Summary
In abstract algebra, a subset of a field is algebraically independent over a subfield if the elements of do not satisfy any non-trivial polynomial equation with coefficients in . In particular, a one element set is algebraically independent over if and only if is transcendental over . In general, all the elements of an algebraically independent set over are by necessity transcendental over , and over all of the field extensions over generated by the remaining elements of . The two real numbers and are each transcendental numbers: they are not the roots of any nontrivial polynomial whose coefficients are rational numbers. Thus, each of the two singleton sets and are algebraically independent over the field of rational numbers. However, the set is not algebraically independent over the rational numbers, because the nontrivial polynomial is zero when and . Although both and e are known to be transcendental, it is not known whether the set of both of them is algebraically independent over . In fact, it is not even known if is irrational. Nesterenko proved in 1996 that: the numbers , , and , where is the gamma function, are algebraically independent over . the numbers and are algebraically independent over . for all positive integers , the number is algebraically independent over . The Lindemann–Weierstrass theorem can often be used to prove that some sets are algebraically independent over . It states that whenever are algebraic numbers that are linearly independent over , then are also algebraically independent over . Algebraic matroid Given a field extension which is not algebraic, Zorn's lemma can be used to show that there always exists a maximal algebraically independent subset of over . Further, all the maximal algebraically independent subsets have the same cardinality, known as the transcendence degree of the extension. For every set of elements of , the algebraically independent subsets of satisfy the axioms that define the independent sets of a matroid.
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