algebraically independent

algebraically independent

[¦al·jə¦brā·ik·lē ‚in·də′pen·dənt]
(mathematics)
A subset S of a commutative ring B is said to be algebraically independent over a subring A of B (or the elements of S are said to be algebraically independent over A) if, whenever a polynominal in elements of S, with coefficients in A, is equal to 0, then all the coefficients in the polynomial equal 0.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
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[8] proved that [[zeta].sub.F](2), [[zeta].sub.F](4) and [[zeta].sub.F](6) are algebraically independent. In 2001, Navas [10] obtained analytic continuation of the Fibonacci Dirichlet series [[zeta].sub.F] (s).
It is easy to see that since A is a set of free generators, the set F is algebraically independent. Hence, Y is a c-generated semigroup.
For example, if [[chi].sub.1] and [[chi].sub.2] are non-equivalent characters, and the numbers [[alpha].sub.1] and [[alpha].sub.2] are algebraically independent over Q, then the function L(s, [[chi].sub.1]) L(s, [[chi].sub.2]) [zeta](s, [[alpha].sub.1]) [zeta](s, [[alpha].sub.2]) satisfies the hypotheses of Theorem 8, since a polynomial p(s) has a preimage (1,1,1,p(s)) [member of] [S.sup.2] x [H.sup.2](D), and the algebraic independence of the numbers [[alpha].sub.1] and [[alpha].sub.2] implies the linear independence of the set L(P, [[alpha].sub.1], [[alpha].sub.2]).
We say that two elements a and b of an algebra B are algebraically independent if P [member of] C[[z.sub.1], [z.sub.2]] is such that P(a, b) = 0 then P [equivalent to] 0.
An extension w, of a valuation v from K to K([X.sub.1],..., [X.sub.n]), symmetric with respect to [X.sub.1],...[X.sub.n], is called symmetrically-open (with respect to [X.sub.1],...[X.sub.n]) if, adding any number of other indeterminates (elements transcendental and algebraically independent over K([X.sub.1],...[X.sub.n])), [X.sub.n+1],...[X.sub.n+r], there exists a symmetric extension of it to K([X.sub.1],..., [X.sub.n+r]) with respect to [X.sub.1],..., [X.sub.n+r].
For instance, [[summation].sub.G](A) is generic when the [a.sub.ij] are algebraically independent. We will use GEN to denote the parameter list of an arbitrary generic bigraphical arrangement.
Let O be an open set in K, and let P [member of] O be a polytope with vertex set V which we suppose to be in general position, for instance the coordinates of the vertices are algebraically independent. (Weaker general position assumptions would also work for our purposes.) For each triple a,b,c [member of] V let R(a, b, c) be the radius of the smallest disk D(a, b, c) in the (unique) plane S(a, b, c) containing a, b, c.
Another approach is to assign algebraically independent values to the nonzeros (i.e., numbers that are not the roots of any multivariate polynomials with integer coefficients, since such roots form a set of measure zero); see Gilbert [12].
Moreover Nori [N] (see also [B] for a different approach) proves that, for A generic, [Gr.sup.2](A) is infinitely generated by showing that the cycles [[C.sub.[alpha]]] - [C??] are algebraically independent, where [C.sub.[alpha]] is the image under an isogeny [r.sub.[alpha]] : j([C'.sub.[alpha]]) [right arrow] A of some Abel-Jacobi embedded curve [C'.sub.[alpha]] in J([C'.sub.[alpha]]).
Let us show that [mathematical expression not reproducible] are algebraically independent on [c.sup.(m').sub.00] ([C.sup.n]) for every m' [greater than or equal to] m.
Suppose that [[chi].sub.1], ..., [[chi].sub.d] are pairwise non-equivalent Dirichlet characters, the numbers [[alpha].sub.1], ..., [[alpha].sub.r] are algebraically independent over the field of rational numbers Q, and that rank ([A.sub.j]) = [l.sub.j], j = 1, ..., r.
[X.sub.10], [Y.sub.12], [X.sub.16] are algebraically independent over F and hence the polynomial Q is uniquely determined.