I am new to Galois theory and while learning I got stuck analysing the following example:
We consider $$f: (x^2-2)(x^2-3) \in \mathbb{Q}[x].$$
The splitting field is $L = \mathbb{Q}(\sqrt{2},\sqrt{3})$ and since $L/\mathbb{Q}$ is a Galois field we know that $$|\operatorname{Gal}(L/\mathbb{Q})| = [L : \mathbb{Q}] = 4.$$
So far so good. Now I'm getting confused: The example states, that each $\phi \in \operatorname{Gal}(L/\mathbb{Q})$ need to permute the zeros of $x^2-2$ and $x^2-3$, that means $\phi(\sqrt{2}) = \pm \sqrt{2}$ and $\phi(\sqrt{3}) = \pm \sqrt{3}$. That yields to four possibilities and we go on showing that $\operatorname{Gal}(L/\mathbb{Q}) \cong Z_2 \times Z_2$.
Why are these the only possibilities for $\phi$. Why - for example - is $$\phi: \sqrt{2} \to -\sqrt{3}, \sqrt{3} \to -\sqrt{2}$$ not allowed? What's the reason for being restricted only on permutations of the zeros of the factors $x^2-2$ and $x^2-3$?
If $\alpha \in L$ and $p(\alpha)=0$ with $p(x) \in \mathbb Q[x]$, then $\phi(\alpha)$ must a root of $p(x)$ because $p(\phi(\alpha))=\phi(p(\alpha))=0$, since $\phi$ is a ring homomorphism.
Take $\alpha=\sqrt2$. Then $\phi(\alpha)$ must be a root of $p(x)=x^2-2$ and so can only be $\pm \sqrt 2$; it cannot be $\pm \sqrt 3$.