proof of characterization of perfect fields
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|
|Date of creation||2013-03-22 14:47:44|
|Last modified on||2013-03-22 14:47:44|
|Last modified by||mclase (549)|