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On an integral scheme, for a Cartier divisor locally represented by $a_i\in K^{\ast }$, its divisor line bundle is the subsheaf of rational functions locally equal to $a_i^{-1}{\mathcal{O}}_X$. Unit ratios glue these free rank-one modules. Given a nonzero rational section of a line bundle $s=a_i{e}_i$, the maps $e_i\mapsto a_i^{-1}$ identify that invertible sheaf with ${\mathcal{O}}_X(D)$. Replacing $s$ by a nonzero rational multiple changes $D$ by a principal Cartier divisor.

If the normalization of an integral scheme is flat, then it is an isomorphism. On an affine chart, let $A\subseteq B\subseteq \mathrm{Frac}(A)$ be the integral closure. Integrality gives surjectivity on spectra, so flatness makes $B$ faithfully flat. For $b=u/v\in B$ with $u,v\in A$ and $v\ne 0$, faithful flatness gives $vB\cap A=vA$. Thus $u=va$ for some $a\in A$, and $b=a$. Consequently $B=A$. This proof does not need finiteness of the normalization.

For a finite surjective map of normal integral schemes with function fields $K\subseteq L$ in characteristic $p$, the condition $L^p\subseteq K$ lets the Absolute Frobenius morphism of the source factor through the map. On an affine chart $A\subseteq B$, send $b$ to $b^p\in A$: it lies in $K$, is integral over $A$, and $A$ is integrally closed. The condition $K\subseteq L^p$ instead lets the map factor after the source Frobenius. Send $a$ to its unique $p$th root in $L$; that root lies in the normal ring $B$ because it satisfies a monic polynomial over $B$. These rules localize and glue; injectivity of Frobenius in the function field supplies uniqueness.

A generic point is a point whose closure is the whole space. An integral scheme has a unique generic point; on an affine scheme associated to an integral domain, it corresponds to the zero prime ideal. Its local ring is the fraction field of the domain. On every nonempty affine chart of an integral scheme these fields identify with the same function field.

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.

For an injective ring homomorphism between nonzero integral domains $A\to B$, the scheme-theoretic fibre over the generic point of $\mathrm{Spec}A$ is $\mathrm{Spec}(S^{-1}B)$ with $S=A\setminus \{0\}$. Localization at these nonzero elements is a nonzero integral domain, so the generic fibre is a nonempty integral scheme. A surjective morphism between these affine schemes necessarily gives an injective ring map: a prime above $(0)$ contains its kernel.

An irreducible scheme is one whose underlying topological space is a nonempty irreducible topological space. It can have nilpotent elements in its structure sheaf; irreducibility concerns only the topology. A reduced scheme is an integral scheme precisely when it is irreducible.

A morphism of schemes between integral schemes can have a scheme-theoretic fibre that is neither reduced nor irreducible. For the map of affine lines given by $t=x^2(x-1{)}^2$, the fibre at $t=0$ has ring $k[x]/(x^2(x-1{)}^2)$. The Chinese remainder theorem identifies it with the product of two dual number rings, so it consists of two nonreduced double points. Its nilpotent structure is invisible to the set-theoretic fibre.

For an integral scheme, take on each affine chart the integral closure of its coordinate ring in its function field. Integral closure commutes with localization, so these rings glue to a normal scheme with an integral birational morphism to the original scheme. This is the normalization. It need not be finite without additional hypotheses; for an algebraic curve of finite type over a field it is finite. It is an isomorphism exactly when the original scheme is normal.

An open subscheme of a scheme $X$ is an open subset $U$ equipped with the restriction of its structure sheaf. Its inclusion is an open immersion. Every scheme is covered by affine open subschemes. Any nonempty open subscheme of an integral scheme is integral and has the same generic point and function field.

An integral scheme is nonempty, reduced and irreducible. For an affine scheme $X=\mathrm{Spec}R$, reducedness is equivalent to $\sqrt{(0)}=0$: nilpotent elements vanish in every stalk, and conversely a nonzero element has nonzero localization somewhere. Irreducibility is equivalent to the nilradical being prime. Indeed, if $ab$ lies in every prime, then $V(ab)=V(a)\cup V(b)=X$; irreducibility makes one of these closed subsets all of $X$, so $a$ or $b$ lies in the nilradical. Conversely, if the nilradical is prime, its point is a generic point of the whole spectrum. Thus reducedness and irreducibility together say that zero is a prime ideal, with $R\ne 0$. Therefore

For a global idempotent $e$, its germ in any local ring is either zero or one. To see this, $e(1-e)=0$, and at least one of $e,1-e$ is a unit. Multiplication by that unit forces the other to vanish. On each affine chart,These equalities also hold globally. Hence $V(1-e)$ is a clopen set. Conversely, on a clopen subset $U$ and its complement, glue the sections one and zero. They define a global idempotent $e_U$, and the two constructions are inverse. This proves the idempotent–clopen correspondence.

Assume each connected component is open. Every clopen subset is a union of connected components, since its intersection with a connected component is clopen in that component. Each individual component is also closed, so it defines a nonzero idempotent. A decomposition into two nonempty disjoint clopen subsets corresponds exactly to a sum of two nonzero orthogonal idempotents. A union containing at least two components admits such a decomposition by separating one component from the rest. A single component admits none. Consequently the precise conclusion is

The printed definition needs both nonzero and orthogonal. Literally forbidding every sum of two nonzero idempotents gives a false statement in characteristic two. For $X=\mathrm{Spec}({\mathbb{F}}_2\times {\mathbb{F}}_2)$, the component idempotent $(1,0)$ decomposes as $(1,1)+(0,1)$; both summands are nonzero idempotents but are not orthogonal. In fact each of the three nonzero idempotents in this ring decomposes into the other two, although $X$ has two open connected components. Excluding zero is also necessary: zero corresponds to the empty subset, not to a component. With the standard primitive idempotent definition the preceding proof establishes the intended bijection in every characteristic.

The affine space $X={\mathbb{A}}_k^n$ is an integral scheme, hence its underlying topological space is irreducible. Every nonempty open subset of an irreducible topological space is irreducible and in particular connected. A section of the constant sheaf with value the function field $K=k(t_1,\dots ,t_n)$ is a locally constant function, so on any such open subset it is constant. Thus its sections are $K$ on each nonempty open subset and zero on the empty set. All restriction maps are either identities on $K$ or maps onto zero.

The constant sheaf is therefore a flasque sheaf. It has its natural ${\mathcal{O}}_X$-module structure because a regular function on a nonempty open subset embeds in the function field and acts by multiplication. Part (i) now givesThis is an instance of constant sheaves on irreducible spaces are flasque; no restriction on $n$ is needed.

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 $\mathbb{C}(t_1,t_2)$.

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.

A fibre product of schemes $Z=X{\times }_{S}Y$ has projection morphisms to $X$ and $Y$ with equal composites to $S$, and for every scheme $T$ the induced map is a bijectionThis universal property determines $Z$ uniquely up to unique isomorphism. On affine charts over $\mathrm{Spec}R$, takeThe tensor product of commutative algebras has the required universal property for ring maps. Localization identifies these constructions on principal-open overlaps. Cover the base and its two inverse images by affine charts and glue the resulting affine products along those overlaps; their local universal properties give the global one.

For a counterexample to preservation of irreducibility, take $X=Y=\mathrm{Spec}\mathbb{C}$ and $S=\mathrm{Spec}\mathbb{R}$ with their usual structure morphisms. All three are integral schemes, but the Chinese remainder theorem givesThe product scheme is the disjoint union of two points, so it is not irreducible.

The rings $A,B$ are nonzero integral domains. Let $\eta$ be the generic point of $Y$, corresponding to $(0)\subseteq A$. Surjectivity provides a prime ideal $\mathfrak{q}\subseteq B$ with ${\alpha }^{-1}(\mathfrak{q})=0$. Since $\ker \alpha \subseteq {\alpha }^{-1}(\mathfrak{q})$, the ring homomorphism $\alpha$ is injective.

Put $K=\mathrm{Frac}A$ and $S=A\setminus \{0\}$. The scheme-theoretic fibre at $\eta$ isEvery element of $\alpha (S)$ is nonzero, so the localization $S^{-1}B$ is a nonzero integral domain. Its zero ideal is prime, ensuring that its spectrum is nonempty; an affine spectrum of an integral domain is an integral scheme. This proves the integrality of the generic fibre of an affine dominant morphism. In fact the proof only needs injectivity of $\alpha$, rather than surjectivity at every point.

Use the ring homomorphism $k[t]\to k[x]$ given by $t\mapsto x^2$. The source scheme $X=\mathrm{Spec}k[x]$ is an integral scheme. At the rational point $y=(t)$, its residue field is $k$, and its scheme-theoretic fibre isThe class of $x$ is nonzero but has square zero, so this is not a reduced scheme. This example works over every field. An integral total space therefore need not have reduced fibres.

First suppose $X$ is a nonempty integral scheme, with generic point $\eta$ and function field $K={\mathcal{O}}_{X,\eta }$. Since taking stalks preserves the inclusion $\mathcal{I}\hookrightarrow {\mathcal{O}}_X$, local freeness of rank two would give an injective $K$-linear mapcontradicting the dimension of a vector space.

For a nonempty Noetherian scheme that is reduced, there are finitely many irreducible components. Choose one, and remove the union of the others. The resulting nonempty open subscheme is irreducible and reduced, hence integral. Restricting the proposed locally free sheaf to it gives the contradiction above. Therefore no locally free ideal of rank two exists on a nonempty reduced Noetherian scheme.

The rank bound for locally free ideals on reduced schemes in fact removes the Noetherian assumption: on an affine open where the rank is fixed, localization at a minimal prime ideal gives a field, so a free ideal has rank at most one. Nonemptiness is necessary for the literal statement: on the empty scheme the zero sheaf is vacuously locally free of every stipulated rank. The question is interpreted with this usual nonempty hypothesis.