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 $\eq 2$ 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 $E_{1}\eq E_{2}$ 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 $K=E_{1}E_{2}=F(\\sqrt{D_{1}},\\sqrt{D_{2}})$ .