2
$\begingroup$

General question: Let $W$ be a proper subvariety of an irreducible affine variety $V/K$. Under what conditions do we know that $W(K)$ is a proper subset of $V(K)$?

If $K$ is finite, then one can bound $|W(K)|$ by $\ll_{\deg(W)} |K|^{\dim(W)}$. (Here we write $|S|$ for the number of elements of a set $S$.) If one can give a bound of the form $|V(K)|\gg |K|^{\dim(V)}$, it then follows that $|W(K)|<|V(K)|$ provided that $K$ is large enough in terms of the degree of $W$.

When can one prove something similar for $K$ infinite? The condition that needs to be replaced here is $|V(K)|\gg |K|^{\dim(V)}$. Things work out if one assumes that $V$ is unirational, but that feels a bit stronger than needed; is a weaker condition enough? What would the proof then be?

(Obviously one needs some sort of condition, as is shown by the example of $V$ given by $x^2+y^2 = 0$, $W$ the zero-dimensional variety $x,y=0$, and $K = \mathbb{R}$.)

Also: is there any sort of model-theoretical transfer principle that one might hope would work here? In which direction would it go - $K$ finite to $K$ infinite, or $K$ infinite to $K$ finite?

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Is it important to you that the variety is irreducible but not necessarily geometrically irreducible? $\endgroup$
    – Arno Fehm
    Commented Feb 24, 2023 at 8:53
  • $\begingroup$ @ArnoFehm I put in the condition of irreducibility just to ensure that $W$ is of positive codimension. Why do you ask? $\endgroup$
    – H A Helfgott
    Commented Feb 24, 2023 at 11:35
  • $\begingroup$ Well, probably this is clear to you, but for geometrically irreducible varieties over finite fields you get a positive answer from the Lang-Weil estimates, which then transfers to ultraproducts, i.e. pseudo-finite fields. More generally, you get a positive answer (for geometrically irreducible varieties) over pseudo-algebraically closed fields. If we add the assumption that $V$ is smooth and $V(K)$ nonempty, then you get a positive answer for many other fields, including $\mathbb{R}$. $\endgroup$
    – Arno Fehm
    Commented Feb 24, 2023 at 18:12
  • $\begingroup$ Right, Lang-Weil is enough over finite fields. How does one get the answer over pseudo-algebraically closed fields? And how does smoothness help? $\endgroup$
    – H A Helfgott
    Commented Feb 24, 2023 at 18:26
  • 1
    $\begingroup$ For $\mathbb{R}$ for example this is just the implicit function theorem: Any smooth rational point on $W$ gives a whole Euclidean neighborhood of rational points, in particular many in $V\setminus W$. $\endgroup$
    – Arno Fehm
    Commented Feb 25, 2023 at 12:14

0

Reset to default

You must log in to answer this question.