英和和英

形容詞

algebraically independent (not comparable)

1. (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.

   The singleton set ${\\displaystyle \\{\\alpha \\}}$ is algebraically independent over ${\\displaystyle K}$ if and only if the element ${\\displaystyle \\alpha }$ is transcendental over ${\\displaystyle K}$.

   A subset ${\\displaystyle S\\subset L}$ is algebraically independent over ${\\displaystyle K}$ if every element of ${\\displaystyle S}$ is transcendental over ${\\displaystyle K}$ and over each of the extension fields over ${\\displaystyle K}$ generated by the remaining elements of ${\\displaystyle S}$.

   • 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 ${\\displaystyle n}$ elements ${\\displaystyle x_{1},\\dots ,x_{n}}$ in ${\\displaystyle R}$ such that their images ${\\displaystyle {\\overline {x}}_{i}}$ in ${\\displaystyle R/P}$ are algebraically independent. Let ${\\displaystyle 0\\neq p\\in P}$. Then ${\\displaystyle p,x_{1},\\dots \\,x_{n}}$ cannot be algebraically independent over ${\\displaystyle k}$, so there is a polynomial ${\\displaystyle P(Y,X_{,}\\dots ,X_{n})}$ over ${\\displaystyle k}$ such that ${\\displaystyle P(p,x_{,}\\dots ,x_{n})=0}$.

   • 2006, Alexander B. Levin, Difference algebra, M. Hazewinkel (editor), Handbook of Algebra, Volume 4, Elsevier (North-Holland), page 251,

     Setting ${\\displaystyle y_{i}=y_{i,1}}$ (where 1 denotes the identity of the semigroup ${\\displaystyle T}$) we obtain a ${\\displaystyle \\sigma }$-algebraically independent over ${\\displaystyle R}$ set ${\\displaystyle \\{y_{i}\\vert i\\in I\\}}$ such that ${\\displaystyle S=R\\{(y_{i})_{i\\in I}\\}}$.

   • 2014, M. Ram Murty, Purusottam Rath, Transcendental Numbers, Springer, page 138,

     Let us begin with the following conjecture of Schneider:

     If ${\\displaystyle \\alpha \\neq 0,1}$ is algebraic and ${\\displaystyle \\beta }$ is an algebraic irrational of degree ${\\displaystyle d\\geq 2}$, then

     ${\\displaystyle \\alpha ^{\\beta },\\dots ,\\alpha ^{\\beta ^{d-1}}}$

     are algebraically independent.