Galois group of a biquadratic extension
This article proves that biquadratic extensions correspond precisely to Galois extensions with Galois group isomorphic to the Klein -group (at least if the characteristic of the base field is not ). More precisely,
Proof. Suppose first that condition (1) holds. Then since neither nor is a square in . Now obviously
and so . If , then , so and . Thus or . If , then is a square. If , then is a square. In any case, this is a contradiction. Thus is a quadratic extension of . So . But is the splitting field for , since the splitting field must contain both square roots, and the polynomial obviously splits in , so has four elements
and is thus isomorphic to .
Now assume that condition (2) holds. Since , there must be three intermediate subfields between and of degree over corresponding to the three subgroups of of order . Thus each of these is a quadratic extension. Suppose where neither nor is a square in . The fact that implies as above that is also not a square in (in fact . Thus , and is of degree over , so .
|Title||Galois group of a biquadratic extension|
|Date of creation||2013-03-22 17:44:06|
|Last modified on||2013-03-22 17:44:06|
|Last modified by||rm50 (10146)|