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# Transcendence basis

Suppose that L has transcendence degree n over K and that L is algebraic over K(&#945;1, . . . , &#945;n). Show that &#945;1, . . . , &#945;n is a transcendence basis for L over K.

Might help:
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)&#61472;-> (3) Any proper subset of A is missing some &#945; in&#61472;A, which by definition is not algebraic over A {a}.
(3)&#61472;-> (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.