4

Can someone give me a starting point for the following question? I don't know where to begin!

Let $C \subset \mathbb{P}^n$ be an irreducible curve of degree $d$. Show that $C$ is contained in a linear subspace of $\mathbb{P}^n$ of dimension $d$.

TheBeiram
  • 435

2 Answers2

5

Arbitrarily choose $d$ distinct points $p_1,p_2,\cdots p_d\in C$ and then some $(d-1)$-dimensional linear space $P^{d-1}\subset \mathbb P^n$ containing all of them: $ p_1,p_2,\cdots p_d\in P^{d-1} $.
Choose a new point $p_0\in C$ and some $d$-dimensional linear space $P^d\subset \mathbb P^n$ containing both $p_0$ and $P^{d-1}$.
Since $P^d$ contains the $d+1$ points $p_0, p_1,p_2,\cdots p_d$ of the curve $C$ of degree $d$ we must have $C\subset P^d$ and the result is proved.

Georges Elencwajg
  • 150,790
  • Thank you for your answer! I get it, the only thing is that I assume you use that $deg(P^d)=deg(P^{d-1})=1$? Is this obvious? – TheBeiram May 29 '15 at 16:20
  • 1
    We have $\deg P^d=1$ because the number of points of intersection of $P^d$ with a generic $P^{n-d}$ is $1$. In general the degre of a variety $X_d$ of dimension $d$ is the number of points of intersection of that variety with a generic linear variety of complementary dimension : $deg X_d= card(X_d \cap P^{n-d})$ . – Georges Elencwajg May 29 '15 at 16:57
1

Another proof!

Let $H$ be a hyperplane of $\mathbb{P}^n$; if $$ X\not\subset H,\,X\cap H\neq\emptyset $$ then by Bézout's theorem $$ |X\cap H|\leq\deg X\deg H=d; $$ that is there are at most $d+1$ distinct points of $X$ in general position, in other words, $X$ is contained in a linear subspace of $\mathbb{P}^n$ of dimention at most $d$.

Armando j18eos
  • 3,907