All fields of characteristic 0 are perfect, so in particular the fields , and are perfect. If is a field of characteristic (with a prime number), then is perfect if and only if the Frobenius endomorphism on , defined by
is an automorphism of . Since the Frobenius map is always injective, it is sufficient to check whether is surjective. In particular, all finite fields are perfect (any injective endomorphism is also surjective). Moreover, any field whose characteristic is nonzero that is algebraic (http://planetmath.org/AlgebraicExtension) over its prime subfield is perfect. Thus, the only fields that are not perfect are those whose characteristic is nonzero and are transcendental over their prime subfield.
Similarly, a ring of characteristic is perfect if the endomorphism of is an automorphism (i.e., is surjective).
|Date of creation||2013-03-22 13:08:23|
|Last modified on||2013-03-22 13:08:23|
|Last modified by||sleske (997)|