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].

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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.