Are all Finite Subsets of Affine n-space Algebraic sets, and related question For an algebraicaly closed field $k$ are all finite subsets of Affine $n$-space $A^{n}\left(k\right)$
algebraic sets (here for $n>1$), and if so, for a given finite set $X\subset A^{n}\left(k\right)$
what set of polynomials in $k\left[x_{1},\dots,x_{n}\right]$ has
$X$ as its common zero set? This probably has an answer, I just don't
know how to phrase the question so as to get an answer by internet
search engine.
Also if I have a set of polynomials $P_{1},\dots,P_{k}\in\mathbb{Q}\left[x_{1},\dots,x_{n}\right]$
such that the common zero set $Z(P_{1},\dots,P_{k})$
is a finite set of size $m$, i.e. $X=\{(z_{11},\dots,z_{1n}),\dots,(z_{m1},\dots,z_{mn})\}$,
does this imply that each coordinate $z_{ij}$ is algebraic over $\mathbb{Q}$.
I asked someone this question, and he felt that each $z_{ij}$ should
turn out to be algebraic, but I can't see how this is proved?
Algebraic geometry is not my area, I just was wondering about these questions and am having a hard time tracking down an answer on the internet.
 A: Here is a simple argument for the second question. I renamed the field $k$ to $K$ for clarity. 
Observe that $K^{\mathrm{Aut}(K/\mathbb{Q})}=\mathbb{Q}$, because $K$ is algebraically closed.
For any $g\in\mathrm{Aut}(K/\mathbb{Q})$ and any $x\in X$ we have $x^g\in X$, since
$$ P_l(x^g) = P_l^g(x^g) = P_l(x)^g = 0^g=0,\qquad 1\leq l\leq k,$$
using that the coefficients of $P_l$ are rational.
This shows that the polynomials 
$$ Q_j(z):=\prod_{i=1}^m(z-z_{ij}),\qquad 1\leq j\leq n, $$
are invariant under $\mathrm{Aut}(K/\mathbb{Q})$, i.e. they also
have rational coefficients. It follows that the coordinates $z_{ij}\in K$ are algebraic over $\mathbb{Q}$.
P.S. Thanks to Qiaochu Yuan and Kevin Ventullo.
A: Let $I\subset \mathbb Q[x_1,...,x_n]$ be the ideal generated by the polynomials $P_1,...,P_k$   and $A$ the $\mathbb Q$-algebra $A=\mathbb Q[x_1,...,x_n]/I$.
You are interested in the scheme $V=Spec(A)\subset \mathbb A^n_\mathbb Q= Spec(\mathbb Q[x_1,...,x_n])$ and its $k$-points for $k$ an extension field of $\mathbb Q$.  
A $ \: k$-point is a    point in $z=(z_1,...,z_n) \in k^n$ such that for all $P\in I$ we have $P(z)=0$ or equivalently a morphism of $\mathbb Q$-algebras $\phi : A\to k$ ( the equivalence is given by the formula $\phi (\bar x_i)=z_i$).  The set of $k$-points of $V$ is denoted  by $V(k)$.   
Now if $k$ is algebraically closed and if $V(k)$ is finite, it follows (from Noether's normalization theorem for example), that $A$ has Krull dimension zero and that it is finite dimensional over $\mathbb Q$ .
Any $\mathbb Q$-algebra morphism $\phi : A\to k$ then satisfies $\phi(A)\subset \overline {\mathbb Q}\subset k$ and the corresponding point $z=(z_1,...,z_n) \in k^n$ thus satisfies $z=(z_1,...,z_n) \in {\overline {\mathbb Q}}^n \subset k^n$
A: 1) The point $(p_1, ... p_n)$ is the vanishing set of the polynomials $x_i - p_i$, and a finite union of algebraic sets is algebraic. 
2) Yes. This should follow concretely from results in elimination theory of which I am totally unaware, and abstractly from Chevalley's theorem. 
