How does the order of a pole of a zeta function indicate any geometric information?

Question

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

$$Z(u)=\frac{1}{(1-uq^2)(1-uq)^2(1-u)}.$$

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

$$Z(u)=\frac{1}{(1-uq^2)(1-uq)(1+uq)(1-u)}.$$

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

How does orders of poles indicate any geometric information?

Answer 1

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.

Answer 2

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 $-q$.