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.

Hartshorne 一章, 問題2.10

The Cone Over a Projective Variety. Let Y \subseteq \mathbb{P}^n be a nonempty algebraic set, and let \theta\colon \mathbb{A}^{n+1} - \{(0,\ldots,0)\} \to \mathbb{P}^n be the map which sends the point with affine coordinates (a_0,\ldots,a_n) to the point with homogeneous coordinates (a_0,\ldots,a_n). We define the affine cone over Y to be

C(Y) = \theta^{-1} \cup \{(0,\ldots,0)\}.

  1. Show that C(Y) is an algebraic set in \mathbb{A}^{n+1}, whose ideal is equal to I(Y), considered as an ordinary ideal in k[x_0,\ldots,x_n].
  2. C(Y) is irreducible if and only if Y is.
  3. \dim C(Y) = \dim Y + 1.

Sometimes we consider the projective closure \overline{C(Y)} of C(Y) in \mathbb{P}^{n+1}. This is called the projective cone over Y.

Proof of a. I(C(Y)) = I(\theta^{-1}(Y) \cup \{(0,\ldots,0)\}) = I(Y) since \theta^{-1}(Y) \cup \{(0,\ldots,0)\} consists of all scalar multiples of points in Y, i.e., all points in \mathbb{A}^{n+1} that are zeros of the generators of I(Y).

Proof of b. C(Y) is irreducible if and only I(C(Y)) is prime if and only if I(Y) is prime (by a) if and only if Y is irreducible.

Proof of c. \dim C(Y) = \dim k[x_0,\ldots,x_n]/I(Y) = \dim S(Y) = \dim Y + 1 by Problem 2.6.

Hartshorne 一章, 問題2.9

Projective Closure of an Affine Variety. If Y \subseteq \mathbb{A}^n is an affine variety, we identify \mathbb{A}^n with an open set U_0 \subseteq \mathbb{P}^n by the homeomorphism \varphi_0. Then we can speak of \overline{Y}, the closure of Y in \mathbb{P}^n, which is called the projective closure of Y.

  1. Show that I(\overline{Y}) is the ideal generated by \beta(I(Y)), using the notation of the proof of (2.2).
  2. Let Y \subseteq \mathbb{A}^3 be the twisted cubic of (Ex. 1.2). Its projective closure \overline{Y} \subseteq \mathbb{P}^3 is called the twisted cubic curve in \mathbb{P}^3. Find generators for I(Y) and I(\overline{Y}), and use this example to show that if f_1,\ldots,f_r generate I(Y), then \beta(f_1),\ldots,\beta(f_r) do not necessarily generate I(\overline{Y}).

Proof of a. Suppose F \in I(\overline{Y}). Then f = \varphi_0(F) = F(1,x_1,\ldots,x_n) vanishes on Y, so f \in I(Y). Since \beta(f) = F, we have that I(\overline{Y}) \subseteq \langle \beta(I(Y))\rangle. Similarly, if f \in \langle \beta(I(Y)) \rangle, then it is the sum of polynomials of the form \beta(F), each one of which is in I(\overline{Y}) after homogenization, and so \langle \beta(I(Y))\rangle \subseteq I(\overline{Y}).

Proof of b. Recall I(Y) = \langle z-x^3,y-x^2 \rangle. Now \overline{Y} = \{(1,s/t,s^2/t^2,s^3/t^3)\} = \{(t^3,st^2,s^2t,s^3)\}, and so I(\overline{Y}) = \langle x_0x_2 - x_1^2,x_0x_3 - x_1x_2,x_1x_3 - x_2^2\rangle, but this is not equal to \langle \beta(z-x^3),\beta(y-x^2) \rangle.

Hartshorne 一章, 問題2.8

A projective variety Y \subseteq \mathbb{P}^n has dimension n-1 if and only if it is the zero set of a single irreducible homogeneous polynomial f of positive degree. Y is called a hypersurface in \mathbb{P}^n.

Proof. \dim Y = n-1 \iff \dim S(Y) = n by Problem 2.6. By Proposition 1.13, this is true if and only if the affine cone is defined by as the zero set of a nonconstant irreducible polynomial f\in S, which is true if and only if Y is defined by the homogenization F of f.