Home > algebraic geometry > The affine line over a finite field

## The affine line over a finite field

Exercise II.2.11 in Hartshorne’s book asks for a description of Spec $\mathbb{F}_p[x]$, where $\mathbb{F}_p$ denotes the field with $p$ elements, including the number of points with a given residue field. I’m going to discuss the affine line over an arbitrary finite field $\mathbb{F}_q$, since this is no more difficult than the case when $q = p$ is prime.

As is the situation over any field, the closed points of the affine line correspond to monic irreducible polynomials, and there is a unique non-closed point, namely the zero ideal, which plays the role of generic point. The nonempty open sets are just the cofinite subsets, and the $\mathbb{F}_q$-algebra over such an open set is obtained by inverting the irreducible polynomials in $\mathbb{F}_q[x]$ corresponding to the removed points. In particular, every open subset of Spec $\mathbb{F}_q[x]$ is affine. The local ring at a closed point $(f)$ consists of all rational functions whose denominator does not divide $f$, and if $f$ has degree $n$ the residue field is $\mathbb{F}_{q^n}$. The local ring at $(0)$ is the function field $\mathbb{F}_q(x)$, which in some respects looks like a number field of positive characteristic: the completions of the various local rings play the role of $p$-adic number rings.

Now for the interesting part: one can count the number of points with a given residue field by using Möbius inversion from elementary number theory. First we must define the Möbius $\mu$-function $\mathbb{N} \to \{ -1,0,1 \}$, which is given by $\mu(1) = 1$, $\mu(n) = (-1)^r$ if $n$ is squarefree with $r$ prime factors, and $\mu(n) = 0$ if $n$ is not squarefree. Then Möbius inversion can be stated as follows: if $f,g : \mathbb{N} \to \mathbb{C}$ satisfy

$g(n) = \sum_{d | n} f(d)$,

then one can solve for $f$ by the formula

$f(n) = \sum_{d | n} \mu (\frac{n}{d}) f(d)$.

Let’s apply this to our situation: let $h(n)$ be the number of points of Spec $\mathbb{F}_q[x]$ with residue field $\mathbb{F}_{q^n}$, which is the same as the number of monic irreducible polynomials in $\mathbb{F}_q[x]$ of degree $n$. Recall that the polynomial $x^{q^n} - x$ is the product of all irreducible monic polynomials of degree dividing $n$, so upon counting degrees we obtain the formula

$q^n = \sum_{d | n} d \cdot h(d)$.

Now apply Möbius inversion with $f(n) = n \cdot h(n)$ and $g(n) = q^n$, so we get

$h(n) = \frac{1}{n} \sum_{d | n} \mu (\frac{n}{d}) q^d$.

This relates to the Hasse-Weil zeta function for Spec $\mathbb{F}_q[x]$ in the following way. Recall that an $\mathbb{F}_{q^n}$-point of a variety $X$ over $\mathbb{F}_q$ is a morphism Spec $\mathbb{F}_{q^n} \to X$ of $\mathbb{F}_q$-schemes, which corresponds to a point $p \in X$ and an $\mathbb{F}_q$-homomorphism $k(p) \to \mathbb{F}_{q^n}$, where $k(p)$ denotes the residue field of $X$ at $p$. Clearly such a homomorphism exists if and only if $p$ is closed with a residue field whose degree over $\mathbb{F}_q$ divides $n$, so if we write $X(\mathbb{F}_{q^n})$ for the set of $\mathbb{F}_{q^n}$-points of $X$ and deg$(p) = [k(p) : \mathbb{F}_{q}]$ we see that

$X(\mathbb{F}_{q^n}) = \coprod_{\text{deg}(p) | n}$ Hom$(k(p),\mathbb{F}_{q^n})$.

When counting the elements of Hom$(k(p),\mathbb{F}_{q^n})$ we must take into account the action of the Frobenius: this set carries a transitive action by Gal$(\mathbb{F}_{q^n} / \mathbb{F}_q)$, which is canonically identified with $\mathbb{Z}/n\mathbb{Z}$, and if deg$(p) = d$ the stabilizer of any point is just Gal$(\mathbb{F}_{q^n} / \mathbb{F}_{q^d}) \cong d\mathbb{Z}/n\mathbb{Z}$. Thus by the point-stabilizer theorem we have $|\text{Hom}(k(p),\mathbb{F}_{q^n})| = d$ and consequently

$|X(\mathbb{F}_{q^n})| = \sum_{d | n} d \cdot | \{ p \in X \ | \ \text{deg}(p) = d \}|$.

If we write $N_n = |X(\mathbb{F}_{q^n})|$ then the Hasse-Weil zeta function of $X$ is the element of $\mathbb{Q}[[t]]$ defined by

$Z(X,t) = \text{exp}(\sum_{n \geq 1} \frac{N_n}{n} t^n)$.

In the case $X = \text{Spec } \mathbb{F}_q[x]$, by the previous observation we have $N_n = q^n$, so

$Z(X,t) = \text{exp}(\sum_{n \geq 1} \frac{q^n}{n} t^n) = \text{exp}(-\text{log}(1-qt)) = \frac{1}{1-qt}$.

So in our situation the zeta function is actually rational! The miracle is that this is true for any variety over $\mathbb{F}_q$ whatsoever. Several proofs are known, all of which are quite hard, although it is possible to give a relatively elementary proof for curves.

Categories: algebraic geometry
1. 13 May 2011 at 6:02 pm

Are you saying that for every finite field, there is a line connecting all the points? Then all functions (line, curve, …) derived from this would be rational? This doesn’t work for infinite fields, huh. Can you put this in laymen terms for me?

• 19 May 2011 at 10:09 pm

I’m not really making a geometrical statement. If we’re working over an algebraically closed field such as the complex numbers, points of the affine line correspond to elements of the field, and it is common to think of the field as a line. This shouldn’t be taken too seriously: it’s a matter of intuition. Also, I’m glossing over a small point, which is that the affine line has one more “generic” point, whose closure is the entire space.

Over a more general field which is not necessarily algebraically closed, e.g. a finite field, the (closed) points of the affine line correspond to elements of the algebraic closure modulo the action of the absolute Galois group, whatever that means.

2. 3 December 2015 at 6:27 pm

This is such a wonderful post; thank you! I have one question, which is standard, I’m sure. Why must a F_{q^n} point of a scheme X over F_q be closed?

• 8 December 2015 at 12:09 am

Thanks for reading. A point of an affine scheme X = Spec R is a prime ideal p of R. The point is closed in X if and only if p is maximal, i.e. R/p is a field.