# 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 $\neq 2$ and $K$ a finite extension of $F$. Then the following are equivalent:

1. 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. 2.

$K$ is a Galois extension of $F$ with $\operatorname{Gal}(K/F)\cong V_{4}$;

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

 $[K:F]=[F(\sqrt{D_{1}},\sqrt{D_{2}}):F(\sqrt{D_{1}})][F(\sqrt{D_{1}}):F]\leq 4$

and so $[K:F(\sqrt{D_{1}})]\leq 2$. 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 $G=\operatorname{Gal}(K/F)$ has four elements

 $id$ $\sigma:\begin{cases}\sqrt{D_{1}}\mapsto-\sqrt{D_{1}}\\ \sqrt{D_{2}}\mapsto\sqrt{D_{2}}\end{cases}$ $\tau:\begin{cases}\sqrt{D_{1}}\mapsto\sqrt{D_{1}}\\ \sqrt{D_{2}}\mapsto-\sqrt{D_{2}}\end{cases}$ $\sigma\tau:\begin{cases}\sqrt{D_{1}}\mapsto-\sqrt{D_{1}}\\ \sqrt{D_{2}}\mapsto-\sqrt{D_{2}}\end{cases}$

and is thus isomorphic to $V_{4}$.

Now assume that condition (2) holds. Since $\operatorname{Gal}(K/F)\cong V_{4}$, 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 $E_{1}\neq E_{2}$ 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 $E_{1}E_{2}\supsetneq E_{1},E_{2}$, 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 GaloisGroupOfABiquadraticExtension 2013-03-22 17:44:06 2013-03-22 17:44:06 rm50 (10146) rm50 (10146) 5 rm50 (10146) Theorem msc 11R16