Of course, I may cite more references as I go.
Goal. The goal is to follow the first chapter of the first reference.
Globalizing unramified maps. We say that a map $\pi : X \rightarrow Y$ of schemes is unramified if
Globalizing unramified maps. We say that a map $\pi : X \rightarrow Y$ of schemes is unramified if
- it is locally of finitely presentation, and
- for every $x \in X,$ the induced map $\kappa_{Y, \pi(x)} = \mathscr{O}_{Y, \pi(x)}/\mathfrak{m}_{Y, \pi(x)} \rightarrow \mathscr{O}_{X, x}/\mathfrak{m}_{Y, \pi(x)}\mathscr{O}_{X, x}$ is a finite separable field extension.
Remark. We note that requiring that $\mathscr{O}_{X, x}/\mathfrak{m}_{Y, \pi(x)}\mathscr{O}_{X, x}$ is a field necessarily implies that $\mathfrak{m}_{Y, \pi(x)}\mathscr{O}_{X, x} = \mathfrak{m}_{X, x},$ which means that $$\mathscr{O}_{X, x}/\mathfrak{m}_{Y, \pi(x)}\mathscr{O}_{X, x} = \mathscr{O}_{X, x}/\mathfrak{m}_{X,x} = \kappa_{X,x}.$$ We note that a ring map $A \rightarrow B$ is unramified in the sense of our first lecture if and only if its induced scheme map $\mathrm{Spec}(B) \rightarrow \mathrm{Spec}(A)$ is unramified.
Globalizing flat maps. We say that a map $\pi : X \rightarrow Y$ of schemes is flat if the induced map $\mathscr{O}_{Y, \pi(x)} \rightarrow \mathscr{O}_{X, x}$ is flat for every $x \in X.$
In the first lecture, we have introduced another notion of flatness. This notion corresponds to the (scheme-theoretic) flatness above although it is not that obvious:
Globalizing flat maps. We say that a map $\pi : X \rightarrow Y$ of schemes is flat if the induced map $\mathscr{O}_{Y, \pi(x)} \rightarrow \mathscr{O}_{X, x}$ is flat for every $x \in X.$
In the first lecture, we have introduced another notion of flatness. This notion corresponds to the (scheme-theoretic) flatness above although it is not that obvious:
Proposition (cf. Vakil 24.F.2). A ring map $\phi : B \rightarrow A$ is flat if and only if the induced map $B_{\phi^{-1}(\mathfrak{p})} \rightarrow A_{\mathfrak{p}}$ is flat for every prime $\mathfrak{p}$ of $A$ (i.e., $\mathrm{Spec}(B) \rightarrow \mathrm{Spec}(A)$ is scheme-theoretically flat).
Proof. Suppose that $\phi : B \rightarrow A$ is flat. To show $B_{\phi^{-1}(\mathfrak{p})} \rightarrow A_{\mathfrak{p}}$ is flat, the key is to notice that $$A_{\phi^{-1}(\mathfrak{p})} \rightarrow A_{\mathfrak{p}}$$ defined by $a/s \mapsto a/\phi(s)$ is a localization of $A_{\phi^{-1}(\mathfrak{p})} = (B \setminus \phi^{-1}(\mathfrak{p}))^{-1}A,$ seen as an $A$-module, at $\mathfrak{p}.$ Since $$(-) \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}} \simeq (-) \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\phi^{-1}(\mathfrak{p})} \otimes_{A_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}},$$ and flatness of $A \rightarrow B$ implies that $A_{\phi^{-1}(\mathfrak{p})} \rightarrow B_{\phi^{-1}(\mathfrak{p})}$ is flat, so the right-hand side is an exact functor, which implies the same for the left-hand side.
Conversely, say $B_{\phi^{-1}(\mathfrak{p})} \rightarrow A_{\mathfrak{p}}$ is flat for all $\mathfrak{p} \in \mathrm{Spec}(A).$ To show that $B \rightarrow A$ is flat, fix any injection $M \hookrightarrow N$ of $B$-modules. Our goal is to show that the induced map $M \otimes_{B} A \rightarrow N \otimes_{B} A$ is injective. Since the last map can be seen as a map of $A$-modules, it is enough to show that $$(M \otimes_{B} A)_{\mathfrak{p}} \rightarrow (N \otimes_{B} A)_{\mathfrak{p}}$$ for a prime $\mathfrak{p}$ of $A.$ It is important to note that this map looks like $$(m \otimes a)/s \mapsto (\phi(m) \otimes a)/s,$$ because then this is reduced to showing the following lemma:
Lemma. Let $M$ be a $B$-module, and say $\phi : B \rightarrow A$ is a ring map. For every prime $\mathfrak{p}$ of $A,$ we have $$(M \otimes_{B} A)_{\mathfrak{p}} \simeq M_{\phi^{-1}(\mathfrak{p})} \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}}$$ given by $$\frac{m \otimes a}{s} \mapsto (m/1) \otimes (a/s).$$
Proof of Lemma. We may define the $B$-module map $$M \otimes_{B} A \rightarrow M_{\phi^{-1}(\mathfrak{p})} \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}}$$ given by $$m \otimes a \mapsto (m/1) \otimes (a/1),$$ using the universal property of the tensor product. Note that this map is also an $A$-module map by the restriction of scalars using $\phi : A \rightarrow B.$ Since the multiplication by any $s \in A \setminus \mathfrak{p}$ on the right-hand side is an isomorphism, we get the $A_{\mathfrak{p}}$-module map $$(M \otimes_{B} A)_{\mathfrak{p}} \rightarrow M_{\phi^{-1}(\mathfrak{p})} \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}}$$ given by $$\frac{m \otimes a}{s} \mapsto (m/1) \otimes (a/s).$$ To show that this is an isomorphism, we only need to show that this is a bijection, so we now construct a set-theoretic inverse.
Consider the $B_{\phi^{-1}(\mathfrak{p})}$-module map $$M_{\phi^{-1}(\mathfrak{p})} \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}} \rightarrow (M \otimes_{B} A)_{\mathfrak{p}}$$ given by $$(m/t) \otimes (a/s) \mapsto \frac{m \otimes a}{\phi(t)s},$$ using the universal property of the tensor product. It is a routine to check that this is the desired inverse. $\Box$
This finishes the proof of Proposition. $\Box$
Globalizing étale maps. We say that a map of schemes is étale if it is flat and unramified.
Remark. Note that a ring map $B \rightarrow A$ is étale in the sense of our first lecture if and only if the induced scheme map $\mathrm{Spec}(A) \rightarrow \mathrm{Spec}(B)$ is étale.
Remark. If $\pi : X \rightarrow Y$ is a scheme map locally of finite presentation, checking that it is étale is checking two conditions on the induced map $\mathscr{O}_{Y, \pi(x)} \rightarrow \mathscr{O}_{X, x}$ for each point $x \in X.$ These two conditions work well with composition of two maps, so we immediately get the following:
Lemma (étaleness is closed under composition). The composition of two étale maps is étale.
The following theorem is an analogous statement to the étale cancellation in our first lecture. The proof takes the same route (although it is not at all obvious to come up with it):
Theorem (étale cancellation). Let $\pi : X \rightarrow Y$ and $\psi : Y \rightarrow Z$ be scheme maps locally of finite presentation. If $\psi$ and $\psi \circ \pi$ are étale, so is $\pi.$
Proposition (étaleness is closed under base change). Given any étale map $X \rightarrow Y$ of schemes, for any scheme map $Z \rightarrow X,$ the base change $Z \times_{Y} X \rightarrow Z$ of the given map is étale.
Proof. It follows from that the following properties are closed under base change:
Proof. Suppose that $\phi : B \rightarrow A$ is flat. To show $B_{\phi^{-1}(\mathfrak{p})} \rightarrow A_{\mathfrak{p}}$ is flat, the key is to notice that $$A_{\phi^{-1}(\mathfrak{p})} \rightarrow A_{\mathfrak{p}}$$ defined by $a/s \mapsto a/\phi(s)$ is a localization of $A_{\phi^{-1}(\mathfrak{p})} = (B \setminus \phi^{-1}(\mathfrak{p}))^{-1}A,$ seen as an $A$-module, at $\mathfrak{p}.$ Since $$(-) \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}} \simeq (-) \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\phi^{-1}(\mathfrak{p})} \otimes_{A_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}},$$ and flatness of $A \rightarrow B$ implies that $A_{\phi^{-1}(\mathfrak{p})} \rightarrow B_{\phi^{-1}(\mathfrak{p})}$ is flat, so the right-hand side is an exact functor, which implies the same for the left-hand side.
Conversely, say $B_{\phi^{-1}(\mathfrak{p})} \rightarrow A_{\mathfrak{p}}$ is flat for all $\mathfrak{p} \in \mathrm{Spec}(A).$ To show that $B \rightarrow A$ is flat, fix any injection $M \hookrightarrow N$ of $B$-modules. Our goal is to show that the induced map $M \otimes_{B} A \rightarrow N \otimes_{B} A$ is injective. Since the last map can be seen as a map of $A$-modules, it is enough to show that $$(M \otimes_{B} A)_{\mathfrak{p}} \rightarrow (N \otimes_{B} A)_{\mathfrak{p}}$$ for a prime $\mathfrak{p}$ of $A.$ It is important to note that this map looks like $$(m \otimes a)/s \mapsto (\phi(m) \otimes a)/s,$$ because then this is reduced to showing the following lemma:
Lemma. Let $M$ be a $B$-module, and say $\phi : B \rightarrow A$ is a ring map. For every prime $\mathfrak{p}$ of $A,$ we have $$(M \otimes_{B} A)_{\mathfrak{p}} \simeq M_{\phi^{-1}(\mathfrak{p})} \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}}$$ given by $$\frac{m \otimes a}{s} \mapsto (m/1) \otimes (a/s).$$
Proof of Lemma. We may define the $B$-module map $$M \otimes_{B} A \rightarrow M_{\phi^{-1}(\mathfrak{p})} \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}}$$ given by $$m \otimes a \mapsto (m/1) \otimes (a/1),$$ using the universal property of the tensor product. Note that this map is also an $A$-module map by the restriction of scalars using $\phi : A \rightarrow B.$ Since the multiplication by any $s \in A \setminus \mathfrak{p}$ on the right-hand side is an isomorphism, we get the $A_{\mathfrak{p}}$-module map $$(M \otimes_{B} A)_{\mathfrak{p}} \rightarrow M_{\phi^{-1}(\mathfrak{p})} \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}}$$ given by $$\frac{m \otimes a}{s} \mapsto (m/1) \otimes (a/s).$$ To show that this is an isomorphism, we only need to show that this is a bijection, so we now construct a set-theoretic inverse.
Consider the $B_{\phi^{-1}(\mathfrak{p})}$-module map $$M_{\phi^{-1}(\mathfrak{p})} \otimes_{B_{\phi^{-1}(\mathfrak{p})}} A_{\mathfrak{p}} \rightarrow (M \otimes_{B} A)_{\mathfrak{p}}$$ given by $$(m/t) \otimes (a/s) \mapsto \frac{m \otimes a}{\phi(t)s},$$ using the universal property of the tensor product. It is a routine to check that this is the desired inverse. $\Box$
This finishes the proof of Proposition. $\Box$
Globalizing étale maps. We say that a map of schemes is étale if it is flat and unramified.
Remark. Note that a ring map $B \rightarrow A$ is étale in the sense of our first lecture if and only if the induced scheme map $\mathrm{Spec}(A) \rightarrow \mathrm{Spec}(B)$ is étale.
Remark. If $\pi : X \rightarrow Y$ is a scheme map locally of finite presentation, checking that it is étale is checking two conditions on the induced map $\mathscr{O}_{Y, \pi(x)} \rightarrow \mathscr{O}_{X, x}$ for each point $x \in X.$ These two conditions work well with composition of two maps, so we immediately get the following:
Lemma (étaleness is closed under composition). The composition of two étale maps is étale.
The following theorem is an analogous statement to the étale cancellation in our first lecture. The proof takes the same route (although it is not at all obvious to come up with it):
Theorem (étale cancellation). Let $\pi : X \rightarrow Y$ and $\psi : Y \rightarrow Z$ be scheme maps locally of finite presentation. If $\psi$ and $\psi \circ \pi$ are étale, so is $\pi.$
Proposition (étaleness is closed under base change). Given any étale map $X \rightarrow Y$ of schemes, for any scheme map $Z \rightarrow X,$ the base change $Z \times_{Y} X \rightarrow Z$ of the given map is étale.
Proof. It follows from that the following properties are closed under base change:
- being locally of finite presentation (Vakil 9.4.B.(h));
- being unramified (Vakil 21.6.C);
- being flat.
We prove the last one. Let $X \rightarrow Y$ be flat. To check $Z \times_{Y} X \rightarrow Z$ is flat, we may check it affine-locally so that we may assume $X, Y, Z$ are affine. Then this is only a consequence of the following lemma:
Lemma (Vakil 24.2.D). Given a ring map $B \rightarrow A$ and an $A$-module $M,$ if $M$ is flat over $A,$ then $M \otimes_{A} B$ is flat over $B.$
Proof of Lemma. This follows from $$M \otimes_{A} B \otimes_{B} (-) \simeq M \otimes_{A} (-),$$ the latter of which is exact using flatness of $M.$ $\Box$
We summarize what we have so far to use it soon. Note that the forth statement below is immediate at this point, even though we have not specifically mentioned it earlier.
Étale cover. Let $X$ be a scheme. Given any scheme map $U \rightarrow X,$ an étale cover of $U$ over $X$ is a collection $\{U_{i} \rightarrow U\}_{i \in I}$ of étale maps over $X$ such that the union of images of $U_{i},$ where $i \in I$ vary, is equal to $U.$ We denote by $\mathrm{Cov}_{X}(U) = \mathrm{Cov}(U)$ to be the set étale covers of $U$ over $X.$
Remark. For any scheme $U$ and a collection $\{U_{i} \rightarrow U\}_{i \in I}$ of scheme maps into $U,$ we have a unique map $$\bigsqcup_{i \in I}U_{i} \rightarrow U$$ such that for each $j \in I,$ we have a unique factoriztion $$U_{j} \hookrightarrow \bigsqcup_{i \in I}U_{i} \rightarrow U$$ given by the universal property of the disjoint union. We note that the following are equivalent:
Proposition (small étale site). The étale coverings over a scheme $X$ give a Grothendieck topology on $X_{ét}.$ When we write $X_{ét},$ calling it the small étale site of $X,$ we not only mean the category $X_{ét}$ but also think of the Grothendieck topology given by étale coverings over $X$ together.
- Any base change of an étale map in the category of schemes is étale.
- A composition of any two étale maps is étale.
- (étale cancellation) Given scheme maps $\pi : X \rightarrow Y$ and $\rho : Y \rightarrow Z,$ if $\rho \circ \pi$ and $\rho$ are étale, then $\pi$ is étale.
- Any isomorphism of schemes is étale.
Étale cover. Let $X$ be a scheme. Given any scheme map $U \rightarrow X,$ an étale cover of $U$ over $X$ is a collection $\{U_{i} \rightarrow U\}_{i \in I}$ of étale maps over $X$ such that the union of images of $U_{i},$ where $i \in I$ vary, is equal to $U.$ We denote by $\mathrm{Cov}_{X}(U) = \mathrm{Cov}(U)$ to be the set étale covers of $U$ over $X.$
Remark. For any scheme $U$ and a collection $\{U_{i} \rightarrow U\}_{i \in I}$ of scheme maps into $U,$ we have a unique map $$\bigsqcup_{i \in I}U_{i} \rightarrow U$$ such that for each $j \in I,$ we have a unique factoriztion $$U_{j} \hookrightarrow \bigsqcup_{i \in I}U_{i} \rightarrow U$$ given by the universal property of the disjoint union. We note that the following are equivalent:
- the union of images of $U_{i},$ where $i \in I$ vary, is equal to $U$;
- the map $\bigsqcup_{i \in I}U_{i} \rightarrow U$ is surjective.
Small/big étale site of a scheme. Given any scheme $X,$ denote by $X_{ét},$ the category whose objects are étale maps $U \rightarrow X$, and whose morphisms are étale maps over $X.$ Thanks to étale cancellation, it is extremely easy to check whether a map belongs to $X_{ét},$ as stated below.
Corollary (to étale cancellation). Given any objects $U, V$ of $X_{ét},$ any scheme map $U \rightarrow V$ over $X$ is étale, so it is a morphism of $X_{ét}.$ In particular, any collection of scheme maps into $U$ whose images cover $U$ is an étale cover.
Remark. Another way to phrase the above statement is to say that $X_{ét}$ is a full subcategory of the category $\textbf{Sch}_{X}$ of schemes over $X.$
Corollary (to étale cancellation). Given any objects $U, V$ of $X_{ét},$ any scheme map $U \rightarrow V$ over $X$ is étale, so it is a morphism of $X_{ét}.$ In particular, any collection of scheme maps into $U$ whose images cover $U$ is an étale cover.
Remark. Another way to phrase the above statement is to say that $X_{ét}$ is a full subcategory of the category $\textbf{Sch}_{X}$ of schemes over $X.$
Proposition (big étale site). The étale coverings over a scheme $X$ give a Grothendieck topology on $\mathrm{Sch}_{X}.$ In other words, we have:
- any isomorphism into $U$ is in $\mathrm{Cov}(U)$ as a singleton;
- given any $\{U_{i} \rightarrow U\}_{i \in I}$ in $\mathrm{Cov}(U),$ for any map $V \rightarrow U$ of $\mathrm{Sch}_{X},$ the fiber product $V \times_{U} U_{i}$ exists in $\mathrm{Sch}_{X}$ for each $i \in I$ and the induced collection $\{V \times_{U} U_{i} \rightarrow V\}_{i \in I}$ is an element of $\mathrm{Cov}(V)$;
- given any $\{U_{i} \rightarrow U\}_{i \in I}$ in $\mathrm{Cov}(U)$ and $\{U_{ij} \rightarrow U_{i}\}_{j \in J_{i}}$ in $\mathrm{Cov}(U_{i})$ for each $i \in I,$ the collection $\{U_{ij} \rightarrow U_{i} \rightarrow U\}_{i \in I, j \in J_{i}}$ of compositions is an element of $\mathrm{Cov}(U).$
We write $(\mathrm{Sch}_{X})_{ét},$ calling it the big étale site of $X,$ to mean the category $\mathrm{Sch}_{X}$ with the Grothendieck topology given by étale coverings over $X.$
Proof. The first condition holds because any isomorphism is an étale map.
To check the second condition, fix any $\{U_{i} \rightarrow U\}_{i \in I}$ in $\mathrm{Cov}(U)$ and $V \rightarrow U$ in $\mathrm{Sch}_{X}.$ Since $U_{i} \rightarrow U$ is étale, so is the base change $V \times_{U} U_{i} \rightarrow V.$ We note that $V \times_{U} U_{i}$ taken over $\mathrm{Spec}(\mathbb{Z})$ is also the fiber product $V \times_{U} U_{i}$ taken over $X$ (i.e., in $\mathrm{Sch}_{X}$).
Remark. If $U \rightarrow X$ is étale, then the above shows that $V \times_{U} U_{i}$ is also the fiber product taken in $X_{ét}.$
If we consider the map $\bigsqcup_{i \in I}U_{i} \rightarrow U$ induced by the maps in the cover $\{U_{i} \rightarrow U\}_{i \in I},$ it is surjective due to the definition of $\mathrm{Cov}(U).$ Thus, the base change of this map under $V \rightarrow U$ in the category $\textbf{Sch}$ of schemes, which is the same as $$\bigsqcup_{i \in I}(V \times_{U} U_{i}) \rightarrow V,$$ is surjective (e.g., Vakil 9.4.D), and this is induced by the collection $\{V \times_{U} U_{i} \rightarrow V\}_{i \in I}$ of maps given by the fiber products. This implies that $\{V \times_{U} U_{i} \rightarrow V\}_{i \in I}$ is an element of $\mathrm{Cov}(V),$ which checks the second condition.
Remark. If $U \rightarrow X$ is étale, then the above shows that $V \times_{U} U_{i}$ is also the fiber product taken in $X_{ét}.$
If we consider the map $\bigsqcup_{i \in I}U_{i} \rightarrow U$ induced by the maps in the cover $\{U_{i} \rightarrow U\}_{i \in I},$ it is surjective due to the definition of $\mathrm{Cov}(U).$ Thus, the base change of this map under $V \rightarrow U$ in the category $\textbf{Sch}$ of schemes, which is the same as $$\bigsqcup_{i \in I}(V \times_{U} U_{i}) \rightarrow V,$$ is surjective (e.g., Vakil 9.4.D), and this is induced by the collection $\{V \times_{U} U_{i} \rightarrow V\}_{i \in I}$ of maps given by the fiber products. This implies that $\{V \times_{U} U_{i} \rightarrow V\}_{i \in I}$ is an element of $\mathrm{Cov}(V),$ which checks the second condition.
For the third condition, let $\{U_{i} \rightarrow U\}_{i \in I} \in \mathrm{Cov}(U)$ and $$\{U_{ij} \rightarrow U_{i}\}_{j \in J_{i}} \in \mathrm{Cov}(U_{i}).$$ By definition of $\mathrm{Cov}(U_{i})$ for each $i \in I,$ the induced map $\bigsqcup_{j \in J_{i}}U_{ij} \rightarrow U_{i}$ is surjective for all $i \in I,$ so we get to induce the map $$\bigsqcup_{i \in I}\bigsqcup_{j \in J_{i}} U_{ij} \rightarrow \bigsqcup_{i \in I} U_{i}$$ that is surjective. Since $\bigsqcup_{i \in I}U_{i} \rightarrow U$ is surjective by definition of $\mathrm{Cov}(U),$ we see that the composition $$\bigsqcup_{i \in I}\bigsqcup_{j \in J_{i}} U_{ij} \rightarrow \bigsqcup_{i \in I} U_{i} \rightarrow U$$ is surjective. This implies that $\{U_{ij} \rightarrow U_{i} \rightarrow U\}_{i \in I, j \in J_{i}} \in \mathrm{Cov}(U).$ This checks the third condition, which finishes the proof $\Box$
Proposition (small étale site). The étale coverings over a scheme $X$ give a Grothendieck topology on $X_{ét}.$ When we write $X_{ét},$ calling it the small étale site of $X,$ we not only mean the category $X_{ét}$ but also think of the Grothendieck topology given by étale coverings over $X$ together.
Proof. The proof for establishing the big étale site works (with Remark within the proof). $\Box$
Convention. Unless mentioned otherwise, we say the étale site of $X$ to mean the small étale site of $X.$
Next time. We will discuss how to define presheaves and sheaves on $X_{ét}.$
No comments:
Post a Comment