Field extension of odd degree whose Galois group is trivial Let $n\geq 1$ be an odd integer. Show there exists a field extension $L$ over $\Bbb{Q}$ of degree $n$ such that $|\mathrm{Gal}(L/\Bbb{Q})|= 1$. 
 A: 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.
A: The answer of Crostul is right only for odd $n$. For an even $n$ it doesn't work, because all the roots are $ \sqrt[n]{2}e^{2 \pi i k/n}$ for $k$ from $0$ to $ n-1.$
So if n is even, in the exponent we get $\frac{ 2 \pi i K}{n}= \frac{ \pi i k}{n/2}$ for $K=0,...,n-1$.
At this moment  for $K_1 = 0,... n/2-1$ and $K_2= n/2 ,...,n$  we realize that $e^{\frac{\pi i k_1}{n/2}}=-e^{\frac{\pi i k_2}{n/2}}$ due to the fact that $e^{\frac{n/2 \pi i}{n/2}}=e^{\pi i}=-1$
Hence $ \sqrt[n]{2}, -\sqrt[n]{2} \in \mathbb{Q}(\sqrt[n]{2})$
and there is no longer a unique root in $\mathbb{Q}(\sqrt[n]{2})$
