Return to Answer

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, we would only get a well defined conjugacy class in $G$.)

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 trivialize when restricted to any fiber of $\mathrm{Sym}^N(X) \to \mathrm{Pic}^N(X)$. So, we restricting to the fiber through $[D]$ and $[E]$, the principal bundle $\psi$ is trivial, and Frobenius must act by permuting the geometric components. In particular, Frobenius acts the same way on $\psi^{-1}([D])$ and $\psi^{-1}([E])$.

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 get a natural splitting sending $d \in \mathbb{Z}$ to $[d x_{\infty}]$, and can thereby write $\mathrm{Pic}(X)(k)$ as $\mathbb{Z} \times \mathrm{Pic}^0(X)(k)$, and we can ask what covering of $X$ corresponds to the quotient $\mathrm{Pic}(X)(k) \to \mathrm{Pic}^0(X)(k)$. (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.)

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.)

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 $\psi^{-1}([E])$.

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.)

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 $\#(\fp)$-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:

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:

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 simply connected! So the $G$-principal bundle $\psi : Z^N \to \mathrm{Sym}^N(X)$ must trivialize when restricted to any fiber of $\mathrm{Sym}^N(X) \to \mathrm{Pic}^N(X)$. So, we restricting to the fiber through $[D]$ and $[E]$, the principal bundle $\psi$ is trivial, and Frobenius must act by permuting the geometric components. In particular, Frobenius acts the same way on $\psi^{-1}([D])$ and $\psi^{-1}([E])$.

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 split, since $\mathbb{Z}$ is projective in the category of abelian groups.) However, if $X$ has a $k$-point $x_{\infty}$, then we get a natural splitting sending $d \in \mathbb{Z}$ to $[d x_{\infty}]$, and can thereby write $\mathrm{Pic}(X)(k)$ as $\mathbb{Z} \times \mathrm{Pic}^0(X)(k)$, and we can ask what covering of $X$ corresponds to the quotient $\mathrm{Pic}(X)(k) \to \mathrm{Pic}^0(X)(k)$. (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.)

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 trivialize when restricted to any fiber of $\mathrm{Sym}^N(X) \to \mathrm{Pic}^N(X)$. So, we restricting to the fiber through $[D]$ and $[E]$, the principal bundle $\psi$ is trivial, and Frobenius must act by permuting the geometric components. In particular, Frobenius acts the same way on $\psi^{-1}([D])$ and $\psi^{-1}([E])$.

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 get a natural splitting sending $d \in \mathbb{Z}$ to $[d x_{\infty}]$, and can thereby write $\mathrm{Pic}(X)(k)$ as $\mathbb{Z} \times \mathrm{Pic}^0(X)(k)$, and we can ask what covering of $X$ corresponds to the quotient $\mathrm{Pic}(X)(k) \to \mathrm{Pic}^0(X)(k)$. (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.)