Fibre Products Base Change and Scheme Theoretic Fibres

Apply the universal property of $P^{\mathrm{\prime }}$ to the compatible maps $p,q$. It supplies a unique map $u:P\to P^{\mathrm{\prime }}$ with the required projections. This works also for $P=\mathrm{\varnothing }$.

Apply the universal property of $P$ to $p^{\mathrm{\prime }},q^{\mathrm{\prime }}$ to obtain $v:P^{\mathrm{\prime }}\to P$. Both $vu$ and ${\mathrm{id}}_P$ have projections $p,q$, so uniqueness gives $vu={\mathrm{id}}_P$; likewise $uv={\mathrm{id}}_{P^{\mathrm{\prime }}}$.

Thus $u$ is an isomorphism, and any projection-compatible isomorphism must equal the map already uniquely obtained. No condition on the number of points or on local nilpotents was used.

For an arbitrary scheme $T$, put $R=\mathrm{\Gamma }(T,{\mathcal{O}}_T)$. Compatible maps from $T$ to the two affine factors are, by the natural bijection in F2, exactly ring maps $B\to R$ and $C\to R$ whose restrictions to $A$ agree.

Use their common restriction to regard $R$ as an $A$-algebra. F3 gives precisely one ring map $B{\otimes }_{A}C\to R$, sending $b\otimes c$ to the product of the two images. F2 converts it to precisely one morphism $T\to \mathrm{Spec}(B{\otimes }_{A}C)$ with the desired projections.

This is the universal property F1, for every $T$, not only affine $T$. The argument permits zero rings: a map to $\mathrm{Spec}0$ is possible precisely for the empty test scheme, whose ring of sections is zero. Tensor-unit and identity cases use the same formula.

F1 identifies the projection ring maps with the two tensor inclusions. Contraction therefore gives the stated primes, and their contractions to $A$ agree because $a\otimes 1=1\otimes a$.

Elements outside the contracted primes map outside $\mathfrak{r}$ and hence become units in $D_{\mathfrak{r}}$. F2 identifies these localizations as stalks, and F3 shows that the displayed maps are local.

Quotient each local map by maximal ideals. The resulting unital maps between fields are injective: their kernels are proper ideals of a field, hence zero. The maps agree on $A$ and then on its residue field after localization and quotient. If $D=0$ there is no prime $\mathfrak{r}$, so the pointwise assertion is vacuous.

Given compatible maps $T\to V,T\to W$ over $U$, their composites to $S$ agree. F1 gives a unique $T\to P$. Its image lies in $Q$, so the morphism factors uniquely through that open subscheme by restriction of its sheaf map.

Conversely a map $T\to Q$ gives maps to $V,W$ agreeing in $S$; they agree in $U$ because $U\to S$ is a monomorphism. The two constructions are inverse, including empty opens and the full opens. F4 supplies the canonical identification with any other product.

For the last assertion, a compatible pair $a:T\to X,b:T\to U$ has $a(T)\subseteq f^{-1}(U)$. Its unique open factorization is a map to $f^{-1}(U)$; its composite to $U$ is $b$ by the monomorphism property. Conversely such a factorization gives the pair. This argument does not assume any general existence theorem.

Let $P_{ij}\subseteq P_i$ be the inverse image of $Y_i\cap Y_j$. By F1 it represents $X{\times }_S(Y_i\cap Y_j)$. F2 identifies it with the corresponding open in $P_j$. The transition maps satisfy identity, inverse and cocycle identities: on each triple overlap both candidate maps have identical projections and are equal by uniqueness.

To use F3 with exactly its affine hypothesis, cover each $P_i$ by affine opens. For two such charts use the open subset on which the preceding transition lands in the second chart; their isomorphisms are restrictions of those transitions. These are open subschemes, even when nonaffine, and their cocycles are already verified. F3 glues the affine charts to a scheme $P$; the charts belonging to $P_i$ glue back to $P_i$. F4 glues the projection maps to $X,Y$.

For a compatible pair $a:T\to X,b:T\to Y$, cover $T$ by $T_i=b^{-1}(Y_i)$. Each pair restricted to $T_i$ gives a unique map to $P_i$. On $T_i\cap T_j$ both factor through $P_{ij}$ and agree by its universal property. F4 glues them to one map $T\to P$. Any other such map has the same restrictions, proving uniqueness for arbitrary, possibly empty, $T$. The empty cover of empty $Y$ gives $P=\mathrm{\varnothing }$.

For a base cover, replace the local factors by $X_i=f^{-1}(S_i)$ and $Y_i=g^{-1}(S_i)$. F1 describes the overlap over $S_i\cap S_j$ as the open inverse image in either local product. The same cocycle and affine refinement construction applies. A compatible pair from $T$ is glued on the inverse images of $S_i$ under its common composite to $S$. A singleton cover changes nothing.

When $X,S$ are affine, cover $Y$ by affines. Each local product exists by F1, so F2 glues them to the desired product, with the displayed affine charts.

When only $S$ is affine, cover $X$ by affine opens. The preceding construction gives each product of one of these opens with $Y$. Interchanging the roles of the two projections in F2 glues them. This interchange is justified directly by the symmetric compatible-pair condition, without needing a later associativity result.

For arbitrary $S$, apply the preceding result over each affine $\mathrm{Spec}{A}_i$, and then use the base-cover clause of F2. Its overlap construction makes the stated tensor spectra an open cover. Empty schemes and zero tensor rings give empty charts, and the empty-base case forces both factors to be empty.

By F2, a map over $S$ is exactly a map commuting with structure maps. Therefore the fibre product supplied by F1 is a categorical product of $S$-schemes. A map from $X$ to the terminal candidate $S$ over $S$ is forced to equal its structure morphism.

The empty scheme has exactly one morphism to every scheme, since both its underlying map and all its sheaf data are unique. Conversely a map into the empty scheme exists only for an empty source. Thus the empty scheme satisfies the initial property, and compatible pairs into $X$ and $\mathrm{\varnothing }$ are represented by $\mathrm{\varnothing }$. This includes $S=\mathrm{\varnothing }$.

The topological disjoint union of schemes, with the structure sheaf specified separately on each component, is a scheme because every component is open and has its original affine charts. Maps out of it are exactly independent component maps, including their sheaf maps, so it is the coproduct over $S$. For zero components it is empty and for one component it is that component.

All products exist by F1. For symmetry swap the two projections; applying this operation twice restores them. For either triple product a map from $T$ is exactly three maps to $X,Y,Z$ with the same composite to $S$. Thus the projections construct mutually inverse maps between the two bracketings.

The pair $({\mathrm{id}}_X,f)$ supplies $X\to X{\times }_{S}S$ and the first projection supplies the inverse. A pair into $X,S$ over $S$ has its second coordinate forced by the first. Every inverse assertion follows by uniqueness, as in F2. This includes empty factors and identity structure maps.

Naturality and all claimed coherence equations are checked after each original projection. Both sides then give exactly the same coordinate maps. Repeated uniqueness in the binary universal property makes the maps equal. This holds for arbitrary test schemes with nilpotents as well as one-point tests.

Using F1, a map $T\to (X{\times }_S{S}^{\mathrm{\prime }}){\times }_{S^{\mathrm{\prime }}}{S}^{{\mathrm{\prime }}^{\mathrm{\prime }}}$ is a triple $(a,b,c)$ into $X,S^{\mathrm{\prime }},S^{{\mathrm{\prime }}^{\mathrm{\prime }}}$ satisfying $fa=hb$ and $b=kc$. Eliminating $b$ gives exactly a pair $(a,c)$ satisfying $fa=hkc$.

The inverse operation is $(a,c)\mapsto (a,kc,c)$. They yield inverse morphisms by the projection-compatible identifications of F2. Empty schemes and identity base maps obey these same formulas.

For $u:X\to Y$ the corresponding pair becomes $(ua,c)$ on both sides. Thus identities and compositions commute with the isomorphism, proving functoriality without a choice of points. More generally, if a property $\mathcal{P}$ is preserved by both base change and composition, the product of two $S$-morphisms $u:X\to Y$ and $v:X^{\mathrm{\prime }}\to Y^{\mathrm{\prime }}$ with $\mathcal{P}$ also has $\mathcal{P}$: factor it as $X{\times }_S{X}^{\mathrm{\prime }}\to Y{\times }_S{X}^{\mathrm{\prime }}\to Y{\times }_S{Y}^{\mathrm{\prime }}$. The first arrow is the pullback of $u$ along $Y{\times }_S{X}^{\mathrm{\prime }}\to Y$, and the second the pullback of $v$ along $Y{\times }_S{Y}^{\mathrm{\prime }}\to Y^{\mathrm{\prime }}$. Their test pairs verify these pullback identifications.

A ring map from $C[t_i]$ to a $C$-algebra $R$ is exactly a choice of elements of $R$ for the variables. Equivalently it is an $A$-algebra map $A[t_i]\to R$ together with the fixed map $C\to R$. F1 therefore identifies $A[t_i]{\otimes }_{A}C$ with $C[t_i]$, fixing coefficients and variables; empty variable sets are included.

The module isomorphism in F3, after swapping tensor factors, sends $(a\mathrm{/}m)\otimes c$ to $ac\mathrm{/}\bar{m}$ and has inverse $c\mathrm{/}\bar{m}\mapsto (1\mathrm{/}m)\otimes c$. These formulas preserve multiplication and 1. They also apply when $0\in \bar{M}$, in which case both rings are zero, and when $M=\{1\}$, in which case both are $C$.

A map out of the quotient must kill each element of $I$, which under the preceding identification means killing $IC[t_i]$. This proves the quotient formula by the same universal property. The module quotient map in F2 is consistent with it: multiplication of pure tensors is sent to the product of their images, so it is a ring map. For $I=0$ this is the polynomial formula and for $I=(1)$ it is the zero ring.

For instance, base change of $k[x,y,t]\mathrm{/}(y^2-x^3-tx)$ along $k[t]\to k[u,v]$, $t\mapsto uv$, gives $k[x,y,u,v]\mathrm{/}(y^2-x^3-uvx)$; along $t\mapsto 3$ it gives $k[x,y]\mathrm{/}(y^2-x^3-3x)$. Finitely many generators and relations remain finite. The substitution follows from the displayed maps and holds in every characteristic.

If $U=\mathrm{Spec}A$ contains $x$, F2 identifies a morphism $\mathrm{Spec}R\to U$ with $\psi :A\to R$. The preimage $\mathfrak{p}={\psi }^{-1}(\mathfrak{m})$ is prime, and every element outside it maps to a unit. Thus $\psi$ factors uniquely through a local map $A_{\mathfrak{p}}\to R$. Conversely any such local map gives $\psi$ and has closed-point image $\mathfrak{p}$.

Every open neighbourhood of the closed point of $\mathrm{Spec}R$ is the whole spectrum: a basic open containing that point is defined by an element outside $\mathfrak{m}$, hence by a unit. Therefore any morphism to $X$ factors through every affine neighbourhood of its closed-point image. The affine constructions agree after shrinking to a common neighbourhood, by uniqueness of the map induced from the stalk. They consequently give inverse constructions globally; for $X=\mathrm{\varnothing }$ both sets are empty.

For $R=K$ a field, locality says precisely that the maximal ideal of ${\mathcal{O}}_{X,x}$ maps to zero. Factoring through the quotient F1 gives a unital field map, necessarily injective. Conversely such an embedding gives a local map. Taking $R={\mathcal{O}}_{X,x}$ and its identity gives the canonical $\mathrm{Spec}{\mathcal{O}}_{X,x}\to X$; composing with its residue-field point gives $\mathrm{Spec}\kappa (x)\to X$. Every field-valued representative at $x$ factors uniquely through this residue-field representative by the specified embedding, so it is the smallest representative in its class. Identity embeddings give the canonical points, and F3 gives their compatibility with a morphism $X\to S$ by composing the residue-field maps.

If two representatives at $x$ use fields $K,L$, their tensor product over $\kappa (x)$ is nonzero: choose bases of these nonzero vector spaces and apply F4. By F5 choose a maximal ideal and take its quotient field $\mathrm{\Omega }$. The unital maps $K,L\to \mathrm{\Omega }$ are injective and agree on $\kappa (x)$, hence give a common representative by step 3.1. Conversely a common representative maps its unique point to both images, forcing those images equal. This is exactly where Choice is used.

F1 defines the fibre as base change by $\mathrm{Spec}\kappa (\mathfrak{p})$. F2 therefore gives its coordinate ring $B{\otimes }_A\kappa (\mathfrak{p})$.

Write $\kappa (\mathfrak{p})=A_{\mathfrak{p}}\mathrm{/}\mathfrak{p}{A}_{\mathfrak{p}}$. Localizing coefficients and then imposing the ideal relations, using F3, identifies the ring with $M^{-1}B\mathrm{/}\mathfrak{p}{M}^{-1}B$. The maps send $b\otimes (a\mathrm{/}s<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>\mathfrak{p})$ to $ab\mathrm{/}s<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>\mathfrak{p}$.

The formula preserves every element, including nilpotents. If the quotient is zero the fibre is empty; otherwise its primes and residue fields are retained. For $\mathfrak{p}=(0)$ in a domain it is extension to the fraction field, and for $B=A$ it is the one-point spectrum of $\kappa (\mathfrak{p})$.

On compatible affine charts put $M=A\setminus \mathfrak{p}$. F1 gives the fibre ring $D=M^{-1}B\mathrm{/}\mathfrak{p}{M}^{-1}B$. By F3 and F4 its primes are exactly primes $\mathfrak{q}$ of $B$ disjoint from $M$ and containing $\mathfrak{p}B$. These two requirements say precisely $\mathfrak{q}\cap A=\mathfrak{p}$. Extension followed by quotient and contraction are inverse.

The basic open $D_D(\overline{b\mathrm{/}m})$ corresponds to $D_B(b)\cap f^{-1}(\mathfrak{p})$, since $m$ is already a unit. Such opens form a basis on both sides, proving the subspace topology assertion, not merely a bijection. Localizing $D$ at this prime and then taking its residue field gives $\mathrm{Frac}(B\mathrm{/}\mathfrak{q})$, the original residue field.