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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.

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 @Jason: To answer your first question quickly: Yes. Any single point is an algebraic set, and any finite union of algebraic sets is algebraic. This is a consequence of the fact that the lattice of algebraic sets is the lattice of closed subsets of a topological space. – Zhen Lin Dec 28 2011 at 7:59
Thank you too everyone who pointed this out. – Jason Suagee Dec 29 2011 at 3:41
@Jason: I responded to your specific question concerning my response, see below. – GH Dec 29 2011 at 3:47

closed as too localized by Dan Petersen, Felipe Voloch, Benjamin Steinberg, Ryan Budney, Andres Caicedo Dec 29 2011 at 6:56

3 Answers

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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 $z=(z_1,...,z_n) \in {\overline {\mathbb Q}}^n \subset k^n$

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As a complement, one can also say that the dimension of $A$ as a $\mathbf{Q}$-vector space is equal to $\# V(k)$. This is because dimension is invariant under base change by $k$, and one can prove (by interpolation, for example) that every function $V(k) \to k$ comes from a polynomial. – François Brunault Dec 28 2011 at 11:00
Dear François, this is not quite correct. For example, if $n=1$ and $A=\mathbb Q[x]/(x^2)$,we have $V(k)=\lbrace 0 \rbrace \subset \mathbb A^1_k$, so that $ \text {card } V(k)=1$ but $dim_\mathbb QA=2$ . What is true is the inequality $\text {card } V(k) \leq dim_\mathbb QA$, which follows from, say, Dedekind's theorem on independance of homomorphisms. – Georges Elencwajg Dec 28 2011 at 12:14
Dear Georges, you're right, thanks. I think I had in mind the case $A$ is the ring of regular functions on some finite set, so that it's automatically reduced. – François Brunault Dec 28 2011 at 13:15
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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.

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 I thought I understood this argument, but now I'm not so sure. If $K = \overline{\mathbb{Q}(t)}$, for example, then $K$ isn't an algebraic extension of $\mathbb{Q}$, so in particular isn't a Galois extension. – Qiaochu Yuan Dec 28 2011 at 20:29
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 @Qiaochu: For this argument to work, you only need $K^{\operatorname{Aut}(K/\mathbb{Q})} = \mathbb{Q}$, which is true for any algebraically closed field $K$. Perhaps GH meant Galois in Pete Clark's sense (mathoverflow.net/questions/11958/…)? – Kevin Ventullo Dec 28 2011 at 21:54
@Qiaochu and Kevin: You are right, I meant $\mathrm{Aut}(K/\mathbb{Q})$ in place of $\mathrm{Gal}(K/\mathbb{Q})$. I will edit and if you like it please give back the lost points. – GH Dec 29 2011 at 0:40
I was confused about this also. Thank you for clarifying this Qiaochu and Kevin. However I am stuck on the last part: You end up with a set of $nm$ symmetric polynomials in $z_{ij}$ each equal to a rational, since the coefficients of $Q_j(z)$ are all rational. This is an improvement since all polynomials are now symmetric, but I can't connect the dots to the conclusion. Is the "it follows.." part a one liner or is there some further manipulation involved? – Jason Suagee Dec 29 2011 at 3:36
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 @Jason: Each $z_{ij}$ is algebraic over $\mathbb{Q}$, because it is the root of $Q_j\in\mathbb{Q}[z]$. Note that there are $n$ polynomials, and they are not symmetric in any sense. – GH Dec 29 2011 at 3:46
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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.

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Qiaochu, please read the comment under my response. – GH Dec 29 2011 at 0:43

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