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For a finite family generating the unit ideal in a commutative ring, this augmented Čech cochain complex of any module is exact. One proof clears the finitely many denominators and cocycle relations by powers $f_i^N$, writes $1={\sum }_i{a}_i{f}_i^N$, and contracts an alternating cocycle by the weighted insertions ${\sum }_i{a}_i{f}_i^N{c}_{iI}$. The argument uses vanishing criterion in a module localization and does not require the module to be finitely generated.

For any multiplicative subset $S$ of a commutative ring $R$, fractions with nilpotent numerators are nilpotent in $S^{-1}R$. Conversely, if $(a/s{)}^n=0$, the vanishing criterion in a module localization gives $t\in S$ with $t{a}^n=0$. Then $(ta{)}^n=t^{n-1}t{a}^n=0$ and $a/s=ta/(ts)$ has a nilpotent numerator. This proves the formula for the nilradical without any finiteness assumption.

Consider the kernel of a ring homomorphismSince $R$ is a Noetherian ring, this ideal has a finite generating set of an ideal, say $K=(a_1,\dots ,a_t)$. For each $i$, choose $s_i\notin \mathfrak{p}$ with $s_i{a}_i=0$, and putThen $f\notin \mathfrak{p}$ by the prime ideal condition, and $fK=0$. If $K=0$, take the empty product $f=1$.

If $a/f^n$ maps to zero under the ring homomorphism ${\theta }_f$ from a, then $a/1=0$ in $R_{\mathfrak{p}}$ because $f/1$ is a unit there. Hence $a\in K$, so $fa=0$. The vanishing criterion in a module localization now gives $a/f^n=0$ already in $R_f$. ThereforeThis is the finite generation clears localization kernels argument: the finite generating set of an ideal permits a single element outside $\mathfrak{p}$ to annihilate the entire kernel of a ring homomorphism.

Choose $f=1$, so that $R_f=R$ and $f\notin \mathfrak{p}$. For an integral domain, the canonical ring homomorphism $R\to R_{\mathfrak{p}}$ is injective: if $a/1=0$, the vanishing criterion in a module localization gives $sa=0$ for some $s\notin \mathfrak{p}$. Such an $s$ is nonzero, so the absence of zero divisors gives $a=0$. Thus

Write $N=\sqrt{(0)}$. Every element of the extension of an ideal $N{R}_{\mathfrak{p}}$ is a fraction $a/s$ with $a\in N$ and $s\notin \mathfrak{p}$: a finite sum of such fractions has a common denominator and numerator still in the ideal $N$. If $a^n=0$, then $(a/s{)}^n=0$, soConversely, let $a/s$ be a nilpotent element of the localization at a prime ideal $R_{\mathfrak{p}}$. For some $n\ge 1$, $a^n/s^n=0$. The vanishing criterion in a module localization supplies $t\notin \mathfrak{p}$ with $t{a}^n=0$. It follows thatso $ta\in N$. Since $ts\notin \mathfrak{p}$,This proves that the nilradical commutes with localization:The multiplication by $t$ is necessary: the numerator $a$ of a nilpotent element in a localization need not itself be nilpotent in $R$.

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.

Let $N$ be the nilradical of $R$. By b, the reduced locus of a ring isThe Noetherian ring hypothesis makes the ideal $N$ finitely generated; write $N=(n_1,\dots ,n_t)$. Fix $\mathfrak{p}\in \mathcal{R}$. For every $i$, the vanishing criterion in a module localization gives $s_i\notin \mathfrak{p}$ with $s_i{n}_i=0$. The product $f=s_1\cdots s_t$ lies outside the prime ideal $\mathfrak{p}$ and satisfies $fN=0$. When $N=0$, take $f=1$.

For every prime ideal $\mathfrak{q}$ in the distinguished open setthe element $f/1$ is a unit in $R_{\mathfrak{q}}$. The equality $fN=0$ therefore implies $N{R}_{\mathfrak{q}}=0$, so $R_{\mathfrak{q}}$ is reduced. Thus $\mathfrak{p}\in D(f)\subseteq \mathcal{R}$, proving that $\mathcal{R}$ is a Zariski-open subset of the spectrum of a ring.

More precisely, let $A={\mathrm{Ann}}_R(N)$, the annihilator of a module $N$. The argument shows that $N{R}_{\mathfrak{p}}=0$ if and only if some $f\in A$ lies outside $\mathfrak{p}$. HenceEquivalently, the complement is the support of a module $N$. This is vanishing of a finitely generated module near a prime applied to $N$: its finite generation lets the separately chosen denominators be combined into one element annihilating all of $N$.

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.

For a commutative ring $R$ and a multiplicative subset $S$ containing $1$, the universal property of localization specifies a pair $(R_S,\iota )$ with $\iota :R\to R_S$ a unital ring homomorphism, such that every $\iota (s)$, $s\in S$, is a unit, and any unital ring homomorphism $f:R\to T$ into a commutative ring that sends $S$ to units factors uniquely through $\iota$. Thus there is a unique $\tilde{f}:R_S\to T$ satisfying $\tilde{f}\iota =f$. The property determines the pair up to a unique compatible isomorphism.

For existence, take pairs $(r,s)\in R\times S$ and declareThe extra multiplier is essential when $R$ has zero divisors. This is an equivalence relation: for transitivity, multiply the two cross-multiplication equations by appropriate denominators and add them, using the product of their annihilating multipliers. Write the class as $r/s$ and defineMultiplying the defining equivalence equations verifies that these operations are well defined. They give a commutative ring, and $s/1$ has inverse $1/s$. The required map is necessarilyIt is well defined because applying $f$ to $u(s^{\prime }r-s{r}^{\prime })=0$ and cancelling the units $f(u),f(s),f(s^{\prime })$ equates the two images. Its addition and multiplication laws follow directly from the fraction formulas. This establishes existence and the universal property of localization. If $0\in S$, this construction gives the zero ring, which is allowed as a localization.

For a prime ideal $P$, its complement $S=R\setminus P$ is multiplicatively closed, and localization at a prime ideal means $R_P=S^{-1}R$. The mapis well defined and surjective, with kernel of a ring homomorphism $P{R}_P$: a numerator has zero image exactly when it belongs to $P$. Since $R/P$ is an integral domain, its field of fractions is a nonzero field, so $P{R}_P$ is a proper maximal ideal. Every fraction $r/s$ with $r\notin P$ is a unit, with inverse $s/r$. Every fraction with $r\in P$ belongs to this proper ideal and cannot be a unit. Thus the nonunits are exactly $P{R}_P$, and every proper ideal lies within it. ConsequentlyNow suppose each $R_P$ is reduced. If $x\in R$ is nilpotent, its image $x/1$ is nilpotent in every $R_P$, hence is zero. If $x\ne 0$, its annihilator ${\mathrm{Ann}}_R(x)$ is a proper ideal, so choose a maximal ideal $P$ containing it. Maximal ideals in a commutative ring are prime ideals. The vanishing criterion in a module localization would give $sx=0$ for some $s\notin P$. But then $s\in {\mathrm{Ann}}_R(x)\subseteq P$, a contradiction. Hence $R$ is reduced. This is reducedness is detected by prime localizations; no finiteness hypothesis is needed.

The stronger local-domain condition does not force $R$ to be an integral domain. Take $R=k\times k$ for a field $k$. Its only prime ideals are $P_1=0\times k$ and $P_2=k\times 0$: a prime ideal must contain one of the orthogonal idempotents $(1,0)$ and $(0,1)$, and the remaining quotient is the field $k$. At $P_1$, inverting $(1,0)$ kills the second component, so $R_{P_1}\cong k$; similarly $R_{P_2}\cong k$. Both localizations are integral domains, butis a product of nonzero elements in $R$. Therefore the answer is no, as illustrated by domain prime localizations do not imply a domain.

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.

Suppose $f$ is injective. If $f_{\mathfrak{m}}(x/s)=0$, some $t\notin \mathfrak{m}$ satisfies $tf(x)=f(tx)=0$. Injectivity gives $tx=0$, and the vanishing criterion in a module localization gives $x/s=0$. Hence every localized module homomorphism is injective.

Conversely, suppose all $f_{\mathfrak{m}}$ are injective. For $x\in \ker f$, its image $x/1$ in $M_{\mathfrak{m}}$ maps to zero and therefore is zero for every maximal ideal $\mathfrak{m}$. The annihilator argument in part (b) implies $x=0$. Thus $\ker f=0$ and $f$ is injective. We have proved that injectivity of a module map is detected at maximal localizations:Equivalently, exactness of localization gives $(\ker f{)}_{\mathfrak{m}}\cong \ker (f_{\mathfrak{m}})$, and part (b) detects whether this kernel is zero.

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