# Finite subset of projective $n$ space is a variety

How do I prove that a finite subset $M$ of the $n$-dimensional projective space is a variety?

I tried finding a homogeneous polynomial of degree one that vanishes at an arbitrary x in M and then go from there, but to no avail. Thanks

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It is easy to come up with an ideal that vanishes at exactly a single point of affine $n$-space: for the point $\mathbf{p}=(a_1,a_2,\ldots,a_n)$, the ideal $I=(x_1-a_1,\ldots,x_n-a_n)$ vanishes at $\mathbf{p}$ and only $\mathbf{p}$. That means that the homogeneous ideal $(x_1-a_1x_0,x_2-a_2x_0,\ldots,x_n-a_nx_0)$ of $K[x_0,\ldots,x_n]$ vanishes exactly at the projective point $(1:a_1\colon a_2\colon\cdots \colon a_n)$ of projective $n$-space.

So a single point (whether projective or affine) is an algebraic set. The union of two algebraic sets is algebraic, since if $Y_1$ is the zero set of $I$ and $Y_2$ is the zero set of $J$, then $Y_1\cup Y_2$ is the zero set of $IJ$. Inductively, the union of any finite collection of algebraic sets is algebraic. In particular, any finite subset of $n$-dimensional (projective or affine) space is algebraic.

A single point is thus an algebraic variety (in the sense given by Hartshorne, which requires it to be an irreducible algebraic set). But a finite collection of points that has more than one point is not irreducible.

If your definition of variety requires irreducibility, then $M$ is not a variety unless $M$ is a singleton or empty. If it does not require irreducibility, then the above works.

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Thanks! I just started learning AG today so some terms are unfamiliar to me, but I think I got it. – Paul Sep 9 '11 at 17:30

It suffices to show that a point is a variety. Call that point $x$. Take any other point $y$. $x = [x_0, \cdots, x_n]$ being different from $y = [y_0, \cdots, y_n]$ means that some ratio $x_i/x_j \neq y_i/y_j$. Take the linear form $x_jz_i - x_i z_j$. This vanishes on $x$ not $y$. Now vary $y$ (and the corresponding $i,j$), and put all such linear forms together.

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Thanks to you, too! – Paul Sep 9 '11 at 17:30
Dear Soarer, the equations $x_jz_i - x_i z_j=0$ you write are correct but have nothing to do with $y$ and they shouldn't. What you have to vary is the pair i,j and not an irrelevant $y$. For example three (redundant) equations define $(a_0:a_1:a_2)\in \mathbb P^2$. One of them is $a_0z_1-a_1z_0=0$. – Georges Elencwajg Sep 9 '11 at 19:52
Dear Georges, I meant to say vary $y$ and the pair $i,j$. The answer is now edited. (Edit: Now I see what you are saying. When I was writing the answer I was having in mind to have one linear form separating $x$ and each point different from $x$, but of course, although to each $y$ corresponds to at least one linear form, these linear forms put together are really just dependent on $x$) – Soarer Sep 9 '11 at 20:12
Dear Soarer, consider the point (4:2:3). My point of view yields the (redundant) three equations $2z_0-4z_1=0,\quad 3z_0-4z_2=0$ and $\; 3z_1-2z_2=0$. What equations do you get by your procedure? – Georges Elencwajg Sep 10 '11 at 10:16
Dear Georges, what we do are essentially the same I think, though the way I thought about it when writing the answer was more convoluted. – Soarer Sep 10 '11 at 15:25