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A hyperplane divisor on projective space is the effective Cartier divisor cut out by one nonzero linear homogeneous polynomial. Its class generates $\mathrm{Cl}({\mathbb{P}}_k^n)\cong \mathbb{Z}$.

The Iitaka dimension of a Cartier divisor $D$ is the maximum dimension of the images of the rational maps given by its nonempty multiples $|mD|$. It is $-\infty$ if all positive multiples have no sections. For an integral $n$-dimensional projective variety it lies between zero and $n$ when nonnegative. The Kodaira dimension of the variety is the special case $D=K_X$.

Successive intersection with Cartier divisors gives a cycle class. On an $n$-dimensional complete scheme, its degree is multilinear and depends only on numerical divisor classes; the top product with one divisor is its self-intersection number.

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.

A Cartier divisor $D$ on an integral projective variety is big exactly when, for every ample divisor $A$, some positive integer $j$ satisfies $jD\sim A+E$ with $E$ effective. To prove the forward direction, subtract an effective high multiple of $A$ using the section subtraction lemma for big divisors, then add an effective representative of the remaining multiple. For the reverse direction, multiply sections of $qA$ by the section of $qE$ and use the ample Hilbert polynomial.

Two Cartier divisors are linearly equivalent when their difference is a principal Cartier divisor. Their associated line bundles are then isomorphic, so their global section spaces have the same dimension. This definition works in every dimension.

A linear system of divisors is a projective family of effective Weil divisors linearly equivalent to a fixed divisor. A complete linear system of a divisor contains every effective representative of that class; a projective subspace gives a subsystem. For a Cartier divisor, the family is obtained from global sections of its associated line bundle. The algebraic curve case recovers the usual moving family of effective divisors.

For a Cartier divisor $D$, the associated line bundle $\mathcal{O}(D)$ has local sections given by meromorphic functions $f$ satisfying $(f)+D\ge 0$. Its canonical meromorphic section has divisor $D$.

An integral Cartier divisor $D$ on a projective scheme is ample if and only if $D^{\dim Y}\cdot Y>0$ for every positive-dimensional integral closed subvariety $Y$. Testing only curves is sufficient in dimension one, but not in higher dimensions.

A line bundle is nef when its degree on every integral complete curve is nonnegative. Equivalently a Cartier divisor $D$ is nef when $D\cdot C\ge 0$ for every such curve. On a projective scheme, adding a positive ample class to a nef class makes it ample; taking limits in the intersection product shows that a nef divisor has nonnegative top self-intersection number.

Write a big real Cartier divisor on an integral projective variety as $D{\sim }_{\mathbb{R}}B+E$, with $B$ ample and $E$ effective real Cartier. Every curve outside $\mathrm{Supp}E$ has $D\cdot C=B\cdot C+E\cdot C>0$. Therefore curves negative against $D$ lie in finitely many support components of codimension one. On a projective scheme use componentwise bigness on a projective scheme and collect the supports on its finitely many integral components; codimension one is taken in the relevant component.

The normalization of an integral algebraic variety is obtained by taking the integral closure of its affine coordinate rings in its function field. Its natural morphism is finite and birational, and its source is normal. For Cartier divisors, bigness under finite normalization compares section growth on the two varieties.

Two Cartier divisors are numerically equivalent when they have equal intersection numbers with every integral complete curve. A numerically trivial divisor has zero intersection with every such curve. Unlike linear equivalence of Cartier divisors, numerical equivalence does not require the same line bundle.

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.

Embed $Y$ in projective space and let $H$ be its hyperplane Cartier divisor. Since $f$ is surjective between integral surfaces, it is generically finite, with degree $r>0$. Its pullback of a Cartier divisor $L=f^{\ast }H$ is a nef divisor, and the projection formula for algebraic cycles gives $L^2=r{H}^2>0$. The same formula gives $L\cdot C=H\cdot f_{\ast }C=0$, because the curve is contracted to a point.

The Hodge index theorem for algebraic surfaces makes the intersection pairing on the Picard group of a surface have signature $(1,\rho -1)$ on $N^1(X{)}_{\mathbb{R}}$. The orthogonal complement of any positive-square vector, including $L$, is negative definite. The numerical class of $C$ is nonzero, since an ample divisor on $X$ intersects every nonzero effective divisor positively. Thus the contracted curve has negative self-intersection conclusion applies:

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.