Editing Fundamental theorem of algebra (section)

====From Galois theory====
Another algebraic proof of the fundamental theorem can be given using [[Galois theory]]. It suffices to show that '''C''' has no proper finite [[field extension]].<ref>A proof of the fact that this suffices can be seen [[Algebraically closed field#The field has no proper finite extension|here]].</ref> Let ''K''/'''C''' be a finite extension. Since the [[Normal extension#Normal closure|normal closure]] of ''K'' over '''R''' still has a finite degree over '''C''' (or '''R'''), we may assume [[without loss of generality]] that ''K'' is a [[normal extension]] of '''R''' (hence it is a [[Galois extension]], as every algebraic extension of a field of [[characteristic (algebra)|characteristic]] 0 is [[separable extension|separable]]). Let ''G'' be the [[Galois group]] of this extension, and let ''H'' be a [[Sylow theorems|Sylow]] 2-subgroup of ''G'', so that the [[order (group theory)|order]] of ''H'' is a power of 2, and the [[index of a subgroup|index]] of ''H'' in ''G'' is odd. By the [[fundamental theorem of Galois theory]], there exists a subextension ''L'' of ''K''/'''R''' such that Gal(''K''/''L'')&nbsp;=&nbsp;''H''. As [''L'':'''R''']&nbsp;=&nbsp;[''G'':''H''] is odd, and there are no nonlinear irreducible real polynomials of odd degree, we must have ''L''&nbsp;= '''R''', thus [''K'':'''R'''] and [''K'':'''C'''] are powers of 2. Assuming by way of contradiction that [''K'':'''C''']&nbsp;>&nbsp;1, we conclude that the [[p-group|2-group]] Gal(''K''/'''C''') contains a subgroup of index 2, so there exists a subextension ''M'' of '''C''' of degree&nbsp;2. However, '''C''' has no extension of degree&nbsp;2, because every quadratic complex polynomial has a complex root, as mentioned above. This shows that [''K'':'''C'''] = 1, and therefore ''K'' = '''C''', which completes the proof.