Galois group of a biquadratic extension
This article proves that biquadratic extensions correspond precisely to Galois extensions^{} with Galois group^{} isomorphic^{} to the Klein $4$group ${V}_{4}$ (at least if the characteristic^{} of the base field^{} is not $2$). More precisely,
Theorem 1.
Let $F$ be a field of characteristic $\mathrm{\ne}\mathrm{2}$ and $K$ a finite extension^{} of $F$. Then the following are equivalent^{}:

1.
$K=F(\sqrt{{D}_{1}},\sqrt{{D}_{2}})$ for some ${D}_{1},{D}_{2}\in F$ such that none of ${D}_{1},{D}_{2}$, or ${D}_{1}{D}_{2}$ is a square in $F$.

2.
$K$ is a Galois extension of $F$ with $\mathrm{Gal}(K/F)\cong {V}_{4}$;
Proof. Suppose first that condition (1) holds. Then $[F(\sqrt{{D}_{1}}):F]=[F(\sqrt{{D}_{2}}):F]=2$ since neither ${D}_{1}$ nor ${D}_{2}$ is a square in $F$. Now obviously
$$[K:F]=[F(\sqrt{{D}_{1}},\sqrt{{D}_{2}}):F(\sqrt{{D}_{1}})][F(\sqrt{{D}_{1}}):F]\le 4$$ 
and so $[K:F(\sqrt{{D}_{1}})]\le 2$. If $K=F(\sqrt{{D}_{1}})$, then $\sqrt{{D}_{2}}\in F(\sqrt{{D}_{1}})$, so $\sqrt{{D}_{2}}=a+b\sqrt{{D}_{1}}$ and ${D}_{2}={a}^{2}+{b}^{2}{D}_{1}+2ab\sqrt{{D}_{1}}$. Thus $a=0$ or $b=0$. If $b=0$, then ${D}_{2}$ is a square. If $a=0$, then ${D}_{1}{D}_{2}={b}^{2}{D}_{1}^{2}$ is a square. In any case, this is a contradiction^{}. Thus $K$ is a quadratic extension of $F(\sqrt{{D}_{1}})$. So $[K:F]=4$. But $K$ is the splitting field^{} for $({x}^{2}{D}_{1})({x}^{2}{D}_{2})$, since the splitting field must contain both square roots, and the polynomial^{} obviously splits in $K$, so $G=\mathrm{Gal}(K/F)$ has four elements
$id$  $\sigma :\{\begin{array}{cc}\sqrt{{D}_{1}}\mapsto \sqrt{{D}_{1}}\hfill & \\ \sqrt{{D}_{2}}\mapsto \sqrt{{D}_{2}}\hfill & \end{array}$  $\tau :\{\begin{array}{cc}\sqrt{{D}_{1}}\mapsto \sqrt{{D}_{1}}\hfill & \\ \sqrt{{D}_{2}}\mapsto \sqrt{{D}_{2}}\hfill & \end{array}$  $\sigma \tau :\{\begin{array}{cc}\sqrt{{D}_{1}}\mapsto \sqrt{{D}_{1}}\hfill & \\ \sqrt{{D}_{2}}\mapsto \sqrt{{D}_{2}}\hfill & \end{array}$ 
and is thus isomorphic to ${V}_{4}$.
Now assume that condition (2) holds. Since $\mathrm{Gal}(K/F)\cong {V}_{4}$, there must be three intermediate subfields^{} ${E}_{1},{E}_{2},{E}_{3}$ between $F$ and $K$ of degree $2$ over $F$ corresponding to the three subgroups^{} of ${V}_{4}$ of order $2$. Thus each of these is a quadratic extension. Suppose ${E}_{1}=F(\sqrt{{D}_{1}}),{E}_{2}=F(\sqrt{{D}_{2}})$ where neither ${D}_{1}$ nor ${D}_{2}$ is a square in $F$. The fact that ${E}_{1}\ne {E}_{2}$ implies as above that ${D}_{1}{D}_{2}$ is also not a square in $F$ (in fact ${E}_{3}=F(\sqrt{{D}_{1}{D}_{2}})$. Thus ${E}_{1}{E}_{2}\u228b{E}_{1},{E}_{2}$, and is of degree $4$ over $F$, so $K={E}_{1}{E}_{2}=F(\sqrt{{D}_{1}},\sqrt{{D}_{2}})$.
Title  Galois group of a biquadratic extension 

Canonical name  GaloisGroupOfABiquadraticExtension 
Date of creation  20130322 17:44:06 
Last modified on  20130322 17:44:06 
Owner  rm50 (10146) 
Last modified by  rm50 (10146) 
Numerical id  5 
Author  rm50 (10146) 
Entry type  Theorem 
Classification  msc 11R16 