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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.

Let $S$ be a multiplicative subset, containing one and closed under products. On $A\times S$, useThe relation is reflexive and symmetric. For transitivity, if $u(ta-sb)=0$ and $v(rb-tc)=0$, then multiplication and addition give $uvt(ra-sc)=0$. Thus it is an equivalence relation. Write $a/s$ for the equivalence class. The localization of a ring $A_S=S^{-1}A$ has operationsThey are well defined. For example, replacing $a/s$ by $b/t$ in its sum with $c/r$ gives cross-numerator difference $r^2(ta-sb)$, and in its product with $c/r$ gives $rc(ta-sb)$. The same annihilating element $u$ kills both differences. Replacing the other input works identically. Associativity and distributivity then follow by putting all terms over the product of their denominators and using the ring identities in $A$. Addition and multiplication are commutative, with identities $0/1$ and $1/1$, and the displayed negative is an additive inverse. Hence $S^{-1}A$ is a commutative ring. If $0\in S$, the fraction relation identifies all elements, producing the zero ring.

The canonical ring homomorphism $\iota :A\to S^{-1}A$, $a\mapsto a/1$, makes each $s\in S$ a unit, with inverse $1/s$. The universal property of localization states that any ring homomorphism $h:A\to C$ sending $S$ into the units factors uniquely through $\iota$. Define the factor byIf $u(ta-sb)=0$, then $h(u)$ is invertible, so $h(t)h(a)=h(s)h(b)$, proving well-definedness. The fraction operations prove that $\tilde{h}$ is a ring homomorphism. Its value is forced by $a/s=\iota (a)\iota (s{)}^{-1}$, so it is unique. This property characterizes the localization up to a unique compatible isomorphism.

For a prime ideal $\mathfrak{p}$, its complement is a multiplicative subset, and the localization at a prime ideal is $A_{\mathfrak{p}}=(A\setminus \mathfrak{p}{)}^{-1}A$. Fractions with numerator outside $\mathfrak{p}$ are units; those with numerator in $\mathfrak{p}$ form the unique maximal ideal $\mathfrak{p}{A}_{\mathfrak{p}}$. Thus it is a local ring.

For the nilpotent question, suppose $a\ne 0$ and $a^n=0$. Its annihilator is a proper ideal, so it lies in some maximal ideal $\mathfrak{m}$. The fraction criterion says $a/1=0$ in $A_{\mathfrak{m}}$ only if $sa=0$ for some $s\notin \mathfrak{m}$, which is impossible by the choice of $\mathfrak{m}$. Thus $a/1$ is a nonzero nilpotent element in that localization. This proves reducedness is detected by prime localizations: if every prime localization is reduced, $A$ has no nonzero nilpotent elements.

For the domain question take $A=k\times k$ with $k$ a field. Its only prime ideals are ${\mathfrak{p}}_1=0\times k$ and ${\mathfrak{p}}_2=k\times 0$: a prime ideal must contain one of the orthogonal idempotents $(1,0),(0,1)$, and the corresponding quotient is $k$. Localizing at ${\mathfrak{p}}_1$ makes $(1,0)$ invertible and kills $(0,1)$, giving $A_{{\mathfrak{p}}_1}\cong k$; similarly $A_{{\mathfrak{p}}_2}\cong k$. But $(1,0)(0,1)=0$ with both factors nonzero. Therefore domain prime localizations do not force $A$ to be an integral domain, as expressed by domain prime localizations do not imply a domain.

Put $A=k[X]$. There is an explicit isomorphism from the principal open subset $U_f$ to the affine algebraic setIts inverse is projection, and both maps are regular maps. The coordinate ring isby the universal property of localization. Since regular functions on an affine variety are its coordinate ring, this proves the natural $k$-algebra isomorphism

Now $W=D(x)\cup D(y)$. A regular function on $W$ determines one element of the function field $k(x,y)$, and its restrictions belong to $k[x,y{]}_x$ and $k[x,y{]}_y$. Conversely, elements of their intersection glue because they agree on $D(xy)$. ThusIf a reduced fraction has denominator dividing a power of $x$ and also a power of $y$, unique factorization makes its denominator a unit. Therefore puncturing the plane does not change its ring of global regular functions:If $W$ were an affine variety, its inclusion $j:W\hookrightarrow {\mathbb{A}}^2$ would correspond to the identity isomorphism $k[x,y]\to k[W]$. The equivalence between affine varieties and their coordinate rings would make $j$ an isomorphism, contradicting the missing origin. Hence $W$ is not affine. This illustrates regular functions on the punctured affine plane: the coordinate ring alone does not recover a nonaffine variety.

A ring homomorphism is understood to preserve $1$. Start with $\phi :A\to B$. Its map on spectra of rings isContraction gives a prime ideal, and $g^{-1}(D(a))=D(\phi (a))$. Thus $g$ is continuous for the Zariski topology. On each principal open subscheme, the structure sheaf map is the ring homomorphismThese maps commute with restrictions, so they define a sheaf morphism ${\mathcal{O}}_X\to g_{\ast }{\mathcal{O}}_Y$. At $\mathfrak{q}$, with $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, its stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. The inverse image of the maximal ideal $\mathfrak{q}{B}_{\mathfrak{q}}$ is $\mathfrak{p}{A}_{\mathfrak{p}}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, let $(g,g^{\#})$ be a morphism of locally ringed spaces. Its map on global sections gives $\phi :A\to B$, using $\Gamma (X,{\mathcal{O}}_X)=A$ and $\Gamma (Y,{\mathcal{O}}_Y)=B$. Fix $\mathfrak{q}\in Y$ and put $\mathfrak{p}=g(\mathfrak{q})$. Compatibility with the stalk maps and the local homomorphism property giveConsequently $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, so the underlying map is forced. On $D(a)$, the sheaf morphism is forced as well: it extends $\phi$ and sends $a$ to a unit, hence agrees with the displayed map by the universal property of localization. The principal open subschemes form a basis, so the entire sheaf morphism is determined. Taking global sections of the construction recovers $\phi$. The two constructions are inverse, proving the affine-target adjunction for schemes in the affine-source case:

To describe the real affine plane scheme points, write $R=\mathbb{R}[t_1,t_2]$. It is a unique factorization domain of Krull dimension two. Its points are exactly the following prime ideals:

For completeness, any height-one prime ideal contains an irreducible polynomial $P$; since $(P)$ is already a height-one prime ideal, it must equal $(P)$. Every remaining nonzero prime ideal has height two and is maximal, by Krull dimension. For a nonreal pair, evaluation generates all of $\mathbb{C}$ over $\mathbb{R}$, so its kernel is maximal. Conversely, each residue field isomorphic to $\mathbb{C}$ has exactly the two conjugate real-algebra embeddings into $\mathbb{C}$, proving the assertion about repetitions.

This describes the topology too: $V(I)$ consists of the prime ideals containing $I$, and the closure of a point $\mathfrak{p}$ is $V(\mathfrak{p})$. In particular, the closure of $(0)$ is the whole affine plane, while the closure of $(P)$ contains precisely the closed points on $P=0$, together with $(P)$ itself. The spectrum is much larger than the set ${\mathbb{R}}^2$. For example, $(t_1^2+1)$ is a height-one point although its curve has no real points. Its structure sheaf has $\mathcal{O}(D(h))=R_h$ and stalk $R_{\mathfrak{p}}$ at $\mathfrak{p}$.

The induced morphism of schemesis contraction of prime ideals, with the structure sheaf maps given above. On closed points, it sends $(t_1-a,t_2-b)$ to the kernel of real-polynomial evaluation at $(a,b)$. A real closed point has one complex point above it; a nonreal closed point has two, interchanged by complex conjugation. The source generic point maps to the target generic point. The source height-one points are generated by irreducible complex polynomials $Q$. Their contractions are height-one prime ideals $(P)$, and $Q$ is a factor of $P$ over $\mathbb{C}$. An irreducible real $P$ either stays irreducible over $\mathbb{C}$ or splits into two distinct conjugate irreducible factors. Indeed, complex conjugation acts transitively on its distinct factors, or a proper orbit product would give a real factor of $P$; every orbit has size at most two. Repeated factors are excluded by separability in characteristic zero. Thus one or two height-one points lie above $(P)$.

The complexification fibres of a real scheme give a uniform description of all scheme-theoretic fibres, including the nonclosed points, is especially useful. The extension of coordinate rings isa free $R$-module with basis $1,s$. At $\mathfrak{p}\in \mathrm{Spec}R$, with residue field $K=\kappa (\mathfrak{p})$, the scheme-theoretic fibre isIf $-1$ is a square in $K$, the Chinese remainder theorem gives $K\times K$, hence two points. Otherwise it is a quadratic field extension, hence one point. The polynomial has no repeated root in characteristic zero, so all these scheme-theoretic fibres are reduced schemes. This also proves surjectivity. Conjugation acts on each two-point fibre by exchanging its points and fixes each one-point fibre. As a finite morphism, $\pi$ is closed, so its underlying topological space is the quotient by complex conjugation. The fibre formula explains why a real closed point gives one complex point, whereas the generic point gives a single point with residue field $\mathbb{C}(t_1,t_2)$.

Let $T=R\setminus P$. Since $P$ is a prime ideal, $T$ is a multiplicative subset containing $1$ and excluding $0$. The localization at a prime ideal isConcretely, its elements are equivalence classes of pairs $(r,s)\in R\times T$, written $r/s$, withThis extra multiplier is essential when $R$ has zero divisors. Addition and multiplication areThe equivalence criterion is compatible with these operations: after multiplying by the multipliers witnessing changes of representatives, the corresponding numerators agree. The commutative ring axioms follow from those of $R$; the identity is $1/1$.

The canonical ring homomorphism $\iota :R\to R_P$, $r\mapsto r/1$, supplies the algebra over a commutative ring structure, with scalar action $a(r/s)=ar/s$. It also has the universal property of localization: whenever a ring homomorphism $\psi :R\to A$ sends $T$ to units, the unique extension is $r/s\mapsto \psi (r)\psi (s{)}^{-1}$.

For later use, $R_P$ is a local ring with maximal ideal $P{R}_P$. Reduction of numerators and denominators gives a surjective ring homomorphism $R_P\to \mathrm{Frac}(R/P)$ with kernel $P{R}_P$, so this ideal is proper and maximal. Every fraction outside it has numerator outside $P$ and is a unit, with inverse $s/r$. Thus it is the unique maximal ideal.

A multiplicative subset $S\subseteq R$ contains $1$ and satisfies $s,t\in S\Rightarrow st\in S$. The localization of a ring consists of fractions $r/s$ modulo the relationIts structure map $\iota :R\to S^{-1}R$ makes every $s\in S$ invertible. The universal property of localization says that if $f:R\to T$ is any ring homomorphism for which every $f(s)$ is a unit, there is one unique homomorphism $\tilde{f}:S^{-1}R\to T$ satisfying $\tilde{f}\iota =f$, namely