Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 18 (Curves and Abelian Varieties)

Throughout the algebraic questions, $C$ is a smooth projective connected algebraic curve over the algebraically closed field $k$. This is the curve convention needed for the Jacobian variety; for an arbitrary singular or nonproper curve the assertions need not hold. Choose $P_0\in C(k)$ and write $\pi :C\times T\to T$.

The degree-$d$ Picard functor of a curve iswith pullback as its action on morphisms. Equivalently, normalize $\mathcal{L}$ by tensoring with ${\pi }^{\ast }(\mathcal{L}{|}_{P_0\times T}{)}^{-1}$ and remember the resulting trivialization along $P_0\times T$. Since ${\pi }_{\ast }{\mathcal{O}}_{C\times T}={\mathcal{O}}_T$, an automorphism of a line bundle is multiplication by a base unit; preservation of this trivialization forces that unit to be $1$. Thus normalized line bundles have effective descent and the displayed functor is already the appropriate sheaf, rather than just a list of classes over fields.

Here is a construction of its representing scheme. The symmetric product of a curve $C^{(r)}$ represents relative effective degree-$r$ Cartier divisors. It is projective, and its completed local ring at $\sum m_i{Q}_i$ is a power-series ring in the elementary symmetric coefficients of $m_i$ local parameters at each $Q_i$. Thus it is smooth of dimension $r$, in every characteristic. PutThis is open by the semicontinuity theorem for coherent cohomology. It is nonempty: choose $g$ points successively imposing independent vanishing conditions on $H^0(C,K_C)$, so that $H^0(C,K_C-D)=0$, and use Serre duality. The Riemann-Roch theorem gives $h^0({\mathcal{O}}_C(D))=1$ on $U$.

For any fixed line bundle $B$ of degree $g-d$, use a copy $U_B$ of $U$ with the universal normalized line bundle $\mathcal{O}(D)\otimes B^{-1}$. This represents exactly the open subfunctor on which $H^1({\mathcal{L}}_t\otimes B)=0$. Indeed, cohomology and base change for line bundles on a curve makes ${\pi }_{\ast }(\mathcal{L}\otimes B)$ a rank-one vector bundle. The evaluation morphismhas a unique relative zero Cartier divisor of degree $g$ and recovers the point of $U_B$. No choice of a global generator on $T$ is needed. Conversely that zero Cartier divisor recovers $\mathcal{L}$ modulo a bundle from $T$.

These charts cover: given any degree-$d$ line bundle $L$, choose a nonspecial effective degree-$g$ Cartier divisor $D$ and take $B=\mathcal{O}(D)\otimes L^{-1}$. On the overlap of $U_B$ and $U_{B^{\prime }}$, the unique effective representative of $\mathcal{O}(D)\otimes B^{-1}\otimes B^{\prime }$ is constructed by the same rank-one evaluation map. It therefore gives a regular transition morphism; uniqueness gives both the inverse and the cocycle identity. Gluing gives a scheme $J^d$ representing ${\mathcal{P}}^d$, with a universal normalized line bundle.

The construction is of finite type, not an infinite union of indispensable charts. Choose $m\ge 2g-1$ and a fixed line bundle $F$ of degree $m-d$. Every $L\otimes F$ has a nonzero section by the Riemann-Roch theorem, so the universal effective Cartier divisor gives a surjection $C^{(m)}\to J^d$. The inverse images of the charts cover the projective, hence quasi-compact, symmetric product of a curve $C^{(m)}$. A finite subcover gives finitely many charts for $J^d$. This surjection also makes $J^d$ irreducible. The charts show that it is smooth of dimension $g$.

For properness, apply the valuative criterion for properness. Let $R$ be a discrete valuation $k$-algebra with fraction field $K$, and let $L_K$ be a degree-$d$ line bundle on $C_K$. A rational section represents it by a Cartier divisor. Its closure in the regular surface $C_R$ is again a Cartier divisor, since regular local rings are factorial. Thus $L_K$ extends to a line bundle on $C_R$, whose fibre degree remains $d$ by constancy of line bundle degree in a family. Two extensions with the same generic class differ by a vertical Cartier divisor. The special fibre is irreducible, so that Cartier divisor is an integer multiple of the entire fibre, the principal Cartier divisor of a power of a uniformizer. The extensions therefore give the same normalized Picard class. This proves existence and uniqueness in the valuative criterion, hence separatedness and properness of the finite-type scheme $J^d$.

Consequently ${\mathrm{Pic}}_C^d={\mathrm{Jac}}_C^d$ is a smooth proper $k$-variety of dimension $g$. Tensoring by ${\mathcal{O}}_C(-d{P}_0)$ identifies it with $J^0$. On $J^0$ the tensor product and inverse of line bundles induce the algebraic group operations; it is the Jacobian variety, an abelian variety.

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.

Let $P$ now denote the degree-zero torsion line bundle represented by the point of $A[N]$, and put $L_P={\mathcal{O}}_C(2K_C)\otimes P$. Again the Riemann-Roch theorem gives $h^0(L_P)=n$ and $h^1(L_P)=0$. For translation, use ${\Theta }_P=\Theta +P$ as a translated subset. With ordinary $T_P(x)=x+P$, this is $T_{-P}^{\ast }\Theta$; the notation $t_P^{\ast }\Theta =\Theta +P$ therefore uses $t_P=T_{-P}$.

Let $s_P$ and $s_0$ be the evaluation determinants for the two chosen bases, interpreted as sections of $\det E_{L_P}$ and $\det E_L$. The kernel for the first determinant is $H^0(C,2K_C+P-\sum z_j)$. By Serre duality and the Riemann-Roch theorem its dimension is $h^0(C,\sum z_j-K_C-P)$. Thus the determinant vanishes along ${\alpha }^{\ast }{\Theta }_P$, including multiplicities by the same determinant-complex argument as in part (i). The quotient is a rational sectionwith zero-minus-pole Cartier divisor ${\alpha }^{\ast }({\Theta }_P-\Theta )$. The diagonal factors in the two determinants cancel.

Choose a trivialization $u:P^{\otimes N}\stackrel{\sim }{\to }{\mathcal{O}}_C$. This is possible because $P\in A[N]$. Taking the $N$th tensor power of $r$ and using ${⨂}_j{\mathrm{pr}}_j^{\ast }u$ produces a scalar rational function $R$ on $C^n$ withOn $X$, the Theorem of the square and $NP=0$ implyConsequently there is a rational function $f_P\in k(X{)}^{\times }$ with $\mathrm{div}(f_P)=N{\Theta }_P-N\Theta$. The map $\alpha$ is surjective, since every degree-$3g-3$ line bundle has a section by the Riemann-Roch theorem. The two rational functions ${\alpha }^{\ast }{f}_P$ and $R$ have the same Cartier divisor on the smooth projective connected product $C^n$; their quotient is an everywhere invertible regular function and hence a constant. Rescale $f_P$ to remove that constant. Away from the diagonals, ordinary evaluation giveswhere the $N$th power is interpreted with the chosen torsion trivialization. Equality as rational functions then extends from this dense open set to the whole product.

The torsion trivialization in a determinant quotient cannot be omitted if one uses raw meromorphic representatives. Explicitly, write $P={\mathcal{O}}_C(p)$ and choose $h$ with $\mathrm{div}(h)=Np$. If $r_{\mathrm{raw}}$ is the quotient of the evaluation matrices in these meromorphic frames, thenThus the displayed formula suppresses bundle trivializations; read as an unframed quotient of arbitrary meromorphic representatives, it needs precisely this extra product. Changes of bases or rescaling $u$ only change the allowable nonzero scalar normalization of $f_P$.

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.

Let $V=H^0(C,{\Omega }_C^1)$, of complex dimension $g$, and choose a symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(C,\mathbb{Z})$, with $a_i\cdot b_j={\delta }_{ij}$. Integration gives a homomorphismWe first prove both injectivity and discreteness, rather than treating periods as a lattice by definition.

Cut the compact Riemann surface along this basis to a polygon. For closed one-forms $\omega ,\eta$, choose a primitive $F$ of $\omega$ on the cut polygon and apply Stokes theorem to $d(F\eta )=\omega \wedge \eta$. Pairing its boundary edges records the jumps of $F$ across the cuts and gives the Riemann bilinear relations for a compact surface:If a holomorphic differential form $\omega$ has all $a$-periods zero, take $\eta =\overline{\omega }$. The right side is zero, while in a holomorphic coordinate, for $\omega =f(z)dz$,Its integral is strictly positive unless $\omega =0$. Thus the $a$-period map $V\to {\mathbb{C}}^g$ is injective, hence an isomorphism. Choose a basis ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_j}{\omega }_i={\delta }_{ij}$, and put ${\tau }_{ij}={\int }_{b_j}{\omega }_i$. Applying the same boundary identity to ${\omega }_i,{\omega }_j$, whose wedge product is zero, gives ${\tau }_{ij}={\tau }_{ji}$. Applying it to $\omega =\sum c_i{\omega }_i$ and its conjugate givesHence $\mathrm{Im}\tau$ is positive definite.

In the dual coordinates, $I(\sum m_i{a}_i+n_i{b}_i)=m+\tau n$. The real-linear map $(x,y)\mapsto x+\tau y$ from ${\mathbb{R}}^{2g}$ to ${\mathbb{C}}^g$ is an isomorphism: taking imaginary parts of a zero gives $(\mathrm{Im}\tau )y=0$, then $y=0$ and $x=0$. Its image of ${\mathbb{Z}}^{2g}$ is therefore discrete and cocompact, and no nonzero integral cycle has zero image. Thus

The Abel-Jacobi map of a compact Riemann surface sends an effective complex analytic divisor $D=\sum Q_i$ to ${\sum }_i{\int }_{P_0}^{Q_i}({\omega }_1,\dots ,{\omega }_g)$ modulo $\Lambda$. Changing a path adds an integral period, so the map is well-defined. For two degree-$d$ complex analytic divisors, let $E=D-D^{\prime }={\sum }_P{n}_{P}P$, with $\sum n_P=0$, and putWe prove that $E$ is principal exactly when $v\in \Lambda$, including the construction of a meromorphic function.

First construct a differential of the third kind $\eta$ with residues $n_P$ at the support of $E$ and no other poles. For the reduced support $S$, the residue exact sequence isOn sheaf cohomology, the obstruction to prescribing residues belongs to $H^1(C,K_C)\cong \mathbb{C}$ by Serre duality; its pairing with the constant section of ${\mathcal{O}}_C$ is the sum of residues. This follows directly by integrating the local simple-pole terms around the boundary circles. Thus the obstruction is zero because $\sum n_P=0$, and the required $\eta$ exists. Subtract a linear combination of the normalized holomorphic differential forms so that all its $a$-periods are zero.

Take its poles and the integration paths inside the same cut polygon. Apply the boundary calculation above to a primitive $F_i(P)={\int }_{P_0}^P{\omega }_i$, removing small circles around the poles of $\eta$. Since $d(F_i\eta )=0$ away from the poles, the paired outer edges contribute $B_i(\eta )$, while the small circles contribute $2\pi i{\sum }_P{n}_P{F}_i(P)$ by the residue theorem. ThereforeThe same argument without normalizing $\eta$ gives the useful reciprocity identityThese identities use compatible cuts and paths; the final conclusions are unchanged by integral periods.

Suppose $v=m+\tau n$ with $m,n\in {\mathbb{Z}}^g$. DefineIts $a$-periods are $-2\pi i{n}_i$, and its $b$-periods are $2\pi i{m}_i$. Around a pole its integral is $2\pi i{n}_P$. These cycles generate the first homology group of the punctured surface, so every period of ${\eta }^{\prime }$ belongs to $2\pi i\mathbb{Z}$. Choosing an auxiliary point $Q$ away from the poles, setIt is single-valued away from the poles. Locally ${\eta }^{\prime }=n_{P}dt/t+\text{a holomorphic one-form}$, so $f=t^{n_P}$ times a nonvanishing holomorphic function. It extends meromorphically with $\mathrm{div}(f)=E$. This proves sufficiency and explicitly produces the required principal complex analytic divisor.

Conversely, if $E=\mathrm{div}(f)$, the meromorphic differential $df/f$ has residues $n_P$ and every period in $2\pi i\mathbb{Z}$, since each integral counts the winding number of $f$ about zero. Write its $a$-period vector as $2\pi ia$ and its $b$-period vector as $2\pi ib$, with $a,b\in {\mathbb{Z}}^g$. The reciprocity identity gives $v=b-\tau a\in \Lambda$. Since $E$ has degree zero, changing the base point cancels in its Abel sum; a base point at the support of $E$ causes no problem, because the Abel integrals use holomorphic forms. We have provedThis establishes the Abel theorem for divisors directly from the period and residue calculations.