Galois-theory - a question about the galois group

Question

I am dealing with galois theory at the moment and I came across with an example in the lecture and I got a question:

Let $K=\mathbb Q$ and $L=\mathbb Q(\sqrt{2},\sqrt{3})\subset \mathbb C$. Lets consider the field-extension $L/K$

Because of $[L:K]=[L:K(\sqrt{2})]\cdot [K(\sqrt{2}):K]=4$

Question 1: How we can argue that the minimal polynomial of $\sqrt{3}$ in $K(\sqrt{2})$ is still $x^2-3?$ I.e. we have to show that $\sqrt{3}\notin K(\sqrt{2})$. It seems trivial, but i would like to see an argument.

Because of $L$ is a splitting field of $f(X)=(x^2-3)(x^2-2)$ $L/K$ we have a galois-extension.

Question 2: Then we concluded that $|G|=|Gal(L/K)|=4$ but why? Does it holds that $|[L:K]|=|G|$ in any case? The definition of $Gal(L/K)$ is $Gal(L/K)=Aut_K(L)$, but i thought that $K$-automorphisms are already defined if we know the images of all generater of the extension, hence $|G|$ must be 2.

The rest is clear to me. We noticed that there are only $2$ groups with order $4$, namely $\mathbb Z_4$ and $\mathbb Z_2 \times \mathbb Z_2$. And we saw that this group cant be cyclic, hence must be $\mathbb Z_2 \times \mathbb Z_2$.

I hope someone can answer my questions. Also if someone knows a useful link to this topic I will be very glad about it.

Answer 1

Actually, you cannot argue that $[L/K(\sqrt 2)] = 2$ without knowing the degree of minimal polynomial of $\sqrt 3$ over $K(\sqrt 2)$. What you do know is that the degree is at most $2$, since it is the root of $x^2-3$. Thus, you can conclude $[L/K]\leq 4$ and that it is divisible by $2$, because $[K(\sqrt 2)/K] = 2$. So, $[L/K]$ is either $2$ or $4$. If it were $2$, it would imply $[L/K(\sqrt 2)] = 1$, thus $\sqrt 3\in\mathbb Q(\sqrt 2)$, and you can proceed as Andrea suggested, deriving contradiction and concluding $[L/K] = 4$.

For the second question, it is one of the first thing one proves in Galois theory that the number of automorphisms coincides with the degree of the extension. It can easily be proved by induction and the fact that $[L/K][K/E] = [L/E]$. But, you are mistaken when you say that there are only $2$ possibilities where automorphism can send generators. $\sqrt 2$ can be sent to $\pm\sqrt 2$ and $\sqrt 3$ to $\pm\sqrt 3$, thus $4$ possibilities, as wanted. This actually directly gives $\operatorname{Gal}(L/K) = \mathbb Z_2\times\mathbb Z_2$, as you stated.

Answer 2

(1) Suppose that $\sqrt{3}=a+b\sqrt{2}$ with $a,b\in\Bbb Q$. Taking squares $$ (a^2+2b^2)+2ab\sqrt{2}=3. $$ Can you derive a contradiction?

(2) The cardinality of the Galois group of a Galois extension always coincides with the degree. What are the generators in this case, and what is the effect of the automorphisms on them?