algebraically independent

== English == === Adjective === algebraically independent (not comparable) (algebra, field theory) (Of a subset S of the extension field L of a given field extension L / K) whose elements do not satisfy any non-trivial polynomial equation with coefficients in K. 1999, David Mumford, The Red Book of Varieties and Schemes: Includes the Michigan Lectures, Springer, Lecture Notes in Mathematics 1358, 2nd Edition, Expanded, page 40, If the statement is false, there are n {\displaystyle n} elements x 1 , … , x n {\displaystyle x_{1},\dots ,x_{n}} in R {\displaystyle R} such that their images x ¯ i {\displaystyle {\overline {x}}_{i}} in R / P {\displaystyle R/P} are algebraically independent. Let 0 ≠ p ∈ P {\displaystyle 0\neq p\in P} . Then p , x 1 , … x n {\displaystyle p,x_{1},\dots \,x_{n}} cannot be algebraically independent over k {\displaystyle k} , so there is a polynomial P ( Y , X , … , X n ) {\displaystyle P(Y,X_{,}\dots ,X_{n})} over k {\displaystyle k} such that P ( p , x , … , x n ) = 0 {\displaystyle P(p,x_{,}\dots ,x_{n})=0} . 2014, M. Ram Murty, Purusottam Rath, Transcendental Numbers, Springer, page 138, Let us begin with the following conjecture of Schneider: If α ≠ 0 , 1 {\displaystyle \alpha \neq 0,1} is algebraic and β {\displaystyle \beta } is an algebraic irrational of degree d ≥ 2 {\displaystyle d\geq 2} , then α β , … , α β d − 1 {\displaystyle \alpha ^{\beta },\dots ,\alpha ^{\beta ^{d-1}}} are algebraically independent. ==== Usage notes ==== Perhaps unexpectedly, a single element of S {\displaystyle S} may be said to be algebraically independent (over K {\displaystyle K} ). ==== Antonyms ==== (antonym(s) of “which does not or whose elements do not satisfy any nontrivial polynomial equation over a given field”): algebraically dependent ==== Translations ==== === Further reading === Algebraic independence on Wikipedia.Wikipedia Algebraic independence on Encyclopedia of Mathematics Algebraically Independent on Wolfram MathWorld