3

I read Shafarevich basic algebraic geometry on page 70 he says for any $X \subseteq \Bbb{P}^N$ can find a linear form $L$ that does not vanish on any component of $X$, I tried to prove this by induction but failed how to prove this?

1 Answers1

6

On each of the finitely many irreducible components $X_1,\cdots, X_r$ of $X$ choose a point $x_i\in X_i$.

The set $\mathcal L_i\subsetneq (\mathbb P^N)^\ast$ of hyperplanes $H \subset \mathbb P^N$ containing $x_i$ is a closed, strict subspace of the space $ (\mathbb P^N)^\ast$ of hyperplanes of $\mathbb P^N$ (this space $ (\mathbb P^N)^\ast$ is known as the dual projective space of $ \mathbb P^N$) .
The complement $ \mathcal U_i=(\mathbb P^N)^\ast \setminus \mathcal L_i \subset (\mathbb P^N)^\ast$ is thus a non-empty Zariski open subset of $(\mathbb P^N)^\ast$: it consists of the hyperplanes of $\mathbb P^N$ not containing $x_i$.
Since $(\mathbb P^N)^\ast$ is irreducible (after all it is isomorphic to $\mathbb P^N$ !), the intersection $\mathcal U=\cap_ {i=1}^n \mathcal U_i\subset (\mathbb P^N)^\ast$ is non-empty too.

And now we are done: any $H\in \mathcal U$ is a hyperplane of $\mathbb P^N$ containing none of the $x_i$'s and thus a fortiori containing none of the irreducible components $X_i$ of $X$.
If $L=\sum_{i=0}^N a_iz_i$ is a linear form defining $H$ (i.e. $H$ is defined by $L=0$), then that linear form is a solution to your problem.

Edit: a technical point and the wisdom of Dieudonné

The dual projective space of $ \mathbb P^N$ is the space $(\mathbb P^N)^\ast$ of hyperplanes of $ \mathbb P^N$.
A hyperplane $H\subset \mathbb P^N$ has an equation $a_0z_0+\cdots +a_Nz_N=0$ and the linear form $a_0z_0+\cdots +a_Nz_N$ gives the point $l_H=[a_0:\cdots :a_N]\in (\mathbb P^N)^\ast$.
If now $P=[p_0:\cdots :p_N]\in \mathbb P^N$ is a fixed point in the original projective space $\mathbb P^N$, the set of hyperplanes of $\mathbb P^N$ containing $P$ is the set $\mathcal L_P\subset (\mathbb P^N)^\ast$ of those $l_H=[a_0:\cdots :a_N]\in (\mathbb P^N)^\ast$ such that $\sum_i p_ia_i=0$, and all those $l_H$ constitute a hyperplane $\mathcal L_P\subset (\mathbb P^N)^\ast$ in the dual projective space.
The subtle point here is that the $p_i$'s must be considered to be constants and the $a_i$'s to be variables!

Dieudonné used to emphasize that a characteristic of modern mathematics is that when mathematicians now write $f(x)$ for a function, it may very well happen that $x$ is fixed whereas $f$ is variable: the delta distribution is an egregious example and here we have another example of Dieudonné's profound observation.

Georges Elencwajg
  • 150,790
  • Thanks for answer Georges. I tried to prove this myself and maybe this is alternatif proof. The number of irreducible components first is at most $N$ say it is $r$. Then for every $x_i$ I find a linear form $f_i$ such that $f_i(x_i) \neq 0$ and $f_i(x_j) = 0$ for all $j \neq i$. Then $f_1 +\ldots + f_r$ is the desired linear form. But I am not sure if we can even find such $f_i$ in the first place.... –  Jun 24 '13 at 06:48
  • Dear Kyle, it is not true at all that $r$ is at most equal to $N$: take for $X$ a billion points in $\mathbb P^1$ so that $r=1,000,000,000$ and $N=1$... – Georges Elencwajg Jun 24 '13 at 06:51
  • Ok what I said about irreducible components is not correct but still for every $x_i$ can we find a linear form $f_i$ such that $f_i(x_i) \neq 0$ and $f_i(x_j)= 0$ for $j \neq i$? –  Jun 24 '13 at 06:58
  • No, we can't even for three points $x_1,x_2,x_3\in \mathbb P^1$. Indeed, any linear form $f=az+bw$ on $\mathbb P^1$ vanishing on $x_1$ and $x_2$ will be the zero form ($a=b=0$) and thus will vanish on $x_3$. – Georges Elencwajg Jun 24 '13 at 07:10
  • @KyleSutherland If you are having trouble seeing that the space of all hyperplanes through $x$ is a closed proper subspace note the following. 1) To see it is a proper subset note we can choose a hyperplane not going through $x$. Indeed we may assume that the first coordinate of $x$ is not zero, and so the hyperplane $x_0 = 0$ does not contain $x$. 2) That it is closed, suppose $x$ has homogeneous coordinates $(a_0,\ldots,a_n)$. Then identifying $(\Bbb{P}^n)^\ast$ with $\Bbb{P}^n$, the set of all hyperplanes through $x$ is the same as the set of all points in $\Bbb{P}^n$ that satisfy the –  Jun 24 '13 at 11:12
  • equation $a_0x_0 + \ldots + a_nx_n = 0$ which is 100% closed in the Zariski topology. Dear @GeorgesElencwajg, +1 for your beautiful answer I am learning so much from reading it right now. –  Jun 24 '13 at 11:12
  • Dear @Benja, thanks and sorry for not answering your former request: I have no interesting insight into the problem :-( – Georges Elencwajg Jun 24 '13 at 12:02