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On an affine variety or affine scheme with coordinate ring $A$, the global section functor and $M\mapsto \tilde{M}$ are inverse equivalences between modules and quasi-coherent sheaves. On every principal open subset, sections are the appropriate localization of a module. Localization detects exactness at maximal ideals, so these functors are exact even for modules which are not finitely generated.

For a module homomorphism $f:M\to N$ over a commutative ring $A$, $f$ is injective if and only if every $f_{\mathfrak{m}}$ is injective, with $\mathfrak{m}$ ranging over maximal ideals. In a localization of a module, $x/s$ maps to zero exactly when some denominator $t$ satisfies $tf(x)=0$. If $f$ is injective this gives $tx=0$, so the original fraction is zero. In general it identifies $\ker (f_{\mathfrak{m}})$ with $(\ker f{)}_{\mathfrak{m}}$. A nonzero element of $\ker f$ has a proper annihilator, contained in some maximal ideal; it remains nonzero in that localization. Thus vanishing of all localized kernels forces $\ker f=0$. No finiteness or Noetherian assumption is required.

An element $a$ of a commutative ring $R$ is zero if and only if its image in every localization at a prime ideal is zero. If $a\ne 0$, its proper annihilator lies in a maximal ideal $\mathfrak{m}$. No element outside $\mathfrak{m}$ can annihilate $a$, so $a/1\ne 0$ in $R_{\mathfrak{m}}$. This also applies to elements of an $R$-module using localization of a module.

Assume $R$ is a nonzero unital ring and all its modules are unital. Order the proper left ideals by inclusion. A chain has an upper bound given by its union, which is still a proper left ideal: if it contained $1$, one member of the chain would contain $1$ and be all of $R$. The family is nonempty because it contains $0$. The Zorn lemma gives a maximal proper left ideal $L$. The nonzero quotient $R/L$ has no proper nonzero submodules, because its submodules correspond to left ideals between $L$ and $R$. Thus $R/L$ is a simple left module. The nonzero-ring qualification is necessary: the zero ring has no nonzero unital simple module.

Define the Jacobson radical $J(R)$ as the intersection of all maximal left ideals. It is equivalently the intersection of the ideals ${\mathrm{Ann}}_R(S)$ over all simple modules, where $\mathrm{Ann}$ is the annihilator of a module. To verify this equivalence, for a nonzero vector $v$ in a simple module $S$, the surjection $R\to S$, $r\mapsto rv$, has a maximal left-ideal kernel. Thus $x\in J(R)$ implies $xv=0$ for every $v$ in every simple module. Conversely, if $x$ annihilates all simple modules, it annihilates $1+L$ in every $R/L$, so $x\in L$. An annihilator of a module is a two-sided ideal; this description therefore proves that $J(R)$ is a two-sided ideal.

For the forward implication of the unit criterion for the Jacobson radical, first let $y\in J(R)$. If $R(1-y)$ were a proper left ideal, it would lie in a maximal one $L$. Both $y$ and $1-y$ would belong to $L$, an impossibility. Hence $c(1-y)=1$ for some $c\in R$. A left inverse alone does not yet establish that $1-y$ is a unit in a general noncommutative ring. However, $c=1+cy$, with $cy\in J(R)$, so the same argument gives $dc=1$ for some $d$. Multiplying $c(1-y)=1$ by $d$ on the left gives $d=1-y$. Therefore $(1-y)c=dc=1$ as well. This proves that $1-y$ is a unit.

For $x\in J(R)$, its two-sided-ideal property gives $axb\in J(R)$ for all $a,b$, so $1-axb$ is a unit. Conversely, if $x\notin J(R)$, choose a maximal left ideal $L$ with $x\notin L$. Then $L+Rx=R$, so $l+ax=1$ for some $l\in L$, $a\in R$. Hence $1-ax=l\in L$ is not a unit: if it were invertible, multiplying it on the left by its inverse would put $1$ in $L$. Taking $b=1$ contradicts the given unit property. We have provedFinally let $R$ be commutative and $I$ an ideal. The set $S=1+I$ contains $1$, andso it is a multiplicative subset. In the localization of a module $I_S\subseteq R_S$, any element is $i/s$ with $i\in I$ and $s\in S$. For any $r/t\in R_S$,Both $ts$ and $ts-ri$ lie in $S$, since $ts\in 1+I$ and $ri\in I$. The fraction is consequently a unit in $R_S$. In a commutative ring, the product of any two coefficients is again a single coefficient, so the proved unit criterion for the Jacobson radical applies to all expressions $1-a(i/s)b$. ThusThis proves localization away from one plus an ideal lies in the Jacobson radical. If $I=R$, then $0\in S$, $R_S=0$, and the inclusion holds trivially.

Set $S=A\setminus \mathfrak{m}$, a multiplicative subset because $\mathfrak{m}$ is a prime ideal. The quotient module $B=A/\mathfrak{a}$ is a cyclic module generated by $\bar{1}$, so its localization of a module $S^{-1}B$ is zero precisely when $\bar{1}/1=0$. By the vanishing criterion in a module localization, this is equivalent to the existence of $s\in S$ with $s\bar{1}=0$ in $B$. The latter condition says $s\in \mathfrak{a}$. ThereforeExplicitly, if $s\in \mathfrak{a}\setminus \mathfrak{m}$, multiplication by $s$ annihilates $B$ but becomes invertible after localization, forcing every localized element to be zero.

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.

A locally ringed space is a topological space $X$ equipped with a sheaf of commutative rings ${\mathcal{O}}_X$ whose stalk ${\mathcal{O}}_{X,x}$ at every point is a local ring. A morphism of locally ringed spaces $(f,f^{\#}):(Y,{\mathcal{O}}_Y)\to (X,{\mathcal{O}}_X)$ consists of a continuous map $f:Y\to X$ and a morphism of sheaves of rings $f^{\#}:{\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$, with every induced stalk map ${\mathcal{O}}_{X,f(y)}\to {\mathcal{O}}_{Y,y}$ a local homomorphism: the inverse image of the target maximal ideal is the source maximal ideal. Here $f_{\ast }{\mathcal{O}}_Y$ is the direct image sheaf, with sections ${\mathcal{O}}_Y(f^{-1}U)$ on $U$.

Let $\varphi :A\to B$ be a ring homomorphism. Define the map between the spectra of rings byThe inverse image is a prime ideal, and $f^{-1}(D(a))=D(\varphi (a))$. Thus $f$ is a continuous map, because the principal open subsets form a basis. On these principal open subsets, define the structure sheaf map using the localizationsThese maps commute with restrictions and consequently give a morphism of sheaves. At $\mathfrak{q}$, putting $\mathfrak{p}={\varphi }^{-1}\mathfrak{q}$, the stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. An element $b/s$ belongs to its inverse image of $\mathfrak{q}{B}_{\mathfrak{q}}$ exactly when $b\in \mathfrak{p}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, take a morphism of locally ringed spaces $f:\mathrm{Spec}B\to \mathrm{Spec}A$. Its map on global sections gives a ring homomorphismLocality of the stalk map implies, for every $a\in A$,Hence the point map must be $\mathfrak{q}\mapsto {\varphi }^{-1}\mathfrak{q}$. On $D(a)$, its structure sheaf map must send $b/1$ to $\varphi (b)/1$ and the inverse of $a$ to the inverse of $\varphi (a)$. Thus it is precisely the localization map constructed above. Agreement on this basis proves agreement of the morphisms of sheaves. In particular, the inducing ring homomorphism exists and is unique.

If $\varphi$ is surjective, write $B=A/I$. The correspondence between prime ideals of $A/I$ and prime ideals of $A$ containing $I$ identifies $\mathrm{Spec}B$ homeomorphically with the closed subset $V(I)$: the principal open subsets correspond on both sides. Moreover, the stalk maps are the surjections $A_{\mathfrak{p}}\to (A/I{)}_{\mathfrak{p}/I}$, so ${\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$ is surjective.

For the converse, it is important not to assume that global sections preserve an arbitrary surjection of sheaves. Here the affine module-sheaf equivalence supplies the extra information: on every $D(a)$,Thus $f_{\ast }{\mathcal{O}}_Y$ is the quasi-coherent sheaf associated with the $A$-module $B$, and the cokernel sheaf of ${\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$ is associated with the module $C=\mathrm{coker}(A\to B)$, since localization of a module is exact. Surjectivity of the morphism of sheaves gives $C_{\mathfrak{p}}=0$ for every prime ideal $\mathfrak{p}$. This implies $C=0$: if $0\ne c\in C$, its proper annihilator lies in a maximal ideal $\mathfrak{m}$, and then $c/1\ne 0$ in $C_{\mathfrak{m}}$. Therefore $A\to B$ is surjective. We have proved the closed immersion criterion for affine schemes:

Use the local-presentation definition of a quasi-coherent sheaf: locally it admits a presentation ${\mathcal{O}}^{(I)}\to {\mathcal{O}}^{(J)}\to \mathcal{F}\to 0$ by direct sums of copies of the structure sheaf, with no finiteness requirement on the index sets. If $M$ is an $A$-module, choose a free module presentation $A^{(I)}\to A^{(J)}\to M\to 0$. Exactness of localization of a module gives the corresponding local presentation of $\tilde{M}$. Hence a sheaf which is associated with a module on every affine open subset is a quasi-coherent sheaf.

For the converse, begin with a quasi-coherent sheaf and a local presentation on an affine open subset $V=\mathrm{Spec}R$. The free sheaves have global sections $R^{(I)}$ and $R^{(J)}$: a section has locally finite support, and quasi-compactness of $V$ makes its support finite globally. Part (i) identifies the presentation map with an $R$-module homomorphism. Exactness of localization then identifies its cokernel sheaf with ${\tilde{M}}_V$ for the module cokernel $M_V$. Thus there is an affine open cover on which $\mathcal{F}$ is associated with modules.

It remains to show this on an arbitrary affine open subset $U=\mathrm{Spec}A$, rather than only the selected cover. Around each point of $U$, choose an affine neighborhood contained in $U$ with such a local presentation. Choose a finite cover $U={\bigcup }_{i=1}^{r}D(s_i)$ refining these neighborhoods. On each $D(s_i)$ the restriction is associated with a module. Its restriction to $D(s_i{s}_j)$ is therefore the corresponding localization, and the same is true after intersecting with $D(a)$. Put $M=\Gamma (U,\mathcal{F})$. The sheaf gluing axiom giveswhere the arrow is the difference of the two restriction maps. Localization of a module commutes with kernels and finite products. Localizing this formula at $a$ therefore gives exactly the gluing formula for $D(a)$, covered by the $D(a{s}_i)$. Consequentlycompatibly with restriction. These isomorphisms on the principal open subsets identify $\tilde{M}$ with $\mathcal{F}{|}_U$. We have proved the global-section localization for quasi-coherent sheaves and, in particular,

An $A$-module homomorphism $h:M\to N$ induces, by localization of a module, compatible mapson all principal open subsets $D(a)$. These glue to an ${\mathcal{O}}_X$-linear morphism of sheaves $\tilde{h}:\tilde{M}\to \tilde{N}$.

Conversely, taking global sections of an ${\mathcal{O}}_X$-linear morphism of sheaves gives an $A$-module homomorphism $h:M\to N$, since $\Gamma (X,\tilde{M})=M$ and $\Gamma (X,\tilde{N})=N$. On $D(a)$, compatibility with restriction determines the image of $m/1$, and ${\mathcal{O}}_X$-linearity determines the image of $a^{-r}m$. It must be $h(m)/a^r$. Thus the morphism of sheaves is uniquely recovered from $h$. Both constructions preserve addition and composition, proving the affine module-sheaf equivalence on morphisms:No finite-generation assumption is used.

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.

Put $u=X_1-1$ and $v=X_2-1$. Evaluation at $(1,1)$ is surjective, andso its kernel is the prime ideal $\boxed{P=(u,v)}$. In particular $P/P^2$ is naturally an $R/P=\mathbb{Z}$-module, generated by the classes of $u,v$.

For integers $a,b$, including negative integers, reduction modulo $P^2$ givesFor negative powers, this follows from $(1+u{)}^{-1}\equiv 1-u$ and $(1+v{)}^{-1}\equiv 1-v$; the mixed product $uv$ vanishes modulo $P^2$. Therefore a Laurent polynomial $f$ satisfiesThe formal derivatives obey the product rule, so $D_i(ab)=\theta (a)D_i(b)+\theta (b)D_i(a)$. If $a,b\in P$, both terms vanish. Hence both $D_i$ vanish on $P^2$ and define a mapIt sends $[u]$ to $(1,0)$ and $[v]$ to $(0,1)$, and the displayed expansion gives its inverse. Thus the conormal module has the explicit basis

Let $A=R_P$ be the localization at a prime ideal, with maximal ideal $\mathfrak{m}=P{R}_P$. Every nonzero integer lies outside $P$ and is therefore inverted. The residue field isIndeed there is an explicit identificationinverting $1+u$ and $1+v$ has no effect in this local ring, and a polynomial denominator lies outside $(u,v)$ precisely when its constant term is nonzero. Conversely, clearing rational denominators turns every such fraction into a fraction from $R_P$.

For the Krull dimension, the prime ideal correspondence for localization preserves the strict chainof prime ideals in $R$. They are prime because the successive quotients are the integral domains $R$, $\mathbb{Z}[X_2,X_2^{-1}]$, and $\mathbb{Z}$. They remain distinct after localization at a prime ideal because they are all contained in $P$. Hence $\dim A\ge 2$.

For the upper bound, $R$ is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring. Since $P$ is generated by two elements, the Krull height theorem gives $\mathrm{ht}P\le 2$. The prime ideal correspondence for localization identifies $\dim R_P$ with $\mathrm{ht}P$, so

A Noetherian local ring $(B,\mathfrak{n})$ with residue field $\kappa$ is a regular local ring ifBy the Nakayama lemma, the dimension on the left is the minimal number of generators of $\mathfrak{n}$; this is also called the embedding dimension.

The localization of a module is exact and commutes with quotients and products of ideals. Applying it to the already computed conormal module gives the cotangent space of a local ringIts dimension is two, equal to $\dim A$. Therefore $\boxed{R_P\text{ is a regular local ring}}$. Notice that $P/P^2$ has a basis over $\mathbb{Z}$, whereas the localized cotangent space of a local ring has a basis over the residue field $\mathbb{Q}$.

The tensor product of sheaves is the sheafification of the presheafwith restriction maps induced by those of $\mathcal{F}$, $\mathcal{G}$, and the structure sheaf of a scheme. Sending the germ of $f\otimes g$ to $f_P\otimes g_P$ gives a canonical mapIt is surjective because a finite sum of tensors of germs has representatives on one common neighborhood. It is injective because an equality in a tensor product of modules is witnessed by finitely many additive and balancing relations; all their germs, including the scalars in ${\mathcal{O}}_{X,P}$, can be represented and their equalities made valid on a common smaller neighborhood. Equivalently, a filtered colimit of modules commutes with these tensor products. Since sheafification preserves the stalk of a sheaf, this proves the asserted isomorphism.

For the pullback of a sheaf of modules, first form the inverse image sheaf ${\varphi }^{-1}\mathcal{F}$: sheafify $U\mapsto {\mathrm{colim}}_{V\supseteq \varphi (U)}\mathcal{F}(V)$. The morphism of sheaves of rings ${\varphi }^{-1}{\mathcal{O}}_Y\to {\mathcal{O}}_X$ then definesExactness of a sequence of sheaf of modules morphisms is checked on every stalk of a sheaf. Applying the right exactness of the tensor product of modules to the stalk sequence over ${\mathcal{O}}_{Y,\varphi (P)}$ gives the exact stalk sequence of the pullbacks. Thus ${\varphi }^{\ast }$ preserves sequences ending in zero, as required.

A quasi-coherent sheaf is a sheaf of modules locally of the form $\tilde{N}$ for a module $N$ over the coordinate ring of an affine neighborhood. On an affine variety $V$ with ring $B$, this associated sheaf is characterized on principal opens by $\tilde{N}(D(b))=N_b$, using localization of a module. Suppose $\mathcal{F}{|}_V\cong \tilde{N}$ and $U\subseteq {\varphi }^{-1}V$ is affine, with ring $C$. The induced algebra map $B\to C$ gives a canonical mapAt a point $P\in U$, both sides identify with ${\mathcal{O}}_{X,P}{\otimes }_{B}N$: on the right use ${\mathcal{F}}_{\varphi (P)}\cong {\mathcal{O}}_{Y,\varphi (P)}{\otimes }_{B}N$ and associativity of the tensor product of modules. The map is therefore an isomorphism on every stalk of a sheaf, hence an isomorphism of sheaves. Such $U$ cover $X$, proving that the pullback is a quasi-coherent sheaf.

Finally, on an affine open $V\subseteq Y$ with ring $B$, the presheaf defining $M_Y$ has, on $D(b)\subseteq V$, the valueConsequently $M_Y{|}_V\cong \widetilde{M{\otimes }_{R}B}$, so $M_Y$ is a quasi-coherent sheaf. The algebra map for $M_X$ is the composite $R\to \Gamma (Y,{\mathcal{O}}_Y)\to \Gamma (X,{\mathcal{O}}_X)$. Restriction of functions and the tensor-product construction give a canonical sheaf morphism ${\varphi }^{\ast }{M}_Y\to M_X$. At $P$, it is the associativity isomorphismIt is therefore a sheaf of modules isomorphism. In particular,

The set $S=1+I$ is a multiplicative subset, since $(1+a)(1+b)=1+(a+b+ab)$. The localization of a ring $S^{-1}R$ consists of fractions $r/s$, whereSimilarly, the localization of a module $S^{-1}M$ consists of fractions $m/s$, with the analogous equivalence relation. The canonical maps areThe Localization of a Noetherian ring is Noetherian. Since a finitely generated module over a Noetherian ring is a Noetherian module, localizing a finite generating set shows that $S^{-1}M$ is Noetherian over $S^{-1}R$.

The prime ideal correspondence for localization identifies $\mathrm{Spec}{S}^{-1}R$ with the primes $P\in \mathrm{Spec}R$ disjoint from $S$. Hereso the spectrum of localization away from one plus an ideal is

An element $m$ maps to zero exactly when $(1+a)m=0$ for some $a\in I$. Such an equation gives $m=(-a{)}^{j}m\in I^{j}M$ for every $j$, proving one inclusion. Conversely, suppose $m\in {\bigcap }_{j\ge 1}{I}^{j}M$. The Artin-Rees lemma applied to $Rm\subseteq M$ says that for some $c$,Taking $n=c+1$ gives $m\in I(Rm)$, so $m=am$ for some $a\in I$. Then $(1-a)m=0$ and $1-a\in S$. Thus

For failure without Noetherianity, takeThe strict chain $(t)⊊(t^{1/2})⊊(t^{1/4})⊊\cdots$ shows that $R$ is not Noetherian. Since $t^q=(t^{q/2}{)}^2$, one has $I^2=I$, and hence ${\bigcap }_j{I}^j=I\ne 0$. But $R$ is an integral domain, so its localization map into $(1+I{)}^{-1}R$ is injective. This is the non-Noetherian failure of the intersection formula for localization.

The fraction relation defining localization of a module gives this criterion. For localization at powers of a single element $f$, a fraction vanishes exactly when some power of $f$ annihilates its numerator. Finitely many such vanishing relations can be cleared with one common exponent. This is distinct from detecting global zero by localizing at every prime ideal.