On Perfect Subfields over Which a Field Is Separably Generated

In this paper, we determine the perfect subfields over which a field is separably generated. Let a field F of characteristic. p^O be an extension of a field £. A separating transcendence basis for the extension £/£ is a transcendence basis B of £/£ such that £ is a separable algebraic extension of £(£). If £/£ has a separating transcendence basis, we say that £ is separably generated over £. A field K of characteristic p^O is perfect if A"=A, that is, if each element of K has a unique pth root in A'. It is clear that any field £ contains a maximal perfect subfield which we denote by £* and that £*= Dn-i Rp"We prove in Theorem 3 that if £is separably generated over a subfield £, then £* is an algebraic extension of £*. As a consequence (Corollary 10), we show that if a field £is separably generated over at least one perfect subfield, then £is separably generated over a perfect subfield A' if and only if £* is algebraic over K. Lemma 1. If F is a pure transcendental extension of E, then £* = £*. Proof. By hypothesis. £=£(£.) where B is an algebraically independent set of elements over £. If y is a nonzero element of £*, then there exist elements u„ in £ such that >' = ir£" for «=1, 2, • • • . Since the polynomial ring E[B] is a unique factorization domain, we may write y—fjg with/and g relatively prime in £[/?]. Similarly, for each n, wn=fjgn with /„ and gn relatively prime in E[B]. Thus fg"n=gfn implies that/and/*" are associates for each «, that is,/=a„/^" where an is in £. As a consequence of the unique factorization in £[5], it follows that/e£ and /„ e £, for all n. Similarly g e E and gn e E, for all n. Thus v e £* = rin-i £'", proving that £* = £*. The following lemma appears as an exercise in [1, Example 16, p. 136]. For the sake of completeness, we provide a proof. Presented to the Society. January 17, 1972; received by the editors September 29, 1971. AMS 1970 subject classifications. Primary 12E05, 12F05, 12F20.