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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 $W_K\longrightarrow K^*$.

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    Thanks, I am seeing this now (see my comment to your comment). I need to consider all abelian extensions.2011-05-13

2 Answers 2

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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.

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    ... the structure of the two sides of the local reciprocity isomorphism well enough. There is nothing more (or less) to it. Regards,2011-05-13
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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 $\mathbb{Z}$

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    @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.2011-05-13