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Theorem 6 If $A$ is abelian, we can take $J$ to be a generalized Jacobian $J(\mathfrak{m}, X)$, for some conductor $\mathfrak{m}$. Moreover, we can take the support of $\mathfrak{m}$ to be the set $S$ of critical points of $Y \to X$, and we can take $X^{\circ} = X \setminus S$. In particular, if $Y \to X$ is unrammified, we can take $J$ to be the Jacobian and $S = \emptyset$.

Let me point out the subtlety of the last sentence of Theorem 6. Let $\mathrm{char}(k) \neq 2$, and let $A = \mathbb{Z}/2$. Then $L = K(\sqrt{\alpha})$ for some $\alpha$. In our high-tech language, we can take $G$ to be the multiplicative group $\mathbb{G}_m$, with $A$ embedded as $\{ \pm 1 \}$. $X^{\circ}$ is the locus where $\alpha$ is nonzero and the map $X \to G/A \cong G$ is given by the function $\alpha$ on $X^{\circ}$. Even if $Y \to X$ is an unbranched cover, the function $\alpha$ will still have zeroes and poles (of even order). We cannot map the entire projective curve $X$ to an affine group like $\mathbb{G}_m$. Theorem 6 is telling us that, by using a very different group, the Jacobian of $X$, we can arrange for the map to $J$ to be defined everywhere on $X$.

Theorem 6 If $A$ is abelian, we can take $J$ to be a generalized Jacobian $J(\mathfrak{m}, X)$, for some conductor $\mathfrak{m}$, and $X \to J$ we can take a translation of the standard map $X \to J(\mathfrak{m}, X)^1$. Moreover, we can take the support of $\mathfrak{m}$ to be the set $S$ of critical points of $Y \to X$, and we can take $X^{\circ} = X \setminus S$. In particular, if $Y \to X$ is unrammified, we can take $J$ to be the Jacobian and $S = \emptyset$.

Let me point out the subtlety of the last sentence of Theorem 6. Let $\mathrm{char}(k) \neq 2$, and let $A = \mathbb{Z}/2$. Then $L = K(\sqrt{\alpha})$ for some $\alpha$. In our high-tech language, we can take $G$ to be the multiplicative group $\mathbb{G}_m$, with $A$ embedded as $\{ \pm 1 \}$. $X^{\circ}$ is the locus where $\alpha$ is nonzero and the map $X \to G/A \cong G$ is given by the function $\alpha$ on $X^{\circ}$. Even if $Y \to X$ is an unbranched cover, the function $\alpha$ will still have zeroes and poles (of even order). We cannot map the entire projective curve $X$ to an affine group like $\mathbb{G}_m$. Theorem 6 is telling us that, by using a very different group, the Jacobian of $X$, we can arrange for the map to $J$ to be defined everywhere on $X$.

Combining Theorems 5 and 6 and the definition of $J(\mathfrak{m}, X)$, we have

Key Theorem Let $k$ be finite and $A$ abelian. Then there is a modulus $\mathfrak{m}$, supported on the ramified primes $S$ of $Y \to X$, such that $\mathrm{Art}$ is trivial on principal ideals whose generators are $1 \bmod \mathfrak{m}$.

Theorem 4 If $A$ is abelian, we can take $J$ to be a generalized Jacobian $J(\mathfrak{m}, X)$, for some conductor $\mathfrak{m}$. Moreover, we can take the support of $\mathfrak{m}$ to be the set $S$ of critical points of $Y \to X$, and we can take $X^{\circ} = X \setminus S$. In particular, if $Y \to X$ is unrammified, we can take $J$ to be the Jacobian and $S = \emptyset$.

This was more or less Theorem 3.2 in Conrad's notes.

Let me point out the subtlety of the last sentence of Theorem 4. Let $\mathrm{char}(k) \neq 2$, and let $A = \mathbb{Z}/2$. Then $L = K(\sqrt{\alpha})$ for some $\alpha$. In our high-tech language, we can take $G$ to be the multiplicative group $\mathbb{G}_m$, with $A$ embedded as $\{ \pm 1 \}$. $X^{\circ}$ is the locus where $\alpha$ is nonzero and the map $X \to G/A \cong G$ is given by the function $\alpha$ on $X^{\circ}$. Even if $Y \to X$ is an unbranched cover, the function $\alpha$ will still have zeroes and poles (of even order). We cannot map the entire projective curve $X$ to an affine group like $\mathbb{G}_m$. Theorem 4 is telling us that, by using a very different group, the Jacobian of $X$, we can arrange for the map to $J$ to be defined everywhere on $X$.

Now let $k= \mathbb{F}_q$ and let $F$ denote the $q$-power Frobenius. As far as I can tell, the following result is independent of Theorem 4, and represents a separate way that the function field case is particularly nice.

Theorem 5 When $A$ is abelian and $k = \mathbb{F}_q$, we can find a map $h: J \to G$ such that the composites $G \to J \to G$ and $J \to G \to J$ are $F-1$.

Warning Again, I suspect that I am missing a technical hypothesis.

Special Cases In an Artin-Schrier extension, $G \cong J$ is the additive group $\mathbb{G}_a$ and the map $G \to J$ is $y \mapsto y^p-y$. If $m | q-1$, then $\mathbb{F}_q$ contains the $m$-th roots of unity. In the Kummer extension, $G \cong J \cong \mathbb{G}_m$, with $G \to J$ being $x \mapsto x^m$ and $J \to G$ being $x \mapsto x^{(q-1)/m}$. So the composite is $x \mapsto x^{q} x^{-1}$.

Generalizing the computation in my other answer shows:

Theorem 6 With the above hypotheses, if $D$ is a divisor supported on $X^{\circ}$, and $\bar{D}$ the divisor which it splits into in $X^{\circ}(\bar{k})$, then $$h \left( \sum_{\pi \in \bar{D}} \pi \right) = \mathrm{Art}(D)$$.

Here the sum $\sum_{\pi \in \bar{D}} \pi$ lives in $J(\bar{k})$, the map $h$ puts it into $G(\bar{k})$, and the assertion is it is equal to $\mathrm{Art}(D)$ under the embedding $A \to G(k)$.

Examples In the Kummer case, Theorem 6 says that $\mathrm{Art}(D) = \prod_{\pi \in \bar{D}} \alpha(\pi)^{(q-1)/m}$. In the Artin-Schrier case, Theorem 6 says that $\mathrm{Art}(D) = \sum_{\pi \in \bar{D}} \beta(\pi)$.

Problem 2 Although the ray class groups of $K$ generalize the groups $J(\mathfrak{m}, X)(k)$, there is no algebraic group which generalizes $J(\mathfrak{m}, X)$, so it is not clear what an analogue of Theorem 4 would look like.

Problem 3 There is no global map $F$, so it is unclear what an analogue of Theorem 5 would look like.

Now let $k= \mathbb{F}_q$ and let $F$ denote the $q$-power Frobenius.

Theorem 4 When $A$ is abelian and $k = \mathbb{F}_q$, we can find a map $h: J \to G$ such that the composites $G \to J \to G$ and $J \to G \to J$ are $F-1$.

Warning I suspect that I am missing a technical hypothesis, perhaps that $L$ and $\bar{k}$ be disjoint.

Special Cases In an Artin-Schrier extension, $G \cong J$ is the additive group $\mathbb{G}_a$ and the map $G \to J$ is $y \mapsto y^p-y$. If $m | q-1$, then $\mathbb{F}_q$ contains the $m$-th roots of unity. In the Kummer extension, $G \cong J \cong \mathbb{G}_m$, with $G \to J$ being $x \mapsto x^m$ and $J \to G$ being $x \mapsto x^{(q-1)/m}$. So the composite is $x \mapsto x^{q} x^{-1}$.

Generalizing the computation in my other answer shows:

Theorem 5 With the above hypotheses, if $D$ is a divisor supported on $X^{\circ}$, and $\bar{D}$ the divisor which it splits into in $X^{\circ}(\bar{k})$, then $$h \left( \sum_{\pi \in \bar{D}} \pi \right) = \mathrm{Art}(D)$$.

Here the sum $\sum_{\pi \in \bar{D}} \pi$ lives in $J(\bar{k})$, the map $h$ puts it into $G(\bar{k})$, and the assertion is it is equal to $\mathrm{Art}(D)$ under the embedding $A \to G(k)$.

Examples In the Kummer case, Theorem 5 says that $\mathrm{Art}(D) = \prod_{\pi \in \bar{D}} \alpha(\pi)^{(q-1)/m}$. In the Artin-Schrier case, Theorem 6 says that $\mathrm{Art}(D) = \sum_{\pi \in \bar{D}} \beta(\pi)$.

Now, for the result which uses the most technical tools. This is basically Theorem 3.2 in Brian Conrad's notes. By the way, this works for any perfect $k$, not just a finite field.

Theorem 6 If $A$ is abelian, we can take $J$ to be a generalized Jacobian $J(\mathfrak{m}, X)$, for some conductor $\mathfrak{m}$. Moreover, we can take the support of $\mathfrak{m}$ to be the set $S$ of critical points of $Y \to X$, and we can take $X^{\circ} = X \setminus S$. In particular, if $Y \to X$ is unrammified, we can take $J$ to be the Jacobian and $S = \emptyset$.

Let me point out the subtlety of the last sentence of Theorem 6. Let $\mathrm{char}(k) \neq 2$, and let $A = \mathbb{Z}/2$. Then $L = K(\sqrt{\alpha})$ for some $\alpha$. In our high-tech language, we can take $G$ to be the multiplicative group $\mathbb{G}_m$, with $A$ embedded as $\{ \pm 1 \}$. $X^{\circ}$ is the locus where $\alpha$ is nonzero and the map $X \to G/A \cong G$ is given by the function $\alpha$ on $X^{\circ}$. Even if $Y \to X$ is an unbranched cover, the function $\alpha$ will still have zeroes and poles (of even order). We cannot map the entire projective curve $X$ to an affine group like $\mathbb{G}_m$. Theorem 6 is telling us that, by using a very different group, the Jacobian of $X$, we can arrange for the map to $J$ to be defined everywhere on $X$.

Problem 2 There is no global map $F$, so it is unclear what an analogue of Theorem 4 would look like.

Problem 3 Although the ray class groups of $K$ generalize the groups $J(\mathfrak{m}, X)(k)$, there is no algebraic group which generalizes $J(\mathfrak{m}, X)$, so it is not clear what an analogue of Theorem 6 would look like.

Examples In the Kummer case, Theorem 6 says that $\mathrm{Art}(D) = \prod_{\pi \in \bar{D}} \alpha(\pi)^{(q-1)/m}$. In the Artin-Schrier case, Theorem 6 says that $\mathrm{Art}(D) = \sum_{\pi \in \bar{D}} \beta(\pi)$.