Algebraically independent
|
|
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 set {√π, 2π + 1} is not algebraically independent, since the non-zero polynomial
- P(x1, x2) = x12 − 1/2 x2 + 1/2
yields zero when √π is substituted for x1 and 2π + 1 is substituted for x2.
It is unknown 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 these maximal algebraically independent subsets have the same cardinality, known as the transcendence degree of the extension.