# Math Genius: When can we have the same number of (general) solution as the projection?

$$newcommandC{mathbb{C}}$$Let $$a_1,dots,a_n$$ be parameters in $$C$$ and let
$$f_1,dots,f_n$$ be non-constant algebraically independent polynomials in $$n$$-variables over $$C$$.

For a general choice of $$a_i$$ (i.e. $$(a_1,dots,a_n)$$ belonging to a Zariski open set in the affine space $$C^n$$) we know that the system

$$f_i = a_i qquad i=1,dots,n$$

is a complete intersection, i.e. there are finite , say $$k$$, solutions in $$C^n$$ (I’m not sure what I should use to justfy this, I guess some argument in ellimination theory would justify this).

My question is the following:

Suppose $$pi: C^n to C^2$$
is a projection on say the first two coordinates.

Can I also claim that for a general choice of $$a_i$$ the projection of the solutions of

$$f_i = a_i qquad i=1,dots,n$$

with respect to $$pi$$ yields $$k$$ points in $$C^2$$? The projection should be the ellimination ideal of the ideal generated by the polynomials $$langle f_i-a_i : i=1,dots,n rangle$$ to polynomials in the first two variables. But can I also have a control on the ellimination
ideal (I want it to have the same number of vanishing points as the original ideal) for general $$a_i$$?

If so, why?

Tagged : /