Editing Fundamental theorem of algebra (section)

====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<sup>''k''</sup> divides the degree ''n'' of ''p''(''z''). Let ''a'' be the coefficient of ''z<sup>n</sup>'' 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''<sub>1</sub>, ''z''<sub>2</sub>, ..., ''z<sub>n</sub>'' in ''F'' such that

:<math>p(z)=a(z-z_1)(z-z_2) \cdots (z-z_n).</math>

If ''k''&nbsp;=&nbsp;0, then ''n'' is odd, and therefore ''p''(''z'') has a real root. Now, suppose that ''n''&nbsp;=&nbsp;2''<sup>k</sup>m'' (with ''m'' odd and ''k''&nbsp;>&nbsp;0) and that the theorem is already proved when the degree of the polynomial has the form 2<sup>''k''&nbsp;−&nbsp;1</sup>''m''′ with ''m''′ odd. For a real number ''t'', define:

:<math>q_t(z)=\prod_{1\le i<j\le n}\left(z-z_i-z_j-tz_iz_j\right).</math>

Then the coefficients of ''q<sub>t</sub>''(''z'') are [[symmetric polynomial]]s in the ''z<sub>i</sub>'' with real coefficients. Therefore, they can be expressed as polynomials with real coefficients in the [[elementary symmetric polynomial]]s, that is, in −''a''<sub>1</sub>, ''a''<sub>2</sub>, ..., (−1)''<sup>n</sup>a<sub>n</sub>''. So ''q<sub>t</sub>''(''z'') has in fact ''real'' coefficients. Furthermore, the degree of ''q<sub>t</sub>''(''z'') is ''n''(''n''&nbsp;−&nbsp;1)/2&nbsp;=&nbsp;2<sup>''k''−1</sup>''m''(''n''&nbsp;−&nbsp;1), and ''m''(''n''&nbsp;−&nbsp;1) is an odd number. So, using the induction hypothesis, ''q<sub>t</sub>'' has at least one complex root; in other words, ''z<sub>i</sub>''&nbsp;+&nbsp;''z<sub>j</sub>''&nbsp;+&nbsp;''tz<sub>i</sub>z<sub>j</sub>'' 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<sub>i</sub>''&nbsp;+&nbsp;''z<sub>j</sub>''&nbsp;+&nbsp;''tz<sub>i</sub>z<sub>j</sub>'' and ''z<sub>i</sub>''&nbsp;+&nbsp;''z<sub>j</sub>''&nbsp;+&nbsp;''sz<sub>i</sub>z<sub>j</sub>'' are complex (for the same ''i'' and ''j''). So, both ''z<sub>i</sub>''&nbsp;+&nbsp;''z<sub>j</sub>'' and ''z<sub>i</sub>z<sub>j</sub>'' 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<sub>i</sub>'' and ''z<sub>j</sub>'' are complex numbers, since they are roots of the quadratic polynomial ''z''<sup>2</sup>&nbsp;−&nbsp; (''z<sub>i</sub>''&nbsp;+&nbsp;''z<sub>j</sub>'')''z''&nbsp;+&nbsp;''z<sub>i</sub>z<sub>j</sub>''.

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).<ref>Shipman, J. [http://www.jon-arny.com/httpdocs/Gauss/Shipman%20Intellig%20Mod%20p%20FTA.pdf Improving the Fundamental Theorem of Algebra].  ''The Mathematical Intelligencer'', volume&nbsp;29 (2007), number&nbsp;4, pp.&nbsp;9–14.</ref> 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''&nbsp;∈&nbsp;''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 {{nowrap|(''x''<sup>2</sup> + 1)<sup>''k''</sup>''f''(''x'')}} has a root, where ''k'' is chosen so that {{nowrap|deg(''f'') + 2''k'' ∈ ''I''}}).