Zeta functions

Here’s the rough idea: E is a set of equations in a finite number of variables over the integers.  Set a_p(E) = \# of solutions mod p.  The goal is to determine formulas that calculate a_p(E).

 

Translating this to schemes, let X be a scheme of finite type over \mathbb{Z} .  Then set a_p(X) = \#X(\mathbb{F}_p) = cardinality of maps from \mathbb{F}_p into X. For example, for X = \mbox{Spec} A[x]/(x^2 - \alpha) for some \alpha \in A, then a_p(X) counts the number of square roots for \alpha \in \mathbb{F}_p.  

Here’s an explicit example.  X = \mbox{Spec} \mathbb{Z}[x,y]/(y^2 - (x^3 - x)).  To give a map from, for example, \mathbb{F}_5 into X is equivalent to give a ring map into the field which requires a choice of where to send x, such that f(x) = x^3 -x has a square root in the field.  Now f(x) for x = 0, 1, 2, 3, 4 is 0,0, 1, -1, 0 and these latter numbers have 1,1,2,2,1 square roots respectively.  Thus a_5(X) = 7.  As it happens a_7(X) = 7.

Definitions and Easy properties

Generalizing a little bit, let X be a scheme of finite type over a finite field \mathbb{F}_q.  Then define 

a_n(X/\mathbb{F}_q) = \# X/\mathbb{F}_q(\mathbb{F}_{q^n})

I.e. the number of maps from \mbox{Spec} \mathbb{F}_{q^n} into X in the category of \mathbb{F}_q – schemes.  

Here are some properties (all unions are meant to be disjoint).  If X = X_1 \sqcup X_2 then a_n(X) = a_n(X_1) + a_n(X_2).  In what follows I write \mathbb{A}^n, \mathbb{P}^n for \mathbb{A}^n_{\mathbb{F}_q}, \mathbb{P}^n_{\mathbb{F}_q}.  Its straightforward to see that \mathbb{A}^m(\mathbb{F}_{q^n}) = (\mathbb{F}_{q^n})^m so a_n(\mathbb{A}^m) = q^{nm}.  Now using

\mathbb{P}^m = \mathbb{A}^m \sqcup \mathbb{P}^{m-1} = \mathbb{A}^m \sqcup \mathbb{A}^{m-1} \sqcup \mathbb{P}^{m-2} = ...

I find a_n(\mathbb{P}^m) = \sum_{i = 0}^m a_n(\mathbb{A}^i) = \sum_i q^{ni} = \frac{z^{m+1}}{z-1}, for z = q^n.

 

Some notation.  Let F_n = \mathbb{F}_{q^n}.  Another important example: X = \mbox{Spec } F_e for fixed e.  Now the elements of X(F_n) correspond to ring homomorphism 

F_e \xrightarrow{\phi} F_n

Its easy to see the map must be injective.  If a \ne 0, then a^{q^e - 1} = 1, so phi(a)^{q^e - 1} = 1.  Further, \phi(a) must now be a root of x^{q^n -1 } - 1.  Meaning q^e - 1| q^n - 1 which happens iff e|n.  In this case the number of such \phi is equal to the order of the Galois group G(F_e/F_1) which is e because its a Galois extension.  Putting this all together I have

a_n(F_e) = \begin{cases} 0 \mbox{ if  } n/e \not \in \mathbb{Z}\\ e \mbox{ otherwise } \end{cases}

 From the previous post this is 

a_n(F_e) = \sum_{\eta \in \mu_e} \eta^n

Where \mu_e are the e-th roots of unity.

The (local) zeta function

For a_n := a_n(X/F_1) set

Z(X/F_1, T) = \exp(\sum_n \frac{a_nT^n}{n}

Also introduce the numbers

b_d(X/F_1) = \# of \mbox{closed pts of X} s.t. \deg x = d

c_d(X/F_1) = \# of \eta = \sum_i n_i x_i s.t. \sum_i n_i \deg(x_i) = d

Where x_i are closed points and n_i \ge 0.  

A bunch of results

In the future I might add some of the proofs of these things.

  1. a_n = \sum_{d|n} d\cdot b_d
  2. \sum_d c_dT^d = \prod_d (1 - T^d)^{-b_d}
  3. If X/F_1 is of finite type of dim = m, then there is a constant c s.t a_n(X/F_1) \le c \cdot q^{mn} for all n.
  4. If dim X is positive then \{a_n(X/F_1)\} is unbounded.

Here’s a big theorem I wont prove

Thm: There exists a finite number of \alpha_i , \beta_i \in \mathbb{C} s.t.

a_n(X/F_1) = \sum_i (\beta_i^n - \alpha_i^n)

In view of the last post, this says

Z(X/F_1, T) = \prod_i \frac{1- \alpha_iT}{1 - \beta_i T}

Further, if the hypothesis of #3 are satisfied then Z(X/F_1,T) converges for |T| < 1/q^m

—————————-

The proof of #3 uses Noether Normalization and CRT; #4 uses the Nulstullenzats (spelling?) and the last statement of the theorem follows from #3 and nth root formula for radius of convergence.

More Results

Thm: Let X/F_1 be a smooth geometrically connected projective curve of genus g.

  1. Z(X/F_1, T) = \frac{\prod_{i = 1}^{2g} (1 - \alpha_iT)}{(1 - T)(1 - qT)}, \alpha_i \in \mathbb{C}
  2. \alpha_i \mapsto q/\alpha_i is a permutation of the \alpha_i and \prod_i \alpha_i = g
  3. |\alpha_i| = \sqrt q for all i
  4. The residue of Z(T) at T = 1 is the order of a certain finite group having to do with \mbox{Pic}_0(X) (I might actually explain this later).

Cor. a_n = a_n(X/F_1) = 1 + q^n - \sum_{i = 1}^{2g} \alpha_i^n so that #3 also implies |a_n - (1 + q^n)| \le 2g \cdot \sqrt{q}^n.

One final thing.  There’s a degree homomorphism Div (X) \to \mathbb{Z}.  In general the image is of the form e\mathbb{Z} for e \ge 1.  For X a curve as the thm above Ogus proved that e = 1 as follows:

Lemma1: Z(X/F_1, T) = \frac{f(T^e)}{(1 - q^eT^e)(1 - T^e)} where f is a poly and so the Z(T) has simple poles at \eta/q.

Lemma2: If Z(X/F_1, T) = \prod_i \frac{1 - \alpha_iT}{1 - \beta_i T} and X' = X \times_{F_1} F_r then Z(X'/F_r, T) = \prod_i \frac{1 - \alpha_i^rT}{1 - \beta_i^rT} .

Lemma2 is not so bad to prove, all the work is in understanding why a_n(X'/F_r) = a_{rn}(X/F_1).

In any case, continuing to show e = 1, using lemma 1

Z(X/F_1) = \frac{\prod_i (1 - \alpha_iT)}{(1 - T^e)(1-q^eT^e)}

= \frac{\prod_i (1 - \alpha_iT)}{\prod_\eta (1 - \eta T)(1-\eta qT)} 

So lemma 2 gives

Z(X'/F_e) = \frac{\prod_i (1 - \alpha_i^e T)}{\prod_\eta (1 - \eta^e T)(1 - \eta^eq^eT)}

= \frac{\prod_i (1 - \alpha_i^e T)}{(1 -T)^e(1 - q^eT)^e}

But by lemma 1 the zeta function has to have simple poles, so e = 1.

 

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