Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 18H (Galois Theory)

Let $f$ be the common minimal polynomial. Evaluation givesso composing the two isomorphisms gives $\mathbb{Q}(\alpha )\cong \mathbb{Q}(\beta )$ and sends $\alpha$ to $\beta$.

Take $\alpha =\beta =\sqrt[4]{2}$. The field is real and contains the two roots $\pm \alpha$ of $X^4-2$, but not $\pm i\alpha$. Hence exactly two $\mathbb{Q}$-automorphisms are possible, while the degree is four.

In general, after choosing one isomorphism $\varphi :\mathbb{Q}(\alpha )\to \mathbb{Q}(\beta )$, every other one is uniquely $\varphi \circ \sigma$ for a $\mathbb{Q}$-automorphism $\sigma$ of $\mathbb{Q}(\alpha )$. ThusBy the automorphism-count divisibility theorem, this order divides $[\mathbb{Q}(\alpha ):\mathbb{Q}]$. Therefore $k$ always divides the degree.

Put $d=[L:K]$. If $\alpha =0$ the result is immediate, so suppose $\alpha =x^n\ne 0$ with $x\in L$. The field norm is multiplicative, sends an element of $K$ to its $d$th power, and commutes with powers. ThereforeChoose integers $r,s$ with $rd+sn=1$. ThenThe expression in parentheses belongs to $K$, so $\alpha$ is an $n$th power in $K$. This is coprime-degree descent for powers.

Let $\beta$ satisfy ${\beta }^p=\alpha$. Since $K=\mathbb{Q}(\alpha )\subseteq \mathbb{Q}(\beta )$,The polynomial $f(X^p)$ has degree $p[K:\mathbb{Q}]$ and has root $\beta$. It is irreducible exactly when $[K(\beta ):K]=p$.

If $\alpha ={\gamma }^p$ for some $\gamma \in K$, then $\gamma$ is a root of $f(X^p)$ of degree at most $[K:\mathbb{Q}]$, so $f(X^p)$ is reducible. Conversely, if it is reducible, thenSince $p$ is prime, $1\le d<p$ implies $\gcd (d,p)=1$. Applying coprime-degree descent for powers to $K(\beta )/K$ and ${\beta }^p=\alpha$ shows that $\alpha$ is a $p$th power in $K$. Hence