What is the automorphism group $Aut_{\mathbb{Q}} \mathbb{Q}(\sqrt{3}+\sqrt{2})$? I am trying to understand the automorphism group $Aut_{\mathbb{Q}} \mathbb{Q}(\sqrt{3}+\sqrt{2})$, aka the the automorphism group of $\mathbb{Q}(\sqrt{3}+\sqrt{2})$ over the rationals $\mathbb{Q}$. I know the following:


*

*$\mathbb{Q}(\sqrt{3}+\sqrt{2}) =\mathbb{Q}(\sqrt{3},\sqrt{2})$ 

*the minimal Polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $x^2-2$ and the minimal Polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$ is $x^2-3$, hence the degree of  $\mathbb{Q}(\sqrt{3}+\sqrt{2}) =\mathbb{Q}(\sqrt{3},\sqrt{2})$ over the rationals is 4.

*the minimal polynomial of $\mathbb{Q}(\sqrt{3}+\sqrt{2})$ is $x^4-10x^2+1$. 

*The automorphism group of $\mathbb{Q}(\sqrt{3},\sqrt{2})$ has $4$ elements, namely the obvious ones. (positive root to negative root, all $4$ possibilities.)


Now my questions: Does $Aut_{\mathbb{Q}} \mathbb{Q}(\sqrt{3}+\sqrt{2})$ also have four elements? It must have, right? But what are these elements? how do I find four of them? I either have all 4 roots in $\mathbb{Q}(\sqrt{3}+\sqrt{2})$in which case I'd have $4!$ automorphism, or $3$ roots giving me $3!$ automorphisms, and so on. What am I missing here?
 A: The extension is degree $4$ and Galois, as you can see from your polynomial. So the order of the Galois group is indeed 4. By the tower law, you also know that 
$$
[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{3})]=
[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})]=2
$$
By the Galois correspondence theorem, this gives you two distinct subgroups of index 2, differentiating the Galois group as $\mathbb{Z}_2\times \mathbb{Z}_2$, rather than $\mathbb{Z}_4$, the other option.
A: What you are missing is the fact that a field automorphism for an algebraic extension always permutes the zeros of the minimal polynomial - but the permutations allowed are not arbitrary, they are those which always yield consistent results when applied to ANY sum or product of field elements.  For example, in $\mathbb{Q}(\sqrt{3},\sqrt{2})$, consider the product expression (1 + $\sqrt{2}$) * $\sqrt{2}$.  If your permutation maps from $\sqrt{2}$ to -$\sqrt{2}$, then you may apply this map to both sides of the product before multiplying, OR multiply and apply the map to the product, and the result will be the same.  This is definitely not true if your permutation maps from $\sqrt{2}$ to $\sqrt{3}$, for example; therefore that permutation does NOT extend to a field automorphism.
For your extension the four roots of the minimal polynomial are $\sqrt{3}+\sqrt{2}$, $\sqrt{3}-\sqrt{2}$, $-\sqrt{3}+\sqrt{2}$, and $-\sqrt{3}-\sqrt{2}$.  What permutations of these four values induce field automorphisms?  Once you know that, there's your group.
