# Galois group of a biquadratic extension

```Galois group of a biquadratic extension∗
rm50†
2013-03-22 0:13:35
extensions with Galois group isomorphic to the Klein 4-group V4 (at least if the
characteristic of the base field is not 2). More precisely,
Theorem 1. Let F be a field of characteristic 6= 2 and K a finite extension of
F . Then the following are equivalent:
√
√
1. K = F ( D1 , D2 ) for some D1 , D2 ∈ F such that none of D1 , D2 , or
D1 D2 is a square in F .
2. K is a Galois extension of F with Gal(K/F ) ∼
= V4 ;
√
Proof.
Suppose first that condition (1) holds. Then [F ( D1 ) : F ] =
√
[F ( D2 ) : F ] = 2 since neither D1 nor D2 is a square in F . Now obviously
p
p
p
p
[K : F ] = [F ( D1 , D2 ) : F ( D1 )][F ( D1 ) : F ] ≤ 4
√
√
√
√
√
and so
2. If K = F √
( D1 ), then D2 ∈ F ( D1 ), so D2 =
√ [K : F ( D1 )] ≤
a + b D1 and D2 = a2 + b2 D1 + 2ab D1 . Thus a = 0 or b = 0. If b = 0, then
D2 is a square. If a = 0, then D1 D2 = b2 D12 is a square.
√ In any case, this is
a contradiction. Thus K is a quadratic extension of F ( D1 ). So [K : F ] = 4.
But K is the splitting field for (x2 − D1 )(x2 − D2 ), since the splitting field
must contain both square roots, and the polynomial obviously splits in K, so
G = Gal(K/F ) has four elements
(√
id
σ:
√
D1 →
7 − D1
√
√
D2 →
7
D2
(√
√
D1 →
7
D1
√
τ: √
D2 →
7 − D2
(√
στ :
√
√
D1 →
7 − D1
√
D2 →
7 − D2
and is thus isomorphic to V4 .
Now assume that condition (2) holds. Since Gal(K/F ) ∼
= V4 , there must
be three intermediate subfields E1 , E2 , E3 between F and K of degree 2 over F
created: h2013-03-2i by: hrm50i version:
h40184i Privacy setting: h1i hTheoremi h11R16i
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