Editing Algebraically closed field (section)

==Examples==
As an example, the field of [[real number]]s is not algebraically closed, because the polynomial equation <math>x^2+1=0</math> has no solution in real numbers, even though all its coefficients (1 and 0) are real.  The same argument proves that no subfield of the real field is algebraically closed; in particular, the field of [[rational number]]s is not algebraically closed. By contrast, the [[fundamental theorem of algebra]] states that the field of [[complex number]]s is algebraically closed. Another example of an algebraically closed field is the field of (complex) [[algebraic number]]s.

No [[finite field]] ''F'' is algebraically closed, because if ''a''<sub>1</sub>, ''a''<sub>2</sub>, ..., ''a<sub>n</sub>'' are the elements of ''F'', then the polynomial (''x''&nbsp;&minus;&nbsp;''a''<sub>1</sub>)(''x''&nbsp;&minus;&nbsp;''a''<sub>2</sub>)&nbsp;⋯&nbsp;(''x''&nbsp;&minus;&nbsp;''a''<sub>''n''</sub>)&nbsp;+&nbsp;1
has no zero in ''F''. However, the union of all finite fields of a fixed characteristic ''p'' (''p'' prime) is an algebraically closed field, which is, in fact, the [[algebraic closure]] of the field <math>\mathbb F_p</math> with ''p'' elements.

The field <math>\mathbb{C}(x)</math> of rational functions with complex coefficients is not closed; for example, the polynomial <math>y^2 - x</math> has roots <math>\pm\sqrt{x}</math>, which are not elements of <math>\mathbb{C}(x)</math>.