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Here, I'm primarily concerced about zeta functions of hypersurfaces over fields of finite characteristic.

Assume $F_q$ to be a finite field with q elements. Consider the zeta function of the hypersurface defined by $-y_0^2+y_1^2+y_2^2+y_3^2=0$ in $\mathbb{P}^3$.
If $-1$ is a square in $F_q$, the zeta function is


It has a pole of order $2$ at $1/q$. If not, it's


It has a pole of order $1$ at $1/q$.

How does orders of poles indicate any geometric information?

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Are you asking for an explanation specific to this case or an explanation in general? –  Qiaochu Yuan Jan 9 '10 at 3:27
Have you already seen the statement of the Weil conjectures? If not, you might want to check this first: en.wikipedia.org/wiki/… –  Steven Sam Jan 9 '10 at 3:51
Yinbang, TeX works here! Cleaned it up for you. Now, I don't do much with Zeta Functions, else I'd have an answer, too. –  Charles Siegel Jan 9 '10 at 3:58

2 Answers 2

up vote 13 down vote accepted

For a smooth projective surface, the order of the pole at 1/q is conjectured to be the rank of the Neron-Severi group of the surface. That's a conjecture of Tate and is an analog of the Birch and Swinnerton-Dyer conjecture. Tate has formulated a more general conjecture for higher dimensional varieties too. For the case of quadrics, as in your example, these conjectures are known.

Edit: maybe you don't want a fancy answer. In the first case, the quadric contains lines and in the second, it doesn't.

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The article is called The Birch and Swinnerton-Dyer conjecture and a Geometric Analogue. –  Timo Keller Jan 25 '10 at 7:50

This is an expository note filling in the background between Steven Sam's comment and Felipe Voloch's answer.

If $X$ is a smooth projective variety, then the Weil conjectures (now theorems) describe the zeroes and poles of the zeta function in terms of the cohomology of $X$, and the action of Frobenius on it. In particular, the poles on the circle $|u|=1/q$ are the reciprocals of the eigenvalues of Frobenius acting on $H^2(X, \mathbb{Q}_{\ell})$.

In your example, $H^2$ is two dimensional. Over the algebraic closure $\overline{F_q}$, your variety is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$. $H^2$ is spanned by the two classes $\mathbb{P}^1 \times \{ \mbox{point} \}$ and $ \{ \mbox{point} \} \times \mathbb{P}^1$.

If $-1$ is a square, Frobenius acts on this two dimensional vector space by multiplication by $q$, so you get a double pole at $1/q$. If $-1$ is not a square, then Frobenius multiplies by $q$ and switches the two generators. So the eigenvalues are $q$ and $-q$.

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