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.



Are all Finite Subsets of Affine n-space Algebraic sets, and related question
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 $z=(z\_1,...,z\_n) \in {\overline {\mathbb Q}}^n \subset k^n$


