Kollár/Smith/Corti 一章, 問題1.8

Let k'/k be a finite Galois extension. The Galois group \mathrm{Gal}(k'/k) acts on \mathbb{A}_{k'}^n = \mathrm{Spec}k'[x_1,\ldots,x_n] coordinatewise. Let X be a closed algebraic set of \mathbb{A}_{k'}^n. Prove that the following are equivalent:

  1. the set X can be defined by polynomials in k[x_1,\ldots,x_n];
  2. the set X is invariant under the \mathrm{Gal}(k'/k)-action.

Proof. Suppose f = \sum a_Ix^I \in k'[x_1,\ldots,x_n] vanishes on X. Let \sigma \in \mathrm{Gal}(k'/k), and set f^\sigma = \sum \sigma^{-1}(a_I)x^I. If X is defined by polynomials in k[x_1,\ldots,x_n] then f = f^\sigma and so X is Galois-invariant.
Conversely, if X is Galois-invariant, then if f vanishes on X then f^\sigma does also. Denoting E_1(f),\ldots,E_d(f) to be the elementary symmetric polynomials of f^\sigma for \sigma \in \mathrm{Gal}(k'/k), each E_i(f) \in k[x_1,\ldots,x_n] and they vanish on X. We claim that these polynomials define X. For, if every E_i(f) vanished at a point P, then every f^\sigma(P) is zero for they are the roots of the polynomial z^d = z^d + \sum_{i=1}^d(-1)^iE_i(f)(P)z^{d-i}. So the E_i(f) define X.

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.

Hartshorne 一章, 問題2.13

Let Y be the image of the 2-uple embedding of \mathbb{P}^2 in \mathbb{P}^5. This is the Veronese surface. If Z \subseteq Y is a closed curve (a curve is a variety of dimension 1), show that there exists a hypersurface V \subseteq \mathbb{P}^5 such that V \cap Y = Z.

Proof. The Veronese embedding is defined by v_2\colon \mathbb{P}^2 \hookrightarrow \mathbb{P}^5, (x_0:x_1:x_2) \mapsto (x_0^2,x_1^2,x_2^2,x_0x_1,x_1x_2,x_2x_0). If Z is defined by f(x_0,x_1,x_2) = 0, then f^2 \in k[x_0^2,x_1^2,x_2^2,x_0x_1,x_1x_2,x_2x_0], and so defines a hypersurface V \subseteq \mathbb{P}^5. Thus Z = v_2(Y) = V \cap Y.

Hartshorne 一章, 問題2.12

The d-Uple Embedding. For given n,d > 0, let M_0,M_1,\ldots,M_N be all the monomials of degree d in the n+1 variables x_0,\ldots,x_n, where N = \binom{n+d}{n} - 1. We define a mapping \rho_d\colon \mathbb{P}^n \to \mathbb{P}^N by sending the point P = (a_0,\ldots,a_n) to the point \rho_d(P) = (M_0(a),\ldots,M_N(a)) obtained by substituting the a_i in the monomials M_j. This is called the d-uple embedding of \mathbb{P}^n in \mathbb{P}^N. For example, if n = 1, d = 2, then N = 2, and the image Y of the 2-uple embedding of \mathbb{P}^n in \mathbb{P}^N is a conic.

  1. Let \theta\colon k[y_0,\ldots,y_N] \to k[x_0,\ldots,x_n] be the homomorphism defined by sending y_i to M_i, and let \mathfrak{a} be the kernel of \theta. Then \mathfrak{a} is a homogeneous prime ideal, and so Z(\mathfrak{a}) is a projective variety in \mathbb{P}^N.
  2. Show that the image of \rho_d is exactly Z(\mathfrak{a}). (One inclusion is easy. The other will require some calculation).
  3. Now show that \rho_d is a homeomorphism of \mathbb{P}^n onto the projective variety Z(\mathfrak{a}).
  4. Show that the twisted cubic curve in \mathbb{P}^3 (Ex. 2.9) is equal to the 3-uple embedding of \mathbb{P}^1 in \mathbb{P}^3, for suitable choice of coordinates.

Proof of a. \mathfrak{a} is homogeneous since it is generated by homogeneous elements in k[y_0,\ldots,y_N], for each y_i is sent to an element of degree d. \mathfrak{a} is prime since \theta maps onto a subring of k[x_0,\ldots,x_n], which is an integral domain, so \mathfrak{a} is prime by the first isomorphism theorem.

Proof of b. f \in \mathfrak{a} \implies f(M_0,\ldots,M_N) = 0, so \mathrm{Im}\:\rho_d \subseteq Z(\mathfrak{a}). Conversely, Z(\mathfrak{a}) \subseteq \mathrm{Im}\:\rho_d \iff \mathfrak{a} \supseteq I(\mathrm{Im}\:\rho_d). If f \in I(\mathrm{Im}\:\rho_d), then it is identically zero for all Q \in \mathrm{Im}\:\rho_d, so f(M_0,\ldots,M_N) = 0, i.e. f \in \ker\theta.

Proof of c. By b it suffices to show that the map \rho_d is continuous with continuous inverse.

Proof of d. \mathbb{P}^1 \hookrightarrow \mathbb{P}^3 maps (t : s) \mapsto (t^3,t^2s,ts^2,s^3), which is the twisted cubic curve by Ex. 2.9b.

Hartshorne 一章, 問題2.11

Linear Varieties in \mathbb{P}^n. A hypersurface defined by a linear polynomial is called a hyperplane.

  1. Show that the following two conditions are equivalent for a variety Y in \mathbb{P}^n:
    1. I(Y) can be generated by linear polynomials.
    2. Y can be written as an intersection of hyperplanes.

    In this case we say that Y is a linear variety in \mathbb{P}^n.

  2. If Y is a linear variety of dimension r in \mathbb{P}^n, show that I(Y) is minimally generated by n - r linear polynomials.
  3. Let Y,Z be linear varieties in \mathbb{P}^n, with \dim Y = r, \dim Z = s. If r+s-n\ge0, then Y \cap Z \ne \emptyset. Furthermore, if Y \cap Z \ne \emptyset, then Y \cap Z is a linear variety of dimension \ge r + s - n.

Proof of a. I(Y) = \langle f_1,\ldots,f_r \rangle for f_i linear implies Y = Z(f_1) \cap \cdots \cap Z(f_r) for Z(f_i) hyperplanes. If Y = H_1 \cap \cdots \cap H_r, then I(Y) = I(H_1 \cap \cdots \cap H_r) = \langle I(H_1),\ldots,I(H_r) \rangle, where we note each I(H_i) is a linear polynomial. Note we can drop the radical since our generators are linear.

Proof of b. Suppose I(Y) can be generated by r+1, i.e., $\dim S(Y) > r+1$. By Problem 2.6 this implies $\dim Y > r$, a contradiction.

Proof of c. Consider their respective affine cones C(Y),C(Z); they have dimension r+1,s+1 respectively by Problem 2.10. Then, their intersection in \mathbb{A}^{n+1} has dimension \ge (r+1) + (s+1) - (n+1) = r + s - n + 1 \ge 1 by the hypothesis r + s - n \ge 0. Since C(Y) \cap C(Z) = C(Y \cap Z), we finally have that \dim Y \cap Z \ge r + s - n by Problem 2.10. Y \cap Z is linear by a.