On Tuesday 29 August 2006 01:59, Daniel Asimov wrote:

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 ???

Obviously not as it stands. P1(z1,z2) = z1^2 - z2^2 P2(z1,z2) = z1^3 - z2^3 Z = (pi,pi)

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 ???

The coordinates must then be algebraic. Here's a proof, which is probably unnecessarily clumsy although its basic idea is very simple: if there's a non-algebraic root then you can tweak it a little and still have a root. Let K be the field generated by the coordinates of Z, and let A be any field homomorphism from K to any subfield of C. Then P(A(Z)) = A(P)(Z) = P(Z) = 0, where A(P) means "P with A applied to its coefficients". Suppose K isn't algebraic. Then it's generated by some (finite, but nonempty) set of independent transcendentals y1, ..., yk, plus perhaps something x algebraic over y1,...,yk; a homomorphism A : K -> C is defined by choosing A(y1), ..., A(yk), A(x) such that q(y1,...,yk,x) = 0 where q is the minimal polynomial for x. Since q, being a minimal polynomial, is irreducible, we have dq/dx(y1,...,yk,x) nonzero. Hence, by the implicit function theorem, there's a neighbourhood of x where we can write x as a smooth function of y1,...,yk. So now choose A(y1), ..., A(yk) to be arbitrarily close to y1,...,yk; and define A(x) according to that smooth function (so it's also close to x). Then A(Z) is arbitrarily close to Z. In other words, there are points where P(_)=0 arbitrarily close to Z. But this is impossible, since the Jacobian of P is nonzero at Z. Contradiction! -- g