Free Resolutions and Hilbert Polynomials

This post is about Hilbert Polynomials and how they can be computed using free resolutions, and is based on a talk I gave on 7/4. There is a pdf version: Free Resolutions and Hilbert Polynomials. For more detail and historical background, see [Eis05, Ch. 1] (this was also used as the basis for this talk); for general facts about commutative algebra that are used, see [AM69, Eis95].

Hartshorne 一章, 問題2.14

The Segre Embedding. Let $\psi \colon \mathbb{P}^r \times \mathbb{P}^s \to \mathbb{P}^N$ be the map defined by sending the ordered pair $(a_0,\ldots,a_r) \times (b_0,\ldots,b_s)$ to $(\ldots,a_ib_j,\ldots)$ in lexicographic order, where $N = rs + r + s$. Note that $\psi$ is well-defined and injective. It is called the Segre embedding. Show that the image of $\psi$ is a subvariety of $\mathbb{P}^N$.

Proof. Let $\mathbb{P}^N$ have coordinates $\{z_{ij}:i=0,\ldots,r,~j=0,\ldots,s\}$, and let $\mathfrak{a} = \ker(k[\{z_{ij}\}] \to k[x_0,\ldots,x_r,y_0,\ldots,y_s])$ where $z_{ij} \mapsto x_iy_j$. Then, $\mathfrak{a} = \langle w_{ij}w_{k\ell} - w_{kj}w_{i\ell}\rangle$. We claim $\mathrm{Im}\:\psi = Z(\mathfrak{a})$. But any $P \in \mathrm{Im}\:\psi$ has coordinates $(\ldots,a_ib_j,\ldots)$, and clearly any $f \in \mathfrak{a}$ vanishes on $P$, and so $\mathrm{Im}\:\psi \subset Z(\mathfrak{a})$. Conversely, note that any point $P \in Z(\mathfrak{a})$ satisfies $w_{ij}w_{k\ell} - w_{kj}w_{i\ell}$, and if $w_{00} \ne 0$, say, then setting $i=j=0$ gives $(x,y),~x = (w_{00},\ldots,w_{r0}), y = (w_{00},\ldots,w_{0s})$ maps to $P$.