Zeta function of smooth conic?

Question

Let $k$ be a finite field of characteristic different from $2$. Let $X = V(x^2 + y^2 + z^2) \subset \mathbb{P}_k^2$ be the smooth conic. What is the zeta function of $X$?

Answer 1

This is easy, but maybe not for the reason that you may be thinking. Namely, if $X/\mathbb{F}_q$ is a smooth conic then $X\cong\mathbb{P}^1_{\mathbb{F}_q}$. There are multiple ways to see this.

They are all predicated upon the following observation. If $X/k$ (any field) is a smooth conic then $X\cong\mathbb{P}^1_k$ if and only if $X(k)\ne \varnothing$. Indeed, if $x\in X(k)$ then $\ell(\mathcal{O}(x))=1+1-0=2$. Thus, there is some $f\in K(X)$ such that the only pole of $f$ is $x$ and thus $f$ defines a degree $1$ map $X\to\mathbb{P}^1_k$ which must be an isomorphism.

So we want to show that any smooth conic over $\mathbb{F}_q$ has a rational point:

1) Try and show that any conic over $\mathbb{F}_q$ has a rational point by first principles.

2) For any conic $ax^2+by^2=z^2$ we get an associated quaternion algebra $Q(a,b)$ over $\mathbb{F}_q$ which, obviously, is split if and only if $ax^2+by^2=z^2$ has a point. But, $\text{Br}(\mathbb{F}_q)=0$ and so this is automatic.

(More generally $\text{Br}(k)$ classifies Brauer-Severi schemes--varieties over $k$ isomorphic over $\overline{k}$ (or $k^\text{sep}$) isomorphic to some projective space--smooth conics are evidently geometrically isomorphic to the projective line)

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So, of course, we get that

$$\begin{aligned}Z(X,t) &=Z(\mathbb{P}^1_{\mathbb{F}_q},t)\\ &=Z(\mathbb{A}^1_{\mathbb{F}_q}+\ast,t)\\ &=Z(\mathbb{A}^1_{\mathbb{F}_q},t)\cdot Z(\ast,t)\\ &=\frac{1}{1-qt}\cdot\frac{1}{1-t}\\ &=\frac{1}{(1-qt)(1-t)}\end{aligned}$$