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29), and prove rigorously that the Axiom of Choice, Zorn’s Lemma, the Principle of Cardinal Comparability and the Well-Ordering Theorem are all equivalent, relative to basic set theory. In fact, these are just a few examples out of many: by now, there is a multitude of statements and theorems for which it has been shown that they are equivalent to the Axiom of Choice (you may have a look at the book by Rubin and Rubin, Equivalents of the Axiom of Choice). t. inclusion of chains). It is actually rather easy to show that this is equivalent to Zorn’s lemma.

Vn ) be the set of those quantifier-free formulas ψ(v 1 , . . , vn ) for which T |= ∀v1 · · · vn (ϕ(v1 · · · vn ) → ψ(v1 · · · vn )) Now work in an extension L of L with n new constants d1 , . . , dn . Let Γ(d) = {ψ(d1 , . . , dn ) | ψ(v1 , . . , vn ) ∈ Γ(v)}. Claim: T ∪ Γ(d) |= ϕ(d) Suppose the Claim holds. Then by the Compactness Theorem there are ψ1 (d), . . , ψk (d) ∈ Γ(d) such that T ∪ {ψ1 (d), . . , ψk (d)} |= ϕ(d) 54 Since the constants d do not occur in T , it follows that T |= ∀v(ψ1 (v) ∧ · · · ∧ ψk (v) → ϕ(v)) But the converse implication follows from the fact that the ψ i are in Γ.

Proof. Suppose 1 ∈ I. Then I is contained in a maximal ideal M of K[X1 , . . , Xn ]. Let K be the algebraic closure of the field K[X 1 , . . , Xn ]/M . In K , the elements X1 , . . , Xn have the property that g(X1 , . . , Xn ) = 0 for every g ∈ I. Now K is a subring of K and both are algebraically closed fields; by quantifier elimination, K is an elementary substructure of K . It follows that for any finite number g1 , . . 16) every ideal of K[X 1 , . . , Xn ] is finitely generated, so we are done.

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