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Galois group of a biquadratic extension∗ rm50† 2013-03-22 0:13:35 This article proves that biquadratic extensions correspond precisely to Galois 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 ∗ hGaloisGroupOfABiquadraticExtensioni created: h2013-03-2i by: hrm50i version: h40184i Privacy setting: h1i hTheoremi h11R16i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 corresponding to the three subgroups of√V4 of order 2. √Thus each of these is a quadratic extension. Suppose E1 = F ( D1 ), E2 = F ( D2 ) where neither D1 nor D2 is a square in F . The fact that E√1 6= E2 implies as above that D1 D2 is also not a square in F (in fact E3 = F√ ( D1√ D2 ). Thus E1 E2 ) E1 , E2 , and is of degree 4 over F , so K = E1 E2 = F ( D1 , D2 ). 2