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The goal of this question is to find a "geometric" definition of Frobenius element in $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.

Here are two definitions that don't work, but that should help explain what I mean. Fix an algebraic closure of $\mathbb{Q}$ and, for a prime $p$, fix an algebraic closure of $\mathbb{F}_p$. If there were a canonical morphism $f : \mathbb{Q} \to \mathbb{F}\_p$, then the induced map $f\_{\ast} : \pi_1(\text{Spec } \mathbb{F}_p) \to \pi_1(\text{Spec } \mathbb{Q})$ of étale fundamental groups (basepoints the algebraic closures above, suppressed) would be an obvious way to define Frobenius elements: we would just take the image of a generator of $\pi_1(\text{Spec } \mathbb{F}_p) \simeq \hat{\mathbb{Z}}$. (The general motivation being: "gee, if I had a category $C$ and a functor, let's call it $\pi_1 : C \to \text{Grp}$, wouldn't it be nice if I could find an object, let's call it $S^1$, with $\pi_1(S^1) \simeq \mathbb{Z}$, so I could probe elements of $\pi_1(X)$ using morphisms $S^1 \to X$," and it seems for schemes that $\pi_1(S^1) \simeq \hat{\mathbb{Z}}$ is the next best thing.)

But of course this is nonsense since no such $f$ exists. The next obvious thing (from my perspective, being someone who knows nothing about all this étale stuff) is to look at the canonical morphism $f : \mathbb{Z} \to \mathbb{F}\_p$, which induces a map $f\_{\ast} : \pi_1(\text{Spec } \mathbb{F}_p) \to \pi_1(\text{Spec } \mathbb{Z})$, and then to look at the canonical morphism $g : \mathbb{Z} \to \mathbb{Q}$ and the induced map $g\_{\ast} : \pi_1(\text{Spec } \mathbb{Q}) \to \pi_1(\text{Spec } \mathbb{Z})$. Perhaps the Frobenius elements at $p$ are nothing more than the inverse image under $g\_{\ast}$ of the image of a generator of $\pi_1(\text{Spec } \mathbb{F}_p)$ under $f\_{\ast}$ (up to conjugacy to account for changes in basepoint).

But this is also nonsense since $\pi_1(\text{Spec } \mathbb{Z})$ is trivial.

So what is the correct version of this construction?

Here's my guess: instead of using $\mathbb{Z}$, we have to use the localization $R = \mathbb{Z}_{(p)}$. As before there are morphisms $f : R \to \mathbb{F}_p, g : R \to \mathbb{Q}$ inducing maps $f\_{\ast} : \pi_1(\text{Spec } \mathbb{F}_p) \to \pi_1(\text{Spec } R)$ and $g\_{\ast} : \pi_1(\text{Spec } \mathbb{Q}) \to \pi_1(\text{Spec } R)$, and maybe now something like the statement "the Frobenius elements in $\pi_1(\text{Spec } \mathbb{Q})$ at $p$ are the inverse image under $g\_{\ast}$ of the image of a generator of $\hat{\mathbb{Z}}$ under $f\_{\ast}$ (up to conjugacy)" is finally true. Is it? And what does $g\_{\ast}$ look like?

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3  
Qiaochu: is it helpful to remark that Gal(Q-bar/Q) doesn't have a Frob_p because Q-bar is ramified at p? Usually one looks at the quotient Gal(Q_S/Q)---the maximal extension unramified outside some set of primes S---and then (and only then) you have conj classes Frob_p for all p not in S. –  Kevin Buzzard Feb 2 '11 at 0:09
    
@Kevin: yes, that is more or less what I'm getting from the other answers. I have read a few expositions which vaguely say things like "well, there are Frobenius elements - only they're not elements, they're conjugacy classes - only they're not conjugacy classes, they're unions of conjugacy classes - but their trace in a Galois representation still makes sense!" and trying to figure out what kind of construction could account for this, and it seems clear-ish now. –  Qiaochu Yuan Feb 2 '11 at 0:35

3 Answers 3

up vote 6 down vote accepted

Your guess is right. $\pi_1(Spec(R))$ is the automorphism group of the maximal extension $\mathbb{Q}^{p\textrm{-}ur} \subset \overline{\mathbb{Q}}$ unramified at $p$. (This is a special case of SGA 1 Exp. V Prop 8.2 about the fundamental group of normal schemes). The morphism $g_*$ is simply the restriction map $Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to Gal(\mathbb{Q}^{p\textrm{-}ur}/\mathbb{Q})$.

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I think you must write $\mathbf{Q}$ as $\mathrm{colim}_n \mathbf{Z}[\frac{1}{n}]$.

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5  
I don't understand why this was down-voted. This answer is correct, if somewhat abbreviated: if $p \nmid n$ then there is a map $\mathbb Z[1/n] \to \mathbb F_p$, as well as a map $\mathbb Z[1/n] \to \mathbb Q$. These induces maps $G_{\mathbb F_p} \to \pi_1(\mathbb Z[1/n])$ and $G_{\mathbb Q} \to \pi_1(\mathbb Z[1/n])$ (the former being injective and the latter surjective), and these are precisely the maps giving rise to the Frobenius elements. –  Emerton Feb 2 '11 at 2:02

Your picture is right. The one thing I'll say (and this is somewhat tangential to your question) is that, for me, the analogue of $\pi_1(S^1)$ viewed as $\pi_1(\mathbb{C}\setminus\{0\})$ is not quite $\pi_1(\operatorname{Spec}\mathbb{F}_p)$, but rather the tame fundamental group of $\operatorname{Spec}\mathbb{Z}_p^{nr}$, where $\mathbb{Z}_p^{nr}$ is the maximal unramified extension of $\mathbb{Z}_p$ (i.e. its universal cover: it's simply connected and therefore analogous to a disk around the origin in $\mathbb{C}$). In other words, $\pi_1(S^1)$ classifies $n$-fold covers of the circle (or of a punctured disk around the origin). Similarly, the tame fundamental group classifies $n$-fold covers of the 'punctured disk' $\operatorname{Spec}\mathbb{Z}_p^{nr}\setminus{p}$. The only hitch is that $n$ has to be prime to $p$ in this setting. This part of the Galois group is in the `geometric' direction (nothing's happening to the residue field of the point), while the part where the Frobenii live is in the arithmetic direction (all about extensions of $\mathbb{F}_p$).

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