proof of fundamental theorem of algebra
If let be a root of in some extension of . Let be a Galois closure of over and set . Let be a Sylow 2-subgroup of and let (the fixed field of in ). By the Fundamental Theorem of Galois Theory we have , an odd number. We may write for some , so the minimal polynomial is irreducible over and of odd degree. That degree must be 1, and hence , which means that , a 2-group. Thus is also a 2-group. If choose such that , and set , so that . But any polynomial of degree 2 over has roots in by the quadratic formula, so such a field cannot exist. This contradiction shows that . Hence and , completing the proof.
|Title||proof of fundamental theorem of algebra|
|Date of creation||2013-03-22 13:09:39|
|Last modified on||2013-03-22 13:09:39|
|Last modified by||scanez (1021)|