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Identify ${\mathrm{Pic}}^{g-1}(X)$ with the Jacobian variety using the base point and the translation of Solution 4. A line bundle $L$ of degree $g-1$ corresponds to the point $z=u(D)-\kappa$ for any divisor representation $L\simeq {\mathcal{O}}_X(D)$. The Riemann singularity theorem isIn particular, the theta divisor is singular exactly at the degree-$g-1$ line bundles with at least two independent sections. We prove the equality, including the upper bound on multiplicity.

Choose a fixed effective divisor $A$ of degree $N\ge g$, and a local holomorphic family $L_{\eta }$ representing nearby points of ${\mathrm{Pic}}^{g-1}(X)$. Such a family can be constructed locally from the holomorphic exponential sequence, by varying line-bundle transition functions through exponentials of Čech cohomology representatives. By Serre duality, $H^1(L_{\eta }(A))$ is dual to $H^0(K_X\otimes L_{\eta }^{-1}(-A))$, whose degree is $g-1-N<0$, so it vanishes. The Riemann-Roch theorem gives $h^0(L_{\eta }(A))=N$. By cohomology and base change for line bundles on a curve, these spaces form a holomorphic vector bundle $E$ of rank $N$. The spaces $H^0(L_{\eta }(A){|}_A)$ form another holomorphic vector bundle $F$ of rank $N$. Evaluation gives a holomorphic square matrix $T(\eta )$ locally, with exact sequenceConsequently $\det T(\eta )=0$ exactly when $L_{\eta }$ has a nonzero section, namely on the support of the theta divisor.

At the point $L$, put $r=h^0(L)=h^1(L)$, the equality being Riemann-Roch theorem. An invertible minor of size $N-r$ persists near this point. Holomorphic row and column operations, or the Schur complement, givewhere $M$ is an $r\times r$ holomorphic matrix. This is the Schur complement presentation of a theta singularity. Its linear term in a tangent direction $\xi \in H^1({\mathcal{O}}_X)$ is the cup product mapTo see this concretely, vary transition functions $g_{ij}$ of $L$ to $g_{ij}(1+ϵ{\xi }_{ij})$. A section $s_i$ lifts to $s_i+ϵs_i^{\prime }$ precisely when its first-order compatibility equation can be solved; the obstruction is the cocycle ${\xi }_{ij}{s}_j$. This is $\xi ⌣s$. Passing to the kernel and cokernel of $T(0)$ gives exactly the derivative of the reduced matrix $M$, proving the assertion without assuming a multiplicity formula.

Before extracting the order of $\det M$, we verify that $\det T$ is a reduced equation of the theta divisor. Generic points of $W_{g-1}$ have $h^0(L)=1$ by Solution 3. At such a point, both $H^0(L)$ and $H^0(K_X\otimes L^{-1})$ are one-dimensional. Their nonzero sections $s,t$ have nonzero product $st\in H^0(K_X)$. Choose $\xi \in H^1({\mathcal{O}}_X)$ with $⟨\xi ,st⟩\ne 0$ under Serre duality. The scalar $d{M}_0(\xi )$ is nonzero, so $\det T$ has a simple zero generically on the irreducible theta divisor. Since its support is exactly that irreducible hypersurface, it defines the reduced divisor. Solution 4 independently proved the same reduced structure for the Riemann theta function. On the smooth ambient Jacobian variety, local equations of the same reduced divisor differ by a nonvanishing holomorphic unit. Hence the orders of the local theta equation and $\det M$ agree everywhere.

Every entry of $M$ vanishes at the point, so each term in its determinant has order at least $r$. This proves ${\mathrm{mult}}_z\Theta \ge r$. To obtain equality, we exhibit an invertible cup-product direction for a line bundle on a curve. Set $V=H^0(L)$ and $W=H^0(K_X\otimes L^{-1})$, both of dimension $r$. There are $r$ distinct points $p_1,\dots ,p_r$ at which evaluation is an isomorphism for $V$: choose them inductively, since a surviving nonzero section has only finitely many zeros. The same holds for $W$. The nonvanishing of each evaluation determinant defines a nonempty open subset of the irreducible variety $X^r$. Their intersection, also avoiding the diagonals, is therefore nonempty. Choose points in it and local coordinates and compatible trivializations of $L$ and $K_X\otimes L^{-1}$ at each point.

Let $\xi$ be the connecting homomorphism class of the meromorphic principal parts ${\lambda }_i/t_i$ at these points, with all ${\lambda }_i\ne 0$. The residue form of Serre duality givesIf $S$ and $U$ are the two invertible evaluation matrices, the matrix of this pairing is $S^t\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_r)U$, and is invertible. Thus the cup product map $\xi ⌣-:H^0(L)\to H^1(L)$ is an isomorphism. Along the corresponding analytic parameter line,with nonzero leading coefficient. The order is at most $r$, completing the proof of the Riemann singularity theorem.

This argument also identifies the tangent cone to a theta divisor: its degree-$r$ equation is $\det (\xi ⌣-)$, which is not the zero polynomial. For $r=1$, the tangent hyperplane is $⟨\xi ,st⟩=0$, in agreement with the annihilator description of the derivative of the Abelian sum map. For $r\ge 2$ all first derivatives vanish, proving singularity. In genus $1$ the theta divisor is a single smooth point of multiplicity $1$. In genus $0$ the effective degree-$-1$ locus is empty, so there is no singularity statement at a point of it.

Assume $X$ is a connected compact Riemann surface of geometric genus $g\ge 1$. Work first in ${\mathrm{Pic}}^{g-1}(X)$, translating back to the Jacobian variety by tensoring with ${\mathcal{O}}_X(-(g-1)p_0)$ when desired. PutThis is the image of the Abelian sum map from $X^{(g-1)}$. The Riemann singularity theorem asserts that this image is a reduced Cartier divisor and thatHere multiplicity means the order of a local defining equation in the maximal ideal of the smooth ambient Jacobian variety. Consequently the singular points are precisely the classes with at least two independent sections. We will prove the multiplicity equality, including the nonvanishing of its proposed leading term.

Choose an effective divisor $E$ of degree $n\ge g$, and a local holomorphic family $\mathcal{L}$ of degree-$g-1$ line bundles about $L$ on $X\times U$, where $U$ is a small neighborhood in ${\mathrm{Pic}}^{g-1}(X)$. Such a family is obtained by varying the Čech cocycles of a line bundle. Since $\mathrm{deg}{L}^{\prime }(E)>2g-2$ for all $L^{\prime }\in U$, Serre duality gives $H^1(L^{\prime }(E))=0$; the Riemann-Roch theorem gives $h^0(L^{\prime }(E))=n$. The cohomology and base change for line bundles on a curve then identifiesas vector bundles of rank $n$. Evaluation defines a matrix $M:A\to B$. For each $L^{\prime }$, its exact sequence isThus $f=\det M$ vanishes exactly on $\Theta$.

The function $f$ is not identically zero. To exhibit a degree-$g-1$ line bundle with no section, choose $g$ distinct points $p_1,\dots ,p_g$ giving independent evaluation functionals on $H^0(K_X)$. Such points can be chosen successively because a nonzero holomorphic differential form cannot vanish everywhere. For $F=\sum p_i$, the Riemann-Roch theorem gives $h^0({\mathcal{O}}_X(F))=1$. Choose $q$ outside $F$. The unique section of ${\mathcal{O}}_X(F)$ does not vanish at $q$, so ${\mathcal{O}}_X(F-q)$ has no section. Therefore the evaluation determinant genuinely cuts out a hypersurface. Its support is irreducible, since it is the image of the irreducible symmetric product of a curve $X^{(g-1)}$.

Fix now $L\in \Theta$, and put $r=h^0(L)=h^1(L)$, the equality following from the Riemann-Roch theorem. The matrix $M(L)$ has an invertible $(n-r)$ by $(n-r)$ block. After invertible holomorphic row and column operations, its local form iswhere $N$ is $r$ by $r$. More explicitly, eliminate the invertible block by its Schur complement; the discarded determinant is a holomorphic unit. Hence $f$ is a unit times $\det N$. Every entry of $N$ lies in the maximal ideal, so its order is at least $r$.

The linear term of $N$ is the cup-product mapTo verify this, deform the transition functions as $g_{ij}(1+\epsilon {\xi }_{ij})$ and try to lift a section $s$ to $s+\epsilon s^{\prime }$. Its first-order gluing equation has right side ${\xi }_{ij}{s}_j$, whose Čech cohomology class is exactly the obstruction to the lift. The kernel-to-cokernel derivative of the evaluation matrix gives that same obstruction. Under Serre duality, if $s_1,\dots ,s_r$ and $t_1,\dots ,t_r$ are bases of $H^0(L)$ and $H^0(K_X\otimes L^{-1})$, thenWe must show that this matrix is invertible for some $\xi$; otherwise the lower bound on multiplicity would not be an equality.

For each of these two $r$-dimensional section spaces, there are $r$ distinct points with independent evaluations. Choose them successively, using a section in the remaining kernel to find the next point. The set of such $r$-tuples is a nonempty open subset of $X^r$. The two open sets intersect because $X^r$ is irreducible, and we may avoid their diagonals. Choose $q_1,\dots ,q_r$ in that intersection and trivialize $L$ and $K_X$ there. Both matricesare invertible. Let $\xi \in H^0(K_X{)}^{\ast }=H^1({\mathcal{O}}_X)$ be the sum of the evaluation functionals at these points, with any nonzero weights $c_k$. Thenis invertible. It follows that the degree-$r$ polynomial $\det N_1$ is nonzero, so ${\mathrm{ord}}_{L}f=r$. This proves the needed invertible cup-product direction for a line bundle on a curve by an explicit construction.

We still need to identify this determinant hypersurface with the reduced image, rather than a multiple of it. Choose $g-1$ distinct points imposing independent conditions on $H^0(K_X)$. For their sum $D$, the Riemann-Roch theorem gives $h^0(D)=h^0(K_X-D)=1$. At its class, the calculation above gives order one. The determinant hypersurface has irreducible support, so its sole divisor on an algebraic curve coefficient is one everywhere. A local hypersurface in a smooth space with these divisor on an algebraic curve coefficients is reduced. Thus it is exactly the reduced image $W_{g-1}$, and the multiplicity formula holds for that image itself.

The argument gives the useful additional descriptionFor $r=1$, its nonzero linear equation is $⟨\xi ,st⟩=0$. This agrees with the derivative of the Abelian sum map: if $\mathrm{div}(s)=D$, then $st$ is the unique holomorphic differential form vanishing along $D$. For $r\ge 2$, the local equation has no linear term and the point is singular. When $g=1$, $W_0$ is the single trivial line bundle in ${\mathrm{Pic}}^0(X)$, with multiplicity one; the argument includes this case using the empty effective divisor. In genus zero there is no $W_{g-1}$ of effective divisors on an algebraic curve, so that hypothesis is necessary.

Here $k$ is algebraically closed and $C$ is a connected smooth projective curve. For a scheme $T$ over $k$, let $\pi :C_T=C{\times }_{k}T\to T$. The degree-$d$ Picard functor of a curve assignsOne can equivalently take the associated faithfully flat sheaf. It is essential to divide out by line bundles from the base: a parameter space should remember the family on $C$, not an independent base line bundle.

Choose $p_0\in C(k)$. A class can be normalized as $L\otimes {\pi }^{\ast }(L{|}_{p_0\times T}{)}^{-1}$, with its canonical trivialization along $p_0\times T$. Since ${\pi }_{\ast }{\mathcal{O}}_{C_T}={\mathcal{O}}_T$, an automorphism of a line bundle is multiplication by a base unit, and preserving this normalization forces that unit to be $1$. Descent of line bundles is effective, so normalized bundles satisfy the faithfully flat sheaf condition with no automorphism ambiguity. Thus this normalized description works on families, including nonreduced bases, rather than only on $k$-points.

We construct a representing scheme by Picard charts from nonspecial divisors. Fix $N\ge 2g$ and put $r=N-g$. The symmetric product of a curve $C^{(g)}$ represents relative effective Cartier divisors on a curve of degree $g$ and has a universal divisor $\Delta$. This uses the scheme symmetric product, not the naive set of unordered $T$-points. Locally on a smooth curve, a divisor of length $m$ is described by a monic polynomial in a smooth coordinate; its coefficients are elementary symmetric functions, so this construction includes coincident points in any characteristic. These local descriptions give the usual effective-divisor interpretation of the symmetric product.

Let $W\subset C^{(g)}$ be the open set where $H^1(C,\mathcal{O}(E))=0$. It is open by semicontinuity theorem for coherent cohomology. The Riemann-Roch theorem gives $h^0(\mathcal{O}(E))=1$ there, so $E$ is the unique effective divisor in its line-bundle class. For every fixed effective divisor $B$ of degree $r$, take a copy $W_B$ of $W$, with familyof degree $N$. Normalize it along $p_0$ as above.

This chart represents the open subfunctor defined byIndeed, for a family satisfying that condition, cohomology and base change for line bundles on a curve makes $V={\pi }_{\ast }L(-B)$ a line bundle on $T$, with formation commuting with base change. The rank is one by Riemann-Roch theorem. The evaluation mapis a nonzero section on each fibre, up to the base twist. Its zero scheme is a relative effective Cartier divisor on a curve $E$ of degree $g$. To check the family assertion, source and target are flat over $T$ and the fibre maps are injections of line bundles on an integral curve, so their cokernel is flat over $T$. Their zeros are Cartier, and the zero scheme is proper with finite fibres of length $g$, hence finite flat of degree $g$. It therefore defines a unique map $T\to W_B$. Conversely this construction recovers the family $\mathcal{O}(B+E)$ modulo a base line bundle.

The base-change step controls infinitesimal families as well. Locally, a finite free complex computes the two cohomology groups. Vanishing of its degree-one cokernel on fibres makes the last differential surjective by the Nakayama lemma. Splitting that differential leaves a locally free degree-zero kernel of rank one and commutes with base change. This is the standard finite-complex form of the permitted cohomology semicontinuity and base-change results.

These chart subfunctors cover all degree-$N$ families. For a geometric-fibre line bundle $L$ of degree $N$, the preceding solution gives $H^1(L)=0$ and $h^0(L)=N+1-g=r+1$. Choose $r$ points successively so that each imposes one independent condition on the current section space. At each stage that space is nonzero; a nonzero section has finitely many zeros, so a point outside its zero set can be chosen, also outside the finitely many previously chosen points. For their sum $B$,and Riemann-Roch theorem forces $h^1(L(-B))=0$. The points may be chosen in $C(k)$ even when the fibre field extends $k$, since $C(k)$ is infinite and a nonzero section on the base-changed curve has only finitely many zeros. Thus fixed divisors $B$ defined over $k$ suffice. Upper semicontinuity gives an open neighbourhood on the base for each successful $B$.

It remains to glue the charts explicitly. On $W_B$, the locusis open. The line bundle inside the braces has degree $g$; its unique section up to scale has an effective zero divisor $E^{\prime }$. Base change and the evaluation construction make $E\mapsto E^{\prime }$ a morphism $W_{B{B}^{\prime }}\to W_{B^{\prime }}$. It satisfies $\mathcal{O}(B+E)\cong \mathcal{O}(B^{\prime }+E^{\prime })$. Reversing $B,B^{\prime }$ gives its inverse. On triple overlaps the transition maps compose correctly because the residual effective divisor is unique. Hence the copies $W_B$ glue along these open isomorphisms to a scheme $P^N$.

The normalized bundles $L_B$ glue as well: on overlaps their normalized isomorphisms exist and are unique, so they satisfy the cocycle condition. Let ${\mathcal{P}}_N$ denote the resulting universal normalized line bundle on $C\times P^N$. For any $T$, the open cover where the conditions for some $B$ hold gives compatible maps $T\to W_B$, and thus a unique map $T\to P^N$. Pulling back ${\mathcal{P}}_N$ reverses this procedure. We have established a natural bijectionThis proves representability of the full functor, not just a parametrization of individual line bundles. The nonspecial-divisor construction is also recorded in Stacks Project, Picard scheme of a curve.

Finally tensoring by ${\mathcal{O}}_C((N-d)p_0)$ is a natural isomorphism ${\mathrm{Pic}}_C^d\to {\mathrm{Pic}}_C^N$ for every integer $d$. Transporting the universal bundle back constructs its degree-$d$ representative $P^d$, usually denoted ${\mathrm{Pic}}_{C/k}^d$. Thus every degree component is represented. These identifications depend on the chosen point and twist; intrinsically the degree-zero component is a group scheme and each degree component is its torsor.

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.

A degree-$g$ line bundle with vanishing first sheaf cohomology has exactly one section up to scale, by Riemann-Roch theorem. Its zero divisor is unique and varies in families by cohomology and base change for line bundles on a curve. For a fixed divisor $B$ of degree $N-g$, the open subset of $C^{(g)}$ where $\mathcal{O}(E)$ has vanishing first cohomology gives a chart in degree $N$, by $E\mapsto \mathcal{O}(B+E)$. Different charts are glued by the unique residual divisor of $\mathcal{O}(B+E-B^{\prime })$ wherever that line bundle is again nonspecial. Normalizing the universal bundles along a fixed rational point gives their canonical compatible gluing and a representing scheme for the Picard functor of a curve.