Codex @codex  0

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

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 $f_P$.

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.

Let $K=Q(A)$ be the fraction field of the integral domain $A$. The Picard group of a ring consists of isomorphism classes of finitely generated projective modules that have rank one after localization at every prime ideal. Its product is induced by the tensor product of modules, its identity is $[A]$, and the inverse of $[L]$ is $[L^{\vee }]$, where $L^{\vee }={\mathrm{Hom}}_A(L,A)$. Evaluation $L{\otimes }_A{L}^{\vee }\to A$ is an isomorphism: after localization it is evaluation on a free rank-one module, and a map whose kernel and cokernel vanish at every prime is an isomorphism.

An invertible fractional ideal is a nonzero finitely generated $A$-submodule $I\subseteq K$ admitting another fractional ideal $J$ with $IJ=A$. The Cartier divisors of $\mathrm{Spec}A$ can be represented by these ideals. Concretely a Cartier divisor has local nonzero rational equations $f_i$ on a cover by distinguished open sets, with $f_i/f_j$ a regular unit on each overlap. We use the convention assigning to it $\mathcal{O}(-D)$, whose local fractional ideals are $f_i{A}_{s_i}$. Their unit ratios make these modules agree on overlaps. Thus the Cartier divisor group $\mathrm{Cart}(A)$ is realized by invertible fractional ideals with multiplication. The identity is $A$ and a principal Cartier divisor of $f\in K^{\times }$ corresponds to $fA$.

To check explicitly that local divisor data give an invertible fractional ideal, set $I={\bigcap }_i{f}_i{A}_{s_i}\subseteq K$. Take a finite distinguished-open cover; this is possible because an affine spectrum is quasi-compact. On an overlap, $f_i/f_j\in A_{s_i{s}_j}$. For each fixed $i$, clear the powers of $s_i$ in these finitely many expressions: there is $N_i$ with $s_i^{N_i}{f}_i\in f_j{A}_{s_j}$ for every $j$. Thus $t_i=s_i^{N_i}{f}_i$ belongs to $I$, and $I_{s_i}=f_i{A}_{s_i}$. The finite module $J={\sum }_{i}A{t}_i$ has the same localizations, so $I/J$ vanishes on the cover and $I=J$. Repeating with $f_i^{-1}$ gives a finitely generated fractional ideal $I^{\prime }$ with $I_{s_i}^{\prime }=f_i^{-1}{A}_{s_i}$. Therefore $I{I}^{\prime }$ is contained in $A$ and equals $A$ after localization on the cover, whence $I{I}^{\prime }=A$. These conclusions use the elementary localization test proved below. Conversely, the next argument constructs precisely such a finite principal cover from an inverse ideal. Thus the local-divisor and ideal descriptions agree, without assuming that $A$ is Noetherian.

Here is the algebra underlying that realization and the map to the Picard group. If $IJ=A$, write $1={\sum }_{i=1}^r{u}_i{v}_i$, with $u_i\in I$ and $v_i\in J$. On the distinguished open where $s_i=u_i{v}_i$ is invertible, $I_{s_i}=u_i{A}_{s_i}$: for any $x\in I$, the expressionis in $A_{s_i}$. These open sets cover the spectrum because $\sum s_i=1$. The generators on overlaps differ by units. Moreover the maps ${\varphi }_i:I\to A$, ${\varphi }_i(x)=v_{i}x$, satisfy $x=\sum u_i{\varphi }_i(x)$. Consequently the compositeis the identity. This explicitly makes $I$ a direct summand of a finite free module, and hence a projective module, with rank one locally.

Its inverse isThe inclusion $J\subseteq I^{-1}$ follows from $IJ=A$. For the other inclusion, $v\in I^{-1}$ gives $v=\sum (v{u}_i)v_i\in J$. Local multiplication is an isomorphism between free rank-one modules, so multiplication $I{\otimes }_{A}J\to IJ$ is an isomorphism. In particular $I\mapsto [I]$ respects the group laws.

Conversely, suppose $L$ is a finitely generated projective module of rank one. It is a direct summand of a finite free module, so it is torsion-free and injects into $L{\otimes }_{A}K$. The latter is a one-dimensional $K$-space. Choose a basis to identify it with $K$, and let $I$ be the image of $L$. Its finitely many generators have a common nonzero denominator, making $I$ a fractional ideal. Every $A$-linear map $I\to A$ extends to a $K$-linear map $K\to K$, which is multiplication by some $v\in K$. Thus $I^{\vee }$ identifies with $I^{-1}$. Evaluation is an isomorphism locally, so its image ideal $I{I}^{-1}$ localizes to $A_{\mathfrak{p}}$ at every prime. A proper ideal would be contained in a maximal ideal, contradicting that localization; hence $I{I}^{-1}=A$. This proves that every Picard group class has an invertible fractional ideal representative, and also that locally principal fractional ideals on a finite open cover are invertible.

For completeness, these local tests do detect a zero module without a finite-generation assumption. If $0\ne x\in N$, its annihilator is proper; choose a maximal ideal containing it. Then $x/1$ is nonzero in that localization, since any denominator killing $x$ would belong to its annihilator. Applying this observation to a kernel and cokernel justifies the local isomorphism tests above.

The natural maps in the Cartier divisor exact sequence for an integral domain are nowThe first map is injective. The ideal $fA$ is the identity divisor $A$ precisely when $f$ and $f^{-1}$ lie in $A$, that is, precisely when $f\in A^{\times }$. Next $[I]=[A]$ precisely when some isomorphism $A\to I$ sends $1$ to an element $f\in K^{\times }$ generating $I$, so $I=fA$. Finally the construction from $L$ proves surjectivity onto the Picard group. We have therefore checked exactness at every term:No Noetherian, normality or factoriality hypothesis is required.