Are all Finite Subsets of Affine n-space Algebraic sets, and related question - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-25T04:23:17Z http://mathoverflow.net/feeds/question/84433 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84433/are-all-finite-subsets-of-affine-n-space-algebraic-sets-and-related-question Are all Finite Subsets of Affine n-space Algebraic sets, and related question Jason Suagee 2011-12-28T07:54:42Z 2011-12-29T00:42:20Z <p>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.</p> <p>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?</p> <p>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.</p> http://mathoverflow.net/questions/84433/are-all-finite-subsets-of-affine-n-space-algebraic-sets-and-related-question/84434#84434 Answer by Qiaochu Yuan for Are all Finite Subsets of Affine n-space Algebraic sets, and related question Qiaochu Yuan 2011-12-28T08:05:45Z 2011-12-28T08:05:45Z <p>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. </p> <p>2) Yes. This should follow concretely from results in <a href="http://en.wikipedia.org/wiki/Elimination_theory" rel="nofollow">elimination theory</a> of which I am totally unaware, and abstractly from <a href="http://en.wikipedia.org/wiki/Constructible_set_(topology)" rel="nofollow">Chevalley's theorem</a>. </p> http://mathoverflow.net/questions/84433/are-all-finite-subsets-of-affine-n-space-algebraic-sets-and-related-question/84436#84436 Answer by GH for Are all Finite Subsets of Affine n-space Algebraic sets, and related question GH 2011-12-28T08:59:43Z 2011-12-29T00:42:20Z <p>Here is a simple argument for the second question. I renamed the field $k$ to $K$ for clarity. </p> <p>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> <p>P.S. Thanks to Qiaochu Yuan and Kevin Ventullo.</p> http://mathoverflow.net/questions/84433/are-all-finite-subsets-of-affine-n-space-algebraic-sets-and-related-question/84439#84439 Answer by Georges Elencwajg for Are all Finite Subsets of Affine n-space Algebraic sets, and related question Georges Elencwajg 2011-12-28T09:44:18Z 2011-12-28T09:44:18Z <p>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 $A=\mathbb Q[x_1,...,x_n]/I$.<br> 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 $\mathbb Q$. </p> <p>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 $V(k)$. </p> <p>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 $\mathbb Q$ .<br> 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 $z=(z_1,...,z_n) \in {\overline {\mathbb Q}}^n \subset k^n$</p>