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Theorem. (Fundamental Theorem of Algebra) C is algebraically closed.

Proof) We note the two results lying outside the algebra:

• Every polynomial with odd degree and real coefficients has some real root. (by the intermediate value theorem)
• Every complex number has a square root in C. (due to the quadratic formula )

These results are translated into the following statements in algebraical terminology:

• [A] R is the unique, odd-dimensional field extension over R.
• [B] C is the the unique, 2-dimensional field extension over C.

To prove the theorem, suppose there is a finite field extension K over C. We shall show K = C. Since char R = 0 and any Galois field extension is a splitting field , one may assume K/R is Galois without loss of generality. 

Let G be the Galois group of K over R. Since C/R is 2-dimensional and [K:R] = [K:C][C:R], 2 divides [K:R] = |G|. Thus G has a 2-Sylow subgroup, say H. Then [G:H] is odd. By the fundamental theorem of Galois theory, there is a subextension F in K/R such that H = Gal(K/F) and [K:F] = |H|. Since [G:H]|H| = |G| = [K:R] = [K:F][F:R] =|H|[F:R], we have [F:R] =  [G:H]. By [A], it must be the case that F = R. Hence the degree of the extension K/R is a power of 2, and so is that of K/C since Gal(K/C) < G.

Now, we prove [K:C] = 1 by contradiction. Otherwise, Gal(K/C) has a subgroup of index 2 by the first Sylow theorem. Since K is Galois over C, too, we have a 2-dimensional extension field over C, say F, by the fundamental theorem of Galois theory. However, it contradicts [B]. Therefore [K:C] = 1 and so K = C. □