Certainly (*) is not sufficient. If all the polynomials are divisible by (z_1 - z_2), then any Z_0 with z_1 = z_2 will satisfy it. Franklin T. Adams-Watters -----Original Message----- From: dasimov@earthlink.net Suppose we have not one integer polynomial in one complex variable, but n > 1 integer polynomials P_1, . . . P_n in n complex variables z_1, . . . , z_n. Set Z = (z_1, . . ., z_n) and P(Z) = (P_1(Z), . . ., P_n(Z)) in C^n, for any z_1, . . ., z_n in C. Suppose (*) Z_0 in C^n satisfies P(Z_0) = 0. ------------------------------------------------------------------- QUESTIONS: 1. Does (*) imply that the complex coordinates of Z_0 are all algebraic numbers ??? 2, If the answer to this is "not necessarily", what if we add the condition (**) det(JP(Z_0)) is non-zero, where JP is the n x n Jacobian matrix (df_j/dz_k (Z_0)), 1 <= i, j <= n ??? 3. If still no, then what kind of numbers do we get, anyway, esp. if we assume both (*) and (**) -- supposing we toss in all complex coordinates of all such "algebraic vectors" Z_0 for all n >= 1 ??? Or for a fixed n ??? Or for all n <= N, for some fixed N ??? Just curious. --Dan ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.