algebraically independent

A subset S#92;subsetL is algebraically independent over K if every element of S is transcendental over K and over each of the extension fields over K generated by the remaining elements of 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 n elements x_1,…,x_n in R such that their images ◌̅x_i in R/P are algebraically independent. Let 0 ne p∈P. Then p,x_1,…,x_n cannot be algebraically independent over k, so there is a polynomial P(Y,X_,…,X_n) over k such that P(p,x_,…,x_n)=0.

Setting y#95;i#61;y#95;#123;i,1#125; (where 1 denotes the identity of the semigroup T) we obtain a #92;sigma-algebraically independent over R set #92;#123;y#95;i#92;verti#92;inI#92;#125; such that S#61;R#92;#123;(y#95;i)#95;#123;i#92;inI#125;#92;#125;.

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

If α ne 0,1 is algebraic and β is an algebraic irrational of degree d>2, then αᵝ,…,α are algebraically independent.

2014, M. Ram Murty, Purusottam Rath, Transcendental Numbers, Springer, page 138, Let us begin with the following conjecture of Schneider