Past exam of the introduction to schemes course of the University of Oxford / 2023 / 1

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.