$\def\cO{\mathcal{O}}\def\FF{\mathbb{F}}\def\fq{\mathfrak{q}}\def\fp{\mathfrak{p}}\def\Gal{\mathrm{Gal}}\def\Frob{\mathrm{Frob}}$I'm going to try to write up two insights that we get from the geometric picture. A basic result of class field theory is that, for $K$ a number field, the Galois group of the maximal unramified abelian extension of $K$ is isomorphic to the class group of $K$. (And then there are generalizations of this to the ramified case, but the unramified case illustrates the points I want to make.)

More precisely, let $L/K$ be an unramified abelian extension. Let $\fq$ be a prime of $\cO_L$ and let $\fp$ be $\cO_K \cap \fq$. There is a unique element $\Frob(\fq)$ of $\Gal(L/K)$ such that $\Frob(\fq)$ fixes $\fq$ and the induced action on $\cO_L/\fq$ is the $\#(\cO_K/\fp)$-power Frobenius. Using that $\Gal(L/K)$ is abelian, $\Frob(\fp)$ depends only on $\fp$, not on $\fq$. (In an nonabelian extension, changing $\fq$ to $\fq'$ would conjugate the Frobenius element.) The Artin map, from the ideal group of $K$ to $\Gal(L/K)$, sends $\prod \fp^{a_p}$ to $\prod \Frob(\fp)^{a_p}$. Then class field theory says two things:

(1) The Artin map factors through the class group and

(2) There is an unramified abelian extension, the class field, for which the Artin map is an isomorphism.

Let's see what each of these mean geometrically.

Let $k$ be the finite field with $q$ elements, let $X$ be a smooth, projective, geometrically irreducible curve over $k$, and let $K$ be the field of meromorphic functions on $X$. Then unramified extensions of $K$ correspond to unbranched covers $Y$ of $X$; such extensions are linearly disjoint from $k^{\text{alg}}$ if and only if $Y$ is geometrically connected.

**Understanding (1):** Let $\pi: Y \to X$ be an abelian unbranched cover with covering group $G$. We can think of $Y$ to $X$ as a principal bundle with group $G$. Let $\mathfrak{p}$ be a closed point of $X$, with residue field of size $q^f$. Geometrically, we can think of this as $f$ points $x_1$, $x_2$, ..., $x_f$ of $X(k^{\text{alg}})$ which are permuted cyclically by the $q$-power Frobenius, say $x_1 \mapsto x_2 \mapsto \dots \mapsto x_f \mapsto x_1$. Then $q$-power Frobenius must also cycle the fibers $\pi^{-1}(x_1) \to \pi^{-1}(x_2) \to \cdots \to \pi^{-1}(x_f) \to \pi^{-1}(x_1)$. So $q^f$ power Frobenius maps each $\pi^{-1}(x_i)$ to itself. This map is multiplication by some element of $G$, and this element is $\Frob(\fp)$. (If $G$ were not abelian, then the action of $\Frob(\fp)$ on $\pi^{-1}(x_1)$ is a permutation which commutes with the $G$-action on $\pi^{-1}(x_1)$. The group of such permutations is isomorphic to $G$, but the isomorphism is only natural up to inner conjugacy, so we only get a conjugacy class in $G$. I learned about this idea from John Baez, search for "dual groups" on the linked page.)

This gives us a geometric way to understand the Frobenius element, but it doesn't help us prove (1) yet. We can do that using an idea I learned from Ben-Zvi's lecture. For any positive integer $N$, we can look at $Y^N \to X^N$ as a principal bundle with group $G^N$ and can then quotient by $\{ (g_1, g_2, \ldots, g_N) : \prod g_N = 1 \}$ to get a principal $G$-bundle over $X^N$. (Note that what we are quotienting by is a subgroup because $G$ is abelian.) Moreover, this quotient descends to a principle $G$-bundle $\psi: Z^N \to \mathrm{Sym}^N(X)$.

If $x_1$, $x_2$, ..., $x_f$ are as two paragraphs ago, then $(x_1, x_2, \ldots, x_f)$ is a fixed point of the $q$-power Frobenius action on $\mathrm{Sym}^f(X)$ and the action of $q$-power Frobenius on $\psi^{-1} (x_1, x_2, \ldots, x_f)$ is $\Frob(\fp)$. More generally, if $D$ is an effective element of the divisor class group, with degree $N$, then $D$ gives a Frobenius-fixed point $[D]$ of $\mathrm{Sym}^N(X)$, and the Artin map describes the action of $q$-power Frobenius action on $\psi^{-1}([D])$.

This gives us a geometrically natural way to think of the Artin map, but can it prove (1)? Let $D$ and $E$ be effective divisors both of degree $N$ which represent the same class in $\mathrm{Pic}^N(X)$. So $[D]$ and $[E]$ are in the same fiber of $\mathrm{Sym}^N(X) \to \mathrm{Pic}^N(X)$. How can we see that the Frobenius actions on $\psi^{-1}([D])$ and $\psi^{-1}([E])$ coincide?

For $N$ large enough, all the fibers of $\mathrm{Sym}^N(X) \to \mathrm{Pic}^N(X)$ are projective spaces! (And, if $N$ isn't large enough, we can add the same high degree divisor to $D$ and $E$ to make it large enough.) And projective space is geometrically simply connected! So the $G$-principal bundle $\psi : Z^N \to \mathrm{Sym}^N(X)$ must be geometrically trivial when restricted to any fiber of $\mathrm{Sym}^N(X) \to \mathrm{Pic}^N(X)$. ("Geometrically" meaning after we base change to $k^{\text{alg}}$.) So, restricting to the fiber through $[D]$ and $[E]$, the principal bundle $\psi$ is geometrically trivial, and Frobenius must act by permuting the geometric components. In particular, Frobenius acts the same way on $\psi^{-1}([D])$ and $\psi^{-1}([E])$.

**Understanding (2).** In number fields, the class group is finite. In function fields, the analogue of the class group is $\mathrm{Pic}(X)(k)$, which is not quite finite, it sits in a short exact sequence
$$0 \to \mathrm{Pic}^0(X)(k) \to \mathrm{Pic}(X)(k) \to \mathbb{Z} \to 0. \quad (\ast)$$
So we are not going to ask for the Artin map to be an isomorphism, but we can still take finite quotients of $\mathrm{Pic}(X)(k)$ and ask what covers they correspond to.

The quotient $\mathrm{Pic}(X)(k) \to \mathbb{Z} \to \mathbb{Z}/d \mathbb{Z}$ corresponds to extending the ground field to have $q^d$ elements; it is a good exercise to check that the Artin map does fit into this story.

The more interesting thing is covers that come from $\mathrm{Pic}^0(X)(k)$. However, these are harder to talk about because the extension $(\ast)$ doesn't come with a natural splitting. (It is splittable, since $\mathbb{Z}$ is projective in the category of abelian groups.) However, if $X$ has a $k$-point $x_{\infty}$, then we can make a splitting $\mathrm{Pic}(X)(k) \longrightarrow \mathrm{Pic}^0(X)(k)$ by $D \mapsto D - \mathrm{deg}(D) x_{\infty}$, and we can ask what covering of $X$ corresponds to this quotient. (If you want to see the theory without the assumption of a $k$-point on $X$, I recall that Serre has a rather difficult chapter on this point.)

Let's abbreviate the Jacobian $\mathrm{Pic}^0(X)$ to $J$, and lets abbreviate the finite group $J(k) = \mathrm{Pic}^0(X)(k)$ to $G$. Frobenius is an addditive map $F:J \to J$ and, using the group structure on $J$, we can talk about the isogeny $F-\mathrm{Id} : J \to J$. The kernel of $F - \mathrm{Id}$ is $G$, so $F-\mathrm{Id} : J \to J$ is a principal $G$-bundle.

Using the point $x_{\infty}$, we can embed $X$ into $J$ by $x \mapsto x-x_{\infty}$, so we can pull the $G$-bundle $F-\mathrm{Id}: J \to J$ back to $X$; let's call this $\pi : Y \to X$.

Let us understand how the Artin map works for $\pi : Y \to X$. Again, let $\fp$ be a closed point of $X$ corresponding to a Frobenius orbit of geometric points $x_1 \mapsto x_2 \mapsto \cdots \mapsto x_f \mapsto x_1$. Choose a geometric point $y_1$ over $x_1$ in $Y$, and let $y_j = F^{j-1}(y_1)$. So $\pi(y_j) = x_j$. Writing out the defnition of $\pi$, we have $F(y_j) - y_j = x_j$ or, in other words, $F^{j+1}(y_1) - F^j(y_1) = x_j$. This telescopes to give $F^f(y_1) - y_1 = \sum x_i$.
But $\Frob(\fp)$ is the element of $G$ by which $F^f$ acts on $\pi^{-1}(x_1)$. So we have just computed that $\Frob(\fp)=\sum x_i$.

More generally, if $D$ is any effective divisor on $X$, we have just computed that the Artin map takes $D$ to the sum of the geometric points of $D$, computed in $G = J(k)$. So the Artin map is the isomorphism to the class group that we want.

heuristicfor the number field case "becomes obvious" after the (etale cohomology...!) of the function-field/geometric case is understood. Lang did use hyperbole now and then. :) $\endgroup$ – paul garrett Mar 24 at 21:591more comment