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In [[abstract algebra]], a [[subset]] ''S'' of a [[field (mathematics)|field]] ''L'' is '''algebraically independent''' over a [[field (mathematics)|subfield]] ''K'' if the elements of ''S'' do not satisfy any non-[[trivial (mathematics)|trivial]] [[polynomial]] equation with coefficients in ''K''.  
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In particular, a one element set {''α''} is algebraically independent over ''K'' [[if and only if]] α is [[Transcendental element|transcendental]] over ''K''. In general, all the elements of an algebraically independent set ''S'' over ''K'' are by necessity transcendental over ''K'', and over all of the [[field extension]]s over ''K'' generated by the remaining elements of ''S''.
 
==Example==
The two [[real number]]s <math>\sqrt{\pi}</math> and <math>2\pi+1</math> are each [[transcendental number]]s: they are not the roots of any nontrivial polynomial whose coefficients are [[rational number]]s. Thus, each of the two [[singleton set]]s <math>\{\sqrt{\pi}\}</math> and <math>\{2\pi+1\}</math> are algebraically independent over the field <math>\mathbb{Q}</math> of rational numbers.
 
However, the set <math>\{ \sqrt{\pi}, 2\pi+1 \}</math> is ''not'' algebraically independent over the rational numbers, because the nontrivial polynomial
:<math>P(x,y)=2x^2-y+1</math>
is zero when <math>x=\sqrt{\pi}</math> and <math>y=2\pi+1</math>.
 
==Algebraic independence of known constants==
Although both <math>\pi</math> and [[E (mathematical constant)|''e'']] are known to be transcendental,
it is not known whether the set of both of them is algebraically independent over <math>\mathbb{Q}</math>.<ref>{{cite book
| url = http://books.google.com/?id=jQ7c8Xqpqk0C
| title = Field and Galois Theory
| author = Patrick Morandi
| publisher = Springer
| year = 1996
| pages = 174
| accessdate = 2008-04-11
| isbn = 978-0-387-94753-2
}}</ref> In fact, it is not even known if <math>\pi+e</math> is irrational.<ref>{{Citation |last=Green |first=Ben |authorlink=Ben J. Green|chapter=III.41 Irrational and Transcendental Numbers |editor-last=Gowers |editor-first=Timothy |year=2008 |title=[[The Princeton Companion to Mathematics]] |page=222 |publisher=Princeton University Press}}</ref>
[[Yuri Valentinovich Nesterenko|Nesterenko]] proved in 1996 that:
* the numbers π, ''e''<sup>π</sup>, and [[Gamma function|Γ]](1/4) are algebraically independent over '''Q'''.<ref name=MP61>{{cite book | first1=Yu. I. | last1=Manin | authorlink1=Yuri I. Manin | first2=A. A. | last2=Panchishkin | title=Introduction to Modern Number Theory | series=Encyclopaedia of Mathematical Sciences | volume=49 | edition=Second | year=2007 | isbn=978-3-540-20364-3 | issn=0938-0396 | zbl=1079.11002 | page=61 }}</ref>
* the numbers π, ''e''<sup>π√<span style="text-decoration: overline">3</span></sup>, and Γ(1/3) are algebraically independent over '''Q'''.
* for all positive integers n, the numbers π, ''e''<sup>π√<span style="text-decoration: overline">n</span></sup> are algebraically independent over '''Q'''.<ref>{{cite journal|author=Nesterenko, Yuri V|authorlink=Yuri Valentinovich Nesterenko|title=Modular Functions and Transcendence Problems|journal=[[Comptes rendus de l'Académie des sciences|Comptes Rendus de l'Académie des Sciences. Série I. Mathématique]]|volume=322|pages=909–914|year=1996|issue=10}}</ref>
 
==Lindemann–Weierstrass theorem==
The [[Lindemann–Weierstrass theorem]] can often be used to prove that some sets are algebraically independent over '''Q'''. It states that whenever α<sub>1</sub>,...,α<sub>''n''</sub> are [[algebraic number]]s that are [[linearly independent]] over '''Q''', then ''e''<sup>α<sub>1</sub></sup>,...,''e''<sup>α<sub>''n''</sub></sup> are algebraically independent over '''Q'''.
 
==Algebraic matroids==
{{main|Algebraic matroid}}
 
Given a [[field extension]] ''L''/''K'', [[Zorn's lemma]] can be used 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.
 
For every set ''S'' of elements of ''L'', the algebraically independent subsets of ''S'' satisfy the axioms that define the independent sets of a [[matroid]]. In this matroid, the rank of a set of elements is its transcendence degree, and the flat generated by a set ''T'' of elements is the intersection of ''L'' with the field ''K''[''T'']. A matroid that can be generated in this way is called an '''algebraic matroid'''. No good characterization of algebraic matroids is known, but certain matroids are known to be non-algebraic; the smallest is the [[Vámos matroid]].<ref>{{citation
| last1 = Ingleton | first1 = A. W.
| last2 = Main | first2 = R. A.
| doi = 10.1112/blms/7.2.144
| journal = [[Bulletin of the London Mathematical Society]]
| mr = 0369110
| pages = 144–146
| title = Non-algebraic matroids exist
| volume = 7
| year = 1975}}.</ref>
 
Many finite matroids may be [[Matroid representation|represented]] by a [[matrix (mathematics)|matrix]] over a field ''K'', in which the matroid elements correspond to matrix columns, and a set of elements is independent if the corresponding set of columns is [[linear independence|linearly independent]]. Every matroid with a linear representation of this type may also be represented as an algebraic matroid, by choosing an [[Indeterminate (variable)|indeterminate]] for each row of the matrix, and by using the matrix coefficients within each column to assign each matroid element a linear combination of these transcendentals. The converse is false: not every algebraic matroid has a linear representation.<ref>{{citation|title=Applied Discrete Structures|first=K. D.|last=Joshi|publisher=New Age International|year=1997|isbn=9788122408263|page=909|url=http://books.google.com/books?id=lxIgGGJXacoC&pg=PA909&lpg=PA909}}.</ref>
 
==References==
{{reflist}}
 
==External links==
*{{MathWorld|urlname=AlgebraicallyIndependent|title=Algebraically Independent|author=Chen, Johnny}}
 
[[Category:Abstract algebra]]
[[Category:Matroid theory]]

Latest revision as of 00:10, 23 December 2014

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