algebraically independent

Adjective

 * 1)  (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.
 * 2) * 1999,, The Red Book of Varieties and Schemes: Includes the Michigan Lectures, Springer, 1358, 2nd Edition, Expanded, page 40,
 * If the statement is false, there are $$n$$ elements $$x_1,\dots, x_n$$ in $$R$$ such that their images $$\overline x_i$$ in $$R/P$$ are algebraically independent. Let $$0\ne p\in P$$. Then $$p, x_1,\dots\,x_n$$ cannot be algebraically independent over $$k$$, so there is a polynomial $$P(Y, X_,\dots, X_n)$$ over $$k$$ such that $$P(p, x_,\dots, x_n)=0$$.
 * 1) * 2014, M. Ram Murty, Purusottam Rath, Transcendental Numbers, Springer, page 138,
 * Let us begin with the following conjecture of Schneider:
 * If $$\alpha\ne 0,1$$ is algebraic and $$\beta$$ is an algebraic irrational of degree $$d \ge 2$$, then
 * $$\alpha^\beta,\dots,\alpha^{\beta^{d-1}}$$
 * are algebraically independent.
 * If $$\alpha\ne 0,1$$ is algebraic and $$\beta$$ is an algebraic irrational of degree $$d \ge 2$$, then
 * $$\alpha^\beta,\dots,\alpha^{\beta^{d-1}}$$
 * are algebraically independent.

Usage notes

 * Perhaps unexpectedly, a single element of $$S$$ may be said to be algebraically independent (over $$K$$).

Translations

 * Italian: algebricamente indipendente