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Pete L. Clark
Pete L. Clark
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Charles Siegel
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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 $$Z(u)=\frac{1}{(1-uq^2)(1-uq)^2(1-u)}$$.

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

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 $F_q$ 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?

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?

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

Assume $F_q$ 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?

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?

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?

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Yinbang Lin
Yinbang Lin
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