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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.

For a Noetherian ring, injectivity of a module can be tested at all prime localizations. By the Baer criterion, test ${\mathrm{Ext}}_R^1(R/J,M)$ for every ideal $J$. Localization of Ext over a Noetherian ring identifies its localization with the corresponding ideal test over $R_P$. Every ideal of $R_P$ is extended from its contraction, and localization detects zero elements applies to the resulting Ext module even when it is not finitely generated. These observations prove both directions.

If $R$ is reduced, its nilradical $N$ is zero. By b, every localization at a prime ideal has zero nilradical, so in particular $R_{\mathfrak{m}}$ is reduced for every maximal ideal $\mathfrak{m}$.

Conversely, suppose every such localization is reduced. Take a nilpotent element $x\in R$. Then $x/1$ is nilpotent in every $R_{\mathfrak{m}}$, hence zero there. To see directly that $x=0$, suppose otherwise and consider the proper annihilatorChoose a maximal ideal $\mathfrak{m}$ containing this annihilator. The equality $x/1=0$ in $R_{\mathfrak{m}}$ would imply $sx=0$ for some $s\notin \mathfrak{m}$, by the vanishing criterion in a module localization. But then $s\in {\mathrm{Ann}}_R(x)\subseteq \mathfrak{m}$, a contradiction. This is the principle that localization detects zero elements. ThereforeThe argument also includes the zero ring, which is reduced and has no maximal ideals.

Set $L=M/\mathfrak{a}M$, a quotient module. The exactness of localization gives, for every maximal ideal $\mathfrak{m}$,If $\mathfrak{m}\supseteq \mathfrak{a}$, the hypothesis $M_{\mathfrak{m}}=0$ makes this quotient module zero. If $\mathfrak{m}⊉\mathfrak{a}$, choose $s\in \mathfrak{a}\setminus \mathfrak{m}$. Its image in $R_{\mathfrak{m}}$ is a unit, so $\mathfrak{a}{R}_{\mathfrak{m}}=R_{\mathfrak{m}}$ and the displayed quotient module is again zero. ThusFor completeness, if $0\ne \ell \in L$, its annihilator is a proper ideal, and hence lies in a maximal ideal $\mathfrak{m}$. The vanishing criterion in a module localization shows $\ell /1\ne 0$ in $L_{\mathfrak{m}}$: any element annihilating $\ell$ lies inside $\mathfrak{m}$. This contradicts the vanishing just proved. Hence $L=0$, andThe use of localization detects zero elements is valid for every module; neither finite generation nor a Noetherian hypothesis is needed.

The forward implication is immediate: the localization of a module that is zero remains zero. Conversely, suppose $M\ne 0$ and choose $0\ne x\in M$. Its annihilatoris a proper ideal, since $1\notin J$. Choose a maximal ideal $\mathfrak{m}$ containing $J$. Then $x/1$ cannot be zero in $M_{\mathfrak{m}}$: otherwise the vanishing criterion in a module localization would give $sx=0$ for some $s\notin \mathfrak{m}$, contradicting $s\in J\subseteq \mathfrak{m}$. Thus some maximal localization is nonzero, provingIn the suggested cyclic description, $Ax\cong A/J$ and part (a) gives $(Ax{)}_{\mathfrak{m}}\ne 0$; the same fraction argument embeds it in $M_{\mathfrak{m}}$. This is localization detects zero elements, applied to an arbitrary module; finite generation is unnecessary.

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

A sheaf of modules $\mathcal{F}$ over ${\mathcal{O}}_X$ is quasi-coherent when locally it admits a presentationwith arbitrary indexing sets $I,J$. On an affine chart, sheafifying the corresponding module presentation and using exactness of localization identifies its cokernel with a sheaf $\tilde{N}$. Thus one may equivalently require that locally on affine charts $\mathcal{F}$ is associated with a module. This does not impose finite generation, which belongs to the stronger coherent condition.

On the affine variety $X$ with coordinate ring $A$, put $M=\Gamma (X,\mathcal{F})$. Refine local module charts to a finite principal open cover $X={\bigcup }_{j}D(g_j)$ such that $\mathcal{F}{|}_{D(g_j)}\cong {\tilde{N}}_j$ for an $A_{g_j}$-module $N_j$. This is possible because principal opens form an open basis and $X$ is quasi-compact. On the overlap $D(g_j{g}_k)$ the sections are the corresponding further localization of $N_j$ or $N_k$. The sheaf gluing axiom gives an equalizerwhere the last map is the difference of restrictions. For any $f\in A$, localize this sequence. Exactness of localization and commutation with finite products give exactly the equalizer for the cover $D(f)={\bigcup }_{j}D(f{g}_j)$. HenceThe isomorphisms are canonical and commute with all basic-open restrictions. Since the sections of $\tilde{M}$ on these opens are $M_f$, they identify the two sheaves:Conversely $\tilde{M}$ is quasi-coherent for every $A$-module $M$, since a free presentation of $M$ gives the required sheaf presentation. Also $\Gamma (X,\tilde{M})=M$, and module homomorphisms sheafify, while a sheaf morphism is determined on every $D(f)$ by the localization of its map on global sections. This proves the affine module-sheaf equivalence.

For a short exact sequence of quasi-coherent sheaves, the sequence of their stalks is exact. Under this equivalence the stalk at $x$ is the localization of the global-section module at ${\mathfrak{m}}_x$. Exactness of module sequences can be checked at all maximal ideals, by localization detects zero elements applied to their homology, so the global-section modules also form a short exact sequence. Therefore global sections are exact on quasi-coherent sheaves over an affine variety. This is stronger than the left exactness of the global section functor on arbitrary sheaves, and follows from localization, not from assuming the cohomology vanishing still to be proved.

For an open inclusion $j:U\hookrightarrow X$, the requested sheaf isRestriction gives $\mathcal{F}\to {}_U\mathcal{F}$. This direct image from an open restriction is not extension by zero: its stalks outside $U$ may be nonzero.

The locally vanishing principle for sheaf cohomology says that a class $\xi \in H^i(X,\mathcal{F})$, $i>0$, is killed on a suitable neighbourhood of every point. Under the stated hypothesis that basis opens and their finite intersections have zero cohomology in degrees $1,\dots ,i-1$, these neighbourhoods can be chosen in that basis so that the image of $\xi$ underis zero. In particular its restriction in $H^i(U,\mathcal{F}{|}_U)$ is zero. This is a statement about individual classes; it does not assert that every locally vanishing class is already globally zero.

Here is a noncircular proof of vanishing of quasi-coherent cohomology on an affine scheme, applied to the variety. Induct on $i>0$, simultaneously for every affine variety and quasi-coherent sheaf. Assume all lower positive degrees vanish. The principal-open basis is closed under finite intersections, so it satisfies the principle's hypothesis. Given $\xi \in H^i(X,\mathcal{F})$, choose a finite principal cover $\mathcal{U}=(D(f_j))$ on which it restricts to zero.

Applying the Čech cochain complex to a flasque resolution gives the Čech lifting below the first possible local cohomology degree comparison segment, under lower-degree vanishing on all intersectionsFor completeness, the double complex has terms ${\check{C}}^p(\mathcal{U},{\mathcal{I}}^q)$. Its augmented rows are exact because each ${\mathcal{I}}^q$ is flasque, so its total cohomology is the cohomology of global sections of the resolution. The vertical cohomology on intersections in degrees $0<q<i$ is zero. Equivalently, start with local primitives of a cocycle representing $\xi$, take their differences on pairwise overlaps, and solve successively for primitives of those differences in degrees $i-1,i-2,\dots ,1$. The last difference is a Čech $i$-cocycle with values in $\mathcal{F}$. This identifies the kernel of the restriction map with the displayed Čech group. No vanishing in degree $i$ on the intersections has been assumed.

It remains a module calculation. Write $\mathcal{F}=\tilde{M}$. The augmented Čech complex isSince the $D(f_j)$ cover $X$, the $f_j$ generate the unit ideal. This complex is exact. To see it algebraically, use alternating cochains, with repeated indices giving zero. For a cocycle $c$, choose $N$ large enough to clear every $f_j$-denominator in $f_j^N{c}_{jI}$, as well as the finitely many cocycle relations; vanishing criterion in a module localization allows a further power to clear relations which initially hold only after localization. Since the ideal $(f_j^N)$ is still the unit ideal, choose $a_j$ with ${\sum }_j{a}_j{f}_j^N=1$. Defineusing those cleared representatives in $M_{f_I}$. The cocycle identity gives $\delta b=({\sum }_j{a}_j{f}_j^N)c=c$. The same argument with the augmentation gives gluing and uniqueness in degree zero. This is exactness of the unit-ideal localization Čech complex.

Thus ${\check{H}}^i(\mathcal{U},\mathcal{F})=0$, so the class $\xi$, whose restrictions were zero, is zero. The induction starts at $i=1$, when the lower-degree condition is empty. We have proved

Set $S=R\setminus P$, a multiplicative subset. The localization at a prime ideal is $R_P=S^{-1}R$, whose elements are fractions $r/s$. Equality of two fractions means that $u(r{s}^{\prime }-r^{\prime }s)=0$ for some $u\in S$. Likewise the localization of a module is $M_P=S^{-1}M$, withAddition uses a common denominator, and the module action isThe equivalence relations make these operations well-defined, so $M_P$ is an $R_P$-module.

If $M\ne 0$, choose $0\ne m\in M$. Its annihilator is proper, so it lies in a maximal ideal $P$. The element $m/1$ cannot vanish in $M_P$: vanishing would mean $sm=0$ for some $s\notin P$, contrary to ${\mathrm{Ann}}_R(m)\subseteq P$. The converse is immediate by localizing the zero module. ThusThis proof of localization detects zero elements actually applies to arbitrary modules, without finite-generation or Noetherian assumptions. We will use that extra generality for an Ext functor module below.

An injective module $E$ has the extension property: for each inclusion $A\hookrightarrow B$, every map $A\to E$ extends to a map $B\to E$. Equivalently, ${\mathrm{Hom}}_R(-,E)$ is exact. A projective module $Q$ has the lifting property: for each surjection $B\twoheadrightarrow C$, every map $Q\to C$ lifts to $Q\to B$. Equivalently, ${\mathrm{Hom}}_R(Q,-)$ is exact, or $Q$ is a direct summand of a free module.

For the local criterion for injectivity over a Noetherian ring, recall the Baer criterion: $E$ is injective precisely when every map from an ideal $J\subseteq R$ into $E$ extends to $R$. Through the short exact sequence $0\to J\to R\to R/J\to 0$, this is equivalent toWe also need localization of Ext over a Noetherian ring. Because $R$ is Noetherian, the module $R/J$ has a free resolution with every term finitely generated: all successive kernels are finitely generated, so this can be built recursively. For a finite free term $F$, the natural mapis an isomorphism, as is clear from a finite basis. Exactness of localization lets us pass to cohomology of the Hom complex. ThereforeThis explains the finiteness hypothesis needed for the localization argument, rather than assuming that localization preserves injectivity automatically.

If $M$ is injective, the left-hand side vanishes for every $J$ and $P$. Every ideal $L$ of $R_P$ is $J{R}_P$, where $J$ is its contraction to $R$: if $r/s\in L$, then $r/1=s(r/s)\in L$, and conversely localization of a member of the contraction stays in $L$. Hence all ideal tests for $M_P$ vanish, and the Baer criterion over $R_P$ makes $M_P$ injective.

Conversely, suppose every $M_P$ is injective. For each ideal $J$, the displayed Ext functor localization is zero at every prime. The zero-detection argument above gives ${\mathrm{Ext}}_R^1(R/J,M)=0$, without needing this Ext module to be finitely generated. Applying the Baer criterion over $R$ provesIn fact, under the Noetherian ring hypothesis this equivalence holds for arbitrary $M$; the printed finite-generation assumption is more than is needed.

The global dimension iswhere projective dimension is the smallest length of a projective resolution, or infinity if there is no finite one. If $\mathrm{gldim}R=0$, every module is projective. Given an inclusion $A\hookrightarrow B$, the quotient $B/A$ is then projective, so the exact sequence $0\to A\to B\to B/A\to 0$ splits. A retraction $\rho :B\to A$ exists. For any module $E$ and any map $f:A\to E$, the composition $f\rho :B\to E$ extends $f$. Thus every module is also injective. Global dimension zero makes all modules both projective and injective, as recorded by global dimension zero and split exact sequences.

No: $R$ cannot contain a nonzero nilpotent element. We prove the needed principle, localization detects zero elements, rather than assume it.

Let $a\ne 0$ in $R$. Its annihilator $\mathrm{Ann}(a)=\{r:ra=0\}$ is a proper ideal, because $1$ does not annihilate $a$. Every proper ideal is contained in a maximal ideal, by the Zorn lemma; choose such a maximal ideal $\mathfrak{m}$ containing $\mathrm{Ann}(a)$. If $a/1=0$ in $R_{\mathfrak{m}}$, then some $s\notin \mathfrak{m}$ satisfies $sa=0$. This would put $s$ in $\mathrm{Ann}(a)\subseteq \mathfrak{m}$, a contradiction. Thus every nonzero element survives in at least one localization at a prime ideal.

Now suppose $a^n=0$. For every prime ideal $P$, $(a/1{)}^n=0$ in $R_P$. The assumed absence of nonzero nilpotent elements forces $a/1=0$ in every $R_P$. The preceding argument then forces $a=0$. In the terminology of reduced rings, we have proved