S.O.S. Mathematics CyberBoard

[caki]

Good afternoon,

How would I approach the following problem?

"Show that a field F of characteristic p (p a prime number) is perfect if and only if for every we can find a such that .

In Herstein's "Topics in Algebra", I am given the following definitions:

Separable element: An element a in an extension K of F is called separable over F if it satisfies a polynomial over F having no multiple roots.

Separable extension: An extension K of F is called separable over F if all its elements are separable over F.

Perfect field: A field F is called perfect if all finite extensions of F are separable.

Thanks very much,
caki

[caki]

Little update on my progress here.

Suppose that for every a in F there exists a b also in F such that but that F is not perfect. Thus, some finite extension K of F is not separable, or some element in K satisfies a polynomial over F having multiple roots. Hence, a satisfies some irreducible polynomial say f(x) over F having multiple roots in some splitting field. Now, there is a theorem in Herstein that says: If f(x) in F[x] is irreducible with char F = p>0 and f(x) has multiple roots then f(x) = g(x^p) where coeffs of g(x) are also in F.

Say . By our original hypothesis

I want to prove that which would contradict the fact that f(x) = g(x^p) is irreducible over F.

Well, I tried multiplying some test polynomials and it DOES work except it is extremely ugly. (remember char F = p is what helps me)

So, how would I solve that more elegantly?

Thanks,
caki