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Let $X$ be a smooth projective variety over a finite field $k$. We can define the zeta function by $$Z_X(T)=\exp\left(\sum N_m \frac{T^m}{m} \right) $$ where $N_m$ is the number of $K$-points, where $K$ is the unique extension of $k$ of degree $m$.

With some work, you can show the equality of formal power series: $$Z_X(T)=\prod_{x \text{ closed point of }X}\frac{1}{1-T^{\deg(x)}}$$

From this, you can show that $$Z_X(T)=\sum_{\text{effective divisors } D}T^{\deg(x)} $$which seems to me to be a simple argument invovling writing out $$\prod_{x \text{ closed point of }X}\frac{1}{1-T^{\deg(x)}} = \prod_{x \text{ closed point of }X}(1+T^{\deg(x)}+T^{2\deg(x)}+\cdots) $$ and doing some combinatorics.

However, this is where I became confused. It occurred to me that I don't know how to make sense of an infinite product of formal power series because there is no topological structure. (We can induce one, but would it be helpful?) Certainly, we can choose to only look at products where finitely factors many are not 1, but I would consider this some kind of restricted product, as opposed to a full product. If $X$ has only finitely many closed points then there is no problem here, but of course even something as simple as $\text{Spec}k[x]$ has infinitely many closed points.

Is there a way to make sense of an infinite product of formal power series?

  • While I don't know for sure, the answer surely must be that for every $d\ge1$, only finitely many factors in the infinite product have any terms involving $T^d$. That would mean that every individial term in the expanded product can be computed using only finitely many factors. – Greg Martin Jul 19 '21 at 21:31
  • Unless I am mistaken, isn't this assuming that the infinite product is a power series? Certainly, if the infinite product is a power series then the coefficients of $T^d$ can be computed by only finitely many factors. But I don't see why the infinite product must be a well-defined power series to begin with. –  Jul 19 '21 at 22:55

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