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 $g:\mathbb{P}^n\dashrightarrow X\subseteq\mathbb{P}^N$.



Rational points of bounded height on a variety
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.


