# Counting Points

## Grothendieck group

A useful way to study the Grothendieck ring of varieties is via its realisation maps into a group ${\displaystyle A}$, 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 ${\displaystyle K_{0}\to A}$. One such measure is counting points:

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

### Exercise

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

## Square polynomials

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

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

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

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

### Exercise

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