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Let $X$ be a variety defined over the finite field $\mathbb{F}_p$. The following is "well known". Let $a_n$ be the number of $\mathbb{F}_{p^{n}}$-rational points of $X$, and let $b_n$ be the number of $\mathbb{F}_p$-rational points of the $n$-fold symmetric product of $X$, i.e $X^n/S_n$ where $S_n$ is the symmetric group on $n$ letters. Then the following equality of rational power series holds:

$\sum{b_n t^n} = \exp(\sum{\frac{a_n}{n} \cdot t^n})$

The explanation is that there is a correspondence between $\mathbb{F}_p$ rational points on $X^n / S_n$ and degree $n$ effective zero cycles on $X$, and these in turn are related to rational points on $X$ in finite field extensions of $\mathbb{F}_p$.

I am not familiar with intersection theory and with algebraic cycles beyond the definitions in the appendix of Hartshorne, but I suppose that the case of degree zero should be easy to explain. I will be happy with an explanation in the case where $X$ is affine

user115940
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You need to know nothing about intersection theory and very little about algebraic geometry to understand this result; it's almost entirely combinatorial. The combinatorial heart of this result is known as the exponential formula; see this blog post and this blog post for some details.

"Degree $n$ effective zero cycle on $X$" just means a formal sum $\sum n_p p$ of points over $\overline{\mathbb{F}_p}$ where $n_p \ge 0$ are non-negative integers summing to $n$ and the sum is closed under Galois action. Such a thing has a unique "cycle decomposition" into Galois orbits, and that's all the algebraic geometry you need to know; the rest is the exponential formula.

Qiaochu Yuan
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  • Hey. I pondered about your answer but I am still not sure where the algebra is involved here. Surely one must use some non-trivial facts about the nature of finite fields and their Galois theory. For example, if you look at the case of the affine line, then counting rational points on the symmetric product is totally combinatorial, but to get the other side of the equation one must know that there is exactly one finite extension of $Fp$ of each degree – user115940 May 20 '14 at 07:22
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    @user: the algebra you need is that the Galois group of every finite extension of $\mathbb{F}_p$ is generated by Frobenius. That's why I can say "cycle decomposition" above and not "orbit decomposition"; I'm really talking about the cycle decomposition of a permutation, namely Frobenius. – Qiaochu Yuan May 20 '14 at 16:44