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On an affine scheme, a morphism between quasi-coherent sheaves is determined by the induced map on global sections, because its maps on principal open subsets are their localizations. Exactness of localization gives the displayed identity. Thus the kernel sheaf of a morphism between quasi-coherent sheaves is quasi-coherent on an arbitrary scheme, by applying the affine argument locally. No finite generation is needed.

A morphism of schemes $f:X\to S$ is locally of finite type if, over every affine open subscheme $\mathrm{Spec}A\subseteq S$, its inverse image is covered by affine schemes $\mathrm{Spec}B$ with $B$ a finitely generated $A$-algebra. It is of finite type exactly when it is also a quasi-compact morphism.

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 Zariski sheaf $F$ is a functor represented by a scheme if it has a set of open subfunctors represented by affine schemes whose pullbacks cover every test scheme. The overlaps are open subschemes in both charts, and their identifications satisfy the compatibility on triple overlaps. Gluing of schemes along open subschemes produces a representing scheme; the Zariski sheaf condition identifies its functor of points with $F$. Conversely, every affine open cover of a scheme supplies such a presentation. A family of local charts covers in this sense without requiring every global element to belong to a single chart.

No. Let $k={\mathbb{F}}_2$ be the field with two elements, put $A=k[\epsilon ]/({\epsilon }^2)$, and take the closed immersioninduced by the ring homomorphism $q:A\twoheadrightarrow k$ with $q(\epsilon )=0$.

For every field $K$, a ring homomorphism $\alpha :A\to K$ must send $\epsilon$ to zero: $\alpha (\epsilon {)}^2=0$, and a field has no nonzero nilpotent elements. Consequently $\alpha$ factors uniquely through $q$. Passing from ring homomorphisms to morphisms of schemes between affine schemes gives a bijectionThis concerns all morphisms of schemes, without requiring them to be over $k$: if $K$ does not have characteristic two, both sets are empty; otherwise each contains exactly one element.

Nevertheless, $f$ is not an isomorphism of schemes. Its map on global sections is $q$, whose kernel of a ring homomorphism contains the nonzero element $\epsilon$, so $q$ is not a ring isomorphism. The target is the nonreduced scheme defined by the dual numbers, whereas the source is a reduced scheme. This illustrates why field-valued points do not detect nilpotents.

Yes. First, the hypothesis implies that $f(T)$ is injective for every scheme $T$, even when $T$ is not an affine scheme. Indeed, suppose $a,b:T\to X$ satisfy $f\circ a=f\circ b$. Choose an affine open cover $T={\bigcup }_{\lambda }{V}_{\lambda }$. On each $V_{\lambda }$, injectivity of $f(V_{\lambda })$ gives $a{|}_{V_{\lambda }}=b{|}_{V_{\lambda }}$. By gluing morphisms of schemes, $a=b$.

Now choose an affine open cover $Y={\bigcup }_i{U}_i$, and denote its inclusion maps by $j_i:U_i\hookrightarrow Y$. Surjectivity of $f(U_i)$ gives morphisms of schemes $g_i:U_i\to X$ with $f\circ g_i=j_i$. On $U_i\cap U_j$, the two restrictions have the same composite with $f$. By the injectivity just established for arbitrary test schemes,In particular, the overlaps need not be affine schemes. Gluing morphisms of schemes supplies $g:Y\to X$ with $g{|}_{U_i}=g_i$, and $f\circ g=1_Y$ because this holds on every $U_i$.

Finally,Injectivity of $f(X)$, already proved without an affineness assumption, gives $g\circ f=1_X$. ThereforeThis proves the relevant instance of affine test schemes determine scheme morphisms: the restricted functor of points still retains the information needed to construct the inverse.

The localization of a ring $R_r$ is generated over $R$ by the single element $1/r$, since every fraction can be written asThe presentation of a localization at one element isTo verify the presentation even when $R$ has zero divisors, observe that $r$ is invertible in $R[t]/(rt-1)$ with inverse $t$. Hence the map from $R$ to this quotient ring extends uniquely to $R_r$. This extension and the displayed map are inverse ring homomorphisms: they agree on $R$ and send the designated inverse of $r$ to the designated inverse of $r$.

Thus $R_r$ is a finitely generated algebra over a ring, and the criterion for a morphism of finite type between affine schemes applies to $u:\mathrm{Spec}{R}_r\to \mathrm{Spec}R$. There is just one affine scheme in the required cover of the source, with one algebra generator. Therefore $u$ is of finite type. If $r$ is nilpotent, the same presentation is the zero ring, and the source is the empty scheme; the conclusion still holds.

Use homogeneous coordinates $x_0,\dots ,x_r$ on projective space and letThese $r+1$ open subschemes cover $X$. Each is a closed subscheme of the affine scheme $D_{+}(x_i)$, hence is an affine scheme.

Every nonempty finite intersection is affine as well. Indeed, inside $D_{+}(x_{i_0})$ the intersection with $D_{+}(x_{i_1}),\dots ,D_{+}(x_{i_p})$ is the principal open subscheme obtained by invertingIntersecting it with the closed subscheme $X$ gives $U_{i_0}\cap \cdots \cap U_{i_p}$, again an affine scheme.

For any quasi-coherent sheaf $F$, vanishing of quasi-coherent cohomology on an affine scheme makes every such intersection acyclic for $F$. The acyclic cover theorem therefore identifies the sheaf cohomology of $F$ with the Čech cohomology of this cover. Its alternating Čech cochain complex has termswhich vanish for $p>r$: one cannot choose $p+1$ distinct indices from $r+1$ indices. ConsequentlyThe choice $N=\max \{1,r\}$ meets the strict requirement $N>0$ even when $r=0$. It is independent of $F$, and applies to all quasi-coherent sheaves, including those which are not coherent sheaves. This is the cohomological dimension bound from an affine cover.

Let $X=\mathrm{Spec}R$ be an affine scheme, and let $(U_{\lambda }{)}_{\lambda \in \Lambda }$ be an open cover. Write each complement as a Zariski closed subset $X\setminus U_{\lambda }=V(I_{\lambda })$ for an ideal $I_{\lambda }\subseteq R$. The intersection rule for Zariski closed subsets givesA proper ideal lies in a maximal ideal, which is a prime ideal; consequently $V(J)=\varnothing$ forces $J=R$. Thus $1\in {\sum }_{\lambda }{I}_{\lambda }$. By the definition of a sum of ideals, this membership is witnessed by finitely many elements:Hence ${\sum }_{j=1}^r{I}_{{\lambda }_j}=R$, so ${\bigcap }_{j=1}^{r}V(I_{{\lambda }_j})=\varnothing$, and therefore $X={\bigcup }_{j=1}^r{U}_{{\lambda }_j}$. If $R$ is the zero ring, $X$ is empty and the empty subcover suffices. This provesThe proof uses only complements, the intersection rule for Zariski closed subsets, and the finite expression of an element in a sum of ideals.

The immersions and gluing. For $p>0$, the prime ideals of the localization ${\mathbb{Z}}_{(p)}$ are $(0)$ and $(p){\mathbb{Z}}_{(p)}$. Inverting $p$ gives ${\mathbb{Z}}_{(p)}[1/p]=\mathbb{Q}$, sois an open immersion sending the generic point to the generic point. For $p=0$, ${\mathbb{Z}}_{(0)}=\mathbb{Q}$, and take ${\psi }_0$ to be the identity. Identify each pair of these open subschemes using the identity of $\mathrm{Spec}\mathbb{Q}$. The identities compose to the identity on every triple overlap, so gluing of schemes along open subschemes applies. No finiteness of the family is required.

The resulting scheme $Y$ has one common generic point $\eta$ and a distinct closed point $y_p$ for every positive prime number $p$. Its affine open subschemes $U_p=\{\eta ,y_p\}$, for $p>0$, intersect pairwise in $U_0=\{\eta \}$. A subset $W\subseteq Y$ is open precisely when its intersection with every $U_p$ is open. In the two-point topological space $U_p$, the open subsets are $\varnothing$, $\{\eta \}$ and $U_p$. Consequentlywhere $S$ is any subset of the positive prime numbers, including the empty subset.

The canonical morphism. The ring homomorphisms $\mathbb{Z}\to {\mathbb{Z}}_{(p)}$ give morphisms of schemes $U_p\to \mathrm{Spec}\mathbb{Z}$ agreeing on the overlaps, hence a morphism of schemes $f:Y\to \mathrm{Spec}\mathbb{Z}$. By uniqueness of the structure morphism, this is the required map. Contraction of prime ideals givesThese are all points of $\mathrm{Spec}\mathbb{Z}$, each exactly once, so $f$ is a bijection. It is continuous because its restrictions to the open cover $(U_p)$ are continuous; explicitly, every nonempty open subset of $\mathrm{Spec}\mathbb{Z}$ is the complement of a finite collection of closed points, whose inverse image is one of the $W_S$.

Sections on every open. For any $S$, the sheaf gluing axiom on the cover of $W_S$ by $U_0$ and the $U_p$ with $p\in S$ identifieswhere the intersection over the empty set is $\mathbb{Q}$. Indeed, a section on $U_p$ is an element of ${\mathbb{Z}}_{(p)}$; restrictions to every overlap are its inclusion into $\mathbb{Q}$. Compatibility means that all these elements are the same rational number. In lowest terms $a/b$, this rational number lies in the intersection exactly when no prime number in $S$ divides $b$. Equivalently, this is the integer localization at a set of primes obtained by inverting all positive prime numbers outside $S$.

Now let $V\subseteq \mathrm{Spec}\mathbb{Z}$ be nonempty and open. Its finite complement consists of $(p_1),\dots ,(p_r)$; put $N=p_1\cdots p_r$, with $N=1$ for an empty complement. Then $V=D(N)$ andThe equality follows by the denominator criterion above. The map $f^{\#}$ is this same equality of subrings of $\mathbb{Q}$, so it is a ring isomorphism. Over the empty open subset both rings of sections are the zero ring. These identifications respect every restriction map. Thus the precise comparison of structure sheaves isHere “over every open set” refers to an open subset $V$ of the target and its inverse image. The preceding intersection formula additionally describes the sections on every open subset of $Y$, including ones whose images are not open in the target.

Topology and finiteness. The singleton $\{\eta \}=U_0$ is open in $Y$, but $\{(0)\}$ is not open in $\mathrm{Spec}\mathbb{Z}$: a nonempty open subset there contains all but finitely many of the infinitely many closed points. Therefore $f$ is not a homeomorphism, despite being a continuous bijection and giving the indicated isomorphism of structure sheaves after direct image.

The open cover $(U_p{)}_{p>0}$ of $Y$ has no finite subcover, because $y_q$ belongs only to $U_q$. Thus $Y$ is not a quasi-compact scheme, and part (a) rules out an affine scheme. Each ${\mathbb{Z}}_{(p)}$ is a Noetherian ring, being a localization of the Noetherian ring $\mathbb{Z}$. Hence the $U_p$ form an affine open cover by spectra of Noetherian rings, proving that $Y$ is a locally Noetherian scheme. A Noetherian scheme is a quasi-compact scheme, so

The open affine representability criterion says that $F$ is a functor represented by a scheme exactly when $F$ is a Zariski sheaf and admits a family of open subfunctors $F_i\subseteq F$, each represented by an affine scheme, which covers $F$ in the following sense: for every scheme $T$ and every $\xi \in F(T)$, the open subschemes representing $h_T{\times }_F{F}_i$ cover $T$. In particular, the covering condition does not demand that every $\xi$ lie in one $F_i(T)$.

In symbols, for affine schemes $X_i$,The family is indexed by a set. For necessity, take an affine open cover of a representing scheme; morphisms of schemes glue over a Zariski cover. For sufficiency, $F_i{\times }_F{F}_j$ is represented by an open subscheme of each $X_i,X_j$. The two descriptions identify these overlaps and satisfy the compatibility on triple overlaps because all maps come from the inclusions into $F$. Apply gluing of schemes along open subschemes to obtain $X$. Local identification gives $h_X\to F$, and the Zariski sheaf condition makes this a natural isomorphism: its elements can be constructed and compared on the pulled-back open cover.

A morphism of schemes $f:X\to Y$ defines a natural transformation $g_f:g_X\to g_Y$ by $u\mapsto f\circ u$. We construct the inverse correspondence.

Let $\alpha :g_X\to g_Y$ be a natural transformation, and choose an affine open cover $(j_i:U_i\hookrightarrow X)$. Regard $j_i$ as an element of $g_X(U_i)$ and setHere and below the equivalence of categories in the question lets us write the restricted functor of points on affine schemes, contravariantly, rather than on commutative rings, covariantly. If $W\subseteq U_i\cap U_j$ is an affine open subscheme, naturality for its two inclusions givesSuch $W$ cover the overlap even when that overlap is not an affine scheme. Consequently the $f_i$ agree and glue to a morphism of schemes $f:X\to Y$.

It remains to check all elements of the restricted functor of points, not only the chosen inclusions. Given an affine scheme $T$ and $u:T\to X$, cover each $u^{-1}(U_i)$ by affine open subschemes $V$. On such a $V$, factor $u{|}_V=j_i\circ v$. Applying naturality first to $v:V\to U_i$ and then to $V\hookrightarrow T$ givesThese local equalities prove ${\alpha }_T(u)=f\circ u$. The glued $f$ is unique, since its restrictions to all $U_i$ are prescribed. Henceso the assignment is fully faithful.

If $g_X\cong g_Y$, the natural isomorphism and its inverse correspond to morphisms of schemes $f:X\to Y$ and $h:Y\to X$. Their compositions induce the identity natural transformations, and faithfulness implies $hf={\mathrm{id}}_X$, $fh={\mathrm{id}}_Y$. Conversely, an isomorphism of schemes induces a natural isomorphism. Thus

Starting from $(r,e)$, the construction in (iii) preserves the composite to the affine line and sends $Z_A$ to the $A$ zero section and $Z_B$ to the $B$ zero section. Part (ii) therefore recovers $(r,e)$.

Conversely, start with $u:T\to X$ and construct $(r,e)$. The only points of $X$ outside $A$ lie on $B_0$, and the only points outside $B$ lie on $A_0$. Consequently $u^{-1}(A)=T\setminus Z_B=T_A$ and $u^{-1}(B)=T\setminus Z_A=T_B$. On each of these open subsets, $u$ is the unique morphism of schemes to the appropriate affine line with coordinate $r$. Thus part (iii) recovers $u$.

Finally, a ring homomorphism $R\to R^{\prime }$ sends $r$ to $r^{\prime }$ and induces $R/(r)\to R^{\prime }/(r^{\prime })$, sending $e$ to an idempotent $e^{\prime }$. Inverse images of the clopen sets $Z_A,Z_B$ are precisely those determined by $e^{\prime }$. The constructions therefore commute with this base change. We have a natural bijectionso $F$ is a functor represented by a scheme, represented by the affine line with doubled origin. The representing scheme is not required to be an affine scheme.

Let $D$ be a valuation ring, $K$ its fraction field, and let $u:\mathrm{Spec}K\to X$, $v:\mathrm{Spec}D\to S$ be a compatible pair. Since $D$ is a local ring, every open subset of $\mathrm{Spec}D$ containing its unique closed point is the whole space: a principal open subset containing the maximal ideal is defined by a unit. Choose an affine open subscheme $V=\mathrm{Spec}A\subseteq S$ containing the image under $v$ of that closed point. The entire $v$ factors through $V$.

Because $f$ is a finite morphism, $f^{-1}(V)=\mathrm{Spec}B$ with $B$ a finite $A$-module. The valuative problem is the compatible ring homomorphismsEvery $b\in B$ is an integral element over $A$. Mapping its monic polynomial equation to $K$ proves that $\psi (b)$ is an integral element over $D$. Part (c) gives $\psi (b)\in D$. Thus $\psi$ factors uniquely as a ring homomorphism $B\to D\hookrightarrow K$, producing the unique extension $\mathrm{Spec}D\to X$.

For completeness, a finite morphism is a quasi-compact morphism and of finite type: over $V$ its inverse image is an affine scheme, and finite $A$-module generators also generate it as an $A$-algebra. It is also a separated morphism. Affine-locally its diagonal morphism corresponds to the surjective multiplication ring homomorphismand hence is a closed immersion. In particular it is a quasi-separated morphism. The valuative criterion for properness now applies, giving

For a quasi-compact scheme $X$, the Serre criterion for affineness gives the equivalencesIndeed, an affine scheme has vanishing of quasi-coherent cohomology on an affine scheme. Conversely, vanishing of the first sheaf cohomology for every quasi-coherent sheaf includes every quasi-coherent sheaf of ideals, which suffices for the criterion. No separated scheme hypothesis is needed in this formulation. Nor may one test only the structure sheaf.^[ref]

Suppose, to the contrary, that $H^1(X,\mathcal{F})=0$ for every quasi-coherent sheaf $\mathcal{F}$. Since $X$ is a quasi-compact scheme, part (a) makes it an affine scheme.

Every affine scheme $X=\mathrm{Spec}A$ is a separated scheme: its diagonal morphism over $\mathrm{Spec}\mathbb{Z}$ is induced by the surjective ring homomorphism $A{\otimes }_{\mathbb{Z}}A\to A$, $a\otimes b\mapsto ab$, so it is a closed immersion. This contradicts the assumption that $X$ is not a separated scheme. ThereforeIn fact, the version of the Serre criterion for affineness using ideals shows that $\mathcal{F}$ may be chosen to be a quasi-coherent sheaf of ideals. This argument does not use a comparison with Čech cohomology on the nonseparated scheme.

Let $U=\mathrm{Spec}A$ and $V=\mathrm{Spec}B$ be affine open subschemes of the separated scheme $X$. Its diagonal morphism is a closed immersionRestricting this closed immersion to the open subscheme $U{\times }_{\mathrm{Spec}\mathbb{Z}}V$ identifies its inverse image with $U\cap V$. Henceis a closed immersion. A closed subscheme of an affine scheme is an affine scheme, soThe argument iterates to every finite intersection of affine open subschemes. The empty intersection of two disjoint charts causes no exception: the empty scheme is $\mathrm{Spec}(0)$.

Suppose $X=U_1\cup \cdots \cup U_n$ with each $U_i$ an affine open subscheme. Every affine scheme is a quasi-compact topological space: a cover by principal open subsets $D(f_{\lambda })$ means that the $f_{\lambda }$ generate the unit ideal; an expression $1={\sum }_{j=1}^r{a}_j{f}_{{\lambda }_j}$ gives a finite subcover. Every open cover has a refinement by principal open subsets, so this proves the general covering property.

Restrict an arbitrary open cover of $X$ to each $U_i$ and choose a finite subcover there. The union of these finitely many finite families covers $X$. Thus $X$ is a quasi-compact scheme.

Now assume $X$ is a separated scheme. By (c), every finite intersection of the $U_i$ is an affine scheme, and hence has vanishing positive sheaf cohomology for the structure sheaf. This affine open cover therefore computes sheaf cohomology using its Čech cochain complex. Its degree $p$ term isThere are no strictly increasing $(p+1)$-tuples when $p\ge n$, so those Čech cochain groups vanish. ConsequentlyThe same argument works for every quasi-coherent sheaf. The relevant comparison requires the acyclicity of the finite intersections; the bare assertion that arbitrary Čech cohomology equals sheaf cohomology would be too broad.

Interpretation of the removed locus. Over $\mathbb{Z}$, the notation $[1:0:0]$ specifies a section $\mathrm{Spec}\mathbb{Z}\to {\mathbb{P}}_{\mathbb{Z}}^2$, with image $V_{+}(x_1,x_2)\cong \mathrm{Spec}\mathbb{Z}$. We remove this entire closed subscheme. This is the interpretation forced by the asserted open cover $X=U_1\cup U_2$. Removing one closed point in a particular fiber would not give that open cover. The native PDF uses the same notation as the transcription.

Put $a=x_0/x_1$, $b=x_2/x_1$ on $U_1$. On $U_2$ put $c=x_0/x_2$, $d=x_1/x_2$. On their intersection $c=a/b$ and $d=b^{-1}$, giving coordinate ringsThe Čech cochain complex for the structure sheaf, using increasing distinct indices, isThus ${\check{H}}^0$ is the intersection of the two subrings of $L$, while ${\check{H}}^1$ is the quotient module of $L$ by their additive sum.

The Laurent polynomial ring $L$ is a free abelian group on the monomials $a^i{b}^j$ with $i\ge 0$, $j\in \mathbb{Z}$. The first subring has the basis $j\ge 0$. The second has the basisso its allowed exponents satisfy $j\le -i$. A monomial belongs to both precisely when $i=j=0$. The monomials belonging to neither satisfy $i\ge 2$ and $-i<j<0$. Since the subgroups are spanned by subsets of the same basis, their sum introduces no other relations. ThereforeHere the square brackets denote cohomology classes, and the direct sums are as abelian groups. In particular, $[a^2{b}^{-1}]=[x_0^2/(x_1{x}_2)]$ is a nonzero cohomology class. This scheme is a separated scheme, being an open subscheme of projective space over $\mathbb{Z}$, and is a quasi-compact scheme by (d). Both charts and their intersection are affine schemes, so this Čech cohomology equals sheaf cohomology. The nonzero first sheaf cohomology of the quasi-coherent structure sheaf rules out an affine scheme:

Let $\mathcal{F}$ be any quasi-coherent sheaf on $X$, and choose $m$ from (i). Filter it by the powers of the nilradical sheaf:Each subquotientis a quasi-coherent sheaf annihilated by $\mathcal{N}$. Apply (ii) to the quasi-coherent sheaf ${\mathcal{N}}^j\mathcal{F}$. Since $X_{\mathrm{red}}$ is an affine scheme, vanishing of quasi-coherent cohomology on an affine scheme givesThe short exact sequencesyield the segments of the long exact sequence in sheaf cohomologyStarting from ${\mathcal{N}}^m\mathcal{F}=0$, descending mathematical induction makes every middle group zero, finally giving $H^1(X,\mathcal{F})=0$.

This holds for every quasi-coherent sheaf $\mathcal{F}$. Since a Noetherian scheme is a quasi-compact scheme, the Serre criterion for affineness in (a) applies and provesThis is an instance of the general fact that affineness is invariant under nilpotent thickening.