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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!

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  • $\begingroup$ Check this post out for some hints and tips. It is essentially the same question. $\endgroup$ – Islands Oct 4 '13 at 21:20
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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.

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  • $\begingroup$ Got it. I did not think of the idea of using the independence over $\mathbb{Q}$ of $1$ and $\sqrt{2}$, but it is clear. Thanks. $\endgroup$ – mimicat Oct 4 '13 at 21:25

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