Algebraic closure

• The fundamental theorem of algebra states that the algebraic closure of the field of real numbers is the field of complex numbers.
• The algebraic closure of the field of rational numbers is the field of algebraic numbers.
• There are many countable algebraically closed fields within the complex numbers, and strictly containing the field of algebraic numbers; these are the algebraic closures of transcendental extensions of the rational numbers, e.g. the algebraic closure of Q(π).
• For a finite field of prime power order q, the algebraic closure is a countably infinite field that contains a copy of the field of order qⁿ for each positive integer n (and is in fact the union of these copies).[6]

Let ${\displaystyle S=\{f_{\lambda }|\lambda \in \Lambda \}}$ be the set of all monic irreducible polynomials in K[x]. For each ${\displaystyle f_{\lambda }\in S}$, introduce new variables ${\displaystyle u_{\lambda ,1},\ldots ,u_{\lambda ,d}}$ where ${\displaystyle d={\rm {degree}}(f_{\lambda })}$. Let R be the polynomial ring over K generated by ${\displaystyle u_{\lambda ,i}}$ for all ${\displaystyle \lambda \in \Lambda }$ and all ${\displaystyle i\leq {\rm {degree}}(f_{\lambda })}$. Write

${\displaystyle f_{\lambda }-\prod _{i=1}^{d}(x-u_{\lambda ,i})=\sum _{j=0}^{d-1}r_{\lambda ,j}\cdot x^{j}\in R[x]}$

with ${\displaystyle r_{\lambda ,j}\in R}$. Let I be the ideal in R generated by the ${\displaystyle r_{\lambda ,j}}$. Since I is strictly smaller than R, Zorn's lemma implies that there exists a maximal ideal M in R that contains I. The field K₁=R/M has the property that every polynomial ${\displaystyle f_{\lambda }}$ with coefficients in K splits as the product of ${\displaystyle x-(u_{\lambda ,i}+M),}$ and hence has all roots in K₁. In the same way, an extension K₂ of K₁ can be constructed, etc. The union of all these extensions is the algebraic closure of K, because any polynomial with coefficients in this new field has its coefficients in some K_n with sufficiently large n, and then its roots are in K_n+1, and hence in the union itself.

It can be shown along the same lines that for any subset S of K[x], there exists a splitting field of S over K.

An algebraic closure K^alg of K contains a unique separable extension K^sep of K containing all (algebraic) separable extensions of K within K^alg. This subextension is called a separable closure of K. Since a separable extension of a separable extension is again separable, there are no finite separable extensions of K^sep, of degree > 1. Saying this another way, K is contained in a separably-closed algebraic extension field. It is unique (up to isomorphism).[7]

The separable closure is the full algebraic closure if and only if K is a perfect field. For example, if K is a field of characteristic p and if X is transcendental over K, ${\displaystyle K(X)({\sqrt[{p}]{X}})\supset K(X)}$ is a non-separable algebraic field extension.

In general, the absolute Galois group of K is the Galois group of K^sep over K.[8]

• Algebraically closed field
• Algebraic extension
• Puiseux expansion
• Complete field

• Kaplansky, Irving (1972). Fields and rings. Chicago lectures in mathematics (Second ed.). University of Chicago Press. ISBN 0-226-42451-0. Zbl 1001.16500.
• McCarthy, Paul J. (1991). Algebraic extensions of fields (Corrected reprint of the 2nd ed.). New York: Dover Publications. Zbl 0768.12001.