6
$\begingroup$

Let $X$ be a complete intersection in $\mathbb{P}^n$ of multidegree $(d_1,\ldots,d_r)$. If we're working over a finite field $\mathbb{F}_q$, the Ax-Chevalley-Warning theorem says that if $X$ is in the Fano range, i.e., $$ \sum d_i \leq n,$$ then $|X(\mathbb{F}_q)| \equiv 1 \pmod q.$

In the case that $X$ is not in the Fano range, one can cook up examples of such $X$ with no $\mathbb{F}_q$-points at all using the norm form, but these are not smooth.

In SGA 7 II Exposè XXI, Katz shows that for a general complete intersection in the Fano range, this congruence is not satisfied, possibly after extending the ground field.

Given a multidegree outside the Fano range, are there explicit examples of smooth complete intersections which don't satisfy this congruence?

Edit: Sorry, I wasn't sufficiently explicit about what I was asking. Restricting to hypersurfaces, what I would like is for each $d$, $p$, and $n$ with $d \geq n+1$, an explicit example of a smooth hypersurface of degree $d$ in $\mathbb{P}^n$ which doesn't satisfy the Chevalley-Warning congruence.

Thanks for the examples, though; I knew some explicit examples before but no infinite families.

$\endgroup$
  • $\begingroup$ You have to be careful about quantifiers. Do you fix $n$ and $(d_1,\dots,d_r)$, yet allow me to choose $q$ small? If $q$ is large, there are the Lang-Weil estimates. $\endgroup$ – Jason Starr Oct 6 '15 at 13:05
  • $\begingroup$ I see now that you are not asking about existence of a point, but about the congruence. The congruence is much stronger than existence of a rational point. $\endgroup$ – Jason Starr Oct 6 '15 at 13:08
6
$\begingroup$

Extension of Daniel Loughan's example (which is also known, but not as well-known as it should(?) be): if a prime $p$ is of the form $dn+1$ then the Fermat hypersurface $\sum_{i=1}^d x_i^d = 0$ in ${\bf P}^{d-1}({\bf F}_q)$ is smooth and its number of rational points is not congruent to $1 \bmod p$. Indeed the usual argument for Chevalley(-Warning) shows that the number of rational points is congruent mod $p$ to $1 \pm t$ where $t$ is the $(x_1 x_2 \cdots x_d)^{p-1}$ coefficient of $\left(\sum_{i=1}^d x_i^d\right)^{p-1}$, and when $p = dn+1$ this coefficient is $(p-1)!/n!^d$ which is clearly not $0 \bmod p$.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

How about $$x_1^{q-1} + \cdots + x_{q-1}^{q-1} = 0 \subset \mathbb{P}^{q-2} \quad ?$$ This is a fairly well-known example which seems to satisfy your criteria.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Could you please double-check this? If $q$ is an odd prime, I compute that the number of solutions is $(q-1)^{q-1}$, which is congruent to $1$ modulo $q$ (possibly making a mistake, but I did get this also for $q=4$). $\endgroup$ – Jason Starr Oct 6 '15 at 15:02
  • 2
    $\begingroup$ I see what happened. You probably wanted to consider the zero locus of $x_0^{q-1} + \dots + x_{q-2}^{q-1}$ in $\mathbb{P}^{q-2}$. If $q$ is prime, this polynomial has zero solutions in $\mathbb{F}_q$, and zero is not congruent to one. $\endgroup$ – Jason Starr Oct 6 '15 at 15:13
  • $\begingroup$ Yes you are right, thanks for pointing this out. The problem was I started with $x_0$, instead of $x_1$. I will change my answer accordingly. $\endgroup$ – Daniel Loughran Oct 6 '15 at 19:11
0
$\begingroup$

How explicit is "explicit"? Even norms of field extensions are somewhat inexplicit.

The following gives just one example, but the method extends to other examples. If you have a pencil of degree $d$ hypersurfaces in $\mathbb{P}^n$ over $\mathbb{F}_q$, if you can count the number of points in the base locus, and if you can count the number of (rational) singular members of the pencil and the number of points of each singular fiber, then you can sometimes prove that the number of points in some (rational) smooth member of the pencil cannot be congruent to $1$ modulo $q$. For instance, if $q=2^r$ for $r\geq 2$, this applies to the pencil from the following: When is the kernel of the etale fundamental group in a fibration abelian?

That example is a pencil of degree $4$ plane curves, the base locus has $4$ rational points, there are precisely two singular fibers, both rational, one of which has $4q-2 = (4q-6)+4$ rational points, and the second of which has $q+1 = (q-3)+4$ rational points. The union of the singular fibers has $4+(4q-6)+(q-3) = 5q-5$ rational points. Thus, the number of rational points contained in no singular fiber is $$(q^2+q+1)-(5q-5) = q^2-4q+6.$$ There are $q-1$ smooth fibers. If the number of points in a smooth fiber is congruent to $1$ modulo $q$, then the number of non-base points is congruent to $1-4=-3$ modulo $q$. Modulo $q$, the total number of non-base points in smooth fibers is congruent to $$-3(q-1) = -3q+3.$$ Since $q=2^r$, $6$ is not congruent to $3$ modulo $q$. Thus, at least one smooth fiber violates the congruence.

Edit. This method seems to apply easily to other pencils. For instance, for every $q$ it follows that for some $a\in \mathbb{F}_q^\times$, the plane cubic curve $\text{Zero}(z^3-axy(y-x))$ in $\mathbb{P}^2$ violates the congruence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.