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There is a duality between degree two coverings and two-torsion points on the Jacobian - i.e. both form elementary abelian 2-groups, and these groups are naturally dual

This is the Artin-Milne Poincaré duality pairing in fppf cohomology (Corollary 4.9 of Duality in the Flat Cohomology of Curves)

$$ H^1(C_{\overline{k}}, \mathbb Z/2) \times H^1 (C_{\overline{k}},\mu_2) \to H^2(C_{\overline{k}}, \mu_2) \to \mathbb Z/2 $$ with $H^1(C_{\overline{k}}, \mathbb Z/2)$ classifying degree $2$ coverings and $H^1 (C_{\overline{k}},\mu_2)$ classifying 2-torsion points of $J$.

Concretely, we can express this by noting that each covering of $C$ comes from a unique covering $A \to J$ of the same degree, and setting the pairing of this covering with a point in $J[2]$ to be zero if and only if that point lies in the image of $A[2]$.

More generally, for $C$ in characteristic $p$ and a $\mathbb Z/p$-covering of $C$ (and thus an isogeny $0 \to \mathbb Z/p \to A \to J\to 0$), we define the pairing with a $p$-torsion point by taking any lift of that point, multiplying by $p$, and obtaining an element of the kernel $\mathbb Z/p$. This doesn't depend on our choice of preimage because two lifts differ by an element of $\mathbb Z/p$ which is killed by multiplication by $p$.

This is linear in the $A[p]$ variable since every step is compatible with multiplication by $p$. This is linear in the covering variable since the addition map on coverings is fiber product followed by adding the kernels, and the product of two lifts is a lift to the fiber product.

To check this pairing has no kernel in the covering variable, the main point is to check that $A[p]$ and $J[p]$ have the same number of elements (since the rank is the multiplicity of the slope 0 of the Newton polygon, which is an isogeny invariant). Since $A \to J$ is a degree $p$ finite etale covering, its kernel contains a single $p$-torsion point, so the image of $A[p] \to J[p]$ has codimension $1$, forcing the pairing to be nontrivial.

To check it has no kernel in the other variable, factor the multiplication by $p$ map $A \to A$ (which kills all the $p$-torsion) into a totally inseperable morphism (which preserves points and thus $p$-torsion and an etale morphism (which therefore must kill all the $p$-torsion). The etale morphism is a fiber product of $\mathbb Z/p$-coverings, so if a point was in the image of $p$-torsion on each $\mathbb Z/p$-covering it would have to be inteh image of $p$-torsion on this covering, forcing it to be trivial.

You're correct on your interpretation of Artin-Schreier theory. (I think you meant unipotent instead of unitary). I don't know how to connect that perspective to the duality with the Jacobian - in particular, I don't think the Poincaré duality pairing above plays nicely with the Artin-Schreier map $\mathbb Z/2 \to \mathcal O_C$.

There is a duality between degree 2 coverings and two-torsion points on the Jacobian — i.e. both form elementary abelian 2-groups, and these groups are naturally dual.

This is the Artin–Milne Poincaré duality pairing in fppf cohomology (Corollary 4.9 of Duality in the Flat Cohomology of Curves)

$$ H^1(C_{\overline{k}}, \mathbb Z/2) \times H^1 (C_{\overline{k}},\mu_2) \to H^2(C_{\overline{k}}, \mu_2) \to \mathbb Z/2 $$ with $H^1(C_{\overline{k}}, \mathbb Z/2)$ classifying degree 2 coverings and $H^1 (C_{\overline{k}},\mu_2)$ classifying 2-torsion points of $J$.

Concretely, we can express this by noting that each covering of $C$ comes from a unique covering $A \to J$ of the same degree, and setting the pairing of this covering with a point in $J[2]$ to be zero if and only if that point lies in the image of $A[2]$.

More generally, for $C$ in characteristic $p$ and a $\mathbb Z/p$-covering of $C$ (and thus an isogeny $0 \to \mathbb Z/p \to A \to J\to 0$), we define the pairing with a $p$-torsion point by taking any lift of that point, multiplying by $p$, and obtaining an element of the kernel $\mathbb Z/p$. This doesn't depend on our choice of preimage because two lifts differ by an element of $\mathbb Z/p$ which is killed by multiplication by $p$.

This is linear in the $A[p]$ variable since every step is compatible with multiplication by $p$. This is linear in the covering variable since the addition map on coverings is fiber product followed by adding the kernels, and the product of two lifts is a lift to the fiber product.

To check this pairing has no kernel in the covering variable, the main point is to check that $A[p]$ and $J[p]$ have the same number of elements (since the rank is the multiplicity of the slope 0 of the Newton polygon, which is an isogeny invariant). Since $A \to J$ is a degree $p$ finite étale covering, its kernel contains a single $p$-torsion point, so the image of $A[p] \to J[p]$ has codimension $1$, forcing the pairing to be nontrivial.

To check it has no kernel in the other variable, factor the multiplication by $p$ map $A \to A$ (which kills all the $p$-torsion) into a totally inseparable morphism (which preserves points and thus $p$-torsion and an étale morphism (which therefore must kill all the $p$-torsion). The étale morphism is a fiber product of $\mathbb Z/p$-coverings, so if a point was in the image of $p$-torsion on each $\mathbb Z/p$-covering it would have to be in the image of $p$-torsion on this covering, forcing it to be trivial.

You're correct on your interpretation of Artin–Schreier theory. (I think you meant unipotent instead of unitary.) I don't know how to connect that perspective to the duality with the Jacobian — in particular, I don't think the Poincaré duality pairing above plays nicely with the Artin–Schreier map $\mathbb Z/2 \to \mathcal O_C$.

There is a duality between degree two coverings and two-torsion points on the Jacobian - i.e. both form elementary abelian 2-groups, and these groups are naturally dual

Morally, this is a Poincaré duality pairing

$$ H^1(C_{\overline{k}}, \mathbb Z/2) \times H^1 (C_{\overline{k}},\mu_2) \to H^2(C_{\overline{k}}, \mu_2) \to \mathbb Z/2 $$ with $H^1(C_{\overline{k}}, \mathbb Z/2)$ classifying degree $2$ coverings and $H^1 (C_{\overline{k}},\mu_2)$ classifying 2-torsion points of $J$.

Concretely, we can express this by noting that each covering of $C$ comes from a unique covering $A \to J$ of the same degree, and setting the pairing of this covering with a point in $J[2]$ to be zero if and only if that point lies in the image of $A[2]$.

More generally, for $C$ in characteristic $p$ and a $\mathbb Z/p$-covering of $C$ (and thus an isogeny $0 \to \mathbb Z/p \to A \to J\to 0$), we define the pairing with a $p$-torsion point by taking any lift of that point, multiplying by $p$, and obtaining an element of the kernel $\mathbb Z/p$. This doesn't depend on our choice of preimage because two lifts differ by an element of $\mathbb Z/p$ which is killed by multiplication by $p$.

This is linear in the $A[p]$ variable since every step is compatible with multiplication by $p$. This is linear in the covering variable since the addition map on coverings is fiber product followed by adding the kernels, and the product of two lifts is a lift to the fiber product.

To check this pairing has no kernel in the covering variable, the main point is to check that $A[p]$ and $J[p]$ have the same number of elements (since the rank is the multiplicity of the slope 0 of the Newton polygon, which is an isogeny invariant). Since $A \to J$ is a degree $p$ finite etale covering, its kernel contains a single $p$-torsion point, so the image of $A[p] \to J[p]$ has codimension $1$, forcing the pairing to be nontrivial.

To check it has no kernel in the other variable, factor the multiplication by $p$ map $A \to A$ (which kills all the $p$-torsion) into a totally inseperable morphism (which preserves points and thus $p$-torsion and an etale morphism (which therefore must kill all the $p$-torsion). The etale morphism is a fiber product of $\mathbb Z/p$-coverings, so if a point was in the image of $p$-torsion on each $\mathbb Z/p$-covering it would have to be inteh image of $p$-torsion on this covering, forcing it to be trivial.

You're correct on your interpretation of Artin-Schreier theory. (I think you meant unipotent instead of unitary). I don't know how to connect that perspective to the duality with the Jacobian - in particular, I don't think the Poincaré duality pairing above plays nicely with the Artin-Schreier map $\mathbb Z/2 \to \mathcal O_C$.

There is a duality between degree two coverings and two-torsion points on the Jacobian - i.e. both form elementary abelian 2-groups, and these groups are naturally dual

This is the Artin-Milne Poincaré duality pairing in fppf cohomology (Corollary 4.9 of Duality in the Flat Cohomology of Curves)

$$ H^1(C_{\overline{k}}, \mathbb Z/2) \times H^1 (C_{\overline{k}},\mu_2) \to H^2(C_{\overline{k}}, \mu_2) \to \mathbb Z/2 $$ with $H^1(C_{\overline{k}}, \mathbb Z/2)$ classifying degree $2$ coverings and $H^1 (C_{\overline{k}},\mu_2)$ classifying 2-torsion points of $J$.

Concretely, we can express this by noting that each covering of $C$ comes from a unique covering $A \to J$ of the same degree, and setting the pairing of this covering with a point in $J[2]$ to be zero if and only if that point lies in the image of $A[2]$.

More generally, for $C$ in characteristic $p$ and a $\mathbb Z/p$-covering of $C$ (and thus an isogeny $0 \to \mathbb Z/p \to A \to J\to 0$), we define the pairing with a $p$-torsion point by taking any lift of that point, multiplying by $p$, and obtaining an element of the kernel $\mathbb Z/p$. This doesn't depend on our choice of preimage because two lifts differ by an element of $\mathbb Z/p$ which is killed by multiplication by $p$.

This is linear in the $A[p]$ variable since every step is compatible with multiplication by $p$. This is linear in the covering variable since the addition map on coverings is fiber product followed by adding the kernels, and the product of two lifts is a lift to the fiber product.

To check this pairing has no kernel in the covering variable, the main point is to check that $A[p]$ and $J[p]$ have the same number of elements (since the rank is the multiplicity of the slope 0 of the Newton polygon, which is an isogeny invariant). Since $A \to J$ is a degree $p$ finite etale covering, its kernel contains a single $p$-torsion point, so the image of $A[p] \to J[p]$ has codimension $1$, forcing the pairing to be nontrivial.

To check it has no kernel in the other variable, factor the multiplication by $p$ map $A \to A$ (which kills all the $p$-torsion) into a totally inseperable morphism (which preserves points and thus $p$-torsion and an etale morphism (which therefore must kill all the $p$-torsion). The etale morphism is a fiber product of $\mathbb Z/p$-coverings, so if a point was in the image of $p$-torsion on each $\mathbb Z/p$-covering it would have to be inteh image of $p$-torsion on this covering, forcing it to be trivial.

You're correct on your interpretation of Artin-Schreier theory. (I think you meant unipotent instead of unitary). I don't know how to connect that perspective to the duality with the Jacobian - in particular, I don't think the Poincaré duality pairing above plays nicely with the Artin-Schreier map $\mathbb Z/2 \to \mathcal O_C$.

There is a duality between degree two coverings and two-torsion points on the Jacobian - i.e. both form elementary abelian 2-groups, and these groups are naturally dual

Cohomologically, this is the Poincaré duality pairing

$$ H^1(C_{\overline{k}}, \mathbb Z/2) \times H^1 (C_{\overline{k}},\mu_2) \to H^2(C_{\overline{k}}, \mu_2) \to \mathbb Z/2 $$ with $H^1(C_{\overline{k}}, \mathbb Z/2)$ classifying degree $2$ coverings and $H^1 (C_{\overline{k}},\mu_2)$ classifying 2-torsion points of $J$.

Concretely, we can express this by noting that each covering of $C$ comes from a unique covering $A \to J$ of the same degree, and setting the pairing of this covering with a point in $J[2]$ to be zero if and only if that point lies in the image of $A[2]$. To get that this is actually a bilinear pairing, the main point is to check that $A[2]$ and $J[2]$ have the same number of elements.

You're correct on your interpretation of Artin-Schreier theory. (I think you meant unipotent instead of unitary). I don't know how to connect that perspective to the duality with the Jacobian - in particular, I don't think the Poincaré duality pairing above plays nicely with the Artin-Schreier map $\mathbb Z/2 \to \mathcal O_C$.

There is a duality between degree two coverings and two-torsion points on the Jacobian - i.e. both form elementary abelian 2-groups, and these groups are naturally dual

Morally, this is a Poincaré duality pairing

$$ H^1(C_{\overline{k}}, \mathbb Z/2) \times H^1 (C_{\overline{k}},\mu_2) \to H^2(C_{\overline{k}}, \mu_2) \to \mathbb Z/2 $$ with $H^1(C_{\overline{k}}, \mathbb Z/2)$ classifying degree $2$ coverings and $H^1 (C_{\overline{k}},\mu_2)$ classifying 2-torsion points of $J$.

Concretely, we can express this by noting that each covering of $C$ comes from a unique covering $A \to J$ of the same degree, and setting the pairing of this covering with a point in $J[2]$ to be zero if and only if that point lies in the image of $A[2]$.

More generally, for $C$ in characteristic $p$ and a $\mathbb Z/p$-covering of $C$ (and thus an isogeny $0 \to \mathbb Z/p \to A \to J\to 0$), we define the pairing with a $p$-torsion point by taking any lift of that point, multiplying by $p$, and obtaining an element of the kernel $\mathbb Z/p$. This doesn't depend on our choice of preimage because two lifts differ by an element of $\mathbb Z/p$ which is killed by multiplication by $p$.

This is linear in the $A[p]$ variable since every step is compatible with multiplication by $p$. This is linear in the covering variable since the addition map on coverings is fiber product followed by adding the kernels, and the product of two lifts is a lift to the fiber product.

To check this pairing has no kernel in the covering variable, the main point is to check that $A[p]$ and $J[p]$ have the same number of elements (since the rank is the multiplicity of the slope 0 of the Newton polygon, which is an isogeny invariant). Since $A \to J$ is a degree $p$ finite etale covering, its kernel contains a single $p$-torsion point, so the image of $A[p] \to J[p]$ has codimension $1$, forcing the pairing to be nontrivial.

To check it has no kernel in the other variable, factor the multiplication by $p$ map $A \to A$ (which kills all the $p$-torsion) into a totally inseperable morphism (which preserves points and thus $p$-torsion and an etale morphism (which therefore must kill all the $p$-torsion). The etale morphism is a fiber product of $\mathbb Z/p$-coverings, so if a point was in the image of $p$-torsion on each $\mathbb Z/p$-covering it would have to be inteh image of $p$-torsion on this covering, forcing it to be trivial.

You're correct on your interpretation of Artin-Schreier theory. (I think you meant unipotent instead of unitary). I don't know how to connect that perspective to the duality with the Jacobian - in particular, I don't think the Poincaré duality pairing above plays nicely with the Artin-Schreier map $\mathbb Z/2 \to \mathcal O_C$.