TITLE: Calculating automorphism groups
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: 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.