MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question may be too elementary for this forum, but I have asked it on math stackexchange and didn't get an answer. I have now deleted it so there wouldn't be duplicates... Here is the question as it appeared on math stackexchange:

I know that this question is pretty elementary, but I'm still trying to grasp the basics of number theory, and I want to make sure I'm thinking about things correctly. Most likely, I am thinking about things in a way that is slightly off. Since it's so complicated, I would appreciate if you can point out for me any misconceptions that I have.

Let $X_{\mathbb{Q}}$ be a variety over $\mathbb{Q}$. I am trying to understand how to associate to $X_{\mathbb{Q}}$ an $L$-function. As I understand it (please correct any mistakes I make in my narrative) there's an $L$-function for every $i=0,1,2,...$ (for convenience call them $L_0, L_1,L_2, ...$). Starting at $2dim(X)+1$ the $L_i$'s are just $1$. The $L$-function'' associated to all of $X_{\mathbb{Q}}$, by which I mean the Hasse-Weil Zeta Function, is $\prod L_i$.

Let's fix $i$. The function $L_i$ is defined to be the product of $L$-functions $L_{i,p}$, where $p$ runs over all primes in $\mathbb{Z}$. Here things get murky again. We choose a model $X_{\mathbb{Z}}$ of $X_{\mathbb{Q}}$ over $\mathbb{Z}$. (How? It can't be that any model would work, can it?) Now we fix a Weil Cohomology theory (as I understand it, it shouldn't matter which. Is that true? Is it only a conjecture that it's true?). For example we can look at $l$-adic cohomology where $l$ is a prime coprime to $p$. Then we define $L_{i,p}(x)$ to be $P_i(x)^{(-1)^i}$ where $P_i(x)$ is the characteristic polynomial of the action of the Frobenius element on $H^i(X_{\mathbb{Z}/p},\mathbb{Q}_l)$ (or whichever Weil Cohomology theory we chose).


First of all I would like to know if the narrative above is accurate. If not, please tell me where. Particular things I am vague about are:

$1.$ How does one choose the model of $X_{\mathbb{Q}}$ over $\mathbb{Z}$?

$2.$ Is it true that we can choose any Weil Cohomology? Is that conjectural, or proven?

$3.$ As I understand it, it should be true that $\prod_{p \in Spec(\mathbb{Z})} L_{i,p}= e^{\sum_{l=1}^{\infty} |X_{\mathbb{Z}/p}(\mathbb{F}_p)|\frac{x^l}{l}}$. Is that right?

$4.$ Does this construction differ in any substantial way if we defined a Hasse-Weil Zeta function for a variety defined over a number field different from $\mathbb{Q}$?

share|cite|improve this question
I'm not a number theorist, but I don't think your question looks 'too elementary'. -- In any case you don't need to excuse for asking a question! – Stefan Kohl Feb 3 '13 at 20:25
The variety is defined over the rationals, so for any $i$ you can look at the $i$th $\ell$-adic cohomology groups; this will be a finite-dimensional $\ell$-adic representation of the abs Gal gp of the rationals. The $L$-function of this representation is probably your $L_i$. The cohomology vanishes for $i>2 dim(X)$. The Hasse-Weil zeta function is not the product, it's the alternating product $(L_0)^{-1}L_1(L_2)^{-1}\ldots$ -- think about the fixed point formula in topology, it's the same idea. But as for your bad factors, I don't think the picture is clear. There are lots of models over $Z$ – user30035 Feb 3 '13 at 20:46
but there's only supposed to be one $L$-function. As far as I know there is only a conjectural definition at the bad primes -- e.g. take the $\ell$-adic etale cohomology, compute the $L$-function there, and then hope that it's independent of $\ell$; this is not known. I don't think your Q2 makes sense (to me). Your Q3 is meaningless because $p$ is fixed on one side and varying on the other. The true statement is that if you take a prime where the variety has good reduction then there's a relation because the local $L$-factor at $p$, not the global $L$-function (which is their product), and – user30035 Feb 3 '13 at 20:48
the number of points. Finally I think the answer to Q4 is that the basic ideas are just the same, and what is known is still known, and what is conjectured is still conjectured. – user30035 Feb 3 '13 at 20:49
@wccanard: are you sure that it's the alternating product as you say? I defined each $L_{i,p}$ to be $P_i^{(-1)^i}$, and I thought that accounted for the alternating part of the story... – Nicole Feb 3 '13 at 21:14
  1. A naive way of choosing a model is to write down some equations and clear denominators. The L-functions will depend on the model but only finitely many factors coming from "bad" primes. I think you want to assume that your variety is smooth and projective, otherwise some pathologies may happen. Nailing down the factors at the bad primes is trickier and I am not sure there is a nice way to do it, independent of the model, in general.

  2. There are theorems giving comparisons between étale (for any $\ell \ne p$) and crystalline cohomology in char $p$. If you have an abstract Weil cohomology that satisfies the Riemann hypothesis, then the local L-factors are uniquely determined (see 3. below).

  3. Your formula is wrong. What is true is that (for a fixed prime $p$ of good reduction) $\prod_i L_{i,p} = e^{\sum_n |X_{\mathbb{Z}/p}(\mathbb{F}_{p^n})|p^{-ns}/n}$. From this it follows that $\prod_i L_{i,p}$ is independent of the cohomology. RH then says that the reciprocals of the zeros of $L_{i,p}$ (as a polynomial in $t=p^{-s}$) have absolute value $p^{i/2}$. From this, you can extract the individual $L_{i,p}$ from the product over $i$.

  4. No. It's basically the same, using prime ideals instead of prime numbers.

share|cite|improve this answer
Yeah, that was silly of me that I took the product over the primes instead of the $i$'s... It was just a typo. So essentially what I'm taking from this is the following: Let $X_{\mathbb{Z}}$ be any model over $\mathbb{Z}$. Then what you call the "bad primes" are the primes of bad reduction of this model? (Where the fibers are not geometrically irreducible and smooth?) And that for the good primes, you define the $L_{i,p}$ in the way I defined it above, but for the bad primes nobody quite knows what to do. But somehow conjecturally there should be a definition at the bad primes so that – Nicole Feb 3 '13 at 21:27
$L_i$ is the product $\prod_{p \in Spec(\mathbb{Z})} L_{i,p}$? – Nicole Feb 3 '13 at 21:27
By the way, with your definition of $L_i$ is it true that $\prod_i L_i$ is the Hasse Weil zeta function? Or is it an alternating series? And is $L_{i,p}$ (for good $p$'s) the characteristic polynomial itself, or is it $P_{i,p}^{(-1)^i}$ where $P_{i,p}$ is the characteristic polynomial? – Nicole Feb 3 '13 at 21:31
By the way, the Hasse-Weil zeta function of a finite type scheme $X_{\mathbb{Z}}$ over $\mathbb{Z}$ (the integral model, that is), is simply the product $\prod_x (1 - |k(x)|^{-s})^{-1}$ over the closed space. But since there is no $\mathbb{Z}$-model in general, what you do instead is to take your $P_{i,p}$ to be the char. poly. of $\mathrm{Frob}_p$ acting on $H^i(\overline{X},\mathbb{Q}_l)^{I_p}$, the invariants of inertia. This does not refer to a $\mathbb{Z}$-model, and works to define the L-function of an abstract Galois representation. See this: – Vesselin Dimitrov Feb 3 '13 at 21:42
(I meant to write: the product over the closed points; that is to say, those of finite residue field $k(x)$). – Vesselin Dimitrov Feb 3 '13 at 21:43

1 One does not need a model to find the L-function. The Frobenius element is a conjugacy class, up to ramification, in the Galois group of $\mathbb Q$. The Galois group of $\mathbb Q$ acts on the $l$-adic etale cohomology for every prime $l$ not $p$, and for the $p$-adic cohomology. However there is much more ramification for $p$-adic cohomology.

One takes the characteristic polynomial of the action of Frobenius on the invariants of the action of the ramification group. This is well-defined.

2 To prove a relationship between different $l$s, one can use a model over $\mathbb Z$. In particular one wants a model that is smooth and proper at $p$. Then one uses base change theorems to relate etale cohomology of the rational fiber and the characteristic p fiber, and then relates different characteristic $p$ cohomologies via the Lefschetz Trace Formula. However this only works for $l\neq p$, so one chooses an $L$-function via $l$-adic cohomology where $l\neq p$.

share|cite|improve this answer
The second part of your answer makes perfect sense to me. Can you explain a little more (or point me to a place that explains) the first part of your answer? Ramification of what over what? – Nicole Feb 3 '13 at 21:22
The ramification subgroups of the Galois group. The map $Gal(\mathbb Q_p) = Gal(\mathbb Q)$ is defined up to conjugacy. There is an extension $1 \to I\to Gal(\mathbb Q_p) \to \hat{\mathbb Z} \to 1$ for a subgroup $I$ called the inertia or ramification group. The Frobenius element is the generator of that quotient. On any representation on which $I$ acts trivially, there is a well-defined conjugacy class of Frobenius, with a well-defined characteristic polynomial. – Will Sawin Feb 3 '13 at 21:53
"that quotient = $\hat{\mathbb Z}$. – Will Sawin Feb 3 '13 at 21:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.