Field extension of odd degree whose Galois group is trivial

Question

Let $n\geq 1$ be an odd integer. Show there exists a field extension $L$ over $\Bbb{Q}$ of degree $n$ such that $|\mathrm{Gal}(L/\Bbb{Q})|= 1$.

Answer 1

Recall the following theorem:

Let $F/K$ be a finite separable extension of degree $n$. Then $L/K$ is normal if and only if the group of $K$-automorphisms of $L$ has $n$ elements.

In particular, this means that your extension cannot be normal (except for the trivial case $n =1$). Hence, I will use the notation $\mathrm{Gal}(L/K)$ for non-normal extensions, denoting the group of $K$-automorphisms of $L$.

For all odd $n \ge 3$, the extension $L= \Bbb{Q}(\sqrt[n]{2})$ does the job. This is because any $\Bbb{Q}$-automorphism of $L$ must fix $\Bbb{Q}$ (by definition), and must fix $\sqrt[n]{2}$ (it's the unique root of $X^n-2$ belonging to $L$); in particular, identity is the unique $\Bbb{Q}$-automorphism of $L$.

For $n=2$ this is impossible, since any quadratic extension of $\Bbb{Q}$ is normal, and its Galois group has two elements.