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

A splitting field of $f$ over $F$ is a field extension $K/F$ in which $f$ is a product of linear factors and which is generated over $F$ by its roots. The generation condition rules out adjoining unrelated elements.

For existence, induct on the degree of a polynomial. Choose an irreducible polynomial divisor $p$ of $f$. The quotient ring $F[X]/(p)$ is a field because $(p)$ is a maximal ideal; the class $\alpha$ of $X$ is a root of $p$, hence of $f$. Over this field extension, divide out $X-\alpha$ and apply induction to the remaining polynomial. This gives an extension in which $f$ splits. Inside it, the subfield generated by all roots is a splitting field. The induction starts with a linear polynomial, whose splitting field is already $F$. Thus every positive-degree polynomial has a splitting field; being monic is unnecessary for existence.

The cubic has no root in ${\mathbb{F}}_2$, since its values at $0$ and $1$ are both $1$. A cubic over a field is irreducible precisely when it has no root. Consequentlyis a field extension of degree $3$, with eight elements. The Frobenius endomorphism $x\mapsto x^2$ takes a root of a polynomial over ${\mathbb{F}}_2$ to another root, so the roots are $\alpha ,{\alpha }^2,{\alpha }^4$. They are distinct: equality between any two would make the Frobenius endomorphism orbit have length less than $3$, contradicting the minimal polynomial of $\alpha$. They all lie in $K$, and one of them generates $K$. Hence

The roots are $\pm \sqrt{2}$ and $\pm \sqrt{3}$, so the splitting field is $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. The first adjunction has degree $2$. To see that the second does too, suppose $\sqrt{3}=u+v\sqrt{2}$ with $u,v\in \mathbb{Q}$. Squaring and comparing rational and irrational parts gives $uv=0$ and $u^2+2v^2=3$. If $v=0$, then $u^2=3$; if $u=0$, then $v^2=3/2$. Neither is a rational square, as follows by comparing prime exponents in numerator and denominator. Thus $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. The tower law now yieldsA basis is $1,\sqrt{2},\sqrt{3},\sqrt{6}$.