"algebraic geometry is the study of solutions to polynomial equations."
That is, we are supposed to study $k$-points of polynomial equations.
References. This posting is a regurgitation of some parts of the following books:
- Eisenbud-Harris,
- Olsson,
- Poonen, and
- Vakil.
Algebraic ways to think about $k$-points. First, note that a $k$-point $(a_{1}, \dots, a_{n}) \in k^{n}$ of $V(f)$ can be thought as the maximal ideal $$\mathfrak{m} = (x_{1} - a_{1}, \dots, x_{n} - a_{n})/(f)$$ of the ring $R = k[x_{1}, \dots, x_{n}]/(f).$ We have $$R/\mathfrak{m} \simeq \frac{k[x_{1}, \dots, x_{n}]}{(x_{1}-a_{1}, \dots, x_{n}-a_{n})} \simeq k,$$ where the second map is given by the evaluation at $(a_{1}, \dots, a_{n}) \in k^{n}.$ This illustrates that the $k$-points of $R$ are precisely maximal ideals of $R$ (i.e., closed points of $\mathrm{Spec}(R)$) whose residue field is $k.$ Such maximal ideals are precisely kernels of $k$-algebra maps $R \rightarrow k.$ They correspond to $k$-scheme maps $$\mathrm{Spec}(k) \rightarrow \mathrm{Spec}(R).$$ In general, a $k$-point of a $k$-scheme $X$ is a $k$-scheme map $\mathrm{Spec}(k) \rightarrow X,$ and one may check that $k$-points of $X$ correspond to the topological/set-theoretic points of $X$ whose residue fields are $k.$
Remark. Given a field $k,$ we know $\mathrm{Spec}(k)$ has one $k$-point. This is because the only $k$-algebra map $k \rightarrow k$ is the identity. If we just think of ring maps, there are many more in general. For instance, note that complex conjugation $\mathbb{C} \rightarrow \mathbb{C}$ gives a ring map that is not the identity.
Generalization. Fix a (commutative and unital) ring $A.$ Given any $A$-algebra $R$ and an $A$-scheme $X,$ an $R$-point of $X$ over $A$ is an $A$-scheme map $$\mathrm{Spec}(R) \rightarrow X.$$ Given any $A$-scheme $T,$ an $T$-point of $X$ over $A$ is an $A$-scheme map $$T \rightarrow X.$$ Note that an $R$-point on $X$ over $A$ is precisely $\mathrm{Spec}(R)$-point over $A.$ Note that we may generalize this notion further by replacing $A$ (or $\mathrm{Spec}(A)$) with another scheme.
Remark. Note that when we were discussing $k$-point earlier, we really meant $k$-point over $k.$ However, it seems that people do not really mention this out loud in general. That is, given an $S$-schemes $X$ and $T,$ we call an $S$-scheme map $T \rightarrow X$ a $T$-point, even though in principle, we should specify the scheme $S$ that we are over.
Functor of points. Given $S$-schemes $X$ and $T,$ we write $$X(T) := \mathrm{Hom}_{\textbf{Sch}_{S}}(T, X)$$ to mean the set of $T$-points. The functor $$h_{X} : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set},$$ given by $T \mapsto X(T)$ is called the functor of points on $X$ over $S$. By Yoneda lemma, we get an embedding $$\textbf{Sch}_{S}^{\mathrm{op}} \hookrightarrow \mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$$ given by $X \mapsto h_{X}.$
Restriction to affine schemes is enough. We can also have $$\textbf{Sch}_{S} \hookrightarrow \mathrm{Fun}(\textbf{Aff}_{S}^{\mathrm{op}}, \textbf{Set}),$$ where $\textbf{Aff}_{S}$ denotes the category of affine schemes over $S$. The functor is given by $$X \mapsto h_{X}|_{\textbf{Aff}_{S}^{\mathrm{op}}}$$ on the level of objects.
How come? Given any $S$-scheme $X,$ consider its functor of points $h_{X}.$ Given an $S$-scheme $T$, we recall that $T$-points on $X$ are precisely $S$-maps $T \rightarrow X.$ For any affine open subset $U \subset T,$ we have an $S$-map given by $$U \hookrightarrow T \rightarrow X,$$ a $U$-point of $X.$ If we denote by $\{U_{i}\}_{i \in I}$ the set of all affine open subsets of $T,$ then an $S$-map $T \rightarrow X$ is given by a set $\{U_{i} \rightarrow X\}_{i \in I}$ of $S$-maps that are compatible with inclusions among various $U_{i}.$ A clean way of saying this is to note that $X(T)$ is the limit (in $\textbf{Set}$) of the diagram $\{X(U_{i})\}_{i \in I},$ whose maps are induced by inclusions among $U_{i}.$ (This seems quite easy because we are dealing with the category of sets.) Thus, given any two $S$-schemes $X$ and $Y,$ if $$h_{X}|_{\textbf{Aff}_{S}^{\mathrm{op}}} \simeq h_{Y}|_{\textbf{Aff}_{S}^{\mathrm{op}}},$$ then $X \simeq Y,$ as $S$-schemes.
Moral. We may consider schemes as a very specific type of functors! Moreover, given any ring $R,$ since the opposite category $\textbf{Aff}_{R}^{\mathrm{op}}$ of the category affine schemes over $R$ is anti-equivalent to the category $\textbf{Alg}_{R}$ of $R$-algebras, so we actually have $$\textbf{Sch}_{R} \hookrightarrow \mathrm{Fun}(\textbf{Alg}_{R}, \textbf{Set}).$$ In particular, taking $R = \mathbb{Z},$ we have $$\textbf{Sch} \hookrightarrow \mathrm{Fun}(\textbf{Ring}, \textbf{Set}),$$ where $\textbf{Ring}$ is the category of (commutative unital) rings.
This is crazy! Isn't it? Schemes are just some functors from the category of rings to the category of sets, which sounds like scheme theory can be thought as a subfield of commutative algebra. Indeed, we often experience that to prove certain statement about schemes, we deduce something about affine opens and then glue together. Note that inducing the above embedding uses such an argument as well.
Remark. I think saying "scheme theory can be thought as a subfield of commutative algebra" is as philosophical as saying that "group theory can be thought of a subfield of topology" (just because every group is the fundamental group of some topological space).
As a terminology, the functors that are in the image of any of the following (Yoneda) embeddings are called representable by schemes (over the given base):
- $\textbf{Sch}_{S} \hookrightarrow \mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set});$
- $\textbf{Sch}_{R} \hookrightarrow \mathrm{Fun}(\textbf{Aff}_{S}^{\mathrm{op}}, \textbf{Set});$
- $\textbf{Sch}_{R} \hookrightarrow \mathrm{Fun}(\textbf{Alg}_{R}, \textbf{Set}).$
Geometry of functors. We are now going to think about the objects in the category $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ as geometric spaces, which include schemes. This sounds quite ridiculous, but it seems that many people have thought about this for many years.
We will build the geometry of functors from the representable ones (i.e., $S$-schemes), so in some sense our notion of geometry will be still somewhat "concrete". For any representable functor $$h_{X} : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ with respect to an $S$-scheme $X,$ we called $$X(T) = h_{X}(T) = \mathrm{Hom}_{\textbf{Sch}_{S}}(T, X)$$ the set of $T$-points on $X.$ We have two takeaways here.
- For any (not necessarily representable) functor $F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set},$ the set $F(T)$ still makes sense.
- Each $T \rightarrow X$ can be thought as $h_{T} \rightarrow h_{X},$ using the Yoneda correspondence $\mathrm{Hom}_{\textbf{Sch}_{S}}(T, X) = h_{X}(T) \simeq \mathrm{Hom}_{\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})}(h_{T}, h_{X}).$ This correspondence works even when $h_{X}$ is replaced by any other functor $F,$ so we have $F(T) \simeq \mathrm{Hom}_{\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})}(h_{T}, F),$ as sets.
Hence, we may call the set $F(T)$ the set of $T$-points on $F$ (over $S$), while thinking of each element of $F(T)$ as a morphism $h_{T} \rightarrow F$, which we call a $T$-point of $F$ (over $S$).
Reading a bit more from Yoneda's lemma. Yoneda lemma is very simple, but it packs even more than the bijection $$F(T) \simeq \mathrm{Hom}_{\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})}(h_{T}, F).$$ That is, this bijection is functorial both in $F$ and $T.$ The functoriality in $T$ says we may see the isomorphism as the isomorphism $$F \simeq \mathrm{Hom}_{\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})}(h_{-}, F)$$ of functors $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$ With $T$ fixed, the functoriality in $F$ does not seem to say much: it means that given any map $F \rightarrow G$ of functors, we have $$\mathrm{Hom}_{\textbf{Set}}(F(T), G(T)) \simeq \mathrm{Hom}_{\textbf{Set}}(\mathrm{Hom}(h_{T}, F), \mathrm{Hom}(h_{T}, G)),$$ where the hom-sets without subscripts were taken in $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set}).$
Fiber product of functors. Let $\alpha : F \rightarrow H$ and $\beta : G \rightarrow H$ be morphisms in the category $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ of functors. Then the fiber product $F \times_{H} G$ of $F$ and $G$ over $H$ in $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ can be constructed as follows. For each $S$-scheme $T,$ we construct $$(F \times_{H} G)(T) := F(T) \times_{H(T)} G(T),$$ which consists of $(x, y) \in F(T) \times G(T)$ such that $\alpha_{T}(x) = \beta_{T}(y) \in H(T).$ Given any $S$-scheme morphism $\phi : T' \rightarrow T,$ we have $$F(T) \times_{H(T)} G(T) \rightarrow F(T') \times_{H(T')} G(T')$$ given by $(x, y) \mapsto (\phi^{F}(x), \phi^{G}(y)).$ When $\phi$ is the identity map of an object, so is this map. This map also respects the composition with another $S$-scheme morphism, so $$F \times_{H} G : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ is indeed a functor. One may easily check that the set projections $$F(T) \leftarrow F(T) \times_{H(T)} G(T) \rightarrow G(T)$$ for $S$-schemes $T$ give rise to morphisms $$F \leftarrow F \times_{H} G \rightarrow G$$ of functors such that the following two compositions are identical: $$F \times_{H} G \rightarrow F \overset{\alpha}{\longrightarrow} H$$ and $$F \times_{H} G \rightarrow G \overset{\beta}{\longrightarrow} H.$$ It requires a small check to notice that $F \times_{H} G$ is indeed the fiber product of $F$ and $G$ over $H$ in $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set}).$
Open subfunctors. We are still thinking about geometry of functors. How do we think about "open subsets" of a functor $$F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ even when $F$ is not a $S$-scheme (i.e., not representable)?
Given any open subset $U \hookrightarrow X,$ one can easily check that any $S$-scheme map $f : Y \rightarrow X$ yields $$h_{f^{-1}(U)} \simeq h_{U} \times_{h_{X}} h_{Y}$$ in the category $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ with respect to $$h_{U} \leftarrow h_{f^{-1}(U)} \rightarrow h_{Y},$$ induced by the restriction $f^{-1}(U) \rightarrow U$ of $f$ and the open embedding $f^{-1}(U) \hookrightarrow Y.$ In particular $$f^{-1}(U) \simeq U \times_{X} Y$$ in the category $\textbf{Sch}_{S}.$ We mimic this situation for the case when $h_{X}$ is replaced by more general functor $F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$
A morphism $G \rightarrow F$ in $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ is called an open subfunctor if for every $S$-scheme $Y$ and a morphism $h_{Y} \rightarrow F$ of functors, the induced map $$G \times_{F} h_{Y} \rightarrow h_{Y}$$ is given by an open embedding $V \hookrightarrow Y.$ This means that $$h_{V} \simeq G \times_{F} h_{Y} \rightarrow h_{Y},$$ with another structure map $h_{V} \rightarrow G$ to define the fiber product.
Remark. Note that an $S$-scheme map $U \rightarrow X$ is an open embedding if and only if $h_{U} \rightarrow h_{X}$ is an open subfunctor. Since two Cartesian squares make a Cartesian rectangle (1.3.G in Vakil), composition of two open subfunctors is an open subfunctor. Eisenbud-Harris (Definition VI-5) gives a weaker definition of an open subfunctor where $Y$ is only allowed to be affine schemes. This is not the same as the definition we gave (from Section 9.1 of Vakil), but the upshot is that they only used affine schemes instead of using all schemes. (I thank Sridhar Venkatesh for explaining this to me.)
Open cover. Given an $S$-scheme, an open cover $\{U_{i}\}_{i \in I}$ of $X$ can be realized as the family of corresponding open subfunctors $\{h_{U_{i}} \hookrightarrow h_{X}\}_{i \in I}.$ Note that for any $S$-scheme map $f : Y \rightarrow X,$ we have an open cover $\{f^{-1}(U_{i}) \hookrightarrow Y\}_{i \in I}$ of $Y.$
Given any open subset $U \hookrightarrow X,$ one can easily check that any $S$-scheme map $f : Y \rightarrow X$ yields $$h_{f^{-1}(U)} \simeq h_{U} \times_{h_{X}} h_{Y}$$ in the category $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ with respect to $$h_{U} \leftarrow h_{f^{-1}(U)} \rightarrow h_{Y},$$ induced by the restriction $f^{-1}(U) \rightarrow U$ of $f$ and the open embedding $f^{-1}(U) \hookrightarrow Y.$ In particular $$f^{-1}(U) \simeq U \times_{X} Y$$ in the category $\textbf{Sch}_{S}.$ We mimic this situation for the case when $h_{X}$ is replaced by more general functor $F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$
A morphism $G \rightarrow F$ in $\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})$ is called an open subfunctor if for every $S$-scheme $Y$ and a morphism $h_{Y} \rightarrow F$ of functors, the induced map $$G \times_{F} h_{Y} \rightarrow h_{Y}$$ is given by an open embedding $V \hookrightarrow Y.$ This means that $$h_{V} \simeq G \times_{F} h_{Y} \rightarrow h_{Y},$$ with another structure map $h_{V} \rightarrow G$ to define the fiber product.
Remark. Note that an $S$-scheme map $U \rightarrow X$ is an open embedding if and only if $h_{U} \rightarrow h_{X}$ is an open subfunctor. Since two Cartesian squares make a Cartesian rectangle (1.3.G in Vakil), composition of two open subfunctors is an open subfunctor. Eisenbud-Harris (Definition VI-5) gives a weaker definition of an open subfunctor where $Y$ is only allowed to be affine schemes. This is not the same as the definition we gave (from Section 9.1 of Vakil), but the upshot is that they only used affine schemes instead of using all schemes. (I thank Sridhar Venkatesh for explaining this to me.)
Open cover. Given an $S$-scheme, an open cover $\{U_{i}\}_{i \in I}$ of $X$ can be realized as the family of corresponding open subfunctors $\{h_{U_{i}} \hookrightarrow h_{X}\}_{i \in I}.$ Note that for any $S$-scheme map $f : Y \rightarrow X,$ we have an open cover $\{f^{-1}(U_{i}) \hookrightarrow Y\}_{i \in I}$ of $Y.$
More generally, an open cover of a functor $F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$ is a family $\{G_{i} \hookrightarrow F\}_{i \in I}$ of open subfunctors such that for every $S$-scheme $Y$ and $h_{Y} \rightarrow F,$ the fiber products $$G_{i} \times_{F} h_{Y} \rightarrow h_{Y}$$ gives rise to an open covering of $Y.$
Sheaves as functors. Given any topological space $X,$ denote by $\mathscr{C}_{X}$ the category of open subsets of $X,$ whose morphisms are given by inclusions. Recall that a sheaf of sets is a functor $F : \mathscr{C}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}$ such that for any open subset $U$ of $X$ (i.e., an object of $\mathscr{C}_{X}$) and an open cover $U = \bigcup_{i \in I}U_{i}$, we have the following equalizer sequence $$F(U) \rightarrow \prod_{i \in I}F(U_{i}) \rightrightarrows \prod_{i, j \in I}F(U_{i} \cap U_{j}),$$ where the first map is given by $s \mapsto (s|_{U_{i}})_{i \in I}$ and the second (double) map is given as follows: for each ordered pair $(i', j') \in I^{2},$ we have two maps $$\prod_{i \in I}F(U_{i}) \rightrightarrows F(U_{i'} \cap U_{j'}),$$ the first of which is $$(s_{i})_{i \in I} \mapsto (s_{i'}|_{U_{i'} \cap U_{j'}}),$$ and the second of which is $$(s_{i})_{i \in I} \mapsto (s_{j'}|_{U_{i'} \cap U_{j'}}),$$ where we note that that these two maps can be equal. Since there are two maps per $F(U_{i} \cap U_{j}),$ we get the desired double map.
Functorial construction of the equalizer sequence. For our purpose it is important that the maps in the equalizer sequence (i.e., the sequence we check to be an equalizer for the definition of sheaves) can be functorially induced. For the first map, the open embeddings $U_{i} \hookrightarrow U$ gives $F(U) \rightarrow F(U_{i}),$ which gives the first map by the property of the product in $\textbf{Set}$. For the second map, we the open embeddings $U_{i'} \cap U_{j'} \hookrightarrow U_{i'}$ and $U_{i'} \cap U_{j'} \hookrightarrow U_{j'}$ to get $F(U_{i'}) \rightarrow F(U_{i'} \cap U_{j'})$ and $F(U_{j'}) \rightarrow F(U_{i'} \cap U_{j'}),$ and using these two maps we have gotten the following two maps: $$\prod_{i \in I}F(U_{i}) \rightarrow F(U_{i'}) \rightarrow F(U_{i'} \cap U_{j'})$$ and $$\prod_{i \in I}F(U_{i}) \rightarrow F(U_{j'}) \rightarrow F(U_{i'} \cap U_{j'}).$$ We can thus use the equalizer sequence that defines a functor as a sheaf works in more generality, even when we replace $\mathscr{C}_{X}$ with another categoty (with some extra structure).
If we assumed that $X$ is a scheme, then we could consider a sheaf on $X$ valued in the category $\textbf{Ring}$ of rings or the category $\textbf{Ab}$ of abelian groups (or that of $R$-modules with a fixed ring $R$, etc.) However, all such categories we take are in the category $\textbf{Set}$ of sets, so we take the equalizer sequence there to give a definition of a sheaf. The bottom line is that a sheaf is a functor $F : \mathscr{C}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}.$
Remark. A bold idea to replace $\mathscr{C}_{X}$ with $\textbf{Sch}_{X}.$ Objects of either category have the notion of open covers, so we call them "sites" (which we formally define below). More formally, we call $\mathscr{C}_{X}$ the small Zariski site on $X,$ while $\textbf{Sch}_{X}$ the big Zariski site on $X.$ The word "small" and "big" are due to the fact that the objects of $\mathscr{C}_{X}$ are open subsets $U \hookrightarrow X,$ while the objects of $\textbf{Sch}_{X}$ are all morphisms $U \rightarrow X,$ not just open embeddings. The word "Zariski" is due to the fact that the notion of "open cover" is given by the Zariski topology on $X.$
A functor $F : \textbf{Sch}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}$ is called a Zariski sheaf if for any $X$-scheme map $U \rightarrow X$ an open cover $U = \bigcup_{i \in I}U_{i},$ the sequence $$F(U) \rightarrow \prod_{i \in I}F(U_{i}) \rightrightarrows \prod_{i, j \in I}F(U_{i} \cap U_{j})$$ induced by the open inclusions is an equalizer in $\textbf{Set}$.
Remark. Note that any representable functor (i.e., the objects in the image of $\textbf{Sch}_{X} \hookrightarrow \mathrm{Fun}(\textbf{Sch}_{X}^{\mathrm{op}}, \textbf{Set})$) is a Zariski sheaf. This is just a fancier way of saying a morphism $U \rightarrow X$ of schemes can be glued from morphisms $U_{i} \rightarrow X$ that are compatible with an open covering $U = \bigcup_{i \in I} U.$
Theorem (Which functors are representable?). A functor $F : \textbf{Sch}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}$ is representable if and only if it is a Zariski sheaf and has an open cover of open subfunctors represented by schemes over $X$.
Proof. As in the previous remark any representable functor $h_{U} : \textbf{Sch}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}$ is a Zariski sheaf. Moreover, taking any open cover $U = \bigcup_{i \in I}U_{i}$ gives rise to the desired open cover $\{h_{U_{i}} \rightarrow h_{U}\}_{i \in I}$ of the functor $h_{U}.$
Conversely, let $F : \textbf{Sch}_{X}^{\mathrm{op}} \rightarrow \textbf{Set}$ be a Zarisk sheaf with an open cover $$\{h_{U_{i}} \rightarrow F\}_{i \in I}$$ such that each $U_{i}$ is a scheme over $X.$ By definition of open subfunctors for any $i, j \in I$, we have an scheme $U_{ij}$ over $X$ such that $$h_{U_{ij}} = h_{U_{i}} \times_{F} h_{U_{j}} \rightarrow h_{U_{i}}$$ is an open subfunctor in $\mathrm{Fun}(\textbf{Sch}_{X}^{\mathrm{op}}, \textbf{Set}),$ which gives rise to an open embedding $\varphi_{i} : U_{ij} \hookrightarrow U_{i},$ which we may think as an actual inclusion by taking its image. Note that $U_{ii} = U_{i}$ because $h_{U_{i}} = h_{U_{i}} \times_{F} h_{U_{i}}.$ The isomorphim $$h_{U_{ij}} = h_{U_{i}} \times_{F} h_{U_{j}} \simeq h_{U_{j}} \times_{F} h_{U_{i}} = h_{U_{ji}}$$ gives rise to an isomorphism $\varphi_{ij} : U_{ij} \overset{\sim}{\longrightarrow} U_{ji}$ of schemes over $X.$ One may check that $$h_{U_{ij} \cap U_{ik}} = h_{U_{ij}} \times_{h_{U_{i}}} h_{U_{ik}}.$$ We continue to note that $$\begin{align*}h_{U_{ij} \cap U_{ik}} &= h_{U_{ij}} \times_{h_{U_{i}}} h_{U_{ik}} \\ &\simeq h_{U_{ij}} \times_{h_{U_{i}}} (h_{U_{i}} \times_{F} h_{U_{k}}) \\ &\simeq (h_{U_{ij}} \times_{h_{U_{i}}} h_{U_{i}}) \times_{F} h_{U_{k}} \\ &\simeq h_{U_{ij}} \times_{F} h_{U_{k}} \\ &\simeq (h_{U_{ij}} \times_{h_{U_{j}}} h_{U_{j}}) \times_{F} h_{U_{k}} \\ &\simeq h_{U_{ij}} \times_{h_{U_{j}}} (h_{U_{j}} \times_{F} h_{U_{k}}) \\ &= h_{U_{ij}} \times_{h_{U_{j}}} h_{U_{jk}} \\ &\simeq h_{U_{ji}} \times_{h_{U_{j}}} h_{U_{jk}} \\ &= h_{U_{ji} \cap U_{jk}},\end{align*}$$ and when we follow the isomorphisms, we notice that all of them are compatible with $h_{U_{ij}} \simeq h_{U_{ji}}$ we fixed before, so this shows that $\varphi_{ij} : U_{ij} \overset{\sim}{\longrightarrow} U_{ji}$ restircts to $$U_{ij} \cap U_{ik} \simeq U_{ji} \cap U_{jk}.$$ Likewise, the isomorphism $\varphi_{jk} : U_{jk} \overset{\sim}{\longrightarrow} U_{kj}$ restricts to $$U_{ji} \cap U_{jk} \simeq U_{ki} \cap U_{kj}.$$ When we combine the isomorphisms $$h_{U_{ij} \cap U_{ik}} \simeq h_{U_{ji} \cap U_{jk}} \simeq h_{U_{ki} \cap U_{kj}},$$ we may note that all the isomorphisms involved are compatible with $h_{U_{ik}} \simeq h_{U_{ki}}.$ Thus, we have $$\varphi_{ik} = \varphi_{jk} \circ \varphi_{ij},$$ with appropriate restrictions. Hence, we may glue the schemes $U_{i}$ to get a scheme $U$ (4.4.A in Vakil), where we have open embeddings $\phi_{i} : U_{i} \hookrightarrow U,$ such that $U = \bigcup_{i \in I}\phi_{i}(U_{i})$ and $$\phi_{i}(U_{i}) \cap \phi_{j}(U_{j}) = \phi_{i}(U_{ij}) = \phi_{j}(U_{ji})$$ so that we may consider $U_{i}$ as an actual open subset of $U.$
Remark. The proof is not over, but so far we have not used either of the hypotheses that the open subfunctors $h_{U_{i}} \rightarrow F$ form a cover nor the functor $F$ is a Zariski sheaf!
Now, let's use the hypothesis that $F$ is a Zariski sheaf. Write $s_{i} : h_{U_{i}} \rightarrow F$ to mean the open subfunctors. We may use the Yoneda correspondence $$\mathrm{Hom}_{\mathrm{Fun}(\textbf{Sch}_{S}^{\mathrm{op}}, \textbf{Set})}(h_{U_{i}}, F) \simeq F(U_{i}),$$ given by $$\phi \mapsto \phi_{U_{i}}(\mathrm{id}_{U_{i}}),$$ (functorial in $U_{i}$) to think of $s_{i} \in F(U_{i}).$ Thinking all $U_{i}$ in $U,$ we have $$s_{i}|_{U_{i} \cap U_{j}} = s_{j}|_{U_{i} \cap U_{j}}$$ for all $i, j \in I$, so we must have a unique $s \in F(U)$ such that $$s|_{U_{i}} = s_{i}$$ for all $i \in I$ because $F$ is a Zariski sheaf. We may interpret $s : h_{U} \rightarrow F$ as a map of functors. One may check that this is an isomorphism of functors by constructing its inverse as follows: for any scheme $W,$ an element of $F(W)$ correponds to a map $h_{W} \rightarrow F$ of functors. Since $h_{U_{i}} \rightarrow F$ is an open subfunctor, we have $h_{U_{i}} \times_{F} h_{W} = h_{W_{i}} \rightarrow h_{W},$ given by an open embedding $W_{i} \hookrightarrow W.$ We have projections $h_{W_{i}} \rightarrow h_{U_{i}},$ which can be composed with $h_{U_{i}} \rightarrow h_{U}$ given by previously constructed open embeddings $U_{i} \hookrightarrow U$ to give maps $$h_{W_{i}} \rightarrow h_{U_{i}} \rightarrow h_{U}.$$ We are not ready to use the hypothesis that $\{h_{U_{i}} \rightarrow F\}_{i \in I}$ is an open cover, which lets us have $$W = \bigcup_{i} W_{i}.$$ Since $F$ is a Zariski sheaf, the maps $h_{W_{i}} \rightarrow h_{U}$ with all $i \in I$ gives rise to $h_{W} \rightarrow h_{U},$ which can be thought as an element of $h_{U}(W).$ Thus, we have constructed a set map $$F(W) \rightarrow h_{U}(W).$$ Using this map, one can construct a map $F \rightarrow h_{U}$ of functors and can check that it is inverse to the map $h_{U} \rightarrow F$ we have constructed earlier. Therefore, we get $F \simeq h_{U},$ so $F$ is representable, as desired. $\Box$
Language of sites. We revisited the usual notion of sheaves on a scheme $X$ as a functor $\mathscr{C}_{X}^{\mathrm{op}} \rightarrow \textbf{Set},$ with an equalizer sequence attached to every open cover of an object of $\mathscr{C}_{X}^{\mathrm{op}},$ where $\mathscr{C}_{X}$ is the category of open subsets of $X,$ whose morphisms are given by inclusions. Then we replaced $\mathscr{C}_{X}$ with $\textbf{Sch}_{X},$ the category of schemes over $X,$ and called the functors $\textbf{Sch}_{X}^{\mathrm{op}} \rightarrow \textbf{Set},$ with an equalizer sequence attached to every open cover of an object of $\textbf{Sch}_{X}.$ We now introduce a general language that combines these two situations.
Given a category $\mathscr{C},$ a Grothendieck topology on $\mathscr{C}$ is given by the following data: for every object $U$ of $\mathscr{C},$ we have a set $\mathrm{Cov}(U),$ each of whose element is a collection of morphisms into $U,$ called the set of coverings of $U$, with the following properties.
- Any isomorphism into $U$ is in $\mathrm{Cov}(U).$
- Coverings respect base change: if $\{U_{i} \rightarrow U\}_{i \in I} \in \mathrm{Cov}(U),$ for any morphism $V \rightarrow U,$ we have $\{U_{i} \times_{U} V \rightarrow V\} \in \mathrm{Cov}(V),$ where we also require the fiber products exist in $\mathscr{C}.$
- We can take subcoverings: if $\{U_{i} \rightarrow U\}_{i \in I} \in \mathrm{Cov}(U)$ and $\{U_{ij} \rightarrow U_{i}\}_{j \in J_{i}} \in \mathrm{Cov}(U_{i})$ for each $i \in I,$ then $\{U_{ij} \rightarrow U_{i} \rightarrow U\}_{i \in I, j \in J_{i}} \in \mathrm{Cov}(U).$
Another word. If a category $\mathscr{C}$ is equipped with a Grothendieck topology, then we call it a site.
Example. Note that $\mathscr{C}_{X}$ can get a Grothendieck topology and be made into a site. Given any object $U$ (i.e., an open subset $U \hookrightarrow X$), the set $\mathrm{Cov}(U)$ consists of all open covers of $U.$ As we mentioned earlier, this site is called the small Zariski site on $X.$
Example. The category $\textbf{Sch}_{X}$ can also be equipped with a Grothendieck topology and be made into a site. Given any object $U$ (i.e., a scheme map $U \rightarrow X$), the set $\mathrm{Cov}(U)$ consists of all open covers of $U.$ As we mentioned earlier, this site is called the big Zariski site on $X.$
Given a site $\mathscr{C},$ a functor $F : \mathscr{C}^{\mathrm{op}} \rightarrow \textbf{Set}$ is called a presheaf. A presheaf $F$ is a called a sheaf if given every object $U$ of $\mathscr{C}$ and $\{U_{i} \rightarrow U\}_{i \in I},$ the sequence (given similarly from the case of the small/big Zariski site on a scheme) $$F(U) \rightarrow \prod_{i \in I}F(U_{i}) \rightrightarrows \prod_{i \in I}F(U_{i} \times_{U} U_{j})$$ is an equalizer in $\textbf{Set}.$
Remark. We have observed before that given $X,$ all the representable functors in $\mathrm{Fun}(\textbf{Sch}_{X}^{\mathrm{op}}, \textbf{Set})$ are sheaves on the big Zariski site of $X.$ Moreover, we learned that a functor in $\mathrm{Fun}(\textbf{Sch}_{X}^{\mathrm{op}}, \textbf{Set})$ is representable if and only if
- it is a sheaf on Zariski site of $X$ and
- it can be covered by representable open subfunctors.
Construction of fiber product of schemes. We finish the posting by giving a basic application: construction of the fiber product of two schemes over a ring $R.$ I am not sure whether we can do this for more general base instead of $\mathrm{Spec}(R)$, because generally structure maps over a base scheme are not necessarily affine, which we would need in the arguments below.
Consider $R$-schemes $X, Y, Z$ with $R$-scheme maps $X \rightarrow Z$ and $Y \rightarrow Z.$ Using the Yoneda embedding $$\textbf{Sch}_{R} \hookrightarrow \mathrm{Fun}(\textbf{Sch}_{R}^{\mathrm{op}}, \textbf{Set}),$$ we note that if the functor $h_{X} \times_{h_{Z}} h_{Y}$ is representable, then the scheme that represents this functor must be the fiber product $X \times_{Z} Y$ in $\textbf{Sch}_{R}.$ Let $\mathrm{Spec}(A) \subset X, \mathrm{Spec}(B) \subset Y, \mathrm{Spec}(C) \subset Z$ be affine open subsets. We have a canonical map $$h_{\mathrm{Spec}(A)} \times_{h_{\mathrm{Spec}(C)}} h_{\mathrm{Spec}(B)} \rightarrow h_{X} \times_{h_{Z}} h_{Y}$$ given by the fiber product in $\mathrm{Fun}(\textbf{Sch}_{R}^{\mathrm{op}}, \textbf{Set}).$ We claim that these are representable open subfunctors that cover $h_{X} \times_{h_{Z}} h_{Y}$ and the functor $h_{X} \times_{h_{Z}} h_{Y}$ is a sheaf on the big Zariski site on $X.$ This will finish showing that $h_{X} \times_{h_{Z}} h_{Y}$ is representable.
Proof. For representability, one can check that $$h_{\mathrm{Spec}(A \otimes_{C} B)} = h_{\mathrm{Spec}(A)} \times_{h_{\mathrm{Spec}(C)}} h_{\mathrm{Spec}(B)}.$$ To check the functor maps into $h_{X} \times_{h_{Z}} h_{Y},$ we can do a bit more generally. Given any open subsets $U \hookrightarrow X, V \hookrightarrow Y, W \hookrightarrow Z,$ where $U$ and $V$ map into $W$ using the maps $X \rightarrow Z$ and $Y \rightarrow Z$ we fixed in the beginning. Given any $h_{T} \rightarrow h_{X} \times_{h_{Z}} h_{Y},$ we have $R$-scheme maps $\alpha : T \rightarrow X$ and $\beta : T \rightarrow Y,$ that are also compatible with the maps from $X, Y$ to $Z.$ We have an open subset $\alpha^{-1}(U) \cap \beta^{-1}(V) \hookrightarrow T,$ and one may check that the corresponding map $$h_{\alpha^{-1}(U) \cap \beta^{-1}(V)} \rightarrow h_{T}$$ gives the fiber prduct $$h_{\alpha^{-1}(U) \cap \beta^{-1}(V)} = (h_{U} \times_{h_{W}} h_{V}) \times_{h_{X} \times_{h_{Z}} h_{Y}} h_{T}.$$ To show the covering condition, take any affine open $\mathrm{Spec}(C) \subset Z$ and cover its preimage in $X$ by affine opens and the same in $Y.$ Then we may cover $T$ with open subsets of the form $\alpha^{-1}(\mathrm{Spec}(A)) \cap \beta^{-1}(\mathrm{Spec}(B)),$ where $\mathrm{Spec}(A), \mathrm{Spec}(B)$ are affine opens of $X$ and $Y,$ respectively.
Finally, we need to show that $h_{X} \times_{h_{Z}} h_{Y}$ is a sheaf (on the big Zariski site). Let $U$ be an $R$-scheme and say $U = \bigcup_{i \in I}U_{i}$ is an open cover. We use the construction of the fiber product of the functors to show that the following sequence is an equalizer in $\textbf{Set}$: $$h_{X}(U) \times_{h_{Z}(U)} h_{Y}(U) \rightarrow \prod_{i \in I} h_{X}(U_{i}) \times_{h_{Z}(U_{i})} h_{Y}(U_{i}) \rightrightarrows \prod_{i, j \in I} h_{X}(U_{ij}) \times_{h_{Z}(U_{ij})} h_{Y}(U_{ij}),$$ where $U_{ij} := U_{i} \cap U_{j}.$ Given $(s_{i}, t_{i}) \in h_{X}(U_{i}) \times_{h_{Z}(U_{i})} h_{Y}(U_{i})$ for $i \in I,$ if $$(s_{i}, t_{i})|_{U_{ij}} = (s_{j}, t_{j})|_{U_{ij}}$$ for all $i, j \in I,$ then $s_{i}|_{U_{ij}} = s_{j}|_{U_{ij}}$ and $t_{i}|_{U_{ij}} = t_{j}|_{U_{ij}},$ for all $i, j \in I.$ Since $h_{X}$ and $h_{Y}$ are sheaves, we have a unique $s \in h_{X}(U)$ such that $s|_{U_{i}} = s_{i}$ for all $i \in I$ and a unique $t \in h_{Y}(U)$ such that $t|_{U_{i}} = t_{i}$ for all $i \in I.$ The compositions $U \overset{s}{\longrightarrow} X \rightarrow Z$ and $U \overset{t}{\longrightarrow} Y \rightarrow Z$ must be identical because their restrictions on $U_{i} \hookrightarrow U$ are for every $i \in I.$ Thus, we have found a unique $(s, t) \in h_{X}(U) \times_{h_{Z}(U)} h_{Y}(U)$ such that $$(s, t)|_{U_{i}} = (s_{i}, t_{i})$$ for all $i \in I.$ This finishes the proof, so in particular we proved that $h_{X} \times_{h_{Z}} h_{Y}$ is representable by an $R$-scheme $\Box$
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