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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Solution Summary
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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