algebraically independentとは 意味・読み方・使い方

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.