# 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 $f$.

# Hartshorne 一章, 問題2.7

1. $\dim \mathbb{P}^n = n$.
2. If $Y \subseteq \mathbb{P}^n$ is a quasi-projective variety, then $\dim Y = \dim \overline{Y}$.

Proof of a. $S(Y) = k[x_0,\ldots,x_n]$, and so $\dim \mathbb{P}^n = \dim S(Y) - 1 = n$ by Problem 2.6.

Proof of b. $\dim S(Y) = \dim Y + 1$ by Problem 2.6. But $\dim S(Y)$ is equal to the dimension of the affine cone of $Y$, which is equal to $\dim S(\overline{Y})$ by Proposition 1.10, which in turn is equal to $\dim \overline{Y} + 1$ by Problem 2.6.

# Hartshorne 一章, 問題2.6

If $Y$ is a projective variety with homogeneous coordinate ring $S(Y)$, show that $\dim S(Y) = \dim Y + 1$.

Proof. Let $\varphi_i\colon U_i \to \mathbb{A}^n$ be the homeomorphism of Proposition 2.2, let $Y_i$ be the affine variety $\varphi_i(Y \cap U_i)$, and let $A(Y_i)$ be its affine coordinate ring. Assume $i=0$ is such that $\dim Y_i = \dim Y$. Any $F/x_0^n \in S(Y)_{x_0}$ of degree zero can be written as $F(1,x_1/x_0,\ldots,x_n/x_0)$ which is exactly $\alpha(f)$ from Proposition 2.2. On the other hand any polynomial $f \in A(Y_0)$ can be homogenized with $F = \beta(f)$ from Proposition 2.2. If $F$ has degree $d$, we associate $F/x_0^d$ with it. Thus $A(Y_0)$ can be identified with $(S(Y)_{x_0})_0$, and so $S(Y)_{x_0} \cong A(Y_0)[x_0,x_0^{-1}]$. Now the transcendence degree of $K(A(Y_0)[x_0,x_0^{-1}])$ over $k$ is $\dim Y_0 + 1$ by Proposition 1.7 and Theorem 1.8A, and so $\dim S(Y) = \dim S(Y)_{x_0} = \dim Y_0 + 1 = \dim Y + 1$ by Problem 1.10.

# Hartshorne 一章, 問題2.5

1. $\mathbb{P}^n$ is a noetherian topological space.
2. Every algebraic set in $\mathbb{P}^n$ can be written uniquely as a finite union of irreducible algebraic sets, no one containing another. These are called its irreducible components.

Proof. a follows from Problem 2.3 since $k[x_0,\ldots,x_n]$ is noetherian. b is just Proposition 1.5.

# Hartshorne 一章, 問題2.4

1. There is a 1-1 inclusion-reversing correspondence between algebraic sets in $\mathbb{P}^n$, and homogeneous radical ideals of $S$ not equal to $S_+$, given by $Y \mapsto I(Y)$ and $\mathfrak{a} \mapsto Z(\mathfrak{a})$.
2. An algebraic set $Y \subseteq \mathbb{P}^n$ is irreducible if and only if $I(Y)$ is a prime ideal.
3. Show that $\mathbb{P}^n$ itself is irreducible.

Proof. a is just Problem 2.3. b follows from Corollary 1.4 by looking at the affine cone. c follows since $\mathbb{P}^n = Z(0)$.