# What is the shape of the zeta function of a singular hypersurface?

So let $X$ be a projective hypersurface inside $\mathbb{P}_{\mathbb{Z}}^n$ of degree $d$. Assume that

(a) $X(\mathbb{C})$ and $X(\overline{\mathbb{F}}_p)$ are irreducible,

(b) and that $X(\mathbb{C})$ and $X(\overline{\mathbb{F}}_p)$ are smooth varieties (therefore of Zariski dimension $n-1$).

Then we know from the classical Weil conjectures and (Hard Lefschetz) that  $$Z(X/\mathbb{F}_p,T)=\frac{P_{n-1}(T)^{(-1)^n}}{(1-T)(1-pT)\cdots (1-p^{n-1}T)},$$  where $$P_{n-1}(T)=\prod_{j=1}^{b_{n-1}}(1-\alpha_{j} T),$$ with $|\alpha_{j}|=p^{(n-1)/2}$ where the $(n-1)$-th Betti number of $X(\mathbb{C})$ is given explicitly by  $$b_{n-1}=\frac{(d-1)^{n+1}+(-1)^{n+1}(d-1)}{d}.$$  The last formula being a direct consequence of Gauss-Bonnet.

Q: So for a general connected projective hypersurface $X$ of $\mathbb{P}_{\mathbb{Z}}^n$ such that

(i) $\dim_{\overline{\mathbb{F}}_p}(X(\overline{\mathbb{F}}_p))=\dim_{\mathbb{C}}(X(\mathbb{C}))=n-1$,

which is no more assumed to be smooth (over $\overline{\mathbb{F}}_p$ and $\mathbb{C}$), what is the "shape" of $Z(X/\mathbb{F}_p,T)$?

I guess that a precise answer to this question should involve a description of the singular locus of $X(\overline{\mathbb{F}}_p)$ and $X(\mathbb{C})$ (the number of components of the singular locus and the type of singularity for each intersection of two components).

P.S. Note that if one has a precise recipe for the zeta function of such a hypersurface, then by the inclusion-exclusion principle one gets a description of the zeta function of a general (equi-dimensional) projective scheme $X$ of finite type over $Spec(\mathbb{Z})$.

added: By shape I mean an explicit factorization of the numerator and denominator of $Z(T)$ that reflects the geometry of $X(\mathbb{C})$. I know nothing about intersection homology but I guess that the shape of the zeta function should encode some data about $IH^{*}(X)$. For example, in the smooth case, the number of reciprocal roots (or poles depending on the parity of $i$) with complex absolute value $p^{i/2}$ of $Z(T)$ equals the $i$-th Betti number of $X(\mathbb{C})$.

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What exactly do you mean by "shape"? – Igor Rivin Nov 29 '11 at 22:45
For curves, the zeta function of an irreducible curve differs from the zeta function of its normalization by a finite product of Euler factors. See, e.g., ams.org/journals/proc/1998-126-09/S0002-9939-98-04333-0/… eq. (3.2). – Felipe Voloch Nov 30 '11 at 13:34