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

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    $\begingroup$ Have you tried the case when $f_1=x_1, f_2=x_2$, the first two co-ordinates? $\endgroup$ – Mohan May 19 at 16:13
  • $\begingroup$ 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! $\endgroup$ – quantum May 24 at 9:07

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