Describe the Galois Group of a field extension I'm struggling to understand the basics of Galois theory. One of the things I don't understand is how to actually derive automorphisms of a field extension. Let's say you had a simple problem:
$x^2-3$ over $\mathbb{Q}$ has splitting field $\mathbb{Q}(\sqrt{3}$) correct? 
Once I have this how would I go about finding the Galois group in this problem? Isn't there an unlimited number of ways I can assign elements of the field? If the Galois group is the set of ALL automorphisms aren't I just limited by my imagination?

It is important to remember that these automorphisms fix the base field. In particular, they fix the polynomial $x^2 - 3$, so they must send roots of that polynomial to other roots. 
They are also field maps and, since they fix the base field, they are in particular linear over the base field - Q, in your example . Since the splitting field is generated by $1$ and $\sqrt(3)$ over $Q$ (as a vector space), knowing where $1$ and $\sqrt(3)$ are sent determines the entire map. But $1$ must be sent to $1$ and $\sqrt(3)$ to another root of your polynomial. So you can see that there are only a few possibilities.
It is not always the case that you can just send roots to other ones - there may be subtle relationships among the roots that prevent every map that permutes roots from being automorphisms. One trick for determining the automorphisms (which happens frequently enough to be worth remembering) is that Galois theory tells you that in the Galois case there are exactly as many automorphism as the degree of the minimal polynomial ( which is degree 2 in this case, since you can factor 1 out of $x^3 - 1$). If you can show that there are exactly that many possible automorphisms by the consideration in the previous paragraph and exhibit them (here the identity and $\sqrt(3) \to -\sqrt(3)$), then you are done.
