proof of characterization of perfect fields
Proposition 1
The following are equivalent:
-
(a)
Every algebraic extension
of K is separable
.
-
(b)
Either charK=0 or charK=p and the Frobenius map
is surjective
.
Proof.
Suppose (a) and not (b). Then we must have charK=p>0, and there must be a∈K
with no p-th root in K. Let L be a splitting field over K for the polynomial
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 charK=p>0 and f has the form:
f=xpn+an-1xp(n-1)+…+a1xp+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-1xn-1+…+b1x+b0)p. |
Since f(α)=0 and since the Frobenius map x↦xp is injective, we see that
αn+bn-1αn-1+…+b1α+b0=0 |
But then α is a root of the polynomial
xn+bn-1xn-1+…+b1x+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 |