Calculating automorphism groups

Question

I've been given a field $L=\mathbb{Q}(\sqrt[4]{2})$ and I have to find $Aut(\frac{L}{Q})$.

Now i know that $\sqrt[4]{2}$ has minimal polynomial $x^4 -2$ over $\mathbb{Q}$, hence $[L : \mathbb{Q}] = 4$, so a $\mathbb{Q}$ basis for $L$ has 4 elements, with two of them being $1$ and $\sqrt[4]{2}$ but I cannot figure out other two.

In terms of finding the automorphisms $\sigma$, I know that if we let $\sigma \in$ $Aut(\frac{L}{Q})$ then $\sigma$ is determined by $\sigma(\sqrt[4]{2})$, and from here I am not quite sure how to calculate the group of automorphisms.

Answer 1

The important thing to remember here is that every automorphism sends an element to another root of its minimal polynomial.

As you correctly observe, $\sigma$ is determined by where it sends $\sqrt[4]{2}$. But it can only send $\sqrt[4]{2}$ to another root of its minimal polynomial, $x^4 - 2 = 0$. The roots of this are $\pm \sqrt[4]{2}$ and $\pm i\sqrt[4]{2}$. So there are four possible $\sigma$s. Now rule out two of those possibilities by saying that this field is entirely contained in the real numbers.

Answer 2

Ok a few things:

First, you are correct that $[L:\mathbb{Q}] = 4$. You can select any basis you want, so long as the elements you've chosen are linearly independent, so in particular, two of them don't have to be $1$ and $\sqrt[4]{2}$. A simple way of doing this for a simple extension $F[a]/F$ of degree $n$ is to simply take powers of the adjoining element: $\{1, a, a^2, \cdots, a^{n-1}\}$ (why?).

Next, it is a theorem that $[K:\mathbb{Q}] = |\operatorname{Aut}(K/\mathbb{Q})|$ if and only if $K$ is a splitting field of a polynomial over $\mathbb{Q}$. Otherwise, it is a theorem that the order of the automorphism group divides $[K:\mathbb{Q}]$. (These are important results that you should be able to find in your textbook.)

Since $\mathbb{Q}[\sqrt[4]{2}]$ is not a splitting field, then notice that this greatly restricts the possibilities for $\operatorname{Aut}(\mathbb{Q}[\sqrt[4]{2}]/\mathbb{Q})$.

Answer 3

There are only two automorphisms, its not Galois. The nontrivial is $\sigma(\sqrt[4]{2})=-\sqrt[4]{2}$