= Solution
We regard $N$ as positive and invertible in $k$; $N=1$ gives the trivial group and the same arguments apply. Put $A=\operatorname{Pic}^0(C)$, 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 $N\operatorname{id}$, 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 $L=L_0\otimes[-1]^*L_0$, which is symmetric and ample. The <Theorem of the Cube> gives the recurrence
$$
[n+1]^*L\otimes[n-1]^*L\cong([n]^*L)^{\otimes2}\otimes L^{\otimes2}.
$$
After fixing a trivialization at the origin, induction from $n=0,1$ gives $[n]^*L\cong L^{\otimes n^2}$. Comparing top <intersection products> under the finite map $[N]$ gives
$$
\deg[N]\,(c_1(L)^g)=(c_1([N]^*L)^g)=N^{2g}(c_1(L)^g).
$$
The intersection number is positive because $L$ is ample, so $\deg[N]=N^{2g}$. 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> proves
$$
\boxed{A[N]\cong(\mathbb Z/N\mathbb Z)^{2g}.}
$$
The same reasoning works for the <dual abelian variety> $\widehat A$, 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 $\widehat A$. Identify $X=\operatorname{Pic}^{g-1}(C)$ 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>, makes
$$
\lambda(q)=T_q^*\mathcal O_X(\Theta)\otimes\mathcal O_X(\Theta)^{-1}
$$
a <group homomorphism> $A\to\widehat A$. Fix the negative <Abel map of an algebraic curve> $j:C\to A$, $P\mapsto[P_0-P]$, and let $\rho:\widehat A\to\operatorname{Pic}^0(C)=A$ 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 $T_q\phi_D=\phi_{D+q}$, the <restriction of the theta divisor to an Abel curve> proved above gives
$$
\phi_D^*\lambda(q)\cong\mathcal O_C(D+q)\otimes\mathcal O_C(D)^{-1}\cong q.
$$
Each $\phi_D$ differs from $j$ by a translation. Thus $\rho\lambda=\operatorname{id}_A$. 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 $g=\dim\widehat A$, so it is all of the reduced irreducible <abelian variety> $\widehat A$. 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 $M\in\widehat A[N]$, the <Theorem of the Cube> gives $[N]^*M\cong M^{\otimes N}\cong\mathcal O_A$. Choose a trivialization of $[N]^*M$. The finite étale cover $[N]:A\to A$ has deck group $G=A[N]$. Its natural descent action on $[N]^*M$ 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 character
$$
\chi_M:G\longrightarrow k^\times,
$$
whose 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 $\chi_{M\otimes M'}=\chi_M\chi_{M'}$. Define
$$
\boxed{e_N(P,Q)=\chi_{\lambda(Q)}(P)\in\mu_N.}
$$
This 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. Thus
$$
\widehat A[N]\longrightarrow\operatorname{Hom}(A[N],\mu_N),\qquad M\longmapsto\chi_M
$$
is 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 \b[the pairing on $A[N]$ is perfect].
Back to article page
