# Prove that if $\operatorname{Gal}(K/F) \simeq \Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\mathbb{Z}$, then $K = F(\sqrt{a},\sqrt{b})$ for some $a,b \in F$

Let $F$ be a field of characteristic not $2$, and let $K$ be a Galois extension with $[K:F] = 4$. Prove that if $\operatorname{Gal}(K/F) \simeq \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, then $K = F(\sqrt{a},\sqrt{b})$ for some $a,b \in F$

I showed that

If $F$ is a field of characteristic not $2$, and $K$ is an extension of $F$ with $[K: F] = 2$, then $K = F (\sqrt{a})$ for some $a \in F$.

Using an idea like this.

But I couldn't use that same idea to prove this case. Can someone help me?

EDIT. Since $\operatorname{Gal}(K/F)$ is Klein group, there is subgroup $H$ of order $2$ and by the Fundamental Theorem of Galois Theory, there is a subfield $L$ of $K/F$ with $L \leftrightarrow H$ such that $[L:F] = [G:H] = 2$ and so, $[K:F] = [K:L][L:F] = [K:F(\sqrt{b})][F(\sqrt{b}):F]$. Can I to apply this result to $[K:F(\sqrt{b})]$ too?

• @AlexNolte read the question Jun 27, 2018 at 17:24
• @AlexNolte, Is not. By hypothesis, the characteristic is not equal to $2$. Jun 27, 2018 at 17:24

The Galois group is $\left<\alpha,\beta\right>$ where $\alpha$ and $\beta$ have order $2$. The fixed field of $\alpha$ has degree $2$ over $F$, so is $F(\sqrt a)$ for some $a$. Similarly the fixed field of $\beta$ is $F(\sqrt b)$ for some $b$. If follows that $K$ must be $F(\sqrt a,\sqrt b)$.