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

$$\newcommand\C{\mathbb{C}}$$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?

• Have you tried the case when $f_1=x_1, f_2=x_2$, the first two co-ordinates? – Mohan May 19 at 16:13
• You are right. So all I can say is that this is a true for a "generic" projection. Thanks Mohan. You should think of writing this as an answer. All I could do was mark this as +1, but I think this answers it! – quantum May 24 at 9:07