Existence of algebraic closures

Every field admits an algebraic closure.
Existence of algebraic closures

Let KK be a .

Theorem (Existence of algebraic closure). There exists a K/K\overline{K}/K such that:

  1. K\overline{K} is algebraically closed (every nonconstant polynomial in K[x]\overline{K}[x] splits into linear factors), and
  2. K/K\overline{K}/K is an (equivalently, every element of K\overline{K} is an over KK).

Such an extension K\overline{K} is called an of KK. A standard proof uses Zorn’s lemma (hence the ) to build a maximal algebraic extension and then shows it must be algebraically closed.

Moreover, any two algebraic closures of KK are KK-isomorphic; see .

Examples

  1. C\mathbb{C} is an algebraic closure of R\mathbb{R}: it is algebraically closed, and [C:R]=2[\mathbb{C}:\mathbb{R}]=2, so every complex number is algebraic over R\mathbb{R}.

  2. The field Q\overline{\mathbb{Q}} of algebraic numbers (elements algebraic over Q\mathbb{Q}) is an algebraic closure of Q\mathbb{Q}.

  3. An algebraic closure of Fp\mathbb{F}_p can be realized as the union

    Fp=n1Fpn \overline{\mathbb{F}_p}=\bigcup_{n\ge1}\mathbb{F}_{p^n}

    inside a fixed ambient algebraic closure. This viewpoint is compatible with the .