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, ..., α_n of elements of S, no two of which are the same, then if P has coefficients in K and

P(α₁,...,α_n) = 0

then P is the zero polynomial.

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 of the real numbers R is not algebraically independent over the rational numbers Q, since the non-zero polynomial

yields zero when is substituted for and is substituted for .

The Lindemann-Weierstrass theorem can often be used to prove that some sets are algebraically independent over Q. It states that whenever α₁,...,α_n are algebraic numbers that are linearly independent over Q, then e^α₁,...,e^α_n are algebraically independent over Q.

It is not known whether the set {π, e} is algebraically independent over Q.^[1] In fact, it is not even known if π+e is irrational.^[2] Nesterenko proved in 1996 that:

• the numbers π, e^π, and Γ(1/4) are algebraically independent over Q.
• the numbers π, e^π√3, and Γ(1/3) are algebraically independent over Q.
• for all positive integers n, the numbers π, e^π√n are algebraically independent over Q.^[3]

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.

See also

• Matroid

References

External links

• Chen, Johnny, "Algebraically Independent" from MathWorld.