If a parametric polynomial system is zero-dimensional, then is the specialized polynomial system zero-dimensional for almost all parameter values?

Question

If we have a parametric polynomial system in say $Q = \mathbb{Q}[p_1,\ldots,p_m][x_1,\ldots,x_n]$ and $I$ is zero-dimensional over $\overline{\mathbb{Q}[p_1,\ldots,p_m]}$, then is it true that for almost all parameter values $\mathbf{p}=(p_1^*,\ldots,p_m^*)$, the image of $I$ under the map $\phi_{\mathbf{p}}\colon Q\rightarrow \mathbb{Q}[x_1,\ldots,x_n]$ taken by evaluating $p_i$ at $p_i^*$ for $i=1,\ldots,m$ is zero-dimensional over $\overline{\mathbb{Q}}$? How would I prove it?

Answer 1

If "almost all" means "all but finitely many", then the answer is wrong. Take for instance the system $\{p_1x_2,p_2x_2\}$.

If "almost all" has a weaker sense, then one could argue as follows: Since the system has dimension $0$, the finitely man solutions are contained in the set $\{a_1,a_2,\dots,a_n\}$, where the $a_i$'s are the roots of separable univariate polynomials $P_i$. By Galois theory, the coefficients of $P_i$ are in $\mathbb Q(p_1,\dots,p_m)$. So each specialization of the $p_j$'s which is defined for all $P_i$'s and which does not annihilate any $P_i$ yields a system with finitely many roots and therefore of dimension $0$.