| Algebraically Independent |
Article Index for Algebraically |
Website Links For Algebraic |
Information AboutAlgebraically Independent |
| CATEGORIES ABOUT ALGEBRAIC INDEPENDENCE | |
| abstract algebra | |
|
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 the same, and every non-zero polynomial ''P''(''x''1, ..., ''x''''n'') with coefficients in ''K'', we have P 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 Number s R is ''not'' algebraically independent over the Rationals '''Q''', since the non-zero polynomial : yields zero when √π is substituted for ''x''1 and 2π+1 is substituted for ''x''2. 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 Number s 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. Nesterenko proved in 1996 that {π, ''e''π, 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. |
|
|