Show that the Galois extension is isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2$

Question

$E$ is the splitting field of $f(x)\in\mathbb{Q}[x]$ s.t. $[E:\mathbb{Q}] = 4$. $E$ has 2 distinct quadratic subextensions, $\mathbb{Q}(\sqrt{m}),\mathbb{Q}(\sqrt{n})$. Prove that $Gal(E/\mathbb{Q}) \cong \mathbb{Z}_2\times \mathbb{Z}_2$ and that every quadratic subextension of $E$ is either $\mathbb{Q}(\sqrt{m}),\mathbb{Q}(\sqrt{n})$ or $\mathbb{Q}(\sqrt{mn})$.

My attempt:

1) $E \cong \mathbb{Q}(\sqrt{m},\sqrt{n})$ and the minimal polynomial for $\sqrt{m},\sqrt{n}$ is $(x^2-m)(x^2-n)$ the roots of which are $\pm \sqrt{m},\pm \sqrt{n}$. Let $Gal(E/\mathbb{Q}) = \{e,\sigma,\tau,\sigma \tau\}$. Let $\sigma(\sqrt{m}) = -\sqrt{m}$. Now I need to show that $\sigma(\sqrt{n}) = \sqrt{n}$ to conclude that the order of $\sigma = 2$. I don't know how use to use the fact that $\sigma$ is a homomorphism to conclude that?

2) I can see that the only subextensions $K$ of $E$ such that $4 = [E:\mathbb{Q}] = [E:K][K:\mathbb{Q}]$ and $[E:K] = [K:\mathbb{Q}] = 2$ are $K=\mathbb{Q}(\sqrt{m}),\mathbb{Q}(\sqrt{n})$ or $\mathbb{Q}(\sqrt{mn})$ but how can I formally say that? Thanks!

Answer 1

For your attempt, saying $\sigma(\sqrt m) = -\sqrt m$ is fine, but you don't show that $\sigma(\sqrt n) = \sqrt n$, you define $\sigma$ to be that way. Then you define $\tau$ by $\tau(\sqrt m) = \sqrt m, \tau(\sqrt n) = -\sqrt n$. This makes it so that $\sigma\tau(\sqrt x) = \tau\sigma(\sqrt x) = -\sqrt x$ for $x = m, n$.

Note that you can't show that $\sigma(\sqrt n) = \sqrt n$. Because there are two elements $\eta$ in the Galois group which have the property $\eta(\sqrt m) = -\sqrt m$. It is (in theory) arbitrary which one of them you happen to call $\sigma$. So you have to tell us which one you choose. It can't be "proven" in any way.

I think the easiest way to solve this problem is by the fundamental theorem of Galois theory. You know the Galois group has $4$ elements, and there are therefore only two possible groups it could be. Each nontrivial subextension corresponds to a non-trivial, proper subgroup of the Galois group, and since there are two distinct non-trivial subextensions, you have two distinct, nontrivial, proper subgroups. This eliminates $\Bbb Z_4$ as a contender, without even looking at the elements of the Galois group.

Answer 2

For the first part, you could simply argue from the following facts : (i) $\Bbb Z_2 \times \Bbb Z_2$ is the only group of order 4 with two distinct subgroups of index 2 and (ii) the Fundamental Theorem Of Galois Theory.

For the second part, again by the Fundamental Theorem Of Galois Theory, $E| \Bbb Q$ has 3 distinct quadratic subextenions and $\Bbb Q(\sqrt{m}), \Bbb Q(\sqrt{n})$ are two such (distinct), thus all you are left to show that $ \sqrt{mn} \notin\Bbb Q$ ( Since, $\sqrt{mn} \in \Bbb Q(\sqrt{m}) \implies \sqrt{n} \in \Bbb Q(\sqrt{m}) \implies \Bbb Q(\sqrt{n}) \subseteq \Bbb Q(\sqrt{m})$ contradicting that those two are distinct quadratic extensions) . But if $\sqrt{mn} \in \Bbb Q$ then $\sqrt{mn} \in \Bbb Q(\sqrt{m}) \implies \sqrt{n} \in \Bbb Q(\sqrt{m})$ contradiction!

Answer 3

Instead of saying "the minimal polynomial of $\sqrt m,\sqrt n$ is $(x^2-m)(x^2-n)$ (which is not true, you can only have a minimal polynomial for an algebraic element), you should say that the conjugates of $\sqrt m$ are $\{\sqrt m,-\sqrt m\}$ and similarly for $n$, we may define the automorphisms by specifying how it maps the generators. For example, $\sigma(\sqrt m)=-\sqrt m$ and $\sigma(\sqrt n)=\sqrt n$ defines an automorphism because it respects addition and multiplication, indeed, $\sigma(\sqrt n\sqrt n)=\sqrt n\sqrt n=n=\sigma(n)$, and $\sigma(\sqrt m\sqrt n)=-\sqrt m\sqrt n=\sigma(\sqrt m)\sigma(\sqrt n)$.

Since the Galois group is abelian, by Galois correspondence, any normal subgroup $H$ of $\text{Gal}(E/\mathbb Q)$ corresponds uniquely to a Galois subextension $\text{Gal}(E/F)$ with degree equals to the cardinality of $H$. And in fact $\text{Gal}(F/\mathbb Q)$ is also Galois with Galois group isomorphic to $\text{Gal}(E/\mathbb Q)/H$. So to find all degree 2 extension $F/\mathbb Q$, it is equivalent to finding all index 2 subgroup of $\mathbb Z\times\mathbb Z$.