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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.

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