I am trying to understand certain aspects of the Weil group $W_K$ for a $p$-adic field K, in particular how it does interplay with local class field theory.

Let $L/K$ be a finite unramified extension of such fields. Then by local reciprocity we have an isomorphism $$\operatorname{Gal}(L/K)\cong K^*/N_{K/L}(L^*),$$ where $N_{L/K}$ denotes the norm of the extension. Under this isomorphism the Frobenius element is mapped to the element $\pi_K(\mod N_{L/K}(L^*))$, where $\pi_K$ is a prime element of $K$. Note that the class $\pi_K(\mod N_{L/K}(L^*))$ does not depend on the choice of prime element $\pi_K$, because all units in $K$ are some norms coming from $L$ (not trivial to prove!). On the other hand, we also have the projection $K^*\longrightarrow K^*/N_{K/L}(L^*)$.

From here on, my goal is to arrive at the (Artin reciprocity) homomorphism $$W_K\longrightarrow K^*$$

For this purpose I take the projective limit over all finite unramified extensions $L/K$. Thus, we have $$\operatorname{Gal}(K^{nr}/K)\cong\varprojlim_{L/K}K^*/N_{K/L}(L^*),$$ where $K^{nr}$ is the maximal non-ramified extension of $K$. Almost by definition of $W_K$, there is a (continuous) homomorphism $W_K\longrightarrow\operatorname{Gal}(K^{nr}/K)$.

But how do we get a (continuous) canonical homomorphism $$\varprojlim_{L/K}K^*/N_{K/L}(L^*)\longrightarrow K^*$$?

My guess is, there is something easy that I don't know about projective limits...

EDIT: Per Matt's comment and answer, taking the projective limit over only finite unramified extensions does not suffice for constructing the desired homomorphism $W_K\longrightarrow K^*$.

  • $\begingroup$ I should have specified that I am actually not looking for just any homomorphism $\varprojlim_{L/K}K^*/N_{K/L}(L^*)\longrightarrow K^*$, but rather for a canonical one, which should mean a (continuous) homomorphism that completes the other arrow into a homomorphism $W_K\longrightarrow K^*$ in the sense of Artin reciprocity. Strictly speaking, I should have also specified a few conditions on the homomorphism $W_K\longrightarrow K^*$, but I don't want to burden the post with such additional stuff as I expect that there is an 'easy canonical' non-trivial map from the proj limit into $K^*$. $\endgroup$
    – M.G.
    May 13, 2011 at 0:47
  • $\begingroup$ There is no such map of the kind you are asking for, for several different reasons. (See my comment to Mephisto's answer.) I will try to post an answer later if I get a chance. $\endgroup$
    – Matt E
    May 13, 2011 at 1:45
  • $\begingroup$ Thanks, I am seeing this now (see my comment to your comment). I need to consider all abelian extensions. $\endgroup$
    – M.G.
    May 13, 2011 at 12:55

2 Answers 2


Local Artin reciprocity, for a finite abelian extension $L/K$ with $K$ a $p$-adic field, is a specific isomorphism $K^{\times}/N_{L/K}(L^{\times}) \cong Gal(L/K)$. You seem particular interested in the unramified case, but let me treat the arbitrary case first.

Passing to the inverse limit over $L$, one gets an isomorphism $$\varprojlim{} K^{\times}/N_{L/K}(L^{\times}) \cong G_K^{ab}.$$ As Mephisto notes in their answer, the norm groups range over all open subgroups of $K^{\times}$, and so we may rewrite this as an isomorphism $$\widehat{K^{\times}} \cong G_K^{ab},$$ where $\widehat{K^{\times}}$ is the profinite completion of $K^{\times}$.

Recall that, if we choose a uniformizer $\pi$ for $K$, then $K^{\times} \cong \mathcal O^{\times} \times \mathbb Z$, where $\mathcal O$ denotes the ring of integers in $K$, and the isomorphism is given by mapping an element $a \in K^{\times}$ to $\bigl(a/\pi^{v(a)}, v(a) \bigr),$ where $v: K^{\times} \to \mathbb Z$ is the valuation, normalized via $v(\pi) = 1$.

Thus $\widehat{K^{\times}} \cong \mathcal O^{\times} \times \hat{\mathbb Z}$. (Recall that $\mathcal O^{\times}$ is its own profinite completion, but $\mathbb Z$ is not; we let $\hat{\mathbb Z}$ denote the profinite completion of $\mathbb Z$.)

Now we can understand what happens if we restrict to unramified extensions. The valuation $v: K^{\times} \to \mathbb Z$ induces a projection $\widehat{K^{\times}} \to \hat{\mathbb Z}$, which is independent of the choice of $\pi$. Similarly, there is a surjection $G_K^{ab} \to Gal(K^{nr}/K)$. The latter group is isomorphic to $\widehat{Z}$; it is naturally identified with the absolute Galois group of the residue field, and is topologically generated by Frobenius.

Under the reciprocity isomorphism $\widehat{K^{\times}} \cong G_K^{ab}$, the projection to $\widehat{Z}$ on the source and the projection to $Gal(K^{nr}/K)$ on the target are compatible with the isomorphism $\widehat{Z} \cong Gal(K^{nr}/K)$ given by mapping $1$ to Frobenius.

Now we can bring in Weil groups. The general definition of the Weil group is somewhat involved, involving the fundamental classes in $H^2(L/K, L^{\times})$, but after you sort everything out, you find that the Weil group $W_K$ can be identified with a subgroup of $G_K$, namely the preimage under the natural map $G_K \to Gal(K^{nr}/K) \cong \hat{\mathbb Z}$ of the subgroup $\mathbb Z \subset \widehat{\mathbb Z}$. From this, one sees that $W_K^{ab}$ can be identified with the subgroup of $G_K^{ab}$ which again is the preimage under the natural map $G_K \to Gal(K^{nr}/K) \cong \hat{\mathbb Z}$ of the subgroup $\mathbb Z\subset \widehat{\mathbb Z}$.

If we go back to the preceding discussion of our reciprocity map, we see that the isomorphism $\widehat{K^{\times}} \cong G_K^{ab}$ restricts to an isomorphism $K^{\times} \cong W_K^{ab}$. The inverse of this is the isomorphism you are looking for, I would guess.

If one restricts to the unramified case, then we have to pass to the quotient $\mathbb Z$ of $K^{\times}$, and the quotient $\mathbb Z \subset \widehat{\mathbb Z} = Gal(K^{nr}/K)$ of $W_K^{ab}$, and then the isomorphism just becomes $\mathbb Z = \mathbb Z$.

Additional remarks: There are various confusions in your question. Here are some: you write $W_K$ but you mean $W_K^{ab}$. (The Weil group itself is not abelian, just as $G_K$ is not abelian.) You say you want the general Artin reciprocity isomorphism $W_K$ [sic] $\to K^{\times}$, but you then restrict attention to unramified extensions. As noted above, these will only see the quotient $\mathbb Z$ of $W_K^{ab}$ and the corresponding quotient $\mathbb Z$ of $K^{\times}$. You then go on to ask for a continuous homomorphism $\varprojlim{} K^{\times}/N_{L/K}(L^{\times}) \to K^{\times}.$ Leaving aside the issue that the left-hand side should involve all abelian $L$ over $K$, not just the unramified ones (or else the left-hand side will be too small), there is no such map, since the left-hand side is profinite and the right hand side has a discrete factor (as noted above). The correct thing is to profinitely complete $K^{\times}$, and one then has the reciprocity isomorphism discussed above.

Further remarks: Here are some additional remarks, prompted in part by the exchange of comments below.

At a technical level, passing from $G_K$ to $W_K$ simply replaces the $\widehat{\mathbb Z}$ quotient of $G_K$ coming from the map $G_K \to Gal(K^{nr}/K) \cong \widehat{\mathbb Z}$ by $\mathbb Z$, so that instead of writing the reciprocity isomorphism as an isomorphism $\widehat{K^{\times}} \cong G_K^{ab}$ we can rewrite it as an isomorphism $K^{\times} \cong W_K^{ab}$, and so avoid passing from $K^{\times}$ to $\widehat{K^{\times}}$.

As for why we do this: one reason is that $K^{\times}$ is what appears as a local factor in the ideles, not $\widehat{K^{\times}}$. There are additional motivations. One thing to remember is that the original definition of the Weil group is not as the preimage of $\mathbb Z \subset \widehat{\mathbb Z}$ in $G_K$, but in terms of the fundamental classes of local class field theory; so the Weil group arises naturally from this point of view. For further motivations, this MO post might help.

  • $\begingroup$ Thank you for your extensive answer. I will try to address several of your remarks to the best possible way I can. 1) I did not use Tate cohomology to define the Weil group, because it involves $H^2=\operatorname{EXT}$ (group extensions), fundamental classes etc. For the purposes of my question and also for practical purposes it is IMHO much more elegant to define the Weil group $W_K$ as a preimage of the Frobenius dense cyclic subgroup of $\operatorname{Gal}(K^{nr}/K)$ and assign to the unique topology coming from the discrete topology of $\mathbb{Z}$. $\endgroup$
    – M.G.
    May 13, 2011 at 12:16
  • $\begingroup$ 2) It is not that clear (at least not to me) that one can go from $G_K^{ab}$ to $W_K^{ab}$ that easily, since abelinization is in general not inclusion-preserving as far as I know. $\endgroup$
    – M.G.
    May 13, 2011 at 12:17
  • $\begingroup$ 3) I don't think that it is optimal to start with $W_K^{ab}$ and work towards $W_K$. The other way around seems more natural to me, but that is rather a matter of opinion, I guess. $\endgroup$
    – M.G.
    May 13, 2011 at 12:18
  • $\begingroup$ 4) I did not mean $W_K^{ab}$ instead of $W_K$, because I asked about Artin reciprocity homomorphism, not isomorphism as you write. The Artin reciprocity isomorphism $W_K^{ab}\cong K^*$ is then induced by the Artin reciprocity homomorphism, so there is really no confusion about that part. $\endgroup$
    – M.G.
    May 13, 2011 at 12:20
  • $\begingroup$ 5) Restricting the attention to unramified extensions and then taking the projective lim over all finite unramified extensions is really convenient, because this way you get the Frobenius elements. Then working in $G_K\longrightarrow\operatorname{Gal}(K^{nr}/K)\longrightarrow\hat{\mathbb{Z}}$ is easy and canonical IMHO (also in view of my first comment). And it also allows for bringing the inertia group in the game and seeing that its inherited (profinite) topology is the same as the induced topology from $W_K$, even though the topology of $W_K$ is not the induced one from $G_K$ (!). $\endgroup$
    – M.G.
    May 13, 2011 at 12:26

You need to use that the norm groups are exactly the open subgroups of finite index in $K^*$ (existence theorem; the subgroups of finite index are automatically open in the characteristic zero case). Therefore the inverse limit is just the profinite completion of $K^*$, which is $K^*$ itself. [Correction: not quite; the quotient $\mathbb{Z}$ of $K^*$ has to be replaced by its profinite completion.] So the canonical homomorphism you want goes the other way.

I am assuming the limit is over all finite extensions $L$ of $K$. If you only take the unramified finite extensions then the inverse limit is equal to profinite completion of $\mathbb{Z}$

  • 1
    $\begingroup$ Dear Mephisto, This is not correct. Firstly, the profinite completion of $K^{\times}$ is not $K^{\times}$. For example, the valuation induces a surjection $K^{\times} \to \mathbb Z$, and $\mathbb Z$ is certainly not profinite. Secondly, even if one modifies the statement so as to take this into account, one would have to take the limit over all finite extensions, and (for reasons I don't really understand) the OP is only taking the limit over unramified extensions. Regards, $\endgroup$
    – Matt E
    May 13, 2011 at 1:42
  • $\begingroup$ Correction: "all finite extensions" should read "all finite abelian extensions". Regards, $\endgroup$
    – Matt E
    May 13, 2011 at 4:37
  • $\begingroup$ @Matt: I took the projective limit only over the unramified extensions, because I thought it would suffice for the purposes of constructing the reciprocity homomorphism $W_K\rightarrow K^*$ using the homomorphism $W_K\rightarrow\operatorname{Gal}(K^{nr}/K)$. But indeed, the existence theorem delivers all the abelian extensions,not just the unramified ones, so the projective lim, the way I want it, would be too small. $\endgroup$
    – M.G.
    May 13, 2011 at 13:09

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