I posted this on MSE, but didn't get any responses, so I'm reposting here. I tried to write down an example of the main theorem of geometric class field theory, but I must be misunderstanding something, since my example isn't turning out right.

**Main theorem.** Let $K$ be a function field of a curve over a finite field. There is a bijection between unramified $\ell$-adic Galois representations of $G_K := \operatorname{Gal}(K^{sep}/K)$ and $\ell$-adic characters of $K^{\times}\backslash\mathbf{A}_K^{\times}/\mathcal{O}_K^{\times}.$ (Here I mean the ideles modulo the diagonal on the left, and on the right quotiented by the subgroup of ideles which have valuation zero at every place.)

--

I was trying to understand what this means when $K = \mathbb{F}_p(T).$ I think unramified Galois representations are the same as representations of $\operatorname{Gal}(K^{un}/K) = \operatorname{Gal}(\overline{\mathbf{F}_p}(T)/\mathbf{F}_p(T))= \hat{\mathbf{Z}}.$ So on one-side of the bijection, I have continuous $\ell$-adic characters of the profinite integers.

--

On the other side, I think that this double quotient is just $\mathbb{Z}.$ One way is by identifying it with the Picard group of $\mathbb{P}^1$, but I also can do it more directly (thanks to these two 'independent' solutions, I don't think this is the incorrect step).

--

So, I think that in this case, geometric class field theory is telling me that continuous homomorphisms $\rho : \hat{\mathbf{Z}} \to \overline{\mathbf{Q}_{\ell}}^{\times}$ are in bijection with continuous homomorphisms $\mathbf{Z}\to \overline{\mathbf{Q}_{\ell}}^{\times}.$

But this surely is incorrect -- the image of a map out of the profinite integers must be compact, but the image out of $\mathbf{Z}$ is not always precompact (take some $\ell$-adic number of multiplicative norm exceeding 1, for instance; this defines a map out of $\mathbb{Z}$ whose image is not contained in any compact set).

So, where have I made a mistake?

Chtoucas de Drinfeld et correspondance de Langlands), there is always the running assumption that the representation is irreducible and its top exterior power hasfinite order. The final assumption removes the difference between $\mathbf Z$ and $\hat{\mathbf Z}$. In the abelian setting, this is also the version you find in Neukirch'sClass field theory(the Bonn lectures, not his other book with the same name), Chapter III, Existence Theorem 7.8. $\endgroup$2more comments