Rational points of bounded height on a variety

Question

I would like to ask for some clarification on the following argument which I can not quite understand.

There is a variety $X$ of dimension $n$ over a number field with a degree two map $f:X\dashrightarrow \mathbb{P}^{\frac{n}{2}}\times\mathbb{P}^{\frac{n}{2}}$ and a rational map $g:\mathbb{P}^n\dashrightarrow X$. The map $f\circ g$ is given by $n+2$ polynomials of a certain degree, say $b$. Hence, if $p\in\mathbb{P}^n$ has height at most $B^{\frac{1}{b}}$ then its image has height at most $B$. So far so good.

Now comes the part that I do not get. This should imply that there is a $t > 0$ such that for any open subset $U\subset X$ the number of rational points of height at most $B$ in $U$ is at least $\lambda B^{t}$ for $B\gg 0$, where $\lambda > 0$ depends on $X$.

Why does the last statement hold true? Thank you very much.

The map $f\circ g$ is generically $b^n$ to one. So the map $g$ should be $\frac{b^n}{2}$ to one. Then the number of points of height at most $B$ of $X$ is at least the number of points of height at most $B^{\frac{1}{b}}$ of $\mathbb{P}^n$ multiplied by $\frac{2}{b^n}$. I think at this point I should use an estimate on the number of points of height at most $B^{\frac{1}{b}}$ of $\mathbb{P}^n$ which is unknown to me (probably a power of $B$). Even if I could do that I would not know where the $\lambda > 0$ depending on $X$ is coming from.

Also it seems that the map $f$ is not really necessary in this argument. One could reason on the polynomials defining $g:\mathbb{P}^n\dashrightarrow X\subseteq\mathbb{P}^N$.

Answer 1

Given the situation, it's understandable that some stuff is missing from the argument.

One part that's missing is the definition of the height.

In modern language (Weil's height machine), we usually take a height function to be defined from an ordered pair of a variety and a line bundle, well-defined up to a constant factor, by the rule that if your line bundle is the pullback of $\mathcal O(1)$ along a map to $\mathbb P^n$ then the associated height is the pullback of the standard height function on $\mathbb P^n$, and height is multiplicative in tensor products of line bundles.

So to even formulate the question, you need to pick a line bundle.

The obvious line bundle on $X$ to work with is the pullback of $\mathcal O(1,1)$ from $\mathbb P^{n/2} \times \mathbb P^{n/2}$.

In that case, to prove the estimate for $X$, it suffices to prove the estimate for $\mathbb P^{n/2} \times \mathbb P^{n/2}$ since by the definition the height doesn't change when you go there. We only have to multiply by a factor of $2$, which will be independent of $X$.

There is a standard bound for heights in this situation. A point of height $<X$ on $\mathbb P^{n/2} \times \mathbb P^{n/2}$ can be written as $((a_0: \dots : a_{n/2}), (b_0:\dots : b_{n/2}))$ where the $a_i$ and $b_i$ are integers, each tuple is relatively prime, and $\max (a_i) \max(b_i)< X$. Thus we can take $\max(a_i) < 2^m$ and $\max(b_i) < 2^\ell$ where $m+l=\lceil \log_2(X) \rceil+1$. The number of choices of $m$ is $\lceil \log_2(X) \rceil+2$ and then we have at most $2^{ (m+1) ( (n/2)+1)}$ possibilities for the $a_i$ and $2^{ (\ell+1) (n/2+1)}$ possibilities for the $b_i$ for $2^{ (m+\ell+2) ( n/2+1)} \leq X^{n/2+1} 2^{2n+4}$ possibilities in total. Summing over the options for $m$ and $\ell$, we get at most a constant times $X^{n/2+1} \log X$.

So we can take any $t> n/2+1$.

There are many things present that are not needed for this (the constant depending on $X$, the open set $U$, and the map from $\mathbb P^n$) so maybe something different was meant.

It is very hard to use the map from $\mathbb P^n$ to control the number of points of bounded height as, first, not all points need lie in the image of such a map (since it's rational and not necessarily well-defined everywhere), and, second, it's possible that the height of the image under the map could be lower than the height of the original point.