# 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.

# 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 $f$.