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[parent] Galois group of a biquadratic extension (Theorem)

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_1D_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

$\displaystyle [K:F]=[F(\sqrt{D_1},\sqrt{D_2}):F(\sqrt{D_1})][F(\sqrt{D_1}):F]\leq 4$
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^2D_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_1D_2=b^2D_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{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 $ {\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\neq E_2$ implies as above that $ D_1D_2$ is also not a square in $ F$ (in fact $ E_3=F(\sqrt{D_1D_2})$. Thus $ E_1E_2\supsetneq E_1, E_2$, and is of degree $ 4$ over $ F$, so $ K=E_1E_2 =F(\sqrt{D_1},\sqrt{D_2})$.



"Galois group of a biquadratic extension" is owned by rm50.
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Cross-references: implies, order, subgroups, degree, subfields, polynomial, square roots, contain, splitting field, quadratic extension, contradiction, square, the following are equivalent, finite extension, field, base field, characteristic, isomorphic, Galois group, Galois extensions, biquadratic extensions
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This is version 2 of Galois group of a biquadratic extension, born on 2008-01-11, modified 2008-04-22.
Object id is 10184, canonical name is GaloisGroupOfABiquadraticExtension.
Accessed 263 times total.

Classification:
AMS MSC: 11R16 (Number theory :: Algebraic number theory: global fields :: Cubic and quartic extensions)
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