Let $k$ be a field, and $P \in k[X]$. Consider the extensions $k \subset L \subset K$, where $L$ is a splitting field for $P$ over $k$ and $K$ is the algebraic closure of $k$. Then (by definition) all roots of $P$ in $K$ actually belong to the smaller field $L$.
What happens to that picture in several variables ? More precisely, let $P_1, \ldots, P_n \in k[X_1, \ldots, X_n]$. Consider the subset $$Z=\bigcap_i P_i^{-1}(0) \subset K^n.$$
Generically $Z$ is finite. Question : is there an analogue of the splitting field, i.e. a smallest extension $k \subset L$ such that $Z \subset L^n$ ?
I realize this is probably very classical, so please forgive my ignorance. I would be glad to just have a precise reference for this question.
EDIT
I wrote the question for $k^n$ but it might be more relevant for $\mathbb P^n(k)$.
EDIT 2
Thanks to an answer, I realize that as such my question was not properly formulated, because there is a always a smallest field $L$ containing any subset $Z$. What I am interested in is the relationship between $L$ and the $P_i$. Also, are the components of points in $Z$ algebraic over $k$ ? (I would say yes, but I don't know how to prove it). If so, is there a natural construction of a polynomial $Q \in k[X]$ in terms of the $P_i$ annihilating, say, the first component of a given $z \in Z$ ?
I think the polynomials are a bit of a red herring here. Given any subset $Z\subset K^n$, finite or not, define $$L = \bigcap \{M\mid K/M/k\text{ and }Z\subset M^n\}.$$ This will satisfy your "analogue of splitting field" property by definition. Another way to obtain $L$ is as the subextension of $K/k$ generated over $k$ by the coordinates of the elements of $Z$. Thus, if $Z$ is finite, then $L/k$ is a finite extension.