4

I recently read the paper "Compactification of a Drinfeld Period Domain over a Finite Field" by Pink and Schieder. (link to the paper)

I am confused about two statements appearing in this paper:

(1) removing all proper $\mathbb{F}_q$-rational linear subspaces from $\mathbb{P}_{\mathbb{F}_q}^{r-1}$ (line 1~3 on p.202);

(2) $\mathbb{P}_{\mathbb{F}_q}^{r-1}\setminus\{ \text{union of all} \ \mathbb{F}_q\text{-rational hyperplanes} \}$ (the first paragraph under Theorem 1.10, p.205).

About the first statement, I cannot understand what a $\mathbb{F}_q$-rational linear subspace is. Since $\mathbb{P}_{\mathbb{F}_q}^{r-1}$ is not a vector space, it cannot have "linear subspace", right? So, what is the definition of a $\mathbb{F}_q$-rational linear subspace?

As for the second one, I think that $\mathbb{P}_{\mathbb{F}_q}^{r-1}\setminus\{ \text{union of all} \ \mathbb{F}_q\text{-rational hyperplanes} \}=\varnothing$. The definition of a $\mathbb{F}_q$-rational hyperplane in $\mathbb{P}_{\mathbb{F}_q}^{r-1}$ shall be the zero locus of a non-constant homogeneous polynomial $f\in \mathbb{F}_q[X_1,\cdots,X_r]$. However, every point in $\mathbb{P}_{\mathbb{F}_q}^{r-1}$ must lie on some $\mathbb{F}_q$-rational hyperplane. For example, let $[a_1:\cdots :a_r]\in \mathbb{P}_{\mathbb{F}_q}^{r-1}$. If there exits $i\in \{ 1,2,\cdots,r \}$ such that $a_i=0$, then $[a_1:\cdots :a_r]\in \{ X_i=0 \}$; otherwise, $[a_1:\cdots :a_r]\in \{ a_2X_1-a_1X_2=0 \}$.

It seems that I have misunderstood something, but I cannot figure it out.

AAAS
  • 93

1 Answers1

3

  1. A proper $\Bbb F_p$-rational linear subspace is a closed subscheme of the form $\Bbb P^a_{\Bbb F_p}\subset \Bbb P^n_{\Bbb F_p}$ with $a<n$ cut out by linear equations with coefficients in $\Bbb F_p$. They exactly correspond to nonzero proper linear subspaces of $\Bbb F_p^{n+1}$, and it is common to refer to such subschemes as linear, like we sometimes refer to copies of $\Bbb P^1$ as "lines".

  2. It should be remembered that a $\Bbb F_p$-rational hyperplane is cut out by a single linear polynomial with coefficients in $\Bbb F_p$, which immediately implies that every closed point on such a hyperplane may be represented as $[a_0:\cdots:a_n]$ with all $a_i\in\Bbb F_p$. On the other hand, not all points in $\Bbb P^n_{\Bbb F_p}$ are of the form $[a_0:\cdots:a_n]$ for $a_i\in \Bbb F_p$. No nonclosed point is of this form, and every non-$\Bbb F_p$-rational point fails to be of this form as well. For instance, consider $\Bbb P^1_{\Bbb F_3}$ and let $i$ be a root of $x^2+1=0$ over $\Bbb F_3$. Then $(x^2+y^2)\subset \Bbb F_3[x,y]$ defines the point $[\pm i:1]$ which is not contained in any $\Bbb F_p$-rational hyperplane (if it were, then the set $\{1,\pm i\}$ would be linearly dependent over $\Bbb F_3$, contradicting the fact that $\Bbb F_3 \subset \Bbb F_3[i]$ is a field extension of degree two with $1,i$ as a basis for $\Bbb F_3[i]$ as a $\Bbb F_3$-vector space).

KReiser
  • 65,137
  • So the key point is: $\mathbb{P}{\mathbb{F}_q}^n$ is the projective space over $\bar{\mathbb{F}_q}$; that is, $\mathbb{P}{\mathbb{F}_q}^n=\bar{\mathbb{F}_q}^{n+1}/\sim$. Do I get the point? – AAAS Dec 16 '19 at 01:51
  • That interpretation is not correct - you're missing all the non-closed points, and you need to glue together non-rational points in each Galois orbit. What is true is that the scheme $\Bbb P^n_{\Bbb F_p}$ has lots of points that aren't $\Bbb F_p$-rational points. – KReiser Dec 16 '19 at 02:26
  • I think I know what the problem is. The definition of the projective space is $\mathbb{P}{\mathbb{F}_q}^n=\text{Proj} \ \mathbb{F}_q[X_0,\cdots,X_n]$, say by Hartshorne's book. The notational confusion is: $\mathbb{P}{\mathbb{F}_q}^n\neq \mathbb{P}^n(\mathbb{F_q})$, where $\mathbb{P}^n(\mathbb{F_q})$ by convention is $\bar{\mathbb{F}_q}^{n+1}/\sim$. – AAAS Dec 16 '19 at 03:13
  • 1
    Yes, that's one way to put it. ($\Bbb P^n(\Bbb F_q)$ is $\Bbb F_q^{n+1}/\sim$, though, no taking algebraic closures here.) – KReiser Dec 16 '19 at 03:15
  • Thanks a lot!! I am confused with this problem for a week. – AAAS Dec 16 '19 at 03:18