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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.

Use the dense standard affine charts. Their product iswhich is also a dense affine chart of ${\mathbb{P}}^{n+m}$. The identification of these charts gives a birational isomorphism between the two projective varieties.

To distinguish them, calculate their divisor class groups explicitly. Let $H\subset {\mathbb{P}}^r$ be a coordinate hyperplane. Its complement is ${\mathbb{A}}^r$, whose coordinate ring is a unique factorization domain. Any Weil divisor on ${\mathbb{P}}^r$ restricts on this chart to a principal Weil divisor; subtract that principal Weil divisor on ${\mathbb{P}}^r$, leaving a Weil divisor supported on $H$. Thus $[H]$ generates $\mathrm{Cl}({\mathbb{P}}^r)$. If $aH=\mathrm{div}(q)$, then $q$ has zero Weil divisor on ${\mathbb{A}}^r$. A reduced fraction in a unique factorization domain with zero orders at every prime Weil divisor must be a unit, so $q\in k^{\ast }$. It follows that $a=0$. We have proved

In the product, let $H_1$ and $H_2$ be coordinate hyperplanes in the factors and putThe complement of $D_1\cup D_2$ is ${\mathbb{A}}^{n+m}$, again with a unique factorization domain as its coordinate ring. The same restriction-and-subtraction argument says that $[D_1],[D_2]$ generate the divisor class group. Ifthen on that affine chart $q$ has neither zeros nor poles along any prime Weil divisor. Unique factorization makes $q$ a unit of $k[s_1,\dots ,s_n,t_1,\dots ,t_m]$, hence a constant. It follows that $a=b=0$. ThereforeBoth spaces are smooth varieties, so their divisor class groups also equal their Picard groups. An isomorphism preserves the divisor class group; ${\mathbb{Z}}^2$ and $\mathbb{Z}$ are not isomorphic. Thus the spaces are birational but not isomorphic. The assumptions $n,m>0$ ensure that both boundary prime Weil divisors actually occur.

An elliptic curve works over every algebraically closed field, including one algebraic over a finite field. Give it explicitly in ${\mathbb{P}}^2$ byIn the first case the only point at infinity is $O=[0:1:0]$, where the derivative with respect to $z$ is nonzero. In the affine chart $z=1$, a singular point of an algebraic variety would have $y=0$, $x^3-x=0$, and $3x^2-1=0$. The possible roots $x=0,1,-1$ cannot satisfy the last equation in characteristic different from $2$: at $0$ it is $-1$, and at $\pm 1$ it is $2$. In the second case the three homogeneous partial derivatives are $x^2,z^2,y^2$, up to harmless signs, and cannot vanish simultaneously at a projective point. Thus both cubics are smooth varieties. A reducible plane cubic would have intersecting positive-degree components by Bézout theorem, and be singular at an intersection; therefore these smooth cubics are irreducible varieties. The genus-degree formula gives $g=(3-1)(3-2)/2=1$. With the marked point $O$ they are elliptic curves.

For a smooth algebraic curve, its divisor class group is its Picard group. Its degree-zero subgroup isThe map $P\mapsto [P-O]$ gives an isomorphism of abelian groupsTo recall why it is a bijection, Riemann-Roch theorem in geometric genus one says every degree-one divisor class has exactly one independent section: a canonical divisor has degree zero, so the dual degree-$-1$ term has no section. Thus the class contains one effective divisor on an algebraic curve of degree one, namely a point $P$, and that point is unique. Translating by $-O$ gives the claimed bijection; this is precisely the chord-and-tangent group law on the elliptic curve.

Choose a prime integer $\ell$ different from $\mathrm{char}k$. The multiplication-by-n morphism $[\ell ]:E\to E$ has derivative $\ell$ times the identity at $O$, so it is nonconstant. A nonconstant morphism of algebraic varieties between smooth projective curves is surjective: its image is closed by proper morphism and cannot have algebraic dimension zero. Since $k$ is an algebraically closed field, this makes $[\ell ]$ surjective on $k$-points. ThereforeYet $E(k)$ is infinite. For each $x\in k$, the affine equation has a solution for $y$ in the algebraically closed field; since $k$ is infinite, this gives infinitely many points.

By the Fundamental theorem of finitely generated abelian groups, a finitely generated abelian group $A$ satisfying $A=\ell A$ has no free part and is a finite torsion group of order prime to $\ell$. The infinite group $E(k)$ cannot have this property while being finitely generated. Thus ${\mathrm{Pic}}^0(E)$ is not finitely generated. A subgroup of a finitely generated abelian group is finitely generated, soThis proof uses divisibility and infinitude, rather than uncountability, so it remains valid over countable algebraically closed fields.

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.

Pullback and tensor product preserve holomorphic line bundles and their isomorphisms. Interpret negative tensor powers of $L$ as powers of $L^{\ast }$. Thus the formula defines a group homomorphism on isomorphism classes.

To prove injectivity, suppose $p^{\ast }M\otimes L^{\otimes n}$ is trivial. Restrict it to any fibre. A line pulled back from its base point is trivial there, so its fibre restriction is ${\mathcal{O}}_{{\mathbb{P}}^1}(n)$. By the Picard group classification, $n=0$. It remains to show that $p^{\ast }M$ trivial implies $M$ trivial.

Let $\sigma$ be a nowhere-zero holomorphic section of $p^{\ast }M$. On a sufficiently small open set where both $E$ and $M$ are trivial, write $\sigma =a_i(x,[v])p^{\ast }{e}_i$ for a frame $e_i$ of $M$. For each fixed $x$, $a_i(x,\cdot )$ is a holomorphic function on the compact complex projective line, hence is constant. It is therefore $b_i(x)$; evaluating at a constant projective point in the local trivialization shows $b_i$ is holomorphic in $x$. The section is nowhere zero, so each $b_i$ is nowhere zero. Its overlap laws are exactly those of a section of $M$, and $b_i{e}_i$ glue to a global holomorphic frame. Thus $M$ is trivial.

The kernel consists only of $({\mathcal{O}}_X,0)$, provingThis is the Picard injection for holomorphic projective bundles. The argument uses local fibrewise constancy, so no global section of the projective bundle is required.

Let $C$ be the projective cubic of a Weierstrass equation of an elliptic curve, allowing its elliptic-curve discriminant to vanish, and let $O=[0:1:0]$. Use only the smooth locus of a variety $C_{\mathrm{sm}}(K)$: the singular point, if present, is excluded. For $P,Q$, intersect their chord with $C$, using the tangent if $P=Q$ and counting intersection multiplicity. If the third intersection is $R$, define $P+Q=-R$, wherein general Weierstrass equation of an elliptic curve coordinates. The line through $R$ and $O$ gives this reflection. A vertical chord gives $P+(-P)=O$, and the tangent at $O$ meets $C$ three times at $O$, so $O$ is the identity. The construction is symmetric in $P,Q$. It stays in the nonsingular locus: a line through a singular point has intersection multiplicity at least two there, and therefore cannot also contain two smooth intersections counted with multiplicity.

For the nonsingular case, prove associativity by transporting a known abelian group law. The Abel-Jacobi map of a genus-one curveis bijective over an algebraic closure. Indeed, the Riemann-Roch theorem in genus one says that every divisor class of degree one has a unique effective representative consisting of one point: existence follows from $\ell (D)=1$, and uniqueness follows because two distinct representatives would produce a degree-one map to the projective line, impossible for a genus one curve. Subtracting $O$ gives the claimed bijection.

Any line section represents the same divisor class as $3O$, including tangencies. Thus $P+Q+R\sim 3O$ as divisors on an algebraic curve, while the vertical line gives $R+(-R)+O\sim 3O$. HenceAddition in the Picard group is associative, soThe chord-and-tangent group law is defined over $K$, so the group operation restricts to the $K$-rational points.

For completeness, a singular Weierstrass equation of an elliptic curve produces the smooth-locus group of a singular Weierstrass cubic, not an elliptic curve. Over an algebraic closure, its normalization of an algebraic curve is the projective line; deleting the two preimages of a nodal crossing gives the multiplicative algebraic group, while deleting the single preimage of a cusp gives the additive group. For example, on $y^2=x^3$, the coordinate $u=x/y$, with $u(O)=0$, makes the smooth-locus law addition. On $y^2=x^2(x+1)$ in characteristic different from two, put $t=y/x$ and $z=(t+1)/(t-1)$, with $z(O)=1$; the chord relation gives $z(P)z(Q)z(R)=1$, so the group law is multiplication of $z$. A nonsplit node gives the corresponding form of the multiplicative algebraic group over $K$. These descriptions also establish the singular-case group laws.

Put $O=0_E$. Choose rational projective coordinates in which $O=(0:1:0)$ and its tangent is $Z=0$. Because $O$ is an inflection point of a plane cubic, that tangent intersects the cubic in the divisor $3O$. Consequently its cubic equation has the formHere $a\ne 0$, since the restriction to $Z=0$ is a nonzero multiple of $X^3$, and $b\ne 0$, since smoothness at $O$ requires a nonzero $Z$-derivative there. Completing the square in $Y$, by the rational linear change $Y^{\prime }=Y+(cX+eZ)/(2b)$, gives on $Z=1$Set $x^{\prime }=c_{0}x$ and $y^{\prime \prime }=c_0{y}^{\prime }$. ThenThese are invertible rational projective coordinate changes, and $O$ remains the unique point at infinity. Smoothness ensures that the resulting monic cubic has no repeated root. Thus this is a Weierstrass equation of an elliptic curve, without requiring any extraction of a square or cube root in $\mathbb{Q}$.

On this model write $-(x,y)=(x,-y)$ and $-O=O$. To add $P,Q$, take their joining line, or the tangent if $P=Q$, and let $R$ be the third intersection counted with multiplicity. DefineFor a vertical line its third intersection is $O$, so $P+(-P)=O$. The line through $O$ and an affine point is vertical, giving $P+O=P$. The tangent at $O$ has triple contact, giving $O+O=O$. This is the chord-and-tangent group law.

If $P,Q$ are rational, their joining line or tangent is rational. Its residual intersection point is rational: substituting the line in the cubic leaves a cubic whose other two roots, counted with multiplicity, are rational. Thus addition preserves $E(\mathbb{Q})$. The construction is symmetric in $P,Q$, so it is commutative.

For associativity, use divisor classes. A smooth plane cubic has geometric genus $1$, by the genus-degree formula. The Riemann-Roch theorem says that a divisor of degree $1$ on a genus-one curve has a one-dimensional space of sections. Therefore every degree-zero divisor class is uniquely represented by $P-O$: add $O$, take its unique effective degree-one representative, and then subtract $O$. Uniqueness also follows because a nonconstant function with its only pole a simple pole at $Q$ would contradict $\ell (Q)=1$.

Every line section is linearly equivalent to the tangent section $3O$. Thus, whenever $P,Q,R$ are collinear, including tangencies,A vertical section gives $[(-R)-O]=-[R-O]$. Hence the chord-and-tangent group law becomes ordinary addition in the degree-zero Picard group under the injective correspondence $P\mapsto [P-O]$. Associativity follows from associativity in that abelian group. Together with closure, identity and inverses already checked, this proves that $E(\mathbb{Q})$ is an abelian group.

Write $L\cong H_1\otimes H_2^{\ast }$ as in part (d). Each very ample line bundle is the pullback of $\mathcal{O}(1)$ under a projective embedding. A hyperplane section therefore supplies an effective divisor on a complex manifold $D_i$ with $H_i\cong \mathcal{O}(D_i)$. HenceEvery Picard group class is therefore in the image of the divisor-to-Picard map:

The Picard group $\mathrm{Pic}(X)$ is the set of isomorphism classes of line bundles on $X$, with tensor product as addition, the trivial bundle as zero and the dual bundle as inverse. On a smooth projective surface, line bundles can be represented by divisors. The intersection pairing on the Picard group of a surface is the symmetric bilinear mapobtained by moving the divisors into proper position and counting their intersections with multiplicity.

The surfaceis the first Hirzebruch surface ${\mathbb{F}}_1$. If $S$ is its negative section and $F$ a fiber of the ruling, its Picard lattice of a Hirzebruch surface is

Now let $\pi :X\to {\mathbb{P}}^2$ be the given nonisomorphic birational morphism. A birational morphism between smooth projective surfaces factors as a nonempty sequence of point blowups. Let $H={\pi }^{\ast }[\text{line}]$, and take the total transform $E$ on $X$ of the exceptional curve of the first blowup. The intersection formula for blowing up a surface givesTherefore the nonzero Picard group element $H+E$ satisfiesThis is the isotropic divisor from a nontrivial birational morphism to the projective plane.

For a morphism $\varphi :{\mathbb{P}}^2\to {\mathbb{P}}^n$, the pullback of the hyperplane bundle has the formfor an integer $d\ge 0$. If a line $\ell$ is contracted to a point, this bundle restricts trivially to $\ell$, whereasIts degree is therefore zero, so $d=0$. The homogeneous sections defining $\varphi$ are then constants, and $\varphi$ is constant. This proves the morphism from the projective plane contracting a line criterion.

Finally choose an integer $r>C^2$ and blow up $r$ distinct points of the smooth curve $C$. For the resulting morphism $\pi :X^{\prime }\to X$, let $E_1,\dots ,E_r$ be the exceptional curves. The strict transform isand the self-intersection after blowing up points on a smooth curve formula givesBlowing up a smooth point of a smooth curve does not change that curve itself, so $\pi {|}_{C^{\prime }}:C^{\prime }\to C$ is an isomorphism.

The standard affine charts of ${\mathbb{P}}^3$ are copies of ${\mathbb{A}}^3$, and the product charts of ${\mathbb{P}}^1\times {\mathbb{P}}^1$ are copies of ${\mathbb{A}}^2$. Their local rings are localizations of polynomial rings over $k$ and hence are regular local rings. Both schemes are therefore regular.

On a regular integral scheme every Weil divisor is Cartier, so the divisor class group is naturally the Picard group. Pullback of line bundles along the Segre embedding therefore definesThe two groups areThe Segre coordinates are bihomogeneous of bidegree $(1,1)$, so ${\iota }^{\ast }{\mathcal{O}}_{{\mathbb{P}}^3}(1)=\mathcal{O}(1,1)$. In these bases the map is $m\mapsto (m,m)$ and

Define the homomorphism associated to a line bundle on an abelian varietyThe Theorem of the square gives ${\varphi }_L(x+y)={\varphi }_L(x)\otimes {\varphi }_L(y)$, so this is a homomorphism.

Tensor products satisfy ${\varphi }_{L\otimes M}={\varphi }_L\otimes {\varphi }_M$ and ${\varphi }_{L^{-1}}={\varphi }_L^{-1}$. Thereforeis a subgroup of the Picard group. If $M={\varphi }_L(x)$, translations commute and the theorem of the square givesfor every $y$. Hence $M\in {\mathrm{Pic}}^{0}X$, proving $\mathrm{im}{\varphi }_L\subseteq {\mathrm{Pic}}^{0}X$.

Write the invariant prime divisors in the order of the raysas $D_1,\dots ,D_4$. The toric divisor class sequence isand the two characters in the standard basis of $M$ giveThuswhere $S=D_2$, $F=D_1=D_3$, and $D_4=S+kF$. Since the fan is smooth, this is also the Picard group.

The orbit-cone correspondence decomposes ${\mathbb{F}}_k$ into one two-dimensional torus, four one-dimensional torus orbits, and four fixed points. Only the zero-dimensional orbits contribute to the compactly supported Euler characteristic, so

For a smooth integral variety, every Weil divisor is a Cartier divisor, so the Picard group equals the divisor class group. If $U$ is obtained by removing irreducible Weil divisors $D_j$, the localization sequence for the divisor class group gives the displayed exact sequence. A rational trivialization of a line bundle on $U$ has divisor supported on the removed divisors, describing the kernel concretely.

The Picard group of a commutative ring $A$ consists of isomorphism classes of finitely generated projective modules that have rank one at every prime ideal. Tensor product is the group law, $[A]$ is the identity, and the module dual supplies the inverse. For an integral domain, every such module can be represented by an invertible fractional ideal after choosing a basis over the fraction field.

For any field $k$, degree identifies the Picard group of the projective line with $\mathbb{Z}$, generated by $\mathcal{O}(1)$. A closed point on the affine chart is cut out by an irreducible polynomial $f(t)$, whose divisor is that point minus $\mathrm{deg}(f)$ times infinity. Hence every divisor is linearly equivalent to its degree times infinity. Principal divisors have degree zero, so the degree map is injective as well as surjective. Thus a line bundle of degree $d$ is $\mathcal{O}(d)$.

The normalized universal line bundle restricts to the class $\xi$ on $A\times \{\xi \}$ and is trivialized on both zero sections. For a line bundle $L$ on an abelian variety, its homomorphism associated to a line bundle on an abelian variety satisfiesas Picard group classes. A choice of trivialization of $L_0$ normalizes the right side; uniqueness then follows from the Seesaw theorem.