# Counting algebraic points of bounded height

Let $K$ be a number field and $X\hookrightarrow\mathbb P^n_K$ be a projective variety of degree $\delta$ (with respect to universal bundle) and dimension $d$. We denote the set $$S(X;D,B)=\{\xi\in X(\overline K)|[K(\xi):K]\leq D,\;H_{K(\xi)}(\xi)\leq B\}.$$ and $$N(X;D,B)=\#S(X;D,B),$$ where $H_{K(\xi)}(\cdot)$ is the common naive height of an algebraic point over the field $K(\xi)$.

I guess that for all $X$ with degree $\delta$ and dimension $d$, we might have $$N(X;D,B)\ll_{n,K,D}\delta N(\mathbb P^d_K;D,B),$$ since for the case of $D=1$, which is the rational points' case, we can prove the result. But I don't know how to prove it.

If this question is too difficult, at least could we prove the complete intersection case? I can only prove the case of hypersurfaces.

Thank you very much.

PS. In this question, the definition of algebraic points of bounded height is different from the usual definition, which is $$S(X;D,B)=\{\xi\in X(\overline K)|[K(\xi):K]=D,\;H_{K(\xi)}(\xi)\leq B\}.$$

This is true. Choose any linear subspace $Z$ (over $K$) of dimension $n- d-1$ and disjoint from $X$, and take $\pi : \mathbb{P}_K^n \setminus Z \to \mathbb{P}_K^d$ the corresponding linear projection. Next, consider a linear automorphism $g \in \mathrm{PGL}(d,K)$ such that $g(Z)$ is the standard linear subspace $Z_0$ defined by the vanishing of the last $n-d-1$ coordinates; $g$ of course depends on $X$.
You can use the projection $\pi' := \pi \cdot g : \mathbb{P}_K^n \setminus Z_0 \to \mathbb{P}_K^d$. It is independent of $X$, hence all $P \in \mathbb{P}^n(\bar{K})$ satisfy $H_K(\pi'(P)) \leq C H_K(P)$ with a constant $C$ depending just on $n$. The original projection, $\pi = \pi' \cdot g^{-1}$, thus sends $\{ P \in X(\bar{K}) \mid [K(P):K] \leq D, \, H_K(P) \leq B \}$ at most $\delta:1$ into the set $\{ Q \in \mathbb{P}^d(\bar{K}) \mid [K(Q):K] \leq D, \, H_K(g' \cdot Q) \leq C \cdot B \}$, where $g' \in \mathrm{PGL}(d,K)$ is a certain element (induced by $g^{-1}$). But the last set is in bijection $Q \leftrightarrow g'^{-1} \cdot Q$ with $\{ Q \in \mathbb{P}^d(\bar{K}) \mid [K(Q):K] \leq D, \, H_K( Q) \leq C \cdot B \}$, and therefore it has the same cardinality. Finally, the latter cardinality is $$\asymp_C \# \{ Q \in \mathbb{P}^d(\bar{K}) \mid [K(Q):K] \leq D, \, H_K( Q) \leq B \},$$ as follows for instance from a generalization of Northcott's precise asymptotics to points of a bounded degree, as proved in: [D. Masser, J. Vaaler: Counting algebraic numbers with large height II, Trans. Amer. Math. Soc., 2006.]
• I don't think your argument is complete. The key problem is the constant $C$ in your answer. I can only bound $C$ as $O(\delta^{n-d+1})$, which is far from the requirement. For the rational points case, I consider the corresponding affine cone, and count integral points by induction. For the hypersurfaces' case, I consider the polynomial directly. – var Nov 12 '17 at 3:15
• Sorry, I wasn't very clear. The constant $C$ depends only on the center $Z$ of the linear projection: a linear subspace of dimension $n-d-1$ disjoint from $X$. In general, there needn't be such a fixed $Z$ if $\delta$ is unbounded. But choose any $Z$ (depending on $X$) and take $g$ a linear automorphism (over $K$) moving $Z$ to a fixed $\mathbb{P}_K^{n-d-1}$. Then use that $\# \{ H_K(\xi) \leq B \}$ and $\# \{ H_K(g \cdot \xi) \leq B \}$ have the same asymptotic count (where $\xi$ ranges over the points of $\mathbb{P}_K^d \subset \mathbb{P}_K^n$ with $[K(\xi):K] \leq D$). Does it make sense? – Vesselin Dimitrov Nov 12 '17 at 4:34
• Yes, the constant $C$ depends only on the center $Z$ of the projection, and the choosing of $Z$ is essentially the choosing of a rational point in $Gr(d+1, n)$ outside the Chow variety (or called Chow form) of $X$. For a fixed degree $X$, the height of such a rational point is at lest $\delta+1$, and this estimate is almost optimal. So I don't think the linear projection method works for this problem, since by this argument, the height of $Z$ is larger if $\delta$ is larger. For your add information, I don't know whether it works, could you provide me more information? Thank you. – var Nov 12 '17 at 8:32
• @var: I edited the above. In my comment, I was trying to make the point that it is not really a problem that $Z$ can't be taken independently of $X$, due to the asymptotic lemma I formulated. – Vesselin Dimitrov Nov 12 '17 at 9:28