This question was given to us by instructor in class notes of Course on Field theory and there are no questions worked out in his notes which are posted online . So, I am looking for an explanation of his question which is solved( but explanation was not good).
Find group $\operatorname{Aut}_{\mathbb{Q}} { \mathbb{Q}{\bigl(\sqrt2, \sqrt3\,\bigr)}} $ ?
Answer posted : let $\sigma \in \operatorname{Aut}_{\mathbb{Q}} \bigl(\mathbb{Q}\bigl(\sqrt2, \sqrt3\,\bigr)\bigr) $ , then $\sigma$ is completely determined by $\sigma(\sqrt{2}\,)$ and $\sigma(\sqrt{3}\,)$.
Why so? Which result is used?
(solution continued) Only possibility is $\sigma(\sqrt2\,) = \sqrt2, -\sqrt2 \:$ and $\sigma(\sqrt3\,) =\sqrt3, -\sqrt3$.
Again no reasoning is given!!
So, there are 4-distinct $\Bbb Q$-homomorphisms (understood).
And $\operatorname{Aut}_{\mathbb{Q}}\bigl(\mathbb{Q}\bigl(\sqrt2, \sqrt3)\bigr) $ is isomorphic to $\mathbb{Z}_2 × \mathbb{Z}_2$.
Why not isomorphic to $\mathbb{Z}_4$?
I have asked 3 questions but I could really not understand the problem after trying to think a lot and I would be grateful for your help.
First question: $\sigma$ is completely determined by the images of $\sqrt 2$ and $\sqrt 3$ simply because any element in $\mathbf Q\bigl(\sqrt2,\sqrt 3\,\bigr)$ is a polynomial in $\sqrt 2$ and $\sqrt 3$, and $\sigma$ is a ring homomorphism.
Second question: the image of $\sqrt 2$ satisfies the same minimal equation, since $\sigma$ is a $\bf Q$-homomorphism. Same reason for $\sqrt 3$.
Last question: The group of automorphisms can't be $\mathbf Z/4\mathbf Z$ because no automorphism of this extension has order $4$, as the $\sigma^i(\sqrt 2\,)$ and the $\sigma^i(\sqrt 3)$ are roots of the quadratic polynomials $X^2-2 $ and $X^2-3$ respectively, and a quadratic polynomial has at most two roots.