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

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

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

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

Assume F $F_q$ to be a finite field with q elements. Consider the zeta function of the hypersurface defined by -y0^2+y1^2+y2^2+y3^2=0 $-y_0^2+y_1^2+y_2^2+y_3^2=0$ in P^3.$\mathbb{P}^3$.
If -1 $-1$ is a square in F, $F_q$, the zeta function is Z(u)=1/{(1-uq^2)(1-uq)^2(1-u)}. $$Z(u)=\frac{1}{(1-uq^2)(1-uq)^2(1-u)}$$. It has a pole of order 2 $2$ at 1/q. $1/q$. If not, it's Z(u)=1/{(1-uq^2)(1-uq)(1+uq)(1-u)}. $$Z(u)=\frac{1}{(1-uq^2)(1-uq)(1+uq)(1-u)}$$. It has a pole of order 1 $1$ at 1/q. $1/q$.

How does orders of poles indicate any geometric information?

Here, I'm primarily concerced about zeta functions of hypersurfaces over fields of finite characteristic.
Assume F to be a finite field with q elements. Consider the zeta function of the hypersurface defined by -y0^2+y1^2+y2^2+y3^2=0 in P^3.
If -1 is a square in F, the zeta function is Z(u)=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)=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?