# 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:

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?

1. 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.
• I would have prefered this to be a comment. This text contains what you want, if you are comfortable with french. here – A.B. Feb 16 '18 at 11:41

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.

• Thank you! Is there anywhere you know of that I can learn these facts from? Do you know where I can learn about normalisation and compactification, particularly over finite fields and/or fields which are not algebraically closed? I have not been able to find the information you provided anywhere else. Do you know of any resources which contain this information? Thank you for your help. – maddels Feb 20 '18 at 11:43
• Is the compactification of a curve the same as the "projective closure"? Or is it a different thing? – maddels Feb 20 '18 at 11:52
• @maddels Yes, compactification is the same as the projective closure. (technically there may be more than one compactification, of which a projective closure is just one example, but only one compactification is needed for this anyways.) – Will Sawin Feb 20 '18 at 13:19
• Would you be willing to direct me to some resources where I can learn about algebraic curves over finite fields and zeta functions? I am an undergraduate in Australia, and I am struggling to find the information you provided elsewhere, as well as the algebraic geometry I need to know for this project at level that is accessible to me. I would appreciate anything you could point me to look at. Thanks again for your help! – maddels Feb 20 '18 at 23:50
• @maddels I don't know a great answer to your question. It seems likely that the book Algebraic Curves over a Finite Field by Hirschfeld, Korchmaros, and Torres contains a lot of helpful information as it is long and focused on concrete problems. jstor.org/stable/j.ctt1287kdw – Will Sawin Feb 22 '18 at 13:25
1. 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.

2. 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.

• Thank you for your suggestion. I computed the zeta function of $y^2=x^3+x^2$ and $y^2=x^3+ax^2$ when $q=p=5$. I computed $Z(t)=\frac{1}{1−5t}$ for the former, and with $a=3$, I computed $Z(t)=\frac{1+t}{(1-t)(1−5t)}$ for the latter. I would like to rewrite/modify my second claim. The claim should instead be: $Z(C,t)$ can be written as $\frac{f(t)}{(1−t)(1−qt)}$ with $f(t)\in\mathbb{Z}[t]$. Or, more precisely, $\frac{f(t)}{(1−t)(1−qt)}(1-t)^i$ for some i. (So some factors of $(1-t)$ may cancel). The examples you have suggested I try are okay under this claim. – maddels Feb 20 '18 at 11:20
• Would you be able to tell me if the modified claim is true? Is there anywhere I could find a reference for it? – maddels Feb 20 '18 at 11:21