What is Hilbert polynomial of this projective variety?

Suppose you have a map $\varphi\colon\mathbb{C}^m\times\mathbb{C}^n\to\mathrm{Mat}_{m,n}(\mathbb{C})$ defined by sending $(\mathbf{u},\mathbf{v})\mapsto\mathbf{u}\cdot\mathbf{v}^T=(u_i,v_j)$. So $\mathrm{Im}(\varphi)$ is the set of matrices with rank at most $1$.

So I get that $\varphi$ is the morphism of affine spaces corresponding to the morphism $\psi\colon\mathbb{C}[z_{11},\dots, z_{mn}]\to\mathbb{C}[x_1,\dots, x_m,y_1,\dots,y_n]$ sending $z_{ij}\mapsto x_iy_j$. so by the correspondence theorem, the set of polynomials vanishing on $\mathrm{Im}(\varphi)$ is just $\ker\psi$.

Then $\ker\psi$ defines a projective variety, and my question is, what exactly is the Hilbert polynomial of this projective variety?

This is based on something I was reading earlier on this site, so I admit I'm stepping out of my usual area of comfort.

-

The projective variety you are interested in is the image $S$ of the Segre embedding $s:\mathbb P^{m-1}\times \mathbb P^{n-1}\to \mathbb P^{mn-1}:( [\ldots:u_i:\ldots], [\ldots:v_j:\ldots])\mapsto [\ldots:u_iv_j:\ldots]$.
A homogeneous polynomial of degree $d$ on $\mathbb P^{mn-1}$ pulls back under $s$ to a polynomial bihomogeneous of degree $(d,d)$ on $\mathbb P^{m-1}\times \mathbb P^{n-1}$.
So the required Hilbert polynomial is $$p_S(d)= \binom {m-1+d}{m-1} \cdot \binom {n-1+d}{n-1}$$

As an application we can calculate the degree of $S$ : it is $dim(S) !=(m+n-2)!$ times the leading coefficient of $p_S$ (that is, the coefficient of $d^{m+n-2}$ ) and we find $$deg(S)=\frac{(m+n-2)!}{(m-1)!(n-1)!} =\binom {m+n-2}{m-1}$$ The simplest example is for $n=m=2$: you obtain the embedding of $\mathbb P^{1}\times \mathbb P^{1}$ as a quadric $Q\subset P^{3}$, of Hilbert polynomial $p_Q(d)=(d+1)\cdot(d+1)$

-
Hey thanks Georges. May I ask how you identified the projective variety with the Segre embedding? I was not familiar with that map until now. – Hana Bailey May 4 '12 at 7:05
Dear Cowbell, since I know the formula for the Segre embedding, a formula like $z_{ij}=u_iv_j$ automatically makes me think of it. Nothing deep or clever there , I confess ... – Georges Elencwajg May 4 '12 at 8:48

Notice the ideal is generated by all two-by-two minors of the matrix of z's.

You will find in the book Combinatorial commutative algebra by E. Miller and B. Sturmfels a proof that, in fact, the set of all those minors is a Groebner basis. That makes it easy to compute the Hilbert series of your ring, since it is the same as that of the quotient by the initial ideal.

-