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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.

An irreducible complex analytic hypersurface is a closed analytic subset of pure complex codimension one which cannot be expressed as the union of two proper closed analytic subsets. Here a local defining function of a complex analytic hypersurface means a reduced local equation: a holomorphic germ generating its vanishing ideal, with each local branch occurring once. The word reduced matters. Merely prescribing the zero set would allow both $f$ and $f^2$, and the requested uniqueness would then be false.

The local facts used are these: in holomorphic coordinates the local ring ${\mathcal{O}}_{X,p}$ is $\mathbb{C}\{z_1,\dots ,z_n\}$, a regular local ring and a unique factorization domain; its units are exactly the germs nonzero at $p$; and the analytic Nullstellensatz identifies the zero-set ideal of a germ $f$ with $\sqrt{(f)}$. A reduced germ is a product of distinct irreducible factors. Unique factorization makes its principal ideal radical: if $f\mid h^r$, every irreducible factor of $f$ divides $h$, so $f\mid h$. Thus two reduced germs with the same hypersurface germ have $(f)=(g)$. Write $g=uf$, $f=vg$. The local ring is a domain, so $uv=1$. Hence $u$ is a holomorphic unit and $u(p)\ne 0$. This proves reduced local defining functions differ by a holomorphic unit. Global irreducibility does not imply local irreducibility at a singular point; the same argument uses the product of all local branches and therefore covers that case too.

A divisor on a complex manifold is a locally finite integer combination of irreducible analytic hypersurfaces. On a compact manifold only finitely many have nonzero coefficients. Choose local meromorphic equations $f_i$ for a divisor $D$, by products of powers of reduced local equations. Their ratios are holomorphic units on overlaps. For the holomorphic line bundle associated to a divisor, choose frames satisfyingThe ratios obey the cocycle identity and so define a holomorphic line bundle $[D]$. This convention makes the local meromorphic sections $f_i{e}_i$ agree, with divisor $D$. Changing the equations by units changes frames and gives an isomorphic bundle.

For hyperplanes $H_1,H_2$ in Complex projective space, choose nonzero linear equations ${\ell }_1,{\ell }_2$. The degree-zero quotient $r={\ell }_1/{\ell }_2$ is a global meromorphic function and has divisor $H_1-H_2$. A principal divisor on a complex manifold gives a trivial associated bundle: using the same global meromorphic equation on every chart makes all frame transitions equal to one. The allowed tensor-product rule now givesIf the hyperplanes coincide the same conclusion is immediate. Equivalently, their common bundle is the hyperplane bundle $\mathcal{O}(1)$.

Locally at the centre of a blowup of a complex manifold at a point, replace a coordinate neighbourhood in ${\mathbb{C}}^n$ byAway from zero the direction is uniquely $[z]$, so this map is an isomorphism there; glue it to $X\setminus \{p\}$. In the chart ${\ell }_j\ne 0$, write $z_j=u$ and $z_i=u{t}_i$ for $i\ne j$. These are smooth holomorphic coordinates on the blowup. The exceptional divisor is $u=0$, globally $E={\sigma }^{-1}(p)\cong \mathbb{P}(T_{p}X)$.

The canonical bundle formula for a point blowup isThe local chart also explains the exponent: pulling back a coordinate volume form gives $\pm u^{n-1}du\wedge {\bigwedge }_{i\ne j}d{t}_i$, so the Jacobian section has divisor $(n-1)E$.

Since $Y$ is smooth at the centre, choose coordinates with $Y=\{z_1=0\}$. Its strict transform is a genuine analytic hypersurface. In a blowup chart $j\ne 1$, it is the closed set $t_1=0$: off $E$, the original equation $u{t}_1=0$ reduces to $t_1=0$, and taking the closure restores the points with $u=0$. In chart $j=1$, there are no such off-exceptional points, and the strict transform is empty there. These descriptions agree on overlaps and show analyticity, with $\tilde{Y}\cap E\cong {\mathbb{CP}}^{n-2}$. The total transform has the extra factor $u$ exactly once, soUnder the given anticanonical assumption, this yieldsTo prove the only-if direction, one must rule out an accidental trivial power of $[E]$. The local blowup is the tautological line bundle over ${\mathbb{CP}}^{n-1}$, with $E$ its zero section. Its normal bundle is therefore ${\mathcal{O}}_E(-1)$. The normal bundle of a smooth analytic hypersurface identifies $[E]{|}_E$ with this bundle. If $n\ge 2$, restrict to a projective line inside $E$; the restriction of $[E{]}^{\otimes (2-n)}$ has degree $n-2$. A trivial line bundle has degree zero, so it is trivial only if $n=2$. Conversely the displayed tensor factor is literally trivial when $n=2$. This is the anticanonical strict transform under a point blowup criterion:For completeness, dimension one cannot satisfy the premise in this compact setting: an irreducible analytic hypersurface is one point, and $[-Y]$ has degree $-1$ on its curve component, whereas $K_X$ has degree $2g-2$. Thus no exceptional dimension-one case invalidates the equivalence.

A holomorphic line bundle is a complex rank-one bundle locally trivialized over holomorphic coordinate opens, with nowhere-zero holomorphic transition functions. A local holomorphic section has a holomorphic coefficient in a holomorphic local frame. Let $h$ be a smooth positive Hermitian metric on its fibres. A Chern connection is a connection compatible with $h$ whose $(0,1)$ part is the bundle's Dolbeault operator.

Fix the Hermitian-product convention linear in the first variable and a nonvanishing holomorphic local frame $e$. Put $h_e=h(e,e)>0$ and write $\nabla e=Ae$. Compatibility with the holomorphic structure forces $A^{0,1}=0$. Metric compatibility givesso taking its $(1,0)$ part determines $A$ uniquely:Define the connection by this formula. To check it exists globally, change frame to $e^{\prime }=ge$ for a nonzero holomorphic function $g$. Then $h_{e^{\prime }}=|g{|}^2{h}_e$ andwhich is exactly the connection transformation law obtained from $\nabla (ge)=dge+g\nabla e$. Hence the local connections glue. The formula and its conjugate give metric compatibility, and $A^{0,1}=0$ gives the required holomorphic structure. Since either condition already forced this unique formula in every frame, existence and uniqueness are proved.

For a line bundle, the curvature form of a connection is $F={\nabla }^2=dA$, as the scalar one-form satisfies $A\wedge A=0$. The local formula for the Chern connection on a line bundle therefore givesThis formula uses a nowhere-zero local holomorphic section; arbitrary sections with zeros are not frames there. It has type $(1,1)$, is $d$-closed and satisfies ${\bar{F}}_h=-F_h$, so $i{F}_h$ is a real $(1,1)$-form. The transformation calculation makes its frame independence explicit.

For two metrics, their ratio $r=h_1(e,e)/h_2(e,e)$ is a global positive smooth function, since the common frame factors cancel. The curvature difference of two Chern connections isEach curvature is $\bar{\partial }$-closed, and the difference has a global $(1,0)$ primitive for $\bar{\partial }$. ThusThis is equality in Dolbeault cohomology, rather than merely equality in de Rham cohomology; the displayed primitive proves precisely that distinction.

Finally let $s$ be a nowhere-zero smooth section of $\hat{L}$. In this global smooth trivialization write ${\bar{\partial }}_{\hat{L}}s=a\otimes s$ with a global $(0,1)$-form $a$. The holomorphic structure has ${\bar{\partial }}_{\hat{L}}^2=0$. Its Leibniz rule givessince a scalar one-form wedges with itself to zero. The hypothesis $H^{0,1}(X)=0$ makes $a=\bar{\partial }f$ for a global smooth complex function $f$. ThenThe exponential never vanishes. Therefore $e^{-f}s$ is a nowhere-zero holomorphic section, proving smoothly trivial holomorphic line bundles when H01 vanishes constructively, without a compactness assumption.

Here the degree-zero Picard group of a curve uses tensor product as its group operation. The map is the Abel map of a pointed smooth projective curve. If $\alpha (p)=\alpha (q)$, then $\mathcal{O}(p-q)$ is trivial. By the divisor description of line bundles, there is a nonzero rational function $f$ with principal divisor $(f)=p-q$.

If $p\ne q$, this function has exactly one pole, of order one. It defines a nonconstant morphism of algebraic varieties $X\to {\mathbb{P}}^1$: locally at a pole one uses $1/f$ as the target coordinate. The degree of a morphism of curves is the total pole degree, so it is one. The map is finite, since a nonconstant map of projective integral curves has finite fibers and is proper, and a degree-one finite map to the normal projective line is an isomorphism. Algebraically, on a target affine chart the finite coordinate algebra lies inside the common function field and is integral over the integrally closed coordinate ring, so it equals that ring. Thus $X\cong {\mathbb{P}}^1$ and its genus is zero, contradicting $g\ge 1$. ConsequentlyThis spells out why a principal divisor with one simple zero and one simple pole cannot occur on a curve of positive genus.

If $g=0$, the chosen point $p_0$ is enough to identify $X$ with ${\mathbb{P}}^1$, even over a nonclosed field. By the Riemann-Roch theorem, $h^0(\mathcal{O}(p_0))=2$, since the complementary canonical twist has degree $-3$ and no nonzero section. Besides constants there is a rational function with its sole pole a simple pole at $p_0$, giving the same degree-one isomorphism. On ${\mathbb{P}}^1$, every degree-zero divisor is principal: over an algebraically closed field write a product of powers of linear forms for its finite point coefficients, with the degree-zero condition giving the coefficient at infinity. Over a general field the same construction uses irreducible polynomial factors for closed points, weighted by their degrees. Therefore

Let $L$ be any degree-zero line bundle on the genus-one smooth projective curve. The twist $M=L\otimes \mathcal{O}(p_0)$ has degree one. The Riemann-Roch theorem givesThe second line bundle has degree $2g-2-1=-1$, so it has no nonzero section: the zero divisor of such a section would be effective of negative degree. Hence $h^0(M)=1$. A nonzero section has an effective zero divisor of degree one, which is a single point $p$. It is a $k$-rational point even over a general field, since a degree-one effective divisor has residue degree one. Thus $M\cong \mathcal{O}(p)$ and $L\cong \mathcal{O}(p-p_0)=\alpha (p)$.

This proves surjectivity; part (a) gives injectivity. Hence $\alpha$ is a bijection, as in the Abel-Jacobi map of a genus-one curve. DefineTensor product of line bundles is associative and commutative, so this is an abelian group law on $X(k)$. Its identity is $p_0$, since $\alpha (p_0)={\mathcal{O}}_X$. Equivalently, $p\oplus q$ is the unique point $r$ with $\mathcal{O}(r-p_0)\cong \mathcal{O}(p+q-2p_0)$.

First use the algebraically closed field convention. Every line bundle on a smooth projective curve has a nonzero rational section of a line bundle: trivialize it on a nonempty open set and take its generic generator. The orders of its zeros and poles give a finite divisor $D$, and the local trivializations identify the line bundle with $\mathcal{O}(D)$. For a degree-zero line bundle writesince every closed point is a $k$-point. ThereforeOnly finitely many factors occur, and negative tensor powers mean dual line bundles. This provesFor genus zero the group is trivial, so the conclusion remains valid.

For a general field, a closed point need not be rational, and its contribution to degree is weighted by its residue degree. Consequently rational points need not generate the degree-zero Picard group of a curve. Here is an explicit counterexample to that stronger reading. Let $C$ be the smooth projective curve over ${\mathbb{F}}_2$ obtained by completing the affine equationThe polynomial is irreducible over ${\overline{\mathbb{F}}}_2(x)$: a function $r^2+r$ has even pole order at infinity when it has a pole there, while the right side has pole order five. Its affine model is smooth because the partial derivative with respect to $y$ is one. The normalized projective completion is therefore a geometrically integral smooth projective curve.

There is exactly one point $p_{\infty }$ above infinity, and it is rational. Indeed the map $x:C\to {\mathbb{P}}^1$ has degree two. At a point above infinity put $v(x)=-e$, where the ramification index $e$ is at most two. The equation forces $v(y)<0$ and $2v(y)=5v(x)=-5e$. Hence $e=2$ and $v(y)=-5$. The degree formula then leaves exactly one point of residue degree one. Thus $x$ and $y$ have pole orders two and five there. There are no affine rational points: for either $x\in {\mathbb{F}}_2$, the right side is one, while $y^2+y=0$ for every $y\in {\mathbb{F}}_2$. Hence $C({\mathbb{F}}_2)=\{p_{\infty }\}$ and the image of its Abel map of a pointed smooth projective curve is trivial.

The equations $y=0$, $x^2+x+1=0$ define a degree-two closed point $D$, since $x^5+x+1$ vanishes modulo $x^2+x+1$. The divisor $D-2p_{\infty }$ has degree zero but is not principal. To prove this, a function with poles bounded by $2p_{\infty }$ is regular on the affine curve and hence has a unique expression $A(x)+B(x)y$. The possible pole orders of its two terms are $2\mathrm{deg}A$ and $2\mathrm{deg}B+5$, of opposite parity, so they cannot cancel. The pole bound forces $B=0$ and $\mathrm{deg}A\le 1$. But no nonzero $a+bx$ over ${\mathbb{F}}_2$ vanishes at $D$, whose residue class of $x$ has irreducible quadratic minimal polynomial. Thus no function has principal divisor $D-2p_{\infty }$, proving its class is nonzero. This establishes the rational points need not generate the degree-zero Picard group qualification with a complete obstruction, while the generation proof above supplies the intended algebraically closed-field answer.

Take the projective line $X={\mathbb{P}}_k^1$ and the coherent sheaves $\mathcal{F}={\mathcal{O}}_X$ and $\mathcal{G}={\mathcal{O}}_X(-1)$, where the latter is a twisting sheaf on projective space. Both are line bundles, so each stalk is a free rank-one ${\mathcal{O}}_{X,x}$-module and therefore ${\mathcal{F}}_x\simeq {\mathcal{G}}_x$ at every point.

They are nevertheless not isomorphic. On the standard charts with coordinates $t$ and $t^{-1}$, choose frames $e_0,e_1$ of $\mathcal{O}(-1)$ with $e_1=t^{-1}{e}_0$. A global section would give polynomials $a(t)$ and $b(t^{-1})$ satisfyingThe left side has only nonnegative powers of $t$, and the right side only strictly negative powers. Thus both vanish, giving $\Gamma (X,\mathcal{O}(-1))=0$. For the structure sheaf, the corresponding equality is $a(t)=b(t^{-1})$, whose solutions are precisely the constants. HenceA morphism of sheaves which is an isomorphism on all stalks would be an isomorphism; this example only asserts the existence of separate stalk isomorphisms, not that they come from one morphism of sheaves.

The Appell–Humbert theorem identifies the positive Hermitian form $H_1$ with a holomorphic line bundle whose First Chern class is represented, in our linear-first convention, byIts integral over the square fundamental domain is $1$, so $\mathrm{deg}L=1$. In particular the oriented Chern pairing is $-E_1(1,i)=1$; using $E_1(1,i)$ without adjusting the Hermitian convention would give the wrong sign.

The complex torus has a nowhere-vanishing differential $dz$, so its canonical divisor is trivial and its genus is one. A nonzero section of a line bundle of negative degree would have an effective divisor of negative degree, which is impossible. Thus $H^0(X,L^{-1})=0$. The Riemann-Roch theorem and Serre duality giveTherefore $\boxed{h^0(X,L)=1}$. The choice of semicharacter of a complex lattice changes the degree-zero component of the line bundle, but not this calculation.

Pulling back the factors of automorphy in the Appell–Humbert theorem by $z\mapsto cz$, where $c=a+bi$, pulls back both the Hermitian form and the semicharacter of a complex lattice. Thus the explicit answer isWriting $\lambda =m+ni$ makes the second formula completely explicit:The second equality is an equality of signs, obtained by reducing the exponents modulo $2$. The Hermitian form follows from $H_1(cz,cw)=|c{|}^2{H}_1(z,w)$. Its integer parameter equals the norm of the Gaussian integer, agreeing with the degree of this nonzero isogeny of elliptic curves. The formulas also include $c=0$, when the pulled-back line bundle is trivial.

For $M\in {\mathrm{Pic}}^0(B)$, translations preserve its line bundle class, so ${\varphi }_M=0$. Equivalently, the kernel of $L\mapsto {\varphi }_L$ is the Identity component of the Picard group. Applying the preceding Poincaré line bundle formula to $M$ yieldssince the normalized Poincaré line bundle is trivial on $B\times \{0\}$. Pullback and tensor products of sheaves also give $D_{L\otimes M}=D_L\otimes D_M$. Thus $D_L$ depends only on $[L]\in \boxed{\mathrm{NS}(B)}$, the Néron-Severi group.

Let $m:B\times B\to B$ be addition, and let $p_1,p_2$ be the projections. Consider the line bundleIts restriction to $B\times \{y\}$ has Picard group class $[T_y^{\ast }L\otimes L^{-1}]={\varphi }_L(y)$; the factor $L_y^{-1}$ is a constant line and has trivial class. Its restriction to $\{0\}\times B$ is also trivial. Choose a trivialization of the fiber $L_0$ and use it to normalize along the zero sections. The defining universal property of the normalized Poincaré line bundle then identifiesMore explicitly, both sides have the same restrictions to all fibers of the second projection, and both are normalized on $\{0\}\times B$; the Seesaw theorem makes their quotient trivial. Here ${\varphi }_L:B\to \hat{B}$ is the homomorphism associated to a line bundle on an abelian variety, and $\hat{B}$ is the dual abelian variety. Pull back by $(f,g)$ to obtain the required identityThe choice of fiber trivialization is immaterial to this identity in the Picard group.

The bundle is on $A$, so the printed, undefined target $\mathrm{Pic}(X)$ must be read as the Picard group $\mathrm{Pic}(A)$. Write this Picard group additively and set $q(h)=[h^{\ast }L]$. The zero morphism pulls back $L$ to a constant one-dimensional vector space tensored with ${\mathcal{O}}_A$, so $q(0)=0$. The Theorem of the Cube, pulled back along $(f,g,h):A\to B^3$, givesThis is the bilinear cross-effect of a line bundle on an abelian variety: set $d(f,g)=q(f+g)-q(f)-q(g)$. Substitution into this identity givesThe definition gives $d(f,g)=d(g,f)$, so additivity holds in both arguments. Additivity includes negative multiples and the zero morphism. Hence $D_L$ is a symmetric bilinear map with values in $\mathrm{Pic}(A)$. The equalities refer to isomorphism classes of line bundles; without chosen rigidifications they are not asserted to be specified canonical isomorphisms of bundles.

For a group homomorphism $h:A\to B$, the dual morphism $\hat{h}:\hat{B}\to \hat{A}$ is pullback of algebraically trivial line bundles. Translations satisfy $h\circ T_x=T_{h(x)}\circ h$, and thereforeAlso ${\varphi }_{L\otimes M}={\varphi }_L+{\varphi }_M$ and ${\varphi }_{L^{-1}}=-{\varphi }_L$. Dualizing is additive: $\widehat{f+g}=\hat{f}+\hat{g}$, since algebraically trivial line bundles obey the Theorem of the square. Apply these identities to the three factors defining $D_L$. The two diagonal terms cancel in the expansionHence $\boxed{{\varphi }_{D_L(f,g)}=\hat{f}\circ {\varphi }_L\circ g+\hat{g}\circ {\varphi }_L\circ f}$.