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.




WordPress.com ロゴ

WordPress.com アカウントを使ってコメントしています。 ログアウト /  変更 )

Google+ フォト

Google+ アカウントを使ってコメントしています。 ログアウト /  変更 )

Twitter 画像

Twitter アカウントを使ってコメントしています。 ログアウト /  変更 )

Facebook の写真

Facebook アカウントを使ってコメントしています。 ログアウト /  変更 )


%s と連携中