proof of characterization of perfect fields

Proposition 1

The following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath:

  1. (a)
  2. (b)

    Either charK=0 or charK=p and the Frobenius mapPlanetmathPlanetmath is surjectivePlanetmathPlanetmath.

Proof. Suppose (a) and not (b). Then we must have charK=p>0, and there must be aK with no p-th root in K. Let L be a splitting fieldMathworldPlanetmath over K for the polynomialPlanetmathPlanetmath Xp-a, and let αL be a root of this polynomial. Then (X-α)p=Xp-αp=Xp-a, which has coefficients in K. This means that the minimum polynomial for α over K must be a divisor of (X-α)p and so must have repeated roots. This is not possible since L is separable over K.

Conversely, suppose (b) and not (a). Let α be an element which is algebraic over K but not separable. Then its minimum polynomial f must have a repeated root, and by replacing α by this root if necessary, we may assume that α is a repeated root of f. Now, f has coefficients in K and also has α as a root. Since it is of lower degree than f, this is not possible unless f=0, whence charK=p>0 and f has the form:


with ao0. By (b), we may choose elements biK, for 0in-1 such that bip=ai. Then we may write f as:


Since f(α)=0 and since the Frobenius map xxp is injectivePlanetmathPlanetmath, we see that


But then α is a root of the polynomial


which has coefficients in K, is non-zero (since bo0), and has lower degree than f. This contradicts the choice of f as the minimum polynomial of α.

Title proof of characterization of perfect fields
Canonical name ProofOfCharacterizationOfPerfectFields
Date of creation 2013-03-22 14:47:44
Last modified on 2013-03-22 14:47:44
Owner mclase (549)
Last modified by mclase (549)
Numerical id 6
Author mclase (549)
Entry type Proof
Classification msc 12F10