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? 2 cleanup and texification 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\$.