Algebraic independence
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In abstract algebra, a subset S of a field L is algebraically independent over a subfield K if the elements of S do not satisfy any non-trivial polynomial equation with coefficients in K. This means that for every finite sequence α1, ..., αn of elements of S, no two the same, and every non-zero polynomial P(x1, ..., xn) with coefficients in K, we have
- P(α1,...,αn) ≠ 0.
In particular, a one element set {α} is algebraically independent over K if and only if α is transcendental over K. In general, all the elements of an algebraically independent set over K are by necessity transcendental over K, but that is far from being a sufficient condition.
For example, the subset {√π, 2π+1} of the real numbers R is not algebraically independent over the rationals Q, since the non-zero polynomial
yields zero when √π is substituted for x1 and 2π+1 is substituted for x2.
The Lindemann-Weierstrass theorem can often be used to prove that some sets are algebraically independent over 
. It states that whenever α1,...,αn are algebraic numbers that are linearly independent over Q, then eα1,...,eαn are algebraically independent over Q.
It is not known whether the set {π, e} is algebraically independent over Q. Nesterenko proved in 1996 that {π, eπ, Γ(1/4)} is algebraically independent over Q.
Given a field extension L/K, we can use Zorn's lemma to show that there always exists a maximal algebraically independent subset of L over K. Further, all the maximal algebraically independent subsets have the same cardinality, known as the transcendence degree of the extension.
