I am trying to compute the Galois group of $\mathbb{Q}[\sqrt{3},\sqrt{2}]/\mathbb{Q}$ in the following way:
First, $\mathbb{Q}[\sqrt{3},\sqrt{2}]/\mathbb{Q}$ is a Galois extension of the separable polynomial $(x^2-2)(x^2-3)$ (separable because $\text{char}(\mathbb{Q})=0$).
Write $G=\text{Gal}(\mathbb{Q}[\sqrt{3},\sqrt{2}]/\mathbb{Q})$. Every element of $G$ sends roots $x^2-2$ to roots of $x^2-2$ and roots of $x^2-3$ to roots of $x^2-3$ (because both polynomials are irreducible). I would like to show that all combinations are possible.
Take the identity automorphism $\mathbb{Q}\rightarrow\mathbb{Q}$. There are two ways to extend it to an automorphism $\mathbb{Q}[\sqrt{2}]\rightarrow\mathbb{Q}[\sqrt{2}]$. I would now like to extend it to an automorphism of $\mathbb{Q}[\sqrt{3},\sqrt{2}]$. The minimum polynomial of $\sqrt{3}$ over $\mathbb{Q}[\sqrt{2}]$ divides $x^2-3$. If it equals $x^2-3$ then we are done.
How do I show that the minimum polynomial of $x^2-3$ over $\mathbb{Q}[\sqrt{2}]$ is $x^2-3$.
One way to go is to show that $\mathbb{Q}[\sqrt{2}]$ does not contain a square root of $3$ by writing $(a+b\sqrt{2})^2=3$ and deriving a contradiction. I did not manage to do that, and anyway, I am hoping for a cleaner way.
(another approach: if we assume that the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}[\sqrt{2}]$ is quadratic, then $[\mathbb{Q}[\sqrt{2},\sqrt{3}:\mathbb{Q}]=4$, and then by Galois theory we have $|\text{Aut}(\mathbb{Q}[\sqrt{2},\sqrt{3}/\mathbb{Q})|=4$, meaning that all 4 possibilies must define automorphisms. This is the approach suggested in the answers to this related question, but they don't explain the part I have problems with).
Note: If you think that an entirely different approach to compute this Galois group is cleaner, please do show it!
Best Answer
Your approach looks good. And you only have to prove that $\sqrt{3} \notin \mathbb{Q}[\sqrt{2}]$. Well, otherwise $(a+b \sqrt{2})^2=3$ for some rationals $a,b$. Expanding and using that $1,\sqrt{2}$ are linearly independent, it follows that $a^2+2b^2=3$ and $2ab=0$. Hence $a=0$ and $2b^2=3$, or $b=0$ and $a^2=3$. But you probably already know that $\sqrt{3}$ and $\sqrt{3/2}$ are irrational.
More generally, if $m_1,\dotsc,m_r$ are coprime square-free integers, then $\mathbb{Q}[\sqrt{m_1}\,,\,\dotsc,\,\sqrt{m_r}\,]$ has degree $2^r$ over $\mathbb{Q}$, and the Galois group is isomorphic to $(C_2)^r$. See here.