# Counting Points

## Grothendieck group

A useful way to study the Grothendieck ring of varieties is via its realisation maps into a group $A$ , also known as motivic measures or additive invariants. These are maps from isomorphism classes of varieties that behave additive under disjoint unions, i.e. descend to $K_{0}\to A$ . One such measure is counting points:

$\#:K_{0}(\mathrm {Var} /\mathbb {F} _{q})\to \mathbb {Z} ,[X]\to \#X(\mathbb {F} _{q^{m}})$ For example:
$\#\mathbb {A} ^{n}(\mathbb {F} _{q})=q^{m}$ $\#\mathbb {A} ^{n}(\mathbb {F} _{q^{m}})=q^{mn}$ $\#\operatorname {Proj} \,^{n}(\mathbb {F} _{q^{m}})={\frac {q^{(n+1)m}}{q^{m}-1}}=1+q^{m}+\dots +q^{mn}$ $\#\operatorname {Spec} \,\mathbb {F} _{q^{m}}(\mathbb {F} _{q^{r}})=\operatorname {Hom} =\operatorname {Hom} ={\begin{cases}m&m|r\\0&{\text{else}}\end{cases}}$ From the Weil Conjectures, it will follow that if a projective variety $X$ of dimension $n$ is the union of $N$ smooth, geometrically irreducible (i.e. after base change to an algebraic closure) components of maximal dimension, then
$\#X(\mathbb {F} _{q^{m}})=Nq^{mn}+{\text{smaller terms}}$ ### Exercise

Complete this argument to show the claim about $K_{0}$ .

## Square polynomials

We can also tackle the first problem now and prove that $f(x)$ assuming square values for all integers is itself the square of a polynomial in $\mathbb {C} [x]$ (although showing $\mathbb {Z} [x]$ is also possible). For this, observe that $f(x)$ is a square if and only if the hyperelliptic curve $C$ given by

$y^{2}=f(x)$ has two irreducible components. This however can be established over finite fields and then be transferred. Reducing the curve $C$ modulo $q$ (not a power of $2$ ), the number of $\mathbb {F} _{q^{m}}$ -rational points is governed by $2\cdot q^{m}$ : For each $x\in \mathbb {F} _{q^{m}}$ (except for the roots of $f$ whose number is bounded by $\deg f$ ), we get two solutions for $y$ .

Alternatively (or equivalently), one can use the Weil estimate for smooth, projective, geom. irreducible curves of genus $g$ :

$|X(\mathbb {F} _{q})-(q+1)|\leq 2g{\sqrt {q}}$ ### Exercise

Formulate and prove the most general version of this type of claim you can imagine. Possible answer: Let $f({\underline {x}})$ be a multivariate polynomial that assumes perfect power values at all integral points (possibly with different exponents). Then $f$ is a perfect power...