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Let $X$ be a projective variety of dimension $r$ in $\mathbb P^n_k$ , where $n>r+1$ and $k$ is an algebraically closed field. Then, can we always find a projective variety $Y$ (depending on $X$ of course) of dimension $n-r-1$ in $\mathbb P^n_k$ such that $X$ and $Y$ has empty intersection ?

Since it is known that for projective varieties $X,Y$ in $\mathbb P^n_k$ , we always have $\dim X\cap Y\ge \dim X +\dim Y-n$ and $X\cap Y$ is nonempty if this quantity is non-negative, so my question asks if this bound is sharp ...

KReiser
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Louis
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1 Answers1

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Yes, this is true, and you may find a suitable linear $\Bbb P^{n-r-1}$ by repeatedly projecting.

Ingredients:

  • 1 projective variety $X\subset \Bbb P^n$ of dimension $r< n$
  • 1 point $p\in\Bbb P^n$ not in $X$
  • 1 hyperplane $\Bbb P^{n-1}\subset \Bbb P^n$ not containing $\Bbb P^n$.

Recipe:

Project $X$ from $p$ to $\Bbb P^{n-1}$. If the image of $X$ under this projection is $\Bbb P^{n-1}$, then $X$ was of dimension $n-1$ and the point $p$ suffices. If not, we can find another point $p_1$ in $\Bbb P^{n-1}$ not in the projection of $X$ and a $\Bbb P^{n-2}\subset \Bbb P^{n-1}$ not containing $p_1$, and project from $p$ to $\Bbb P^{n-2}$.

If the image of $X$ under this projection is $\Bbb P^{n-2}$, then $X$ was of dimension $n-2$ and the preimage of $p_1$ under the projection to $\Bbb P^{n-1}$ is a line in $\Bbb P^n$ not intersecting $X$, which works. If not, we can find another point $p_2$ in $\Bbb P^{n-2}$ not in the projection of $X$, and repeat the process.

Eventually, when we're projecting on to a linear subspace of dimension $\dim X$ we'll find that we get that the image of $X$ under this composite projection is all of the $\Bbb P^{\dim X}$ and the preimage in $\Bbb P^n$ of the last point we projected from is a linear subspace of dimension $n-r-1$ which doesn't intersect $X$.

KReiser
  • 65,137