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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.

For an integral domain $A$ with fraction field $K$, a fractional ideal is a nonzero $A$-submodule $I\subseteq K$ for which $dI\subseteq A$ for some $0\ne d\in A$. A finitely generated submodule of $K$ has a common denominator and is therefore a fractional ideal. The product is generated by pairwise products of elements. An invertible fractional ideal has a multiplicative inverse and is finitely generated; arbitrary fractional ideals need not be finitely generated over a non-Noetherian domain. The familiar fractional ideals of a number field are the special case $A={\mathcal{O}}_K$. This general definition also applies to the ideals representing Cartier divisors on affine integral schemes.

Over an integral domain $A$ with fraction field $K$, an invertible fractional ideal is a finitely generated nonzero $A$-submodule $I\subseteq K$ for which another fractional ideal $J$ satisfies $IJ=A$. Here $I^{-1}=\{x\in K:xI\subseteq A\}$ equals $J$. Choose $1=\sum u_i{v}_i$ with $u_i\in I$, $v_i\in J$. On the distinguished open where $u_i{v}_i$ is invertible, $I$ is generated by $u_i$, since $x/u_i=(x{v}_i)/(u_i{v}_i)$. The maps $x\mapsto v_{i}x$ form a finite dual basis and prove that $I$ is a rank-one projective module. Conversely, a finitely generated rank-one projective module embeds in $K$ after choosing a basis over $K$; its dual identifies with $I^{-1}$, and local evaluation shows $I{I}^{-1}=A$.

A Dedekind domain is a Noetherian integrally closed domain in which every nonzero prime ideal is maximal. The equivalent local characterization is that its localizations at nonzero prime ideals are discrete valuation rings. A field is harmless as the dimension-zero case.

Write $R$ for the domain, $F$ for its fraction field and $I$ for the nonzero fractional ideal. Its proposed inverse is an $R$-submodule of $F$. Choose $0\ne d\in R$ with $dI\subseteq R$, so $d\in I^{-1}$ and the inverse is nonzero. Choose also $0\ne a\in I$. Every $x\in I^{-1}$ satisfies $xa\in R$, hence $I^{-1}\subseteq a^{-1}R$. Writing $a=r/s$ with $r,s\in R$ gives $r{I}^{-1}\subseteq sR\subseteq R$. Thus $I^{-1}$ is a fractional ideal; the Noetherian hypothesis also makes it finitely generated.

For a nonzero maximal prime ideal $\mathfrak{p}$, the discrete valuation ring $R_{\mathfrak{p}}$ makes $I_{\mathfrak{p}}$ principal, say $I_{\mathfrak{p}}=a_{\mathfrak{p}}{R}_{\mathfrak{p}}$. We need to justify that inversion commutes with this localization. One inclusion is immediate. Conversely, if $x{I}_{\mathfrak{p}}\subseteq R_{\mathfrak{p}}$, finite generation of $I$ supplies a common $s\notin \mathfrak{p}$ such that $sxI\subseteq R$. Therefore $sx\in I^{-1}$, provingIt follows that $(I{I}^{-1}{)}_{\mathfrak{p}}=R_{\mathfrak{p}}$ at every such prime ideal. Globally $I{I}^{-1}\subseteq R$. If this were a proper ideal, it would be contained in a maximal ideal and would remain proper after localizing there, a contradiction. In the field case every nonzero fractional ideal is the whole field. Hence in all casesThis proves that every nonzero fractional ideal of a Dedekind domain is an invertible fractional ideal, rather than presupposing its inverse exists.

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.

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.