proof of characterization of perfect fields


Proposition 1

The following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath:

  1. (a)
  2. (b)

    Either char⁡K=0 or char⁡K=p and the Frobenius mapPlanetmathPlanetmath is surjectivePlanetmathPlanetmath.

Proof. Suppose (a) and not (b). Then we must have char⁡K=p>0, and there must be a∈K with no p-th root in K. Let L be a splitting fieldMathworldPlanetmath over K for the polynomialPlanetmathPlanetmath Xp-a, and let α∈L 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 char⁡K=p>0 and f has the form:

f=xp⁢n+an-1⁢xp⁢(n-1)+…+a1⁢xp+a0.

with ao≠0. By (b), we may choose elements bi∈K, for 0≤i≤n-1 such that bip=ai. Then we may write f as:

f=(xn+bn-1⁢xn-1+…+b1⁢x+b0)p.

Since f⁢(α)=0 and since the Frobenius map x↦xp is injectivePlanetmathPlanetmath, we see that

αn+bn-1⁢αn-1+…+b1⁢α+b0=0

But then α is a root of the polynomial

xn+bn-1⁢xn-1+…+b1⁢x+b0

which has coefficients in K, is non-zero (since bo≠0), 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