proof of characterization of perfect fields
Proposition 1
The following are equivalent:
-
(a)
Every algebraic extension
of is separable
.
-
(b)
Either or and the Frobenius map
is surjective
.
Proof.
Suppose (a) and not (b). Then we must have , and there must be
with no -th root in . Let be a splitting field over for the polynomial
,
and let be a root of this polynomial.
Then , which has coefficients in .
This means that the minimum polynomial for over must be a divisor of
and so must have repeated roots. This is not possible since is separable
over .
Conversely, suppose (b) and not (a). Let be an element which is algebraic over but not separable. Then its minimum polynomial must have a repeated root, and by replacing by this root if necessary, we may assume that is a repeated root of . Now, has coefficients in and also has as a root. Since it is of lower degree than , this is not possible unless , whence and has the form:
with . By (b), we may choose elements , for such that . Then we may write as:
Since and since the Frobenius map is injective, we see that
But then is a root of the polynomial
which has coefficients in , is non-zero (since ), and has lower degree than . This contradicts the choice of as the minimum polynomial of .
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 |