I am working on a project which involves learning about the zeta function (weil zeta function) for plane curves, but I do not know much algebraic geometry. (I do not know anything about schemes).


I am mostly interested in irreducible projective plane curves $C$ over finite fields $\mathbb{F}\_q$. We define the zeta function by


$$Z(C, t) = \exp\left(\sum^{\infty}\_{n = 1}{\frac{\# C(\mathbb{F}\_{q^n})}{n} t^n}\right)$$


My supervisor has told me some facts about the zeta function for plane curves, and I am seeking a reference for these facts. I would also like to know under what conditions these facts hold true, in particular whether they are still true even when the curve is *singular*, or if they are still true if the curve is an affine plane curve. 


The facts I want information about are: 


1. For a plane curve $C \subseteq \mathbb{A}^2$ given by $f(x,y) = 0$ with $f \in \mathbb{F}\_q[x,y]$, the zeta function $Z(C,t)$ is rational. Also, for a plane curve $C \subseteq \mathbb{P}^2$, $Z(C,t)$ is rational.
2. $Z(C,t) = \frac{f(t)}{(1-t)(1-qt)}$, i.e. the denominator has the form $g(t) = (1-t)(1-qt)$. My understanding is that this the case for irreducible projective plane curves, whether or not the curve is singular or non-singular. Is this also true for affine plane curves?


EDIT: I have phrased (2) incorrectly. What I want to know is if $Z(C,t)$ can be *written* in the form $\frac{f(t)}{(1-t)(1-qt)}$ where $f(t) \in \mathbb{Z}[t]$, or more precisely, if it is true that $$Z(C,t) = \frac{f(t)}{(1-t)(1-qt)} (1-t)^i$$ for some $i \geq 0$. Is there anywhere I can find a reference for this?


3. When the (irreducible projective plane) curve is *non-singular*, we have $f(t) = 1 + a\_1 t + \dots + a\_g t^g + q a\_{g-1} t^{g+1} + \dots + q^g t^{2g}$. My understanding is that this is not the case when the curve is *singular*.



Question about Zeta Function of Singular Plane Curve
For a singular curve $C$, you have a compactification $\overline{C}$ and a normalization $\tilde{C}$. Over a perfect field, such as a finite field, the normalization is smooth (and projective).


The best way to study the zeta function of $C$ is by studying the zeta function of $\tilde{C}$ and then adding additional factors to get to the zeta function of $C$. 


Using the product formula for the zeta function, the zeta function of $\overline{C}$ is equal to the zeta function of $C$ times one factor of $(1-t^d)^{-1}$ for each degree $d$ point added to make the compactification. To go from the zeta function of $\overline{C}$ to the zeta function of $\tilde{C}$, we remove the factor of $(1-t^d)^{-1}$ for each singular point of degree $d$ and then add back in a factor of $(1-t^e)^{-1}$ for each degree $e$ point lying over it. Because $e$ is always a multiple of $d$, and there is always at least one point, $(1-t^d)$ always divides $(1-t^e)$. In other words, the zeta function of $C$ is the zeta function of $\tilde{C}$ times a polynomial, as you suspect.


However, this polynomial need not always have roots only at one. Instead, they can be arbitrary roots of unity. So we have


$$Z(C,t) = \frac{ f(t)}{(1-t)(1-qt)} \prod\_{i=1}^n (1- \zeta\_i t) $$ 


where $f(t)$ satisfies the symmetry property you state and $\zeta\_i$ are some roots of unity.


