>>435-436
どうも、コメントありがとう

>“fの次数の2進付値”についての帰納法の話ならあり得ると思うし実際>>416の方法がまさにそれ

へー、発想が違うね

>もちろん代数的にやるなら素直にシローの定理まで使う方がいいと思うけど、数学科の学生向けでないならこういう方法もアリかもしれない

へー、発想が違うね、と、今改めて見ると、別の筋で「From Galois Theory」があって、
”Let G be the Galois group of this extension, and let H be a Sylow 2-subgroup of G, so that the order of H is a power of 2, and the index of H in G is odd. ”
とありますな
へー、すごいね

(>>417より)
https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
Fundamental theorem of algebra
2.3 Algebraic proofs
2.3.2 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.[12] Let K/C be a finite extension. Since the 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 0 is separable). Let G be the Galois group of this extension, and let H be a Sylow 2-subgroup of G, so that the order of H is a power of 2, and the 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) = H. As [L:R] = [G:H] is odd, and there are no nonlinear irreducible real polynomials of odd degree, we must have L = R, thus [K:R] and [K:C] are powers of 2. Assuming by way of contradiction that [K:C] > 1, we conclude that the 2-group Gal(K/C) contains a subgroup of index 2, so there exists a subextension M of C of degree 2. However, C has no extension of degree 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.