Let $P_0,P_1,\ldots,P_r$ be distinct points in $\mathbb{P}^n$. Why there is a hyperplane $H$ in $\mathbb{P}^n$ passing through $P_0$ but not through any of $P_1,\ldots,P_r$?
Hyperplane in projective space
2
$\begingroup$
algebraic-geometry
-
1The result is obviously false for projective space over a finite field $k$ : just take $P_0,P_1,...,P_n$ to be an enumeration of all the points of $\mathbb P^n(k)$ ! – 2012-06-21
1 Answers
5
By projective duality, your question is equivalent to asking why, given a finite collection of distinct hyperplanes in $\mathbb P^n$, there is a point lying on exactly one of them. Does that make it any easier?
-
0@user10: Dear user, No, that wouldn't be valid, because not all hyperplane configurations are projectively equivalent to coordinate hyperplane configurations. (If $r$ is larger then $n+1$, then you can't find $r$ different coordinates in the first place.) Regards, – 2012-06-22