# Pick's Theorem for rational points of bounded height

I wonder if the various lattice-point theorems, such as Pick's Theorem or Minkowski's Lattice Theorem, have been generalized to the collection of points with rational coordinates no more than height $h$?

A rational $a/b$ in lowest terms has height $\max( |a|,|b| )$. For example, here are the positive rationals of height $\le 5$: $$\left\{\frac{1}{5},\frac{1}{4}, \frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac {2}{3},\frac{3}{4},\frac{4} {5},1,\frac{5}{4},\frac{4}{ 3},\frac{3}{2},\frac{5}{3}, 2,\frac{5}{2},3,4,5 \right\}$$ (Incidentally, the length of this list is given by integer sequence A018805.)

Let us call rationals of height $\le h$, $h$-rationals, and points with both coordinates $h$-rationals, $h$-rational points (my own terminology). Generalizations from lattice points to $h$-rational points seems natural, and likely has been explored...? If so, I would appreciate a reference! That's my primary question.

Here is a specific, Pick's Theorem -like question:

Can the number $i$ of $h$-rational points inside a polygon $P$ be expressed in a form that is not tantamount to enumerating each interior point? Assume we are given the vertex coordinates of $P$, and perhaps the number of boundary points on each edge.

If $P$ is an axis-aligned rectangle, then the number of interior points $i$ can be computed from the number of points $b_x$ lying on the bottom side and the number lying on the left side $b_y$: $i = (b_x - 2)(b_y - 2)$. So, for the rectangle with lowerleft corner $(1,1)$ below, $b_x=8$ and $b_y=7$ leads to $i=30$. But just knowing the total number of boundary points $b=26$, a la Pick's Theorem, is inadequate to determine $i$. It seems feasible that there is a nonlinear transformation that maps the $h$-rational points to the integer lattice, allowing Pick's Theorem to be applied there. Answering this question when $P$ is a triangle should lead to a result for arbitrary $P$ via triangulation. • Why do you not include negative rational numbers or $0$ when you illustrate the count on the number of rational points with height at most 5? – KConrad Jul 1 '13 at 12:55
• @KConrad: Just an arbitrary decision, matching A018805. – Joseph O'Rourke Jul 1 '13 at 13:01
• But the count of all rational numbers with height at most $n$ is not that online sequence. – KConrad Jul 1 '13 at 13:05
• @KConrad: You are right. I corrected the sentence to say the positive rational numbers of height at most $h$. Thanks. – Joseph O'Rourke Jul 1 '13 at 13:15
• I doubt that there can be any formula nearly as nice as Pick's here. You're asking for primitive integer points in a 3-dim. polyhedron (intersection of the cube $|x|\leq H, |y|\leq H, |z|\leq H$ with the polyhedral cone $\lbrace x,y,z \in {\bf R}^3 : (x/z,y/z) \in P \rbrace$). Even without the primitivity condition, the formulas are somewhat complicated, even when as here some of the polyhedron's sides are parallel to the coordinate axes; and the primitive subset surely cannot be any easier to enumerate when the asymptotic behavior as $H\rightarrow\infty$ includes a factor $1/\zeta(3)$ ! – Noam D. Elkies Jul 1 '13 at 23:08

Here is a consequence of that theorem, for the case of the projective line $\mathbf P^1_{\mathbf Q}$. Since you normalized the height of a point $p/q\in\mathbf P^1(\mathbf Q)$ to be $\max(|p|,|q|)$, when written in lower terms, let us define a measure on $\mathbf P^1(\mathbf R)=\mathbf R\cup\{\infty\}$ by $\mu=\mathrm dt/\max(1,|t|)^2$. (For an explanation of the exponent $2$ in the definition of the measure $\mu$, observe that the differential form $\mathrm dt$ has a pole of order $2$ at infinity.)
For any subset $U$ of $\mathbf P^1(\mathbf R)$, let $N(U;B)$ be the number of points of $\mathbf P^1(\mathbf Q)$ of height $\leq B$ which belong to $U$. Peyre's Theorem asserts that for any open subset $U$ of $\mathbf P^1(\mathbf R)$ whose boundary has $\mu$-measure $0$, the following limit formula holds: $$\frac{N(U;B)}{N(\mathbf P^1(\mathbf R);B)} \to \frac{\mu(U)}{\mu(\mathbf P^1(\mathbf R))}$$ when $B\to\infty$.