Question about Zeta Function of Singular Plane Curve 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: 


*

*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.

*$Z(C,t) = \frac{f(t)}{(1-t)(1-qt)}$, 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?


*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.

 A: *

*is a general fact for varieties, proved by Dwork. For curves, you argue directly, since there exists a smooth projective curve $C'$ and open sets $U \subset C, U' \subset C'$ with $U,U'$ isomorphic. Then 
$$Z(C,T)=Z(U,T)Z(C-U,T)=Z(C',T)Z(C-U,T)/Z(C'-U',T)$$
and the zeta functions of the zero dimensional schemes $C-U,C'-U'$ are easy to control.

*and 3. are not correct if the hypotheses are removed.
Try the examples of projective curves $y^2=x^3+x^2, y^2=x^3+ax^2$, with $a$ a non-square and of the affine line.
A: 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.
