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
 A: 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. 
A: 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.
