Affine Schemes and the Structure Sheaf

If $D(g)\subseteq D(f)$, localization universality gives the canonical restriction $A_f\to A_g$; for equal opens the two restrictions are inverse.

Under $D(f)\cong \mathrm{Spec}(A_f)$, the cover becomes the distinguished cover by the images of the $g_i$. The spectrum-cover lemma makes those images generate the unit ideal in $A_f$, so a finite subfamily already generates $1$. The standard localization calculation for a finite unit-ideal cover then glues every compatible family in the $(A_f{)}_{g_i}\cong A_{g_i}$ uniquely to an element of $A_f$.

This includes the empty case: if $D(f)=\mathrm{\varnothing }$, then $f$ lies in every prime ideal, so it is nilpotent; hence $A_f$ is the zero ring, and the empty compatible family glues uniquely to its sole element.

Thus gluing and uniqueness hold for every cover of a distinguished open by distinguished opens, not only for finite covers. This is precisely the sheaf axiom for the localization presheaf on the distinguished-open basis, so steps 2.1 and 2.2 prove the claim.

The preceding lemma gives a sheaf of rings $\tilde{A}$ on the distinguished-open basis.

For an open $U$, let $\mathcal{O}(U)$ be the ring of families $(s_x{)}_{x\in U}$ of germs of $\tilde{A}$ such that every $x\in U$ has a distinguished neighborhood $D(f)\subseteq U$ on which the family is represented by one section of $\tilde{A}(D(f))$. Restriction discards the germs outside the smaller open, and ring operations are pointwise. These maps make $\mathcal{O}$ a presheaf of rings; for $U=\mathrm{\varnothing }$, the empty germ family is its unique element.

Local representability is itself local, so compatible families in the rings $\mathcal{O}(U_i)$ glue uniquely by taking their pointwise germs. Hence $\mathcal{O}$ is a sheaf of rings. If $U=D(f)$ is distinguished, the map from $\tilde{A}(D(f))$ to its family of germs is bijective by locality and gluing for the basis sheaf in step 1.1. Thus $\mathcal{O}$ agrees with the localization assignment on every distinguished open.

If $\mathcal{G}$ is any other sheaf of rings with the same basis restriction, its sections on each open $U$ map to the locally representable germ families of step 2.1. The sheaf axiom makes this map bijective and compatible with restrictions, so $\mathcal{G}\cong \mathcal{O}$ uniquely through the prescribed basis identifications.

The extended sheaf agrees with the basis assignment on $D(f)$, so its sections are $A_f$.

Its restriction along $D(g)\subseteq D(f)$ is the prescribed canonical localization map $A_f\to A_g$.

If $D(f)=\mathrm{\varnothing }$, then $A_f=0$, the unique ring of sections on the empty open.

The opens $D(f)$ with $f\mathrm{\notin }\mathfrak{p}$ are cofinal neighborhoods of $\mathfrak{p}$, and their sections are $A_f$.

Hence the stalk is .

This colimit is $A_{\mathfrak{p}}$ by the universal property of localization.

At $\mathfrak{p}$, the stalk is canonically isomorphic to [given] $A_{\mathfrak{p}}$.

The ring $A_{\mathfrak{p}}$ is local, with maximal ideal [step 1.1] $\mathfrak{p}{A}_{\mathfrak{p}}$, so the canonically isomorphic stalk is local as well.

Thus every stalk is local, which is exactly the locally ringed condition.

$D(1)=\mathrm{Spec}A$.

The basic-open calculation gives $\mathrm{\Gamma }(D(1),\mathcal{O})=A_1$.

The canonical map $A\to A_1$ is an isomorphism, also for $A=0$.

By [F1], the underlying map is a homeomorphism from $\mathrm{Spec}(A_f)$ onto $D(f)$.

On $D(g)\subseteq D(f)$ the relevant section rings are $(A_f{)}_{g\mathrm{/}1}$ and $A_g$, canonically isomorphic and compatible with restrictions.

The basic opens cover $D(f)$, so the preceding identifications give an isomorphism of locally ringed spaces.

By [F1], the stalk map sends $a\mathrm{/}s$ to $\phi (a)\mathrm{/}\phi (s)$, for $s\mathrm{\notin }\mathfrak{p}$.

This image lies in $\mathfrak{q}{B}_{\mathfrak{q}}$ exactly when $\phi (a)\in \mathfrak{q}$, because its denominator is outside the prime ideal.

Thus [F2] identifies the inverse image of $\mathfrak{q}{B}_{\mathfrak{q}}$ with $\mathfrak{p}{A}_{\mathfrak{p}}$, so the map is local.

By [F2], every ring map $A\to B$ induces a locally ringed-space morphism $\mathrm{Spec}B\to \mathrm{Spec}A$.

A morphism $u$ gives a global-sections map $u^{\mathrm{♯}}:A\to B$ using [F1].

Locality determines its point map by contraction and localization determines each basic-open section map, so $u=\mathrm{Spec}(u^{\mathrm{♯}})$.

The constructions of steps 1.1--2.1 are inverse and natural; global sections is the quasi-inverse.

If $u$ is an isomorphism, [F1] carries its inverse to an inverse of the associated ring map $A\to B$.

If $A\to B$ is an isomorphism, its inverse ring map induces the inverse of $u$.

These two implications prove the claim.

A closed point is maximal by [F2], and [F1] identifies every such ideal with a kernel $A\to k$.

Every unital $k$-algebra map $A\to k$ is surjective, so its kernel is maximal and hence closed by [F2].

Thus classical $k$-points and closed points agree, but this makes no assertion that every prime is maximal.

The irreducible closed subset $Z$ is nonempty, so [F1] gives a prime $\mathfrak{p}$ generic for $Z$.

Any other generic point has the same closure and is equal to $\mathfrak{p}$ by the uniqueness in [F1].

Therefore every irreducible closed subset has a unique generic point.

By [F1], the natural isomorphism and its inverse are induced by morphisms $X\to Y$ and $Y\to X$.

Their composites induce identity natural transformations, so faithfulness in [F1] makes both composites identity morphisms.

The two morphisms are inverse scheme isomorphisms.

By [F1], $D(f)$ is homeomorphic to $\mathrm{Spec}(A_f)$.

The latter is compact by [F2], including when $A_f$ is the zero ring.

Compactness transfers across the homeomorphism.

Choose an isomorphism $X\cong \mathrm{Spec}A$ from affineness.

The whole spectrum is $D(1)$, which is quasi-compact by [F1] under the assumed Axiom of Choice. This is the only use of AC in this argument.

Its homeomorphic copy $X$ is quasi-compact.

Every element of $R$ has the form $a+bϵ$. If $a\mathrm{\ne }0$, then $a+bϵ$ is a unit with inverse $a^{-1}-a^{-2}bϵ$; while $(bϵ{)}^2=0$. Thus $\mathrm{Nil}(R)=(ϵ)$, and $R\mathrm{/}(ϵ)\cong k$. Since the field $k$ has exactly one prime ideal, $(0)$, [F1] identifies $\mathrm{Spec}R$ with its one-point spectrum; its sole prime is $(ϵ)$.

The nonzero class of $ϵ$ squares to zero, so $R$ is not reduced.

Every element outside $(ϵ)$ is a unit by step 1.1, so $R_{(ϵ)}=R$. The residue-field definition therefore gives $\kappa ((ϵ))=R_{(ϵ)}\mathrm{/}(ϵ)R_{(ϵ)}\cong R\mathrm{/}(ϵ)\cong k$.

This is the asserted one-point nonreduced scheme.

The only proper ideal of a field is $(0)$, and it is prime; hence the spectrum has exactly that point.

By [F1], its local ring is $k_{(0)}=k$, whose residue field is $k$.

By [F2], its global sections are $k$.