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.

$K=F(\sqrt{D_{1}},\sqrt{D_{2}})$ 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 $\operatorname{Gal}(K/F)\cong V_{4}$ ;

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

$i d$

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

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

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

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
Canonical name	GaloisGroupOfABiquadraticExtension
Date of creation	2013-03-22 17:44:06
Last modified on	2013-03-22 17:44:06
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	5
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 11R16