Let F, K, S, P be as in Theorem 6.7 and suppose E is an intermediate field. Then

(a) F is purely inseparable over E if and only if S ⊂ E.

(b) If F is separable over E, then P ⊂ E.

(c) If E ∩ S = K, then E ⊂ P.

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Theorem 6.7. If A is a maximal element of S then A is an algebraic closure of K. (9) Problem Prove this theorem. A sketch of the proof follows. Let A be a maximal element of S and assume we have an algebraic extension ACB. Lemma 1 Refer to a result in BB to show that B is algebraic over K. Lemma 2 Define a function mcC: Mk → P(II) by C((p(x)) = R(p(x))) - A. Since B is algebraic over K we have IB: B → MK). For p(x) = MK we will have IB¹ (p(X) - A|≤|R(p(x) - A| and therefore we have (many) injections from T¹(p(X)) - A to R(p(x)) - A. Let I (p(x) be the non-empty set of injections of IB¹(p(x) - A to R(p(x)) - A Use the Axiom of Choice to get an injection : B→ II such that (a) = a for a E A and (3) € (C) (Irr(3) for 3EB-A. Lemma 3 Let F = (B). Using Definition 3.1 push the field structure on B forward to one on F. Show that F E S, that is, F is special. Conclusion Now use the maximality of A to show F = A and hence B = A. Conclude A is algebraically closed.