# Transcendence basis

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Suppose that L has transcendence degree n over K and that L is algebraic over K(α1, . . . , αn). Show that α1, . . . , αn is a transcendence basis for L over K.

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Theorem - Definition: Let L be an extension of K, A a subset of L. The following are equivalent:

(1) A is a maximal algebraically independent set.

(2) A is algebraically independent and everything in L is algebraic over A.

(3) A is a minimal set such that L is algebraic over K(A)

Such an A is a transcendence basis of L over K.

Proof: (1) <-> (2) is clear.

(1)-> (3) Any proper subset of A is missing some α inA, which by definition is not algebraic over A {a}.

(3)-> (1) Since the union of a chain of algebraically independent sets is algebraically independent, Zorn's lemma implies there is a maximal algebraically independent set B contained in A. K(A) is algebraic over K(B) by maximality, and L is algebraic over K(A), hence L is algebraic over K(B). Since A is minimal, B =A.

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This provides an example of a proof regarding a transcendence basis.

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Proof:

From the condition, we know that is an extension field of that is an extension field of . In another word, . has transcendence degree over and is algebraic over .

First, I claim that is algebraically independent over ...

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