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For a ring homomorphism $A\to B$, extension of scalars takes an invertible module to an invertible $B$-module. A direct-summand presentation in a finite free module remains a direct-summand presentation after tensoring. At a prime $\mathfrak{q}\subset B$ contracting to $\mathfrak{p}$, localization gives $B_{\mathfrak{q}}{\otimes }_{A_{\mathfrak{p}}}{L}_{\mathfrak{p}}\cong B_{\mathfrak{q}}$. Compatibility with tensor products, identities and composition makes the Picard group of a ring a covariant functor.

For an integral domain $A$ with fraction field $K$, realize Cartier divisors as invertible fractional ideals, with local divisor equation $f$ corresponding to the ideal $fA$. This is the $\mathcal{O}(-D)$ convention. The maps in the displayed exact sequence are unit inclusion, $f\mapsto fA$, and $I\mapsto [I]$. A principal fractional ideal is trivial as a divisor exactly when its generator is a unit. An invertible fractional ideal is trivial in the Picard group of a ring exactly when it is principal. Every rank-one projective module embeds in $K$ and produces an invertible fractional ideal, proving surjectivity. Local multiplication identifies $I{\otimes }_{A}J$ with $IJ$, so these are group homomorphisms.

An invertible module over a commutative ring $A$ is a finitely generated module that is a projective module and whose localization at a prime ideal is free of rank one at every prime. Its dual module $L^{\ast }={\mathrm{Hom}}_A(L,A)$ is also finite projective. Evaluation $L{\otimes }_A{L}^{\ast }\to A$ is an isomorphism after every prime localization, hence is an isomorphism globally by localization detects zero elements. The Picard group of a ring consists of the isomorphism classes of these modules under the tensor product of modules.

An algebraic line bundle over $A$ is an invertible module: a finitely generated projective module $L$ such that $L_{\mathfrak{p}}\cong A_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$. Define the Picard group of a ring as the set of its isomorphism classes, withThe tensor product of modules of two finite projective modules is finite projective: write each as a direct summand of a finite free module and tensor the two decompositions. Localization gives $(L{\otimes }_{A}K{)}_{\mathfrak{p}}\cong A_{\mathfrak{p}}$, so this multiplication stays within the set of invertible modules. It is well defined on isomorphism classes. The associativity and symmetry isomorphisms of the tensor product give associativity and commutativity, and $[A]$ is the identity.

Let $L^{\ast }={\mathrm{Hom}}_A(L,A)$ be the dual module. A finite direct-summand presentation shows that $L^{\ast }$ is finite projective and that localization commutes with this dual. Since $L_{\mathfrak{p}}$ is free of rank one, so is $(L^{\ast }{)}_{\mathfrak{p}}$. The evaluation homomorphismis an isomorphism after every prime localization. Its kernel and cokernel are therefore zero, by exactness of localization and localization detects zero elements for modules. Thus $[L{]}^{-1}=[L^{\ast }]$, proving that $\mathrm{Pic}(A)$ is an abelian group.

For a ring homomorphism $f:A\to B$, tensoring a split presentation $L\oplus L^{\prime }\cong A^r$ gives $(B{\otimes }_{A}L)\oplus (B{\otimes }_A{L}^{\prime })\cong B^r$, so $B{\otimes }_{A}L$ is finite projective. For $\mathfrak{q}\in \mathrm{Spec}B$ put $\mathfrak{p}=f^{-1}(\mathfrak{q})$. Every element of $A\setminus \mathfrak{p}$ becomes a unit in $B_{\mathfrak{q}}$, andThis proves base change of invertible modules. Moreover, the canonical mapis an isomorphism, with inverse $b\otimes (\ell \otimes k)\mapsto (b\otimes \ell )\otimes (1\otimes k)$. It shows that $[L]\mapsto [B{\otimes }_{A}L]$ is a group homomorphism. Identity maps induce the identity, and for $A\to B\to C$, the associativity isomorphism $C{\otimes }_B(B{\otimes }_{A}L)\cong C{\otimes }_{A}L$ proves composition compatibility. Therefore $\mathrm{Pic}$ is a covariant functor to abelian groups.

The polynomial ring $\mathbb{C}[x]$ is a principal ideal domain by division with remainder. A submodule $N$ of a finite free module over a principal ideal domain is free: induct on the ambient rank, project onto the first coordinate, and note that the image is a principal ideal, hence either zero or a free rank-one module. This gives a split exact sequence with that image as quotient; the kernel lies in a free module of one smaller rank, completing the induction. A finite projective module is such a submodule, and so is free. An invertible module has rank one after passing to the fraction field, so its free rank is exactly one. Hence every algebraic line bundle over $\mathbb{C}[x]$ is trivial and

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.