Reputation: 11
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.
Upvotes: 1
Views: 167
Reputation: 6257
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.
Upvotes: 0