Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 18H (Galois Theory)

An algebraic element $\alpha$ over $K$ is a root of a nonzero polynomial in $K[T]$. Its degree is the degree of its minimal polynomial, equivalently $[K(\alpha ):K]$. Since $\alpha$ satisfies $T^2-{\alpha }^2$ over $K({\alpha }^2)$, the tower law gives $[K(\alpha ):K({\alpha }^2)]\le 2$ and makes that degree divide the odd integer $[K(\alpha ):K]$. It must therefore be one, proving $\boxed{K(\alpha )=K({\alpha }^2)}$.

For the addition assertion, let polynomial relations of degrees $r,s$ express ${\alpha }^r$ and ${\beta }^s$ in terms of lower powers. The finite-dimensional $K$-space spanned by ${\alpha }^i{\beta }^j$ for $0\le i<r$, $0\le j<s$ contains one and is closed under multiplication by $\alpha +\beta$. Thus $1,\alpha +\beta ,\dots ,(\alpha +\beta {)}^{rs}$ are linearly dependent. Their dependence is a nonzero polynomial relation over $K$, so $\alpha +\beta$ is algebraic.

A separable algebraic element has a minimal polynomial with distinct roots in an algebraic closure. A separable field extension is an algebraic extension all of whose elements are separable. In characteristic $p$, the extension ${\mathbb{F}}_p(t)/{\mathbb{F}}_p(t^p)$ is inseparable: the minimal polynomial of $t$ is $T^p-t^p$, irreducible over the smaller field but having the single root $t$ with multiplicity $p$.

If $L$ is a finite field of order $q$, every element of $L$ is a root of $T^q-T\in K[T]$, because $K$ contains its prime field. This polynomial has derivative $-1$, so it has no repeated roots. Every element's minimal polynomial divides it and is therefore separable. Hence $L/K$ is separable.