# Determining automorphism group

Let $K=\mathbb{Q}[\sqrt{2},\sqrt{3}]$. In the book Abstract Algebra by Dummit and Foote, page 563, the author gave an example of finding the Galois group of $K/\mathbb{Q}$. Here are some arguments that relate to my question :

The extension $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ is Galois over $Q$ since it is the splitting field of $(x^2-2)(x^2-3)$. Any automorphism $\sigma$ is completely determined by its action on the generators $\sqrt{2},\sqrt{3}$, which must be mapped to $\pm\sqrt{2}, \pm\sqrt{3}$, respectively.

Hence the only possibilities for automorphisms are the maps: \begin{cases} \sqrt{2}\mapsto \sqrt{2} \\ \sqrt{3}\mapsto \sqrt{3} \end{cases} \begin{cases} \sqrt{2}\mapsto -\sqrt{2} \\ \sqrt{3}\mapsto \sqrt{3} \end{cases} \begin{cases} \sqrt{2}\mapsto \sqrt{2} \\ \sqrt{3}\mapsto -\sqrt{3} \end{cases} \begin{cases} \sqrt{2}\mapsto -\sqrt{2} \\ \sqrt{3}\mapsto -\sqrt{3} \end{cases}

My question is :

1. Why does the automorphism have to map $\sqrt{2}$ to $\pm\sqrt{2}$, and $\sqrt{3}$ to $\pm\sqrt{3}$ ? Why can't we choose an automorphism like \begin{array}{l l} \sqrt{2}\mapsto \sqrt{3} \\ -\sqrt{2}\mapsto -\sqrt{3} \end{array} I tried to prove that the above map is not an automorphism but my attempt failed. Where am I wrong ?
2. Let $\sigma$ be the 2nd automorphism, $\tau$ be the 3rd automorphism, then what is: $\sigma(-\sqrt{2})$ and $\tau(-\sqrt{3})$ ?

P/S : I do not know the latex code of the bracket, mod please help me. Thanks

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If $\alpha^2=2$ then $\sigma(\alpha)^2=\sigma(\alpha^2)=\sigma(2)=2$. In general, if $f(\alpha)=0$ with $f\in\mathbb Q[X]$, then also $f(\sigma\alpha)=0$ (because $\sigma f=f$).
For your second question note that with $u,v\in \mathbb Q$ you have $\sigma(u\alpha+v\beta)=u\sigma(\alpha)+v\sigma(\beta)$.
Thank @Hagen von Eitzen for your very nice answer and edit. I have one more question : How can we determine the Galois group of the splitting field ofsome seperable polynomial ? For example, determining the Galois group of the splitting field of $x^3-2=0$, there are 9 possibilites for the automorphism but the Galois group is only of order 6 ? –  knot Jan 1 '13 at 15:37