]/(p)$, since an irreducible polynomial generates a maximal ideal in the polynomial ring over a field. The class $\alpha$ of $X$ is a root of $p$ and hence of $f$. By the factor theorem, $f=(X-\alpha )g$ over $F_1$, with $\mathrm{deg}g=\mathrm{deg}f-1$. Induction gives a splitting field of $g$ over $F_1$. It contains all roots of $f$; taking the subfield generated by these roots over $F$ gives exactly a splitting field of $f$. The degree-one case is immediate. Thus

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

All roots are $\pm \sqrt{2}$ and $\pm \sqrt{3}$, so the splitting field is $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. We have $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$.

To show that adjoining $\sqrt{3}$ doubles the degree, suppose $\sqrt{3}=a+b\sqrt{2}$ with rational $a,b$. Squaring and using irrationality of $\sqrt{2}$ gives $ab=0$. If $b=0$, then $a^2=3$; if $a=0$, then $b^2=3/2$. Neither is a rational square, as the prime exponents in a rational square are even. Thus $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. Its quadratic polynomial remains irreducible over that field, and the tower law givesA basis is $1,\sqrt{2},\sqrt{3},\sqrt{6}$.