structural:expansion, information:highlighting
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. 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 = is odd, and there are no nonlinear irreducible real polynomials of odd degree, we must have L = R, thus and are powers of 2. Assuming by way of contradiction that > 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 = 1, and therefore K = C, which completes the proof.