Let $K$ be a number field, $\bar K$ an algebraic closure and $G$ the associated absolute Galois group. How can I define the Frobenius elements of $G$ or at least their conjugacy class?

I know how these elements arise in finite Galois extensions, and I know that $G$ is the inverse limit among this finite extensions, however I don't know how to put together these fact to obtain an answer.


[This answer is an elaboration on what KConrad mentions in his comments below, posted while I was first writing this.]

At finite level one has a perfectly good theory of Frobenius elements attached to maximal ideals upstairs with ambiguity of inertia subgroups (so no ambiguity in the unramified case, but don't get too attached to that). This goes similarly at "infinite level" and one can in fact think about Frobenius elements in a "naive way" for $\overline{K}/K$ by working directly with the integral closure $O_{\overline{K}}$ of $O_K$ in $\overline{K}$.

That integral closure is quite gigantic (and not noetherian) but is really not as frightening as it may seem because it is the directed union of the rings of integers $O_{K'}$ of the subextensions $K'/K$ of finite degree, with those $O_{K'}$ quite concrete things. If you think about it, a prime ideal of $O_{\overline{K}}$ "is" nothing more or less than a compatible system of prime ideals $P_{K'} \subset O_{K'}$ (meaning $P_{K''} \cap O_{K'} = P_{K'}$ whenever $K' \subset K''$).

Let's see how this plays out with maximal ideals in $O_{\overline{K}}$ and $O_K$, and then with that experience under our belts we'll see that a "naive" approach to Frobenius elements works very well. For a maximal ideal $M$ of $O_K$, the set $S_{K'}$ of maximal ideals in $O_{K'}$ over $M$ is a non-empty finite set, and we have the natural maps $S_{K''} \rightarrow S_{K'}$ defined by $\mathfrak{m} \mapsto \mathfrak{m} \cap O_{K'}$ (even surjective, but doesn't matter). In this way $\{S_{K'}\}$ is an inverse system of non-empty finite sets, so the inverse limit $\lim S_{K'}$ is non-empty (no need for Zorn stuff here since these are all countable inverse systems).

But this inverse limit is precisely "the same" as a prime ideal $P$ of $O_{\overline{K}}$ satisfying $P \cap O_K = M$. For any such $P$, each prime ideal $P \cap O_{K'}$ is maximal (i.e., a nonzero prime ideal). Thus, $O_{\overline{K}}/P$ is a field, as it is the direct limit of the quotients $O_{K'}/(P \cap O_{K'})$ that are all fields, so $P$ is maximal. The upshot is that every maximal ideal $M$ in $O_K$ lifts to a maximal ideal $\overline{M}$ of $O_{\overline{K}}$.

Conversely, every maximal ideal of $O_{\overline{K}}$ is nonzero and hence its prime ideal intersection with some $O_{K'}$ is nonzero and thus maximal, whence its intersection with $O_K$ is maximal. So we conclude that $\overline{M} \mapsto \overline{M} \cap O_K$ is a surjection from the set of maximal ideals of $O_{\overline{K}}$ onto the set of maximal ideals of $O_K$. There is an evident action of ${\rm{Gal}}(\overline{K}/K)$ on the set of such $\overline{M}$ which permutes those over a common $M$. Let's next see that this Galois action is transitive.

If $\overline{M}$ and $\overline{N}$ are two maximal ideals of $O_{\overline{K}}$ over the same $M$ then the same goes for the maximal ideals $\overline{M} \cap O_{K'}$ and $\overline{N} \cap O_{K'}$ for each finite $K'/K$, especially those which are Galois. But the usual story for number fields assures that the finite set $T_{K'}$ of elements of ${\rm{Gal}}(K'/K)$ carrying $\overline{M} \cap O_{K'}$ over to $\overline{N} \cap O_{K'}$ is non-empty, and for $K''/K'$ with $K''/K$ also Galois the natural map ${\rm{Gal}}(K''/K) \twoheadrightarrow {\rm{Gal}}(K'/K)$ clearly carries $T_{K''}$ into $T_{K'}$ (surjectively, but doesn't matter). So once again the inverse limit $$\lim T_{K'} \subset \lim {\rm{Gal}}(K'/K) = {\rm{Gal}}(\overline{K}/K)$$ is non-empty. But if you think about it, this inverse limit is precisely the set of $g \in {\rm{Gal}}(\overline{K}/K)$ carrying $\overline{M}$ over to $\overline{N}$. Hence, the fibers of the map $\overline{M} \mapsto \overline{M} \cap O_K$ of sets of maximal ideals consist precisely of ${\rm{Gal}}(\overline{K}/K)$-orbits on maximal ideals of $O_{\overline{K}}$. In that sense, all $\overline{M}$ over a given $M$ are "created equal" from the perspective of the Galois-action.

Now finally (!) we come to your question. For each maximal ideal $M$ of $O_K$ we choose a maximal ideal $\overline{M}$ of $O_{\overline{K}}$ over it, and define the maximal ideal $\overline{M}_{K'} := \overline{M} \cap O_{K'}$ of $O_{K'}$ over $M$ for each $K'/K$. Consider the stabilizer subgroup ("decomposition group") $$D(\overline{M}|M) = \{g \in {\rm{Gal}}(\overline{K}/K)\,|\,g(\overline{M}) = \overline{M}\}.$$ The above transitivity shows that each of the natural surjections ${\rm{Gal}}(\overline{K}/K) \twoheadrightarrow {\rm{Gal}}(K'/K)$ with finite Galois $K'/K$ carries $D(\overline{M}|M)$ onto the decomposition group $D(\overline{M}_{K'}|M)$. But for each $K''/K'$ that is Galois over $K$ the natural map $D(\overline{M}_{K''}|M) \rightarrow D(\overline{M}_{K'}|M)$ is surjective by the usual theory, so in particular for the non-empty finite subset $\phi_{K'} \subset D(\overline{M}_{K'}|M)$ of "Frobenius elements" we get a natural surjection $\phi_{K''} \rightarrow \phi_{K'}$ induced by the restriction map ${\rm{Gal}}(K''/K) \twoheadrightarrow {\rm{Gal}}(K'/K)$.

Finally, consider the inverse limit $\lim \phi_{K'} \subset {\rm{Gal}}(\overline{K}/K)$. This is non-empty and consists of precisely those elements $g \in {\rm{Gal}}(\overline{K}/K)$ such that $g(\overline{M}) = \overline{M}$ and the induced map on the field $O_{\overline{K}}/\overline{M}$ has restriction to each finite subfield $O_{K'}/\overline{M}_{K'}$ is the $q$-power map for $q = \#(O_K/M)$. But that says $g$ induces the $q$-power map on $O_{\overline{K}}/\overline{M}$, so such $g$ certainly deserve to be called "Frobenius elements" (at $\overline{M}$). Note that given one such $g$, all others are related to it through multiplication against the subgroup $I(\overline{M}|M) = \lim I(\overline{M}_{K'}|M)$ ("inertia subgroup") of elements of $D(\overline{M}|M)$ that induces the identity on the residue field $O_{\overline{K}}/\overline{M}$.

So that finally is the definition of a Frobenius element in ${\rm{Gal}}(\overline{K}/K)$ relative to $\overline{M}$ over $M$: it preserves $\overline{M}$ and induces the map $t \mapsto t^{\#(O_K/M)}$ on the residue field $O_{\overline{K}}/\overline{M}$. The above shows that such elements exist, and for any $\overline{N}$ over $M$ conjugation against any choice (which we saw does exist!) of element of ${\rm{Gal}}(\overline{K}/K)$ carrying $\overline{M}$ to $\overline{N}$ clearly carries $D(\overline{M}|M)$ over to $D(\overline{N}|M)$ and carries Frobenius elements over to Frobenius elements. (With a tiny bit of work one sees that $O_{\overline{K}}/\overline{M}$ is algebraically closed, so it is even an algebraic closure of $O_K/M$, but this was not needed above; indeed, $\overline{K}$ everywhere above could have been an arbitrary Galois algebraic extension of $K$.) What could be better than that?

  • $\begingroup$ This has more detail than Washington's appendix, which also argues some things differently, e.g., in it the proof of transitivity of the Galois action on primes over a given prime downstairs appeals to the compactness of the Galois group (together with transitivity at finite levels) rather than appealing to an inverse limit being nonempty. $\endgroup$ – KConrad Oct 10 '15 at 22:42
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    $\begingroup$ @KConrad: Sure, but it is more "elementary" to argue via exhaustion by subrings that are fields than to appeal to integrality arguments, and the point of my answer was to try to make the argument as elementary as possible (given the context of working with an algebraic Galois extension of possibly infinite degree). $\endgroup$ – grghxy Oct 10 '15 at 22:58
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    $\begingroup$ @user75536, there is not a Frobenius conjugacy class in general. Even before you try to vary the prime upstairs, there is an ambiguity by a coset of the inertia subgroup of the prime you pick upstairs, and while in the finite case most such subgroups are trivial, in the infinite case they are potentially all nontrivial. In the finite case you don't have Frobenius conjugacy classes associated to ramified primes, and in the infinite case every prime could be ramified. For an infinite extension $L/K$ unramified at all but finitely many primes you would have many Frob. conj. classes, however. $\endgroup$ – KConrad Oct 10 '15 at 23:23
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    $\begingroup$ @user75536: Definitely no. As KConrad notes, for a Galois extension "unramified" at $M \subset O_K$ (i.e., "inertia group" defined at infinite level as in my answer is trivial, or equivalently all finite-degree subextensions are unramified over $M$) there is a "Frobenius conjugacy class" attached to $M$; otherwise not. By Chebotarev at finite levels, for Galois algebraic extensions unramified at all but finitely $M$ (e.g., splitting fields of $\ell$-adic representations "arising from algebraic geometry") members of Frobenius conjugacy classes at unramified $M$'s form a dense subset. $\endgroup$ – grghxy Oct 11 '15 at 0:12
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    $\begingroup$ @user75536: "Frobenius conjugacy class" in the absolute Galois group is meaningless; no prime of $O_K$ is unramified in that extension. But "Frobenius element" makes sense (conjugacy class of inertial cosets), and these form a dense subset (enough to check at finite level, and there use Chebotarev). This notion of "Frobenius element" is defined in two equivalent ways: element of $\lim \phi_{K'}$, or as at the end of my answer. The whole point of my answer is the latter definition, avoiding explicit appeal to the crutch of finite layers. Embrace infinite-degree extensions on their own terms! $\endgroup$ – grghxy Oct 11 '15 at 0:19

You can easily define it for an infinite extension $\bar{K}|K$, and an unramified maximal ideal $\mathfrak{p} \in O_K$. Take $\mathfrak{P} \in O_\bar{K}$ maximal over $\mathfrak{p}$. The Frobenius element $\mathrm{Frob}(\mathfrak{P}/\mathfrak{p})$ works just fine:

$$\mathrm{Frob}(\mathfrak{P}^\sigma/\mathfrak{p})=\sigma \mathrm{Frob}(\mathfrak{P}/\mathfrak{p}) \sigma^{-1}$$

for $\sigma \in \mathrm{Gal}(\bar{K}/K)$, with the conjugacy classes determined by $\mathfrak{p}$.

You might find useful this older MO question, particularly Kevin Buzzard's answer:

  • $\begingroup$ Here is $\mathrm{Frob}(\mathfrak{P}|\mathfrak{p})$ defined as the generator of the galois group of the residue fields extensions? $\endgroup$ – user75536 Oct 10 '15 at 22:15
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    $\begingroup$ @user75536 Yes. It is usually defined as the generator of the decomposition group of $\mathfrak{P}/\mathfrak{p}$, but since the ideal is unramified, that's the same as the Galois group of the residue field extensions. $\endgroup$ – Myshkin Oct 10 '15 at 22:22
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    $\begingroup$ There is nothing problematic if you start with a prime ideal $\mathfrak P$ upstairs in the ring of integers of $\overline K$. Then define a Frobenius automorphism in $G = {\rm Gal}({\overline K}/K)$ associated to $\mathfrak P$ in a "naive sense" as some element $\sigma$ of $G$ such that $\sigma(\alpha) \equiv \alpha^{{\rm N}\mathfrak p} \bmod \mathfrak P$ for all $\alpha$ in $\mathcal O_{\overline K}$, where $\mathfrak p = \mathfrak P \cap \mathcal O_K$. It exists. See section 2 of the appendix to Washington's book on cyclotomic fields. $\endgroup$ – KConrad Oct 10 '15 at 22:27
  • $\begingroup$ Thanks for the answer. Only one thing, I remember that in infinite Galois extension the decomposition group is the inverse limit of the decomposition group that occur in finite sub-extension. So why should be generated by one element? $\endgroup$ – user75536 Oct 10 '15 at 22:29
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    $\begingroup$ @user75536, the decomposition group is not generated by one element, but the decomposition group modulo the inertia subgroup (both associated to a prime ideal $\mathfrak P$ upstairs) is isomorphic to the Galois group of the residue field extension and that infinite Galois group is pro-cyclic, so it is topologically generated by one element, namely the ${\rm N}\mathfrak p$-th power map on the residue field upstairs, $\mathcal O_{\overline K}/\mathfrak P$. $\endgroup$ – KConrad Oct 10 '15 at 22:32

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