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The Abel map of an algebraic curve sends an effective Cartier divisor $D$ to its line bundle ${\mathcal{O}}_C(D)$. Its scheme fibre over a line bundle $L$ is the projective space of one-dimensional subspaces of $H^0(C,L)$. This follows in families: a relative effective Cartier divisor in the fixed class determines a nowhere fibre-zero section of $L$ up to a line bundle from the parameter space, equivalently a line subbundle of $H^0(C,L)\otimes {\mathcal{O}}_T$. Thus the scheme fibre is smooth, not merely reduced on its points. The dimensions can nevertheless jump, so the Abel map of an algebraic curve need not be a smooth morphism.

The Jacobian variety of a smooth projective algebraic curve parametrizes degree-zero line bundles. Its dimension is the genus of the algebraic curve. An effective degree-$d$ Cartier divisor defines a point in a degree-$d$ torsor through the Abel map of an algebraic curve. The theta divisor induces an isomorphism with the dual abelian variety.

Choose a projective embedding of the abelian surface $X$. By Bertini's theorem, a general hyperplane section $C$ is a smooth projective curve that is connected. It is ample, so $C^2>0$. Choose $p\in C$ and translate the inclusion so it sends $p$ to the identity of $X$. The universal property of the Jacobian variety then gives a homomorphism of abelian varietieswhose composition with the based Abel map of an algebraic curve is this translated inclusion.

The image is a closed connected algebraic subgroup, since $J(C)$ is projective, and it contains a translate of $C$. If the image had dimension one, that irreducible curve would be the image subgroup itself. A distinct coset is disjoint from it and has the same numerical divisor class: translation acts trivially on cohomology, being homotopic to the identity on the complex torus. Its self-intersection would therefore be zero, contradicting $C^2>0$. The image has dimension two and is all of $X$. Thus $\Phi$ is surjective, as expressed by the principle that an ample curve on an abelian surface generates the surface.

Write $X={\mathrm{Pic}}^{g-1}(C)$. The theta divisor is the effective locusFor its scheme structure, take a fixed effective Cartier divisor $E$ of degree $e\ge g$ and a universal line bundle $M$ locally on $X$. Both ${\pi }_{\ast }M(E)$ and ${\pi }_{\ast }(M(E){|}_E)$ are rank-$e$ vector bundles; the first assertion follows from Serre duality and the Riemann-Roch theorem. Define $\Theta$ by the determinant of their evaluation map. Its kernel is $H^0(C,M)$, so this is precisely the required locus. Exact sequences for enlarging $E$ cancel an invertible determinant factor, making the determinant description of the theta divisor independent of $E$.

This is a genuine irreducible reduced Cartier divisor. Choose $g-1$ points imposing independent conditions on $H^0(C,K_C)$. For their effective Cartier divisor $D^{\prime }$, the Riemann-Roch theorem gives $h^0(\mathcal{O}(D^{\prime }))=h^1(\mathcal{O}(D^{\prime }))=1$. These points form a nonempty open set in the irreducible symmetric product of a curve $C^{(g-1)}$; the Abel map of an algebraic curve is generically one-to-one there, so its image has dimension $g-1$. At such a class $M$, the first-order change of the evaluation determinant is the cup-product functionalUnder Serre duality its dual is multiplication $H^0(M)\otimes H^0(K_C\otimes M^{-1})\to H^0(K_C)$. The product of two nonzero sections on an integral algebraic curve is nonzero. Thus the functional is nonzero and the determinant has multiplicity one generically. Its irreducible support is consequently the reduced theta divisor.

Now let $L={\mathcal{O}}_C(D)$ have degree $g$. The Riemann-Roch theorem gives $h^0(L)\ge 1$, while evaluation at $P$ givesIf $h^0(L)\ge 2$, this kernel is nonzero for every $P$, so the entire image of ${\varphi }_D$ lies in $\Theta$. If $h^0(L)=1$, write $s$ for a nonzero section. The kernel is nonzero exactly at the zeros of $s$, so the image is not contained in $\Theta$. Moreover $h^1(L)=0$, and the family $L(-P)$ is computed by the rank-one evaluation map $H^0(L)\otimes {\mathcal{O}}_C\to L$. Its determinant is $s$ itself. This proves the scheme-theoretic restriction of the theta divisor to an Abel curve:Here the intersection is regarded as a Cartier divisor on $C$. Since $D$ was specified as a divisor class, it means its unique effective representative, with all multiplicities. The map ${\varphi }_D$ is an embedding: distinct points with the same image would give a degree-one map $C\to {\mathbb{P}}^1$; its cotangent map is evaluation of canonical differentials, which is surjective because $K_C$ is base-point-free. To see the latter, $h^0({\mathcal{O}}_C(P))=1$ for genus at least one, and the Riemann-Roch theorem gives $h^0(K_C-P)=g-1$. Thus the pullback also describes the intersection on the embedded curve.

It remains to count sections on $X$. Let $s_{\Theta }$ be the section defining the theta divisor, and take any $s\in H^0(X,{\mathcal{O}}_X(\Theta ))$. The quotient $r=s/s_{\Theta }$ is a rational function. Choose a fixed degree-$2g-1$ line bundle $T$ and the morphismIt is surjective: for every $M\in X$, the degree-$g$ bundle $T\otimes M^{-1}$ has a nonzero section by the Riemann-Roch theorem. On a general coordinate curve, put $D=T-{\sum }_{j\ne i}{P}_j$. The $g-1$ general points impose independent conditions on the $g$-dimensional space $H^0(C,T)$, leaving $h^0(\mathcal{O}(D))=1$. This independence follows inductively by choosing each new point away from the common zero set of the remaining nonzero sections. The restriction already proved identifies ${\mathcal{O}}_X(\Theta )$ on that coordinate curve with ${\mathcal{O}}_C(D)$, whose section space is one-dimensional. Hence $r\circ \beta$ is constant along every general coordinate curve.

A rational function on a product of integral curves which is constant in each factor is constant: on a common dense open set, successively varying one coordinate identifies its value with the value at a fixed general tuple. Therefore $r\circ \beta \in k$, and surjectivity of $\beta$ gives $r\in k$. Every $s$ is a scalar multiple of $s_{\Theta }$, proving

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.

Put $n=3g-3$. To specify the Abel map of an algebraic curve into $X={\mathrm{Pic}}^{g-1}(C)$, useThe canonical-class twist is necessary: an untwisted sum has degree $3g-3$, rather than $g-1$. Let $L={\mathcal{O}}_C(2K_C)$. The Riemann-Roch theorem and Serre duality give $h^0(L)=n$ and $h^1(L)=0$.

On $C\times C^n$, write ${\Gamma }_i$ for the graph of the $i$th coordinate, $\mathcal{D}={\sum }_i{\Gamma }_i$, and $\pi$ for projection to $C^n$. The locally free sheaf $E_L={\pi }_{\ast }(L{|}_{\mathcal{D}})$ has rank $n$; at repeated points it records values and jets. Evaluation gives a morphism of rank-$n$ vector bundlesIts kernel at a tuple is $H^0(C,2K_C-\sum z_j)$. By the Riemann-Roch theorem, this space has the same dimension as $H^0(C,\sum z_j-K_C)$, since both line bundles have degree $g-1$. Thus its determinant vanishes exactly along ${\alpha }^{\ast }\Theta$. This is also an equality of Cartier divisors with multiplicities: the evaluation complex computes the sheaf cohomology of $L(-\mathcal{D})$; Serre duality identifies its determinant equation, up to an invertible factor, with the defining theta evaluation equation for $K_C\otimes L^{-1}(\mathcal{D})$.

To calculate the determinant line, set ${\mathcal{D}}_i={\Gamma }_1+\cdots +{\Gamma }_i$. Restriction to the extra graph givesProjection is exact on these finite relative divisors. On ${\Gamma }_i\cong C^n$, the first term is ${\mathrm{pr}}_i^{\ast }L\otimes \mathcal{O}(-{\sum }_{j<i}{\Delta }_{ij})$, where ${\Delta }_{ij}=\{z_i=z_j\}$. Taking determinants inductively yields the evaluation determinant on a product of curves:The determinant of the fixed space $H^0(C,L)$ contributes only a constant line. ThereforeThe diagonal subtraction is essential: ordinary evaluation determinants vanish when columns coincide, whereas restriction to a repeated-point Cartier divisor uses jets. The restriction exact sequences account for those factors in every characteristic.

Fix a point $[L]\in {\mathrm{Pic}}^d(C)$ and put $V=H^0(C,L)$. The set of effective Cartier divisors in its divisor class is the complete linear system of a divisor, but identifying points alone would not prove the requested scheme assertion. We identify its functor on arbitrary parameter schemes, including nonreduced ones.

Let $T$ be a $k$-scheme and let $\mathcal{D}\subset C\times T$ be a relative effective degree-$d$ Cartier divisor whose Abel map of an algebraic curve is the constant morphism $[L]$. The definition of the Picard functor of a curve saysfor a line bundle $M$ on $T$. Its canonical section is nonzero on every geometric fibre. Since the constant family has direct image $V{\otimes }_k{\mathcal{O}}_T$, this section determines a line subbundleIt is a subbundle, rather than merely an injective sheaf map: locally some coordinate is nonzero on each residue field, hence a unit on a neighbourhood, so the quotient is locally free. Such a subbundle is exactly a $T$-point of the projective space ${\mathbb{P}}_{\mathrm{lines}}(V)$.

Conversely, its tautological line subbundle evaluates to a section of ${\mathrm{pr}}_C^{\ast }L\otimes {\pi }^{\ast }M$, nonzero on every geometric fibre. On the smooth integral fibres its zero locus is an effective Cartier divisor. The fibrewise regularity criterion gives a relative Cartier divisor flat over $T$, of degree $d$, and its Picard class is constant $[L]$. These constructions are inverse and commute with arbitrary base change. Thus they identify functors, and therefore the scheme fibre itself:If $h^0(C,L)=0$, the fibre is empty, which is smooth as well. The argument remains valid over a point's residue field, so every scheme fibre, not just each closed geometric fibre, is smooth. Fibre dimensions can jump when $h^0(C,L)$ jumps; this does not assert that the entire Abel map of an algebraic curve is a smooth morphism.

A morphism of schemes is smooth when it is locally of finite presentation, flat and has geometrically regular fibres. Smooth fibres alone do not imply that a morphism is smooth: flatness and the dimension behaviour matter as well. The Abel map of an algebraic curve has smooth projective space fibres whose dimensions can jump.

For $T_q(x)=x+q$, the homomorphism associated to a line bundle on an abelian variety gives ${\lambda }_{\Theta }(q)=T_q^{\ast }\mathcal{O}(\Theta )\otimes \mathcal{O}(\Theta {)}^{-1}$. Restrict to the negative Abel map of an algebraic curve. The restriction of the theta divisor to an Abel curve identifies this restriction with the degree-zero class $q$. Pullback of algebraically trivial line bundles does not change when the embedding is translated. Thus pullback $\rho :\widehat{J(C)}\to J(C)$ satisfies $\rho {\lambda }_{\Theta }=1$. Equal dimensions and properness make ${\lambda }_{\Theta }$ an isomorphism, including on schemes, rather than only on closed points.

For a smooth connected projective curve $C$ with chosen basepoint $p$, the Abel map of an algebraic curve $i_p:C\to J(C)$ sends $q$ to $[q-p]$. Every morphism $g:C\to A$ to an abelian variety with $g(p)=0$ factors uniquely through a homomorphism $J(C)\to A$. This universal property converts a curve embedded in an abelian variety into a homomorphism from its Jacobian.