linguistic:paraphrasing, information:classification, information:highlighting
By induction

As mentioned above, it suffices to check the statement "every non-constant polynomial p (z) with real coefficients has a complex root". This statement can be proved by induction on the greatest non-negative integer k such that 2 divides the degree n of p (z). Let a be the coefficient of z in p (z) and let F be a splitting field of p (z) over C; in other words, the field F contains C and there are elements z, z,..., z in F such that

If k = 0, then n is odd, and therefore p (z) has a real root. Now, suppose that n = 2 m (with m odd and k > 0) and that the theorem is already proved when the degree of the polynomial has the form 2 m ′ with m ′ odd. For a real number t, define:

Then the coefficients of q (z) are symmetric polynomials in the z with real coefficients. Therefore, they can be expressed as polynomials with real coefficients in the elementary symmetric polynomials, that is, in − a, a,..., (−1) a. So q (z) has in fact real coefficients. Furthermore, the degree of q (z) is n (n − 1)/2 = 2 m (n − 1), and m (n − 1) is an odd number. So, using the induction hypothesis, q has at least one complex root; in other words, z + z + tz z is complex for two distinct elements i and j from {1,..., n }. Since there are more real numbers than pairs (i, j), one can find distinct real numbers t and s such that z + z + tz z and z + z + sz z are complex (for the same i and j). So, both z + z and z z are complex numbers. It is easy to check that every complex number has a complex square root, thus every complex polynomial of degree 2 has a complex root by the quadratic formula. It follows that z and z are complex numbers, since they are roots of the quadratic polynomial z − (z + z) z + z z.

Joseph Shipman showed in 2007 that the assumption that odd degree polynomials have roots is stronger than necessary; any field in which polynomials of prime degree have roots is algebraically closed (so "odd" can be replaced by "odd prime" and this holds for fields of all characteristics). For axiomatization of algebraically closed fields, this is the best possible, as there are counterexamples if a single prime is excluded. However, these counterexamples rely on −1 having a square root. If we take a field where −1 has no square root, and every polynomial of degree n ∈ I has a root, where I is any fixed infinite set of odd numbers, then every polynomial f (x) of odd degree has a root (since (x + 1) f (x) has a root, where k is chosen so that deg(f) + 2 k ∈ I).