Introduction to schemes course of the University of Oxford (C2.6)

Solutions to all three questions of the Introduction to schemes course of the University of Oxford, Part C, sat on 8 June 2023. Source: Oxford's released examination PDF, compilation pages 346–348; paper pages 1–3. The original numeric, letter and Roman labels are retained. The compactness hypotheses in the valuative criterion and the interpretation of the removed section in 3(e) are made explicit below.

Write $h_T(S)=\mathrm{Hom}(S,T)$ for the functor of points of a scheme $T$. A subfunctor $H\subseteq F$ is an open subfunctor if every natural transformation $h_T\to F$ has a pullback represented by an open subscheme $U\subseteq T$:The displayed natural isomorphism must identify the projection to $h_T$ with the natural transformation induced by $U\hookrightarrow T$. Equivalently, for each $\xi \in F(T)$, there is an open subscheme $U\subseteq T$ such that a morphism of schemes $a:S\to T$ factors through $U$ exactly when $F(a)(\xi )\in H(S)$. The pullback is taken in the functor category, so this tests arbitrary families, rather than just underlying points.

A Zariski cover of a scheme $T$ is a family of open immersions $(j_i:U_i\hookrightarrow T{)}_{i\in I}$ whose images cover its underlying topological space:There is no finiteness requirement. Intersections used in the sheaf on a site condition are the pullbacks $U_i{\times }_T{U}_j$, identified with the corresponding open subschemes $U_i\cap U_j$.

A presheaf $F:{\mathsf{Sch}}^{\mathrm{op}}\to \mathsf{Set}$ is a Zariski sheaf when every Zariski cover $(U_i\to T)$ gives an equalizer diagramThus compatible local elements glue to a unique element of $F(T)$. Compatibility means that the two restrictions agree on every intersection. This includes the empty Zariski cover of the empty scheme, so $F(\varnothing )$ is a singleton.

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

For an idempotent $e\in R$, define the principal open subset $D(e)\subseteq \mathrm{Spec}R$. Since $e(1-e)=0$, a prime ideal contains at least one of $e,1-e$, and it cannot contain both. ThereforeThus $D(e)$ is a clopen set.

Conversely, let $C$ be a clopen set in $T=\mathrm{Spec}R$. The structure sheaf glues the section $1$ on $C$ and the section $0$ on its complement to a unique global section $e\in \Gamma (T,{\mathcal{O}}_T)=R$. Checking on these two open subsets gives $e^2=e$, and $D(e)=C$. Conversely, the restriction of an idempotent $e$ to $D(e)$ is $1$ because $e$ is a unit there, and its restriction to $D(1-e)$ is $0$. Thus the two constructions are inverse, including for the zero ring:

Label the two charts of the affine line with doubled origin by $A,B$, and use $t$ for their common coordinate. Let $u:T=\mathrm{Spec}R\to X$ be an $R$-valued element of the functor of points. Its composite with $\varphi$ corresponds to a unique $r\in R$ under the affine-target adjunction for schemes:The closed subscheme $Z=V(r)=\mathrm{Spec}(R/(r))$ maps into the disjoint union of the two zero sections,The inverse image $Z_A$ of $A_0$ is a clopen set in $Z$. By the idempotent–clopen correspondence, it determines a unique idempotent $e\in R/(r)$, equal to $1$ on $Z_A$ and $0$ on the inverse image $Z_B$ of $B_0$. HenceOver $\mathbb{Z}$, the two “origins” are entire zero sections, rather than single closed points; this construction uses their full scheme-theoretic fibers.

Let $T=\mathrm{Spec}R$ and $Z=\mathrm{Spec}(R/(r))$. The idempotent $e$ determines the disjoint clopen sets $Z_A=D_Z(e)$ and $Z_B=D_Z(1-e)$. Each is closed in $T$. Define the open subsetsThey cover $T$ and satisfy $T_A\cap T_B=T\setminus Z=D(r)$. The global section $r$ defines morphisms of schemes $T_A\to A$ and $T_B\to B$, both sending the coordinate $t$ to $r$. On their intersection $r$ is a unit, so both maps land in the common open subscheme $A\cap B=\mathrm{Spec}\mathbb{Z}[t,t^{-1}]$, where they agree. They therefore glue uniquely:For an entirely explicit description, choose any lift $a\in R$ of $e$. Then $a(1-a)\in (r)$ andThe equality $a+(1-a)=1$ proves that these open subsets cover. The inclusion $D(a(1-a))\subseteq D(r)$ proves that their intersection is exactly $D(r)$. Their description as complements of $Z_A,Z_B$ proves independence of the chosen lift. In particular, no lift of $e$ to an idempotent of $R$ is required.

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.

For every valuation ring $D$ with fraction field $K$, consider compatible morphisms of schemesAn extension is a morphism of schemes $\ell :\mathrm{Spec}D\to X$ satisfying $\ell {|}_{\mathrm{Spec}K}=u$ and $f\ell =v$. The valuative criterion for universal closedness statesEquivalently, among quasi-compact morphisms, the existence condition characterizes universal closedness. A universally closed morphism is automatically a quasi-compact morphism; the converse direction of the criterion needs this compactness condition. All valuation rings are allowed, with arbitrary fraction fields; there is no Noetherian ring hypothesis and one cannot replace them by discrete valuation rings without additional assumptions.^[ref][ref]

Use the compatible pair $u,v$ and the notion of extension from (i). The valuative criterion for separatedness isA quasi-separated morphism means one whose diagonal morphism is a quasi-compact morphism. A separated morphism automatically has this property because its diagonal morphism is a closed immersion. The quasi-separated morphism hypothesis is needed in the converse valuative test for arbitrary schemes. Existence is not demanded.^[ref]

With the same compatible pair $u,v$, the valuative criterion for properness isEquivalently, one can replace finite type by being a quasi-compact morphism and locally of finite type. Necessity follows from the definition of a proper morphism and the existence and uniqueness criteria in (i) and (ii). Conversely, finite type supplies the compactness needed in (i), while being a quasi-separated morphism supplies the compactness of the diagonal morphism needed in (ii). Thus the extension condition gives both universal closedness and separatedness, and these together with finite type give a proper morphism. These hypotheses are part of the criteria, even though the question is phrased for unrestricted morphisms of schemes.^[ref]

Let $S^{\prime }\to S$ be a morphism of schemes and set $X^{\prime }=X{\times }_S{S}^{\prime }$, with projection $f^{\prime }:X^{\prime }\to S^{\prime }$. Given compatible morphisms of schemes $u^{\prime }:\mathrm{Spec}K\to X^{\prime }$ and $v^{\prime }:\mathrm{Spec}D\to S^{\prime }$, compose them with the projections to obtain a valuative problem for $f$. An extension $\ell :\mathrm{Spec}D\to X$ for that problem and the prescribed $v^{\prime }$ determine a unique morphism of schemesThe universal property of a fibre product shows that ${\ell }^{\prime }$ extends $u^{\prime }$ and lies over $v^{\prime }$. Conversely, every extension ${\ell }^{\prime }$ has such a first component $\ell$. Thus the sets of extensions in the two problems are in bijection.

If $f$ is a universally closed morphism, it is a quasi-compact morphism and satisfies existence. The allowed stability of quasi-compact morphisms under base change and (a)(i) prove that $f^{\prime }$ is a universally closed morphism. If $f$ is a separated morphism, it is a quasi-separated morphism and satisfies uniqueness. Stability of quasi-separated morphisms under base change and (a)(ii) prove that $f^{\prime }$ is a separated morphism.

If $f$ is a proper morphism, the same bijection supplies existence and uniqueness. The permitted stability under base change of quasi-compact morphisms, quasi-separated morphisms and locally of finite type supplies all remaining hypotheses of (a)(iii). Hence

Let $X\stackrel{f}{\to }Y\stackrel{g}{\to }S$ be morphisms of schemes. Start with $u:\mathrm{Spec}K\to X$ and $v:\mathrm{Spec}D\to S$ satisfying $gfu=v{|}_{\mathrm{Spec}K}$.

For existence, first apply the extension condition for $g$ to $fu$ and $v$, obtaining $w:\mathrm{Spec}D\to Y$. Then apply the extension condition for $f$ to $u,w$, obtaining $\ell :\mathrm{Spec}D\to X$. This $\ell$ solves the problem for $gf$. If both $f,g$ are universally closed morphisms, both are quasi-compact morphisms; their composite is a quasi-compact morphism by the allowed stability fact. Part (a)(i) therefore proves that $gf$ is a universally closed morphism.

For uniqueness, suppose ${\ell }_1,{\ell }_2$ are extensions for $gf$. Uniqueness for $g$ gives $f{\ell }_1=f{\ell }_2=w$, and uniqueness for $f$, now over this same $w$, gives ${\ell }_1={\ell }_2$. If $f,g$ are separated morphisms, their composite is a quasi-separated morphism by the allowed stability fact. Part (a)(ii) proves that $gf$ is a separated morphism.

Finally, if $f,g$ are proper morphisms, these arguments give both existence and uniqueness. Stability under composition of quasi-compact morphisms, quasi-separated morphisms and locally of finite type gives the other hypotheses of (a)(iii). Consequently

Let $x\in \mathrm{Frac}(D)$ satisfy a monic polynomial equationIf $x=0$, there is nothing to prove. Suppose $x\notin D$. The defining property of a valuation ring gives $y=x^{-1}\in D$. Multiplying the equation by $y^{n-1}$ yieldsa contradiction. ThereforeThis proof uses only the comparison of an element with its inverse; no discreteness or finite generation of the valuation is needed.

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

Write $X=\mathrm{Spec}B$, $S=\mathrm{Spec}A$ with $A,B$ integral domains, let $\alpha :A\to B$ be the ring homomorphism, and put $K=\mathrm{Frac}(B)$. We do not assume that $\alpha$ is injective. Write $A^{\prime }=\alpha (A)\subseteq B\subseteq K$.

Let $D\subseteq K$ be any valuation ring containing $A^{\prime }$, with fraction field $K$. The canonical morphism of schemes $\mathrm{Spec}K\to \mathrm{Spec}B$ and the ring homomorphism $A\to A^{\prime }\hookrightarrow D$ form a valuative problem for $f$. Since $f$ is a proper morphism, there is an extension. On coordinate rings it is a ring homomorphism $\beta :B\to D$ such thatis the original inclusion of $B$ into $K$. Therefore every element of $B$ lies in $D$. Since $D$ was arbitrary, the stated description of integral closure as an intersection of valuation rings givesThus every element of $B$ is an integral element over $A^{\prime }$, and hence over $A$: lift the coefficients of its monic polynomial along the surjection $A\to A^{\prime }$. The ring homomorphism $\alpha$ is therefore an integral extension in the sense of the question.

A proper morphism is of finite type, so in this affine situation $B=A[b_1,\dots ,b_s]$. If $b_i$ satisfies a monic polynomial of degree $n_i$, repeatedly reducing its powers shows that the finitely many monomialsgenerate $B$ as an $A$-module. Equivalently, apply the allowed equivalence between finite and integral plus finite type. Hence

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:

A Noetherian scheme is a quasi-compact scheme, so choose a finite affine open cover $X={\bigcup }_{i=1}^s\mathrm{Spec}{A}_i$ with all $A_i$ Noetherian rings. The nilradical sheaf restricts on this chart to the quasi-coherent sheaf associated to $\mathrm{Nil}(A_i)$: the nilradical commutes with localization.

By the allowed fact, choose $d_i\ge 1$ such that $\mathrm{Nil}(A_i{)}^{d_i}=0$ and put $m={\max }_i{d}_i$. On every chart, the ideal sheaf ${\mathcal{N}}^m$ is zero, and hence it is zero globally. Higher powers are also zero. ThusIf $X$ is empty, take $m=1$. The finite affine open cover is essential for choosing one bound valid everywhere.

Let $i:X_{\mathrm{red}}\hookrightarrow X$ be the reduced induced subscheme, with structure sheaf ${\mathcal{O}}_{X_{\mathrm{red}}}={\mathcal{O}}_X/\mathcal{N}$. Its underlying map of topological spaces is a homeomorphism, since every prime ideal contains the nilradical.

Put $\mathcal{G}=\mathcal{F}/\mathcal{N}\mathcal{F}$. Multiplication by $\mathcal{N}$ annihilates $\mathcal{G}$, so its action by ${\mathcal{O}}_X$ factors through ${\mathcal{O}}_X/\mathcal{N}$. Thus $\mathcal{G}$ is naturally a sheaf of modules over ${\mathcal{O}}_{X_{\mathrm{red}}}$. On an affine open subscheme $\mathrm{Spec}A$, with $\mathcal{F}=\tilde{M}$, it corresponds to $M/\mathrm{Nil}(A)M$ over $A/\mathrm{Nil}(A)$, so it remains a quasi-coherent sheaf.

The sheaf cohomology of a sheaf of modules is the sheaf cohomology of its underlying sheaf of abelian groups. Under the above homeomorphism, that underlying sheaf is identical on the two spaces: the global sections functor and its flasque resolutions are identical. Equivalently, use sheaf cohomology under a closed inclusion with $i_{\ast }\mathcal{G}$. Therefore there are canonical isomorphisms, in every degree,In particular, this gives the requested equality in degree one. Equality of the underlying topological spaces alone would not compare arbitrary coefficient sheaves; the factorization of the action explains why the indicated coefficients are the same.

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.

Solutions to all three questions of the Introduction to schemes course of the University of Oxford, Part C, sat on 1 June 2021. Source: Oxford's released examination PDF, compilation pages 103–106; paper pages 1–4. The original question and letter labels are retained; the paper has no Roman subparts. The correction to the printed closed-subset description in 1(b), and the meaning of the comparison of structure sheaves in 1(c), are explained where they arise.

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 quotient. Evaluation $\mathbb{Z}[t]\to \mathbb{Z}[i]$, $t\mapsto i$, is a surjective ring homomorphism whose kernel of a ring homomorphism is $(t^2+1)$. Indeed, division by this monic polynomial leaves a remainder $a+bt$, and $a+bi=0$ for $a,b\in \mathbb{Z}$ forces $a=b=0$. Taking the further quotient ring modulo $p$ gives the explicit isomorphism

All fibers, including the generic fiber. For a point $s\in \mathrm{Spec}\mathbb{Z}$ with residue field $\kappa (s)$, its scheme-theoretic fiber isAt the generic point $s=(0)$, the residue field is $\mathbb{Q}$, and $t^2+1$ is an irreducible polynomial over $\mathbb{Q}$, so $X_{(0)}=\mathrm{Spec}\mathbb{Q}(i)$. For a closed point $(p)$ the residue field is ${\mathbb{F}}_p$. The permitted criterion for a square root of $-1$ gives three cases. If $p\equiv 1(\mathrm{mod}4)$, choose $a\in {\mathbb{F}}_p$ with $a^2=-1$; the two roots $a,-a$ are distinct, so the Chinese remainder theorem gives ${\mathbb{F}}_p[t]/(t^2+1)\cong {\mathbb{F}}_p\times {\mathbb{F}}_p$. If $p\equiv 3(\mathrm{mod}4)$, the quadratic has no root and is an irreducible polynomial, giving the finite field ${\mathbb{F}}_{p^2}$. Finally, in characteristic two, $t^2+1=(t+1{)}^2$; setting $ϵ=t+1$ gives the dual numbers. ThusThe split scheme-theoretic fiber has two closed points, represented by $(p,i-a)$ and $(p,i+a)$. The inert scheme-theoretic fiber has one closed point, represented by $(p)$. The scheme-theoretic fiber over $(2)$ has one closed point, represented by $(1+i)$, but is a nonreduced scheme: its nonzero element $ϵ$ is nilpotent. These descriptions retain the residue fields and the nilpotent structure, which a count of points alone would lose.

The topology. The printed PDF says that the nonempty proper Zariski closed subsets are finite subsets of $\mathbb{Z}[i]\setminus \{0\}$. Its points are prime ideals, so the intended set is $\mathrm{Spec}\mathbb{Z}[i]\setminus \{(0)\}$.

Put $R=\mathbb{Z}[i]$. Since $R$ is a principal ideal domain, every nonzero prime ideal is a maximal ideal. Each such prime ideal contains a nonzero Gaussian integer $z$, and therefore contains the positive integer $z\overline{z}$. Its contraction to $\mathbb{Z}$ is consequently a nonzero prime ideal $(p)$, so the scheme-theoretic fibers just computed list all the closed points of $X$. The sole remaining point is the generic point $(0)$.

Every proper Zariski closed subset is $V((a))$ with $a\ne 0$: if its defining ideal were zero, it would be all of $X$. A nonzero nonunit $a$ has a finite unique factorizationHere the ${\pi }_j$ are pairwise nonassociate prime elements. Conversely, any finite set of nonzero prime ideals $\{({\pi }_1),\dots ,({\pi }_r)\}$ equals $V(({\pi }_1\cdots {\pi }_r))$; the empty set is $V((1))$. ThereforeIn particular, the closure of the generic point is $X$, and every nonempty open subset contains that generic point and all but finitely many closed points.

A nonidentity automorphism. Complex conjugationis a ring automorphism fixing $\mathbb{Z}$. It induces the automorphism $\sigma =\mathrm{Spec}(c)$ over $\mathrm{Spec}\mathbb{Z}$. Since $c(i)=-i\ne i$, this is not the identity morphism of schemes. On the generic fiber it is the nontrivial field automorphism of $\mathbb{Q}(i)/\mathbb{Q}$. On a split scheme-theoretic fiber it swaps the two closed points, since $i-a$ maps to $-(i+a)$. On an inert scheme-theoretic fiber it fixes the only point but induces the nonidentity finite-field Frobenius automorphism $z\mapsto z^p$ on its residue field, because $i^p=-i$. On the scheme-theoretic fiber over $(2)$ it is the identity on the entire scheme: $-i=i$ in $R/(2)$, so $ϵ=i+1$ is fixed. Thus