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We regard $N$ as positive and invertible in $k$; $N=1$ gives the trivial group and the same arguments apply. Put $A={\mathrm{Pic}}^0(C)$, an abelian variety of dimension $g$.

First compute the prime-to-characteristic torsion of an abelian variety. The differential of $[N]:A\to A$ at the origin is $N\mathrm{id}$, and translation gives the same assertion everywhere. Consequently $[N]$ is an étale morphism, its fibres have dimension zero, and properness makes it finite. Its image is closed of dimension $g$, hence all of $A$. Choose an ample line bundle $L_0$ on $A$ and put $L=L_0\otimes [-1{]}^{\ast }{L}_0$, which is symmetric and ample. The Theorem of the Cube gives the recurrenceAfter fixing a trivialization at the origin, induction from $n=0,1$ gives $[n{]}^{\ast }L\cong L^{\otimes n^2}$. Comparing top intersection products under the finite map $[N]$ givesThe intersection number is positive because $L$ is ample, so $\mathrm{deg}[N]=N^{2g}$. Since $[N]$ is étale and $k$ is algebraically closed field, its kernel has precisely $N^{2g}$ points.

For a prime power ${\ell }^r\mid N$, write the finite abelian group $A[{\ell }^r]$ as a product of cyclic groups of orders ${\ell }^{a_j}$, with $1\le a_j\le r$. Its subgroup killed by $\ell$ is $A[\ell ]$, so the number of factors is $2g$. Its total order is ${\ell }^{2gr}$, so $\sum a_j=2gr$ forces every $a_j=r$. Combining the prime-power factors by the Chinese remainder theorem provesThe same reasoning works for the dual abelian variety $\hat{A}$, also of dimension $g$.

To define the Weil pairing for an abelian variety on two copies of $A[N]$, we first justify the required theta identification with $\hat{A}$. Identify $X={\mathrm{Pic}}^{g-1}(C)$ with an $A$-torsor and use ordinary translation $T_q(x)=x+q$. The Theorem of the square, a consequence of the Theorem of the Cube, makesa group homomorphism $A\to \hat{A}$. Fix the negative Abel map of an algebraic curve $j:C\to A$, $P\mapsto [P_0-P]$, and let $\rho :\hat{A}\to {\mathrm{Pic}}^0(C)=A$ be pullback by $j$. Pullback by a translate of $j$ gives the same map: algebraically trivial line bundles on an abelian variety are translation-invariant as Picard classes, by the Theorem of the square.

For any $q\in A$, choose a degree-$g$ class $D$ such that both $D$ and $D+q$ have one-dimensional section spaces. This is possible because the nonspecial degree-$g$ classes form a nonempty open subset of the irreducible Picard scheme of a curve, and its translate has a nonempty intersection with it. Since $T_q{\varphi }_D={\varphi }_{D+q}$, the restriction of the theta divisor to an Abel curve proved above givesEach ${\varphi }_D$ differs from $j$ by a translation. Thus $\rho \lambda ={\mathrm{id}}_A$. This equality on closed points is equality of morphisms, since $A$ is reduced and the target is separated. In particular $\lambda$ is a closed immersion, being a section of the separated morphism $\rho$. Its image has dimension $g=\dim \hat{A}$, so it is all of the reduced irreducible abelian variety $\hat{A}$. Therefore $\lambda$ is an isomorphism of schemes. This proves the needed theta autoduality of a Jacobian, including possible infinitesimal issues in positive characteristic.

Here is a definition of the Weil pairing for an abelian variety that also proves nondegeneracy. For $M\in \hat{A}[N]$, the Theorem of the Cube gives $[N{]}^{\ast }M\cong M^{\otimes N}\cong {\mathcal{O}}_A$. Choose a trivialization of $[N{]}^{\ast }M$. The finite étale cover $[N]:A\to A$ has deck group $G=A[N]$. Its natural descent action on $[N{]}^{\ast }M$ becomes multiplication by scalars under this trivialization: these scalars are constant because every invertible regular function on the proper connected abelian variety $A$ is constant. They form a characterwhose values lie in ${\mu }_N$, since $NP=0$ for every $P\in G$. Rescaling the trivialization changes none of these scalars. Translation gives multiplicativity in $P$, and tensor product gives ${\chi }_{M\otimes M^{\prime }}={\chi }_M{\chi }_{M^{\prime }}$. DefineThis fixes a convention for the Weil pairing for an abelian variety; reversing the convention for the descent action inverts all values and changes no nondegeneracy assertion.

If ${\chi }_M=1$, the trivialization is equivariant for the deck group. Descent then makes $M$ the trivial line bundle on the target. Thusis injective. The source and target both have $N^{2g}$ elements, since $k$ contains all $N$ distinct roots of unity and $A[N]\cong (\mathbb{Z}/N{)}^{2g}$. Hence it is an isomorphism. Characters of this group separate its points, so the Weil pairing for an abelian variety is nondegenerate in both variables. Composing with the isomorphism $\lambda$ proves the pairing on $A[N]$ is perfect.

For an abelian variety and an integer $N$ invertible in the base field, a torsion line bundle $M\in \hat{A}[N]$ has trivial pullback by $[N]$. Choose a trivialization. The descent action of the kernel $A[N]$ on that trivial line bundle acts through a character ${\chi }_M:A[N]\to {\mu }_N$. Define $e_N(P,M)={\chi }_M(P)$. This is independent of rescaling the trivialization and bilinear under translation and tensor product. A trivial character descends the trivial bundle, so $M$ is trivial. Since both groups have order $N^{2\dim A}$, the character map is an isomorphism and the Weil pairing for an abelian variety is perfect. The theta autoduality of a Jacobian turns it into a pairing on two copies of the Jacobian variety's torsion group.