Algebraic closure

A field F is algebraically closed if there is a solution to every polynomial over F in one variable. Every field is contained in an algebraically closed field and there is a minimal one, called the algebraic closure of F.

The definition of algebraically closed has several other equivalent forms:

1. For every Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\in F} and every polynomial f(x), there is a $b\in F$ such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(b)=a} .
2. The only irreducible polynomials in F[x] have degree 1.
3. F has no finite extensions other than itself.
4. F is a maximal field of its transcendence degree.
5. F is the union of all its finitely-generated localized subrings.
6. Every algebraic variety defined over F has a point.
7. F is not finite and the automorphism group of F has cardinality strictly greater than the cardinality of F.

The complex numbers Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}} are algebraically closed by the Fundamental Theorem of Algebra, however the rational numbers Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{Q}} are not since x²+1 does not have a root. This same polynomial shows the real numbers $\mathbb {R}$ are also not algebraically closed, yielding as a corollary (by the equivalence) that ${\textrm {Aut}}(\mathbb {R} )$ has cardinality the continuum.