# nLab algebraically closed field

A field $k$ is algebraically closed if every non-constant polynomial (with one variable and coefficients from $k$) has a root in $k$. It follows that every polynomial of degree $n$ can be factored uniquely as

$p=c\prod _{i=1}^{n}\left(\mathrm{x}-{a}_{i}\right),$p = c \prod_{i = 1}^n (\mathrm{x} - a_i) ,

where $c$ and the ${a}_{i}$ are elements of $k$.

The fundamental theorem of algebra is, classically, the statement that the complex numbers form an algebraically closed field $ℂ$. Arguably, this theorem is not entirely algebraic; the algebraic portion is that $R\left[\mathrm{i}\right]$ is algebraically closed whenever $R$ is a real-closed field. Unusually, this algebraic portion is not (as stated) valid in constructive mathematics, while the result the real numbers form a real closed field $ℝ$ is constructively valid with the proper definitions.

An algebraic closure of an arbitrary field $k$ is an algebraically closed field $\overline{k}$ equipped with a field homomorphism (necessarily an injection) $i:k\to \overline{k}$ such that $\overline{k}$ is an algebraic extension? of $k$ (which means that every element of $\overline{k}$ is the root of some non-zero polynomial with coefficients only from $k$). For example, $ℂ$ is an algebraic closure of $ℝ$. The axiom of choice proves the existence of $\overline{k}$ for any field $k$, as well as its uniqueness up to isomorphism over $k$. However, note that $\overline{k}$ need not be unique up to unique isomorphism, so it's not really appropriate to speak of the algebraic closure of $k$. For example, complex conjugation is a nontrivial automorphism of $ℂ$ over $ℝ$.

Without choice, the existence and uniqueness of algebraic closures may fail; see

Revised on September 25, 2012 18:28:25 by Todd Trimble (68.198.214.232)