alternative definition of algebraically closed


Proposition 1.

If K is a field, the following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath:

  1. (1)

    K is algebraically closedMathworldPlanetmath, i.e. every nonconstant polynomialPlanetmathPlanetmath f in K[x] has a root in K.

  2. (2)

    Every nonconstant polynomial f in K[x] splits completely over K.

  3. (3)

    If L|K is an algebraic extensionMathworldPlanetmath then L=K.

Proof.

If (1) is true then we can prove by inductionMathworldPlanetmath on degree of f that every nonconstant polynomial f splits completely over K. Conversely, (2) (1) is trivial.
(2) (3) If L|K is algebraic and αL, then α is a root of a polynomial fK[x]. By (2) f splits over K, which implies that αK. It follows that L=K.
(3) (1) Let fK[x] and α a root of f (in some extensionPlanetmathPlanetmathPlanetmath of K). Then K(α) is an algebraic extension of K, hence αK. ∎

Examples 1) The field of real numbers is not algebraically closed. Consider the equation x2+1=0. The square of a real number is always positive and cannot be -1 so the equation has no roots.
2) The p-adic field p is not algebraically closed because the equation x2-p=0 has no roots. Otherwise x2=p implies 2vpx=1, which is false.

Title alternative definition of algebraically closed
Canonical name AlternativeDefinitionOfAlgebraicallyClosed
Date of creation 2013-03-22 16:53:23
Last modified on 2013-03-22 16:53:23
Owner polarbear (3475)
Last modified by polarbear (3475)
Numerical id 8
Author polarbear (3475)
Entry type DerivationPlanetmathPlanetmath
Classification msc 12F05