I've been reading:
math.stanford.edu/~conrad/249BPage/handouts/geomcft.pdf


in an attempt to shed some geometric light on class field theory. The last paragraph there reads:


*In case the ground field $k$ is perfect, the essential difficulty in the proof of class field theory – proving
that the Artin map kills certain principal ideals – becomes easy to prove geometrically by means of the
interpretation of geometric points of generalized Jacobians in terms of generalized ideal class groups. (More
precisely, one has $J\_m (k) = Cl\_m (K)$ when $Br(k) = 1$, as happens when $k$ is finite but not when $k$ is a number
field.)*


What precisely is he referring to? What particular part of class field theory ("that the Artin map kills certain principal ideals") becomes easier to prove, and why ("by means of the
interpretation of geometric points of generalized Jacobians in terms of generalized ideal class groups.")?



What makes Geometric CFT easier than CFT?
There is a lot here and it is hard to tell where you are confused. I'll try to fill in some steps along the way. For simplicity, I'll present the whole theory only for unramified extensions. I'll assume that you have gone through CFT for number fields in the ideal theoretic (not adelic) presentation, as in Janusz or Cox's books. 


$\def\Art{\mathrm{Art}}$Let $K$ be a number field, and let $L/K$ be an unramified abelian extension with Galois group $A$. Let $I(K)$ be the group of fractional ideals of $K$. So the Artin map is the map $\Art:I(K) \to A$ which sends a prime ideal $\pi$ to the Frobenius lift of $\pi$. 


**Key Theorem:** With the above notation, $\Art$ is trivial on the principal ideals.


**Example:** The extension $\mathbb{Q}(\sqrt{-5}, i)/\mathbb{Q}(\sqrt{-5})$ is unramified (exercise!) with Galois group $\{ \pm 1 \}$. Let $\pi$ be a prime of $\mathbb{Q}(\sqrt{-5})$ and, for simplicity, let $\pi \neq \langle 2, \sqrt{-5}-1 \rangle$, which is the unique prime lying over 2. Then $\mathrm{Frob}(\pi)=1$ if $-1$ is square in $\mathbb{Z}[\sqrt{-5}]/\pi$ and $\mathrm{Frob}(\pi)=-1$ if $-1$ is not square in this residue field. So the above theorem says the following: Take any odd element $\alpha \in \mathbb{Z}[\sqrt{-5}]$ and factor the ideal $(\alpha)$ as $\pi\_1 \pi\_2 \cdots \pi\_r$. Then there is an even number of $\pi\_k$ for which $-1$ is square in $\mathbb{Z}[\sqrt{-5}]$. In fact, the class group of $\mathbb{Z}[\sqrt{-5}]$ is $\mathbb{Z}/2$, and it turns out that $-1$ is a square in $\mathbb{Z}[\sqrt{-5}]/\pi$ if and only if $\pi$ is principal. 
(I encourage you to check a half dozen examples of this!)


I hope you will agree that this theorem is extremely deep. Even the corollary about $\mathbb{Z}[\sqrt{-5}]$ in the above example is not obvious, although it can be proved by elementary means.


It is the Key Theorem which Brian Conrad is calling the "essential difficulty" in CFT. More precisely, you need to prove the generalization of this theorem for ramified extensions, which requires the introduction of an auxiliary modulus $\mathfrak{m}$, in which case the set of all principal ideals is replaced by a certain subgroup of principal ideals. But all the key points I want to make can be seen in the unramified setup.


Now, let's see how this Key Theorem, in the geometric case, is deduced from Theorem 3.2 in Conrad's handout. Theorem 3.2 is itself not obvious, but (in my experience) proofs of it are usually more motivated than direct assaults on the Key Theorem.


$\def\bark{\bar{k}}$Here is a specialization of Theorem 3.2 which contains all the cases I want to talk about. Let $k = \mathbb{F}\_q$ and $\bark$ be the algebraic closure of $k$. The letter $F$ will always denote the $q$-power Frobenius; what $F$ is acting on will (hopefully) be clear from context. Let $f: Y \to X$ be an unbranched abelian cover of $X$. Assume further that $Y \times\_k \bark$ is connected and $Y(k) \neq 0$. These last two conditions aren't really necessary, but they make the exposition simpler. (If you don't want simplifying assumptions, read Conrad!) 


Let $y\_0 \in Y(k)$ and let $x\_0 \in X(k)$ be the image of $y\_0$. Algebraically, $x\_0$ is a prime of the downstairs field and $y\_0$ is a prime above it with residue extension of degree $1$. Because $Y \to X$ is abelian, all the primes in $g^{-1}(x\_0)$ have the same degree; in other words, $g^{-1}(x\_0) \subseteq Y(k)$. This will be relevant later.


Let $J$ be the Jacobian of $X$, and map $X \to J$ by $x \mapsto x-x\_0$.


**Theorem 3.2 specialized:** With the above notation, there is a finite unramified abelian cover $g: G \to J$ such that the diagram
$$\begin{matrix} 
Y & \rightarrow & G \\ \downarrow & & \downarrow \\ X & \rightarrow & J 
\end{matrix}$$
commutes, with $Y = X \times\_J G$. (Compare this with the diagram in Conrad's Theorem 3.2.)


**Sketch of the deduction of the Key Theorem from the above:** The following won't be easy, but it is (I hope) easier than standard proofs of CFT.


**Claim 1:** $G \times\_k \bark$ is connected. **Proof Sketch:** If not, then $Y \times\_k \bark$ would also be disconnected.


From the general theory of abelian varieties, this tells us that $G$ is an abelian variety and $J \times\_k \bark=(G \times\_k \bark)/H$ for a finite subgroup $H$ of $G(\bark)$. 


**Claim 2:** $H \subseteq G(k)$. **Proof sketch** We have $H = g^{-1}(0) = f^{-1}(x\_0) \subseteq Y(k)$, as discussed above. We have $Y(k) \subseteq G(k)$.


Note that $G(k) = \mathrm{Ker}(F-\mathrm{Id})$, where $F - \mathrm{Id}$ maps $G \to G$. So $H$ is contained in the kernel of $F - \mathrm{Id}$. We deduce that we can factor $G \to J \to G$ where the composite is $F - \mathrm{Id}$. This proof is already getting too long, so let me make things simpler by assuming that the second map $J \to G$ is an isomorphism, and leave adapting the more general case as an exercise for you.


Our setting is now the following: We have a curve $X$ embedded in its Jacobian $J$. $Y$ is the curve
$$\{ y \in J : \mathrm{Frob}(y) - y \in X \}$$
The map $Y \to X$ is $y \mapsto F(y) - y$. Note that this is indeed an abelian cover: The abelian group $J(k)$ acts on $Y$ with the element $z \in J(k)$ acting by $y \mapsto y+z$. So we identify $A = J(k)$. The fact that we have been able to reduce from an arbitrary abelian cover to one of this special form is very deep (and, of course, most of the work is inside Theorem 3.2).


Now, let's understand what the Artin map does. Let $\pi$ be a prime in the downstairs field, with residue field $\mathbb{F}\_{q^r}$. Geometrically, $\pi$ corresponds to a set $(\pi\_1, \pi\_2, \ldots, \pi\_r) \in X(\bark)$ such that $F(\pi\_i) = \pi\_{i+1}$ and $F(\pi\_r) = \pi\_1$. Let $y \in Y(\bark)$ lie above $\pi\_1$. Then the Frobenius lift at $\pi$ is required to send $y$ to $F^r(y)$. So $\mathrm{Frob}\_{\pi}$ is the unique element $z$ of $J(k)$ such that $y+z = F^r(y)$. In other words, $z=F^r(y)-y$. 


Now, here comes a key computation:
$$z=F^r(y) - y = \sum\_{i=1}^r \left( F^{i}(y) - F^{i-1}(y) \right) = \sum\_{i=1}^r F^{i-1} \left( F(y)-y \right) = \sum\_{i=1}^r F^{i-1}(\pi\_1) = \sum\_{i=1}^r \pi\_i$$


So $\mathrm{Frob}(\pi)$ is equal, in $J(\bark)$, to the sum of the points into which $\pi$ splits in $X(\bark)$. 


We deduce the following corollary by linearity: Let $D$ be a divisor in $X$. Let $D$ split in $X(\bark)$ as $\sum a\_i \pi\_i$, for various points $\pi\_i \in X(\bark)$. Then $\Art(D) = \sum a\_i \pi\_i$, where the sum takes place in $J(\bark)$.


In particular, if $D$ is principal, then $\Art(D)=0$, as desired. QED.


