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On an integral scheme $X$, every nonempty open $U$ contains its generic point $\eta$. Taking a germ gives an injective ring homomorphism from its regular functions to the function field ${\mathcal{O}}_{X,\eta }$. Injectivity follows on each nonempty affine open subscheme from the injection of an integral domain into its field of fractions, and then from the sheaf gluing axiom. This lets regular functions on different open subsets be compared inside one field.

The punctured affine plane $U={\mathbb{A}}_k^2\setminus \{(0,0)\}$ is not an affine scheme. On its cover $D(x)\cup D(y)$, global sections are $k[x,y{]}_x\cap k[x,y{]}_y=k[x,y]$ inside the field of fractions. Coprimality of $x$ and $y$ proves this equality. If $U$ were affine, its inclusion into ${\mathbb{A}}_k^2$ would be an isomorphism by affine scheme reconstruction from global sections, contradicting the omitted point.

A right Ore localization at $S$ is a unital homomorphism $\iota :R\to Q$ such that every $\iota (s)$, $s\in S$, is a unit, every element of $Q$ is $\iota (r)\iota (s{)}^{-1}$, andEquivalently, the usual fraction construction has this zero criterion and is universal among homomorphisms that invert $S$. Injectivity is required only when the denominators are regular elements of a ring. If $0\in S$, every homomorphism inverting $S$ has zero target; this is the degenerate zero-ring localization. For a nonzero localization we require $0\notin S$.

The necessary and sufficient conditions are that $S$ be a right denominator set:

For necessity, write the element $\iota (s{)}^{-1}\iota (r)$ as $\iota (b)\iota (t{)}^{-1}$. Multiplying by $\iota (s)$ on the left and $\iota (t)$ on the right gives $\iota (rt-sb)=0$. The zero criterion supplies $u\in S$ with $(rt-sb)u=0$. Hence $r(tu)=s(bu)$, the right Ore condition in $R$, not merely an equality of images in $Q$. Also $sr=0$ makes $\iota (r)=0$ because $\iota (s)$ is invertible, and the zero criterion gives $rt=0$. This proves both necessities.

The sufficiency is the Ore theorem for denominator sets. Concretely, right fractions can be represented by $(r,s)$; a common-denominator equality has $su=tv\in S$ and $ru=r^{\prime }v$ for suitable $u,v\in R$. The two conditions make the induced fraction equivalence transitive and make addition and multiplication well defined; the zero fraction is exactly the kernel displayed above. The requested direction of the theorem was necessity, for which the complete argument has just been given.

For a right Noetherian domain, any two nonzero principal right ideals intersect. To see this without using the desired localization, suppose $bR\cap aR=0$ with $a,b\ne 0$. The sumis direct: reduce a finite relation modulo $aR$ to get its first summand zero, then cancel $a$ in the noncommutative domain and repeat. Every summand is nonzero, so the finite partial sums form a strictly ascending chain of right ideals, contradicting the ascending chain condition. A nonzero intersection $rR\cap sR$ thus provides $rt=sb\ne 0$, with $t\ne 0$. This is the right Ore condition for $S=R\setminus \{0\}$. Reversibility is automatic in a noncommutative domain, and the zero criterion makes the map injective. The localization is a division ring, since a nonzero $r{s}^{-1}$ has inverse $s{r}^{-1}$.

The Goldie theorem states that a semiprime ring has a semisimple Artinian classical right ring of quotients if and only if it has finite right uniform dimension and satisfies the ascending chain condition on right annihilators. A right Noetherian ring satisfies the annihilator condition because right annihilators are right ideals. Its right regular module has finite uniform dimension: an infinite direct sum of nonzero submodules would give a strictly ascending chain of finite partial sums. A prime ring is semiprime, so the theorem applies to a prime right Noetherian ring $R$.

To obtain a single matrix-ring factor, not just a finite product, prove that its quotient $Q$ is prime. A nonzero two-sided ideal $I\subseteq Q$ contains a nonzero fraction $r{s}^{-1}$ and hence the nonzero element $r=(r{s}^{-1})s\in I\cap R$. The same holds for any other nonzero ideal $K$. Since $R$ is prime, $(I\cap R)(K\cap R)\ne 0$, so $IK\ne 0$. Thus $Q$ is prime. The Artin–Wedderburn theorem gives a finite product of matrix rings over division rings; primeness excludes two different nonzero factors with zero product. This proves the prime classical quotient of a prime right Goldie ring:For the requested counterexample to Noetherianity, take $R=k[x_1,x_2,\dots ]$. It is a commutative integral domain and has its field of fractions as its classical right ring of quotients. But $(x_1)⊊(x_1,x_2)⊊\cdots$ is a strictly ascending ideal chain: evaluation at $x_1=\cdots =x_m=0$ shows $x_{m+1}$ is not in the preceding ideal. Therefore existence of a classical quotient does not imply right Noetherianity.

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.

An irreducible scheme is a nonempty scheme whose underlying topological space cannot be expressed as the union of two proper closed subsets. Equivalently, any two nonempty open subsets meet. A reduced scheme is one whose local rings have no nonzero nilpotent elements. Equivalently, every affine open subscheme has a reduced ring of regular functions. We may define an integral scheme as a nonempty scheme for which the coordinate ring of every nonempty affine open subscheme is an integral domain. We shall show that this is equivalent to being reduced and irreducible. The nonempty convention matters: the empty scheme is reduced but is not irreducible or integral.

For a commutative ring $A$, the spectrum of a ring is reduced exactly when $A$ is a reduced ring. One direction follows since localization preserves reducedness. Conversely, if $a$ is a nonzero nilpotent element, choose a prime ideal containing its proper annihilator. Then $a/1$ cannot vanish at that localization, contradicting reducedness of its local ring.

The spectrum of a ring is irreducible exactly when its nilradical $N=\sqrt{(0)}$ is a prime ideal. To see the essential implication directly, if $ab\in N$, then $D(a)\cap D(b)=D(ab)$ is empty. Irreducibility forces $D(a)$ or $D(b)$ to be empty, hence $a\in N$ or $b\in N$. Also $N$ is proper because the spectrum is nonempty. Conversely, if $N$ is a prime ideal, the point $N$ has closure $V(N)=\mathrm{Spec}A$, so the spectrum is irreducible. Combining the two criteria gives

Now suppose $X$ is reduced and irreducible. Every nonempty affine open subscheme is also reduced and irreducible: irreducibility passes to nonempty open subsets, because their nonempty open subsets are nonempty opens of $X$. The affine criterion makes each coordinate ring an integral domain, so $X$ is integral. Conversely, suppose all its nonempty affine coordinate rings are integral domains. Their localizations show that $X$ is reduced. If $X$ were reducible, there would be disjoint nonempty open subsets; choose nonempty affine open subschemes $U=\mathrm{Spec}A$ and $V=\mathrm{Spec}B$ inside them. Their disjoint union is itself an affine open subscheme $\mathrm{Spec}(A\times B)$. Since $A,B$ are nonzero, $(1,0)(0,1)=0$ contradicts the domain condition. Therefore

The generic point of an integral scheme $X$ is the unique point $\eta$ with $\overline{\{\eta \}}=X$. For existence, take a nonempty affine open subscheme $V=\mathrm{Spec}A$. Its point $(0)$ has closure containing $V$, which is dense in $X$, so its closure in $X$ is all of $X$. For uniqueness, both proposed generic points lie in every nonempty open subset, hence in $V$, where the only dense point is $(0)$. The function field isHere the stalk description shows that the field of fractions is independent of the choice of nonempty affine open subscheme.

Every nonempty open subscheme $U$ contains $\eta$, so taking a germ there defines $\Gamma (U,{\mathcal{O}}_X)\to K(X)$. If a section has zero germ, restrict it to any nonempty affine open subscheme $V\subseteq U$. Its image in $\mathrm{Frac}\Gamma (V,{\mathcal{O}}_X)$ is zero. Since this coordinate ring is an integral domain, the section is zero on $V$. Such affines cover $U$, and the sheaf gluing axiom makes the section zero on $U$. Thus the generic-point embedding of regular functions is

For a nonreduced reducible fibre between integral schemes, take both source and target to be the affine line over $\mathbb{C}$, and use the ring homomorphismBoth coordinate rings are integral domains, so both schemes are integral. At the target closed point $t=0$, the scheme-theoretic fibre has ringby the Chinese remainder theorem. Its underlying space has two distinct closed points, so it is reducible. The class of $x(x-1)$ is nonzero but has square zero, so it is not reduced. Even a morphism between integral schemes can have a fibre consisting of two nonreduced double points.

The point to retain in the direct-image theorem is quasi-compactness of inverse images of affine opens and of overlaps; no separation assumption has been supplied. The underlying space of a Noetherian scheme is Noetherian, so every open subset of $X$ is quasi-compact. Fix an affine open subscheme $V=\mathrm{Spec}A$ of $Y$, put $W=f^{-1}V$, and choose a finite open cover by affine open subschemes $W={\bigcup }_i{U}_i$. For each pair $i,j$, choose a finite affine cover $(W_{ij\ell }{)}_{\ell }$ of $U_i\cap U_j$. Empty overlaps contribute no terms.

Write $M=\Gamma (W,\mathcal{F})$. The sheaf gluing axiom gives the exact sequencewhere $\delta$ is the difference of the two restrictions to each overlap chart. All terms are $A$-modules through $f$. For $a\in A$, the inverse image of $D(a)$ cuts each $U_i$ and $W_{ij\ell }$ by a principal open subscheme. Because $\mathcal{F}$ is a quasi-coherent sheaf, sections on these smaller affine charts are the corresponding module localizations at $a$. Exactness of localization and its commutation with finite products now identify the localized equalizer with the equalizer for the restricted cover. ThusThe isomorphisms respect restrictions, proving $(f_{\ast }\mathcal{F}){|}_V\cong \tilde{M}$ on the basis of principal open subschemes. Since $V$ was arbitrary, $f_{\ast }\mathcal{F}$ is a quasi-coherent sheaf. The finite covers of overlaps are what allow the proof to work for nonseparated Noetherian schemes; this is the quasi-coherence of direct image under a quasi-compact quasi-separated morphism.

For a quasi-coherent sheaf which is not coherent but has coherent direct image, use $f:{\mathbb{P}}_k^1\to \mathrm{Spec}k$ andThis infinite negative-twist sum with zero global sections is quasi-coherent: on each standard affine chart it is the sheaf associated with a direct sum of free rank-one modules. Its stalk at every point, modulo the maximal ideal, is an infinite-dimensional vector space. A finitely generated module would have a finite-dimensional quotient, so $\mathcal{F}$ is not a coherent sheaf.

Nevertheless, $\Gamma ({\mathbb{P}}^1,\mathcal{O}(-1))=0$. This follows from cohomology of twisting sheaves on projective space, or directly by gluing on the two standard affine charts: if $z=t_1/t_0$, a section is a polynomial $p(z)$ on the first chart and $q(z^{-1})$ on the second with $p(z)=z^{-1}q(z^{-1})$, which forces both to vanish. Global sections commute with this direct sum: they are the kernel of the difference map for the two-chart cover, and direct sums commute with that finite equalizer of modules. Hencewhich is coherent on $\mathrm{Spec}k$. Both schemes in this example are Noetherian.

For a finite morphism, such an example is impossible. On an affine open subscheme $V=\mathrm{Spec}A$ of the target, its inverse image is $\mathrm{Spec}B$, with $B$ a finite $A$-module. Write $\mathcal{F}{|}_{f^{-1}V}=\tilde{M}$. Its direct image corresponds to $M$ considered as an $A$-module. If the direct image is coherent, $M$ is finitely generated over $A$. The same generators also generate it over $B$, because $A$ acts through $B$. Since $B$ is Noetherian, $\tilde{M}$ is coherent. Conversely, a finite set of $B$-generators combined with a finite set of $A$-generators of $B$ gives finitely many $A$-generators of $M$. Thus coherence reflected by finite direct image gives the stronger equivalence

Finally, let $Y=\mathrm{Spec}k[x,y]$, let $X=Y\setminus \{(x,y)\}$ be the punctured affine plane, and let $j:X\hookrightarrow Y$ be the open immersion. Both are integral schemes, but $j$ is not an isomorphism of schemes because it omits a point. The cover $X=D(x)\cup D(y)$ givesinside the field of fractions $k(x,y)$. Indeed, in a reduced fraction, membership in the first localization forces every denominator factor to be associated to $x$, while membership in the second forces it to be associated to $y$. Unique factorization and coprimality force the denominator to be a unit. By the theorem just proved, $j_{\ast }{\mathcal{O}}_X$ is quasi-coherent on the affine $Y$, hence determined by this module of global sections. The natural map ${\mathcal{O}}_Y\to j_{\ast }{\mathcal{O}}_X$ corresponds to the identity of $k[x,y]$, soa coherent sheaf. This is a concrete case of codimension-two extension of regular functions on a normal variety.

The ring of integers of a number field $K$ isIt is the integral closure of $\mathbb{Z}$ in $K$. Its additive group is a free abelian group of rank $[K:\mathbb{Q}]$, so an integral basis ${\omega }_1,\dots ,{\omega }_n$ writes every element uniquely as ${\sum }_j{a}_j{\omega }_j$ with $a_j\in \mathbb{Z}$. Its field of fractions is $K$.^[ref]

For $K=\mathbb{Q}$, this gives ${\mathcal{O}}_K=\mathbb{Z}$; for $K=\mathbb{Q}(\sqrt{2})$, it gives $\mathbb{Z}[\sqrt{2}]$. The Gaussian integers give the corresponding example inside $\mathbb{Q}(i)$. The ring of integers is determined by the field, and need not equal $\mathbb{Z}[\theta ]$ for a chosen algebraic integer generator: in $\mathbb{Q}(\sqrt{5})$, the larger ring $\mathbb{Z}[(1+\sqrt{5})/2]$ contains $\mathbb{Z}[\sqrt{5}]$ with index two.

This is the setting for divisibility, global units, and prime ideal factorization. A ring of integers is a Dedekind domain, so every nonzero proper ideal factors uniquely into nonzero prime ideals, even when unique factorization of elements fails.^[ref]