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.