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Suppose that a finite set of polynomials in C[x,y,z] has a finite number of solutions (i.e. the generated ideal is 0-dimensional).

Suppose also that the Groebner basis with respect to lex order x>y>z is

[f(z), g(y,z), h(y,z), k(x,y,z)]

As well known, the system can be now easily solved: choose a root z0 of f, plug it into g and h and look for a common root (y0) etc.

The question is the following: Is it true that for EVERY root z0 of f there exist y0, z0 such that (x0,y0,z0) satisfy the system?

In all the examples I have seen this is true, but I don't know whether this is true in general or there is a counterexample.

Thank you.

V. Vil
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Yes, any root z0 of f can be extended to a root (x0,y0,z0) of the system f = g = h = k = 0.

To see this consider that Iz = <f>, where Iz is the intersection of the generated zero dimensional ideal I with C[z] and <f> is the ideal generated by f. As can be seen in the proof that non-trivial intersections of I with C[xi] for all variables xi implies a finite zero set (see e.g. here, page 2 bottom and especially page 3 top), <f> contains a polynomial which factors only in (powers of) minimal polynomials of values appearing as z-value in the common roots of I. Since f divides this polynomial, it also has only roots that can be extended to roots of the system.

coproc
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