# Submersion implies many rational points in image?

Let $$A \colon V \to W$$ be a surjective linear map (defined over $$\mathbb{Z}$$), inducing a projection $$\alpha \colon \mathbb{P}(V) \to \mathbb{P}(W)$$. Let $$X \subseteq \mathbb{P}(V)$$ and $$Y \subseteq \mathbb{P}(W)$$ be some absolutely irreducible projective varieties (defined over $$\mathbb{Z}$$) that we know well.

Suppose that when considered over large fields, the restriction of the map $$\alpha$$ is close to a surjection. For example, suppose that $$\alpha$$ restricted to $$X$$ is a submersion almost everywhere when considered as a differentiable function over $$\mathbb{R}$$, $$\mathbb{C}$$, $$\mathbb{Q}_p$$.

My question is whether or not such a behaviour implies that we can deduce that $$\alpha$$ is close to being a surjection also over finite fields.

How to produce a lower bound on the number of $$\mathbb{F}_p$$ points of $$\alpha(X)$$?

Is there a standard procedure of how to pass from information over $$\mathbb{R}$$, $$\mathbb{C}$$, $$\mathbb{Q}_p$$ to answering the above question? Could someone please point out some relevant theorems? Which other properties of $$\alpha$$ restricted to $$X$$ should one prove in order to obtain the desired lower bound?

Remark 1: The concrete example of $$X$$ that I have in mind is a product of Grassmannians.

Remark 2: I am aware of the fibre dimension theorem, saying that over $$\bar{\mathbb{F}}_p$$, the generic fibers are of the same dimension. Together with Lang-Weil estimates, this would bring us close to giving an answer, as long as one could say something about the set over which generic fibres are, the fact that the fibres are absolutely irreducible and have known dimension (that is, the same dimension as they have over $$\mathbb{C}$$). Are any of these known?

• I think it would be good to clarify more the question. By "local submersion", do you mean on the level of $\mathbb{R}/ \mathbb{Q}_p/\mathbb{C}$-points? The analogue of submersion in algebraic geometry is a smooth morphism, so are you asking that $X \to Y$ is a smooth morphism of schemes? This in itself is a very mild assumption, as by generic smoothness it always holds over some open subset of the base (in char $0$). And when counting $\mathbb{F}_p$-points, one is often happy to remove closed subvarieties as these usually contribute negligibly to the count. – Daniel Loughran Oct 1 '18 at 15:08
• Usually $A \colon V \to W$ surjective is considered as inducing an immersion , $\alpha \colon \mathbb{P}(W) \hookrightarrow \mathbb{P}(V)$. Is that a typo or do you want to do it that way. – meh Oct 1 '18 at 15:09
• @DanielLoughran: Indeed, as an analytic function. I have edited the relevant part in the post. Would smoothness be enough to give an answer to the question? How is smoothness over $\mathbb{F}_p$ related to that over $\mathbb{R}/\mathbb{Q}_p/\mathbb{C}$? – darko Oct 1 '18 at 15:41
• @aginensky: I really want to have a rational map $\mathbb{P}(V) \to \mathbb{P}(W)$ coming from a surjective linear map. – darko Oct 1 '18 at 15:42
• I'm still not sure what you are after and I think that the submersion condition you are imposing is not as important as you think it is. For example the map $x \to x^2$ on the affine line is a smooth morphism away from the origin, hence a submersion on points over these fields, but is clearly not surjective on $\mathbb{F}_p$-points. The image contains $(p-1)/2$ many points; is this a "good enough" lower bound for you? – Daniel Loughran Oct 1 '18 at 15:49

This is an attempt at an answer to what I think the question is (please tell me if anything isn't clear).

Theorem

Let $$f:X\to Y$$ be a dominant morphism of finite type schemes over $$\mathbb{Z}$$ with $$X_{\mathbb{Q}}, Y_{\mathbb{Q}}$$ geometrically integral of dimensions $$n,m$$, respectively. Then there exists an absolute constant $$c$$ such that $$\#f(X(\mathbb{F}_p)) \geq c p^{m} + O(p^{m-1/2})$$ for all sufficiently large primes $$p$$, where the implied constant in the big-$$O$$ is independent of $$p$$.

To prove this, we use the following key lemma.

Lemma

There exists an absolute constant $$c'$$ such that for all $$y \in Y(\mathbb{F}_p)$$ we have $$\# f^{-1}(y)(\mathbb{F}_p) \leq c'p^{\mathrm{dim}\, f^{-1}(y)}$$

There are a few ways to prove this lemma. It can be proved using the Lang-Weil estimates, Deligne's proof of the Weil conjectures, or the method given in the answer to Number of solutions to polynomial congruences. (The key point here is that Deligne proved that the resulting sheaves which arise are constructible; this is what allows one to obtain the stated uniformity in the absolute constant $$c'$$).

Proof of the Theorem: There exists a proper closed subset $$Z \subset Y$$ such that every fibre outside of $$Z$$ has dimension $$n-m$$ (this follows from Lemma 36.28.1. of https://stacks.math.columbia.edu/tag/05F6 applied to $$f: X \to Y$$). Moreover the number of $$\mathbb{F}_p$$-points in $$Z$$ is at most $$c''p^{m-1}$$ for some absolute constant $$c''$$. Thus replacing $$Y$$ by $$Y \setminus Z$$, we may assume that the dimension of every fibre of $$f$$ is exactly $$n-m$$. Thus by the key lemma, we have $$\# X(\mathbb{F}_p) = \sum_{y \in f(X(\mathbb{F}_p))} \# \{ x \in f^{-1}(y)(\mathbb{F}_p)\} \leq c'p^{n-m}\#f(X(\mathbb{F}_p))$$ However, by the Lang-Weil estimates, we have $$\# X(\mathbb{F}_p) = p^n + O(p^{n-1/2})$$ as $$p \to \infty$$, since $$X_{\mathbb{Q}}$$ is geometrically integral. The result now easily follows with $$c = 1/c'$$. $$\Box$$

Remark: The constant $$c$$ depends on the number of irreducible components of the generic fibre of $$f$$ over the algebraic closure. In particular, if the generic fibre of $$f$$ is geometrically integral, then you can take $$c = 1$$.

• Great, this is just what I was looking for! Thank you very much! I would be grateful for some additional explanations regarding the key lemma. I will state them as separate comments. – darko Oct 2 '18 at 15:21
• 1. (dimensions) There are several dimensions appearing, over $\mathbb{F}_p$, $\bar{\mathbb{F}}_p$, $\mathbb{Q}$ and $\bar{\mathbb{Q}}$. These all come from Noether normalization of the ring of regular functions of the fibre $f^{-1}(y)$. (I am assuming everything is affine.) Are these dimensions all the same if the prime is large enough? (I understand that as long as $p$ is large enough, algebraic independence of polynomials over $\mathbb{Q}$ implies independence over $\mathbb{F}_p$, so at least these two dimensions eventually agree.) – darko Oct 2 '18 at 15:21
• 2. (the constant) Over a fixed point $y$, we can get the required bound with the constant basically coming from the algebraic part of Noether normalization. Could you please elaborate on why this constant can be chosen to be universal? (In principle it could happen that by varying $y$, the constants might grow.) I have looked at the answer you linked to and I unfortunately do not understand what you mean by "spreading out". Could you please provide me with a reference? – darko Oct 2 '18 at 15:22
• 1. I don't quite understand this comment. If $V$ is a variety over a field $k$ and $\bar{k}$ an algebraic closure of $k$, then $\mathrm{dim} V = \mathrm{dim} V_{\bar{k}}$. Also some of the fibres of $f$ can have larger dimension, but this can only occur on some closed subset $Z$ as in my answer. If you like, I'm putting all the bad primes into the closed subset $Z$. – Daniel Loughran Oct 2 '18 at 15:44
• BTW: In terrytao.wordpress.com/2012/08/31/the-lang-weil-bound, Tao proves a version of the key lemma and calls it a "Schwarz-Zippel type bound" – Daniel Loughran Oct 2 '18 at 15:54