Grassmannians - Part 2

We continue a previous posting following Vakil's notes. We work over a fixed base scheme $S$ and fixed integers $0 \leq r \leq n,$ where $n \geq 1.$ Recall that the Grassmanian is a functor $\mathrm{Gr}_{S}(r, n) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$ that sends an $S$-scheme $Y$ to the isomorphism class of an epimorphism $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q},$ where $\mathscr{Q}$ is a locally free $\mathscr{O}_{Y}$-module of rank $r.$ When $r = 1,$ we saw in the previous posting that $\mathrm{Gr}_{S}(1, n)$ is represented by $\mathbb{P}^{n-1}_{S}.$

Special case: rational points over a field. Let $S = \mathrm{Spec}(k)$ for some field $k.$ Then the elements of $\mathrm{Gr}(r, n)(k) := \mathrm{Gr}_{\mathrm{Spec}(k)}(r, n)(\mathrm{Spec}(k))$ are given by the $k$-linear surjections $k^{n} \twoheadrightarrow k^{r},$ which can be viewed as $r \times n$ matrices with at least one $r \times r$ submatrix having nonzero determinant. (In other words, the matrices have rank $r,$ which is the full rank.) We are these maps up to isomorphisms, which amounts to the action of $\mathrm{GL}_{r}(k)$ on the left. Thus, we have established: $$\mathrm{Gr}(r, n)(k) \leftrightarrow \mathrm{GL}_{r}(k) \backslash \mathrm{Mat}_{r \times n}(k).$$ The elements of the right-hand side can be seen as picking $r$-linearly independent vectors up to $k$-linear isomorphisms of the vector spaces they generate in $k^{n}.$ Thus, we see that $\mathrm{Gr}(r, n)(k)$ parametrizes $r$-dimensional subspaces of $k^{n}.$ The following lemma explains this a bit more clearly:

Lemma. Let $\{v_{1}, \dots, v_{r}\}$ and $\{w_{1}, \dots, w_{r}\}$ two sets of $r$ $k$-linearly independent vectors in $k^{n}.$ Denote by $A$ the $r \times n$ matrix whose $i$-th row is given by $v_{i}$ and define $B$ similarly using $w_{i}.$ Then the following are equivalent:

• $\{v_{1}, \dots, v_{r}\}$ and $\{w_{1}, \dots, w_{r}\}$ generate the same subspace in $k^{n}$;
• $B = gA$ for some $g \in \mathrm{GL}_{r}(k).$

Proof. Write $v_{i} = (v_{i,1}, \dots, v_{i,n})$ and $w_{i} = (w_{i,1}, \dots, w_{i,n})$ with $v_{i,j}, w_{i,j} \in k.$ Then for any $g = (g_{i,j}) \in \mathrm{Mat}_{r}(k),$ saying that $B = gA$ is equivalent to saying $$w_{ij} = g_{i,1}v_{1,j} + \cdots + g_{i,r}v_{r,j}$$ for $1 \leq i \leq r$ and $1 \leq j \leq n.$ This condition can be rewritten as $$w_{i} = g_{i,1}v_{1} + \cdots + g_{i,r}v_{r}$$ for $1 \leq i \leq r.$ Now, note that $\{v_{1}, \dots, v_{r}\}$ and $\{w_{1}, \dots, w_{r}\}$ generate the same subspace in $k^{n}$ if and only if the last condition we mentioned is true for some $g_{i,j} \in k.$ When this happens, it is immediate that $(g_{ij})$ is necessarily invertible, so this finishes the proof. $\Box$

Back to functorial nonsense. We are going to show that $$\mathrm{Gr}_{S}(r, n) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ is representable. Since all representable functor is a Zariski sheaf (i.e., a sheaf on the big Zariski site of $S$), the strategy is to show that $\mathrm{Gr}_{S}(r, n)$ is a Zarski sheaf and then check a certain property on top of that so that we can guarantee that it is representable by some $S$-scheme (which will be necessarily a unique one). That is, after showing $\mathrm{Gr}_{S}(r, n)$ is a Zarski sheaf, we will use the following theorem we studied in a previous posting:

Theorem. A functor $F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$ is representable if and only if the following two conditions are satisfied:

• $F$ is a Zariski sheaf;
• $F$ can be covered by representable open subfunctors.

Checking Grassmanianns are Zariski sheaves. Saying that $\mathrm{Gr}_{S}(r, n)$ is a sheaf means that for any $S$-scheme $U$ and an open cover $U = \bigcup_{i \in I}U_{i},$ the following two conditions are satisfied:

• if $\phi, \psi \in \mathrm{Gr}_{S}(r, n)(U)$ such that $\phi|_{U_{i}} = \psi|_{U_{i}}$ for all $i \in I,$ then $\phi = \psi$;
• if $(\phi_{i})_{i \in I} \in \prod_{i \in I}\mathrm{Gr}_{S}(r, n)(U_{i})$ satisfies that $\phi_{i}|_{U_{i} \cap U_{j}} = \phi_{j}|_{U_{i} \cap U_{j}}$ for all $i, j \in I,$ then there is $\phi \in \mathrm{Gr}_{S}(r, n)(U)$ such that $\phi|_{U_{i}} = \phi_{i}$ for all $i \in I.$

Proof for the first condition. To check the first among the two conditions above, consider $\phi : \mathscr{O}_{U}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ and $\psi : \mathscr{O}_{U}^{\oplus n} \twoheadrightarrow \mathscr{Q}'.$ Since we are working with representatives of isomorphism classes, instead of checking needed equalities, we aim to establish corresponding isomorphisms. Denote by $s_{1}, \dots, s_{n} \in \mathscr{Q}(U)$ and $t_{1}, \dots, t_{n} \in \mathscr{Q}'(U)$ the images of $e_{1}, \dots, e_{n} \in \mathscr{O}(U)^{\oplus n},$ writing $\mathscr{O} = \mathscr{O}_{U}$ for convenience.

Suppose that $\phi|_{U_{i}} \simeq \psi|_{U_{i}}$ for all $i \in I.$ Our goal is to show that $\phi \simeq \psi.$ For this, we may assume that $U_{i}$'s are refined enough so that both $\mathscr{Q}|_{U_{i}}$ and $\mathscr{Q}'|_{U_{i}}$ are free of rank $r$ over $U_{i}.$ The restrictions $\phi|_{U_{i}} : \mathscr{O}_{U_{i}}^{\oplus n} \twoheadrightarrow \mathscr{Q}|_{U_{i}}$ and $\psi|_{U_{i}} : \mathscr{O}_{U_{i}}^{\oplus n} \twoheadrightarrow \mathscr{Q}'|_{U_{i}}$ are isomorphic so that we have $\eta_{i} : \mathscr{Q}|_{U_{i}} \overset{\sim}{\longrightarrow} \mathscr{Q}'|_{U_{i}}$ with $\psi|_{U_{i}} = \eta_{i} \circ \phi|_{U_{i}}$ for $i \in I.$ Given $i \in I,$ the commutativity condition can be written as $$t_{l}|_{U_{i}} = \eta_{i, U_{i}}(s_{l}|_{U_{i}})$$ for $1 \leq l \leq n.$ An important observation is the following:

Claim. We have $\eta_{i, U_{i} \cap U_{j}} = \eta_{j, U_{i} \cap U_{j}}.$

Proof of Claim. Note that $$\eta_{i, U_{i} \cap U_{j}}(s_{l}|_{U_{i} \cap U_{j}}) = t_{l}|_{U_{i} \cap U_{j}} = \eta_{j, U_{i} \cap U_{j}}(s_{l}|_{U_{i} \cap U_{j}})$$ for all $1 \leq l \leq n.$ Since $s_{1}|_{U_{i} \cap U_{j}}, \dots, s_{l}|_{U_{i} \cap U_{j}}$ generate $\mathscr{Q}(U_{i} \cap U_{j}),$ this establishes the identity. $\Box$ (Claim)

From now on let us write $U_{ij} = U_{i} \cap U_{j},$ as the notations get messier. Let $V \subset U$ be any open subset. We write $V_{i} = V \cap U_{i}$ and $V_{ij} = V \cap U_{ij}.$ Given any $s \in \mathscr{Q}(V),$ we have $$\eta_{i, V_{i}}(s|_{V_{ij}})|_{V_{ij}} = \eta_{i, V_{ij}}(s|_{V_{ij}}) = \eta_{j, V_{ij}}(s|_{V_{ij}}) = \eta_{j, V_{j}}(s|_{V_{j}})|_{V_{ij}}$$ for $i, j \in I.$ Since $\mathscr{Q}'$ is a sheaf, there is unique $\eta_{V}(s) \in \mathscr{Q}'(V)$ such that $\eta_{V}(s)|_{V_{i}} = \eta_{i, V_{i}}(s|_{V_{i}})$ for $i \in I.$ This defines a map $\eta_{V} : \mathscr{Q}(V) \rightarrow \mathscr{Q}'(V),$ which can be immediately recognized as a sheaf map $\eta : \mathscr{Q} \rightarrow \mathscr{Q}'$ by intersection with $U_{i}$'s to check some commutative diagrams. $\Box$

Proof for the second condition. Consider an $\mathscr{O}_{U_{i}}$-linear epimorphism $$\phi_{i} : \mathscr{O}_{U_{i}}^{\oplus n} \twoheadrightarrow \mathscr{Q}_{i}$$ for each $i \in I$ such that $\phi_{i}|_{U_{ij}} \simeq \phi_{j}|_{U_{ij}}$ for $i, j \in I.$ This means that, for $i, j \in I,$ we have $\eta_{ij} : \mathscr{Q}_{i}|_{U_{ij}} \overset{\sim}{\longrightarrow} \mathscr{Q}_{j}|_{U_{ij}}$ such that $\phi_{j}|_{U_{ij}} = \eta_{ij} \circ \phi_{i}|_{U_{ij}}.$ Our goal is to find $$\phi : \mathscr{O}_{U}^{\oplus n} \twoheadrightarrow \mathscr{Q}$$ such that $\phi|_{U_{i}} \simeq \phi_{i}$ for $i \in I.$ Writing $U_{ijk} = U_{i} \cap U_{j} \cap U_{k},$ we note that we have $$\eta_{jk}|_{U_{ijk}} \circ \eta_{ij}|_{U_{ijk}} \circ \phi_{i}|_{U_{ijk}} = \eta_{jk}|_{U_{ijk}} \circ \phi_{j}|_{U_{ijk}} = \phi_{k}|_{U_{ijk}} = \eta_{kj}|_{U_{ijk}} \circ \phi_{i}|_{U_{ijk}}.$$ This implies that we have the cocycle condition $$\eta_{jk}|_{U_{ijk}} \circ \eta_{ij}|_{U_{ijk}} = \eta_{kj}|_{U_{ijk}},$$ so there must be a unique sheaf (that is necessarily locally free of rank $r$) $\mathscr{Q}$ gluing $\mathscr{Q}_{i}$'s. (For example, see SP:00AK.) Now, we are given $\phi_{i} : \mathscr{O}_{U_{i}}^{\oplus n} \rightarrow \mathscr{Q}|_{U_{i}}$ with $\phi_{i}|_{U_{ij}} \simeq \phi_{j}|_{U_{ij}}$ for $i, j \in I.$ Working similarly to the last paragraph of the last proof, we are done. $\Box$ 

Hence, we have shown that $\mathrm{Gr}_{S}(r, n)$ is a Zariski sheaf.

What remains to show Grassmannians are representable. Now that we know $\mathrm{Gr}_{S}(r, n)$ is a Zariski sheaf, to show it is representable, it remains to show that $\mathrm{Gr}_{S}(r, n)$ is covered by reprsentable open subfunctors.

Given any subset $I \subset \{1, \dots, n\}$ of size $r,$ and an $S$-scheme $Y,$ denote by $\mathrm{Gr}_{S}(r, n)_{I}(Y)$ the subset of $\mathrm{Gr}_{S}(r, n)$ consisting of $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ such that on every trivializing open subset $U \subset Y$ of $\mathscr{Q}$ (which in particular gives $\mathscr{Q}|_{U} \simeq \mathscr{O}_{U}^{\oplus r}$), the $I$-th minor of the $r \times n$ matrix given by $\mathscr{O}_{U}^{\oplus r} \twoheadrightarrow \mathscr{Q}|_{U}$ is nonzero at every point of $U.$

More concretely, given $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q},$ let $s_{1}, \dots, s_{n} \in \mathscr{Q}(Y)$ be the images of $$e_{1}, \dots, e_{n} \in \mathscr{O}_{Y}(Y)^{\oplus n}.$$ Then $s_{1}|_{U}, \dots, s_{n}|_{U}$ are the columns of the $r \times n$ matrix in discussion. Fix any $S$-scheme map $\pi : X \rightarrow Y.$ We know this induces $$\mathrm{Gr}_{S}(r, n)(Y) \rightarrow \mathrm{Gr}_{S}(r, n)(X)$$ by the pullback $\pi^{*}.$ Moreover, we note that $\pi^{*}s_{1}, \dots, \pi^{*}s_{n} \in \mathscr{O}_{X}(X)^{\oplus n}$ are the images of $e_{1}, \dots, e_{n} \in \mathscr{O}_{X}(X)^{\oplus n},$ so considering trivializing open subsets, we may observe that the above map restricts to $$\mathrm{Gr}_{S}(r, n)_{I}(Y) \rightarrow \mathrm{Gr}_{S}(r, n)_{I}(X)$$ for each $I \subset \{1, \dots, n\}$ of size $r.$ Thus, we have seen that $\mathrm{Gr}_{S}(r, n)_{I}$ is a functor $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$ We have a map $\mathrm{Gr}_{S}(r, n)_{I} \rightarrow \mathrm{Gr}_{S}(r, n)$ of functors given by the relevant inclusions of sets.

Claim. The functor $\mathrm{Gr}_{S}(r, n)_{I} \rightarrow \mathrm{Gr}_{S}(r, n)$ is an open subfunctor. That is, given any $S$-scheme $Y$ and a map $h_{Y} \rightarrow \mathrm{Gr}_{S}(r, n)$ of functors, the base chage $$\mathrm{Gr}_{S}(r, n)_{I} \times_{\mathrm{Gr}_{S}(r, n)} h_{Y} \rightarrow h_{Y}$$ is given by an open embedding $V_{I} \hookrightarrow Y$ (i.e., $h_{V_{I}} \rightarrow h_{Y}$).

Construction of open subsets $V_{I}$. We are given be an $S$-scheme $Y$ and an element of $$\mathrm{Hom}(h_{Y}, \mathrm{Gr}_{S}(r, n)) \simeq \mathrm{Gr}_{S}(r, n)(Y),$$ where the hom-set is taken over the functors $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$ The correspondence is given by the Yoneda lemma, and explicitly, it is given as $\xi \mapsto \xi_{Y}(\mathrm{id}_{Y}).$ We denote by $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ (a reperesentative of) the given element in $\mathrm{Gr}_{S}(r, n)(Y).$ It is convenient to denote by $s_{1}, \dots, s_{n} \in \mathscr{Q}(Y)$ the images of $e_{1}, \dots, e_{n} \in \mathscr{O}_{Y}(Y)^{\oplus n}$ under $\phi_{Y}.$

It necessarily follows that $h_{Y}(X) \rightarrow \mathrm{Gr}_{S}(r, n)(X)$ is given by $$[\pi : X \rightarrow Y] \mapsto [\mathscr{O}_{X}^{\oplus n} \simeq \pi^{*}\mathscr{O}_{Y}^{\oplus n} \xrightarrow{\pi^{*}\phi} \pi^{*}\mathscr{Q}.]$$ Thus, the set $$\mathrm{Gr}_{S}(r, n)_{I}(X) \times_{\mathrm{Gr}_{S}(r, n)(X)} h_{Y}(X)$$ consists of the elements of the form $(\pi^{*}\phi, \pi)$ such that $\pi^{*}\phi \in \mathrm{Gr}_{S}(r, n)(X)_{I}$ (i.e., the $I$-th minor of $[(\pi^{*}s_{1})|_{W} | \cdots | (\pi^{*}s_{1})|_{W}]$ is nowhere zero in any trivializing open $W \subset X$ for $\pi^{*}\mathscr{Q}$). The map $$\mathrm{Gr}_{S}(r, n)_{I}(X) \times_{\mathrm{Gr}_{S}(r, n)(X)} h_{Y}(X) \rightarrow h_{Y}(X)$$ is simply given by $(\pi^{*}\phi, \pi) \mapsto \pi.$ Of course, we note that $\phi$ is only given up to an isomorphism, but the above discussion still makes sense even if we just consider isomorphism classes.

Let's construct $V_{I}$'s now. Again, note that $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ is fixed. Let $U \subset Y$ be any trivializing open subset for $\mathscr{Q}.$ Denote by $f_{I} \in \mathscr{O}_{Y}(U)$ the $I$-th minor (i.e., the determinant of the $I$-th $r \times r$ submatrix of $[s_{1}|_{U} | \cdots | s_{n}|_{U}]$). We define $V_{I}$ to be the union of $D_{U}(f_{I})$ where we vary trivializing open subsets $U \subset Y$ for $\mathscr{Q}.$

Proof of Claim. For any $\pi : X \rightarrow Y$ in $h_{Y}(X) = \mathrm{Hom}_{S}(X, Y),$ we have $\pi^{*}\phi : \mathscr{O}_{X}^{\oplus n} \twoheadrightarrow \pi^{*}\mathscr{Q}$ in $\mathrm{Gr}_{S}(r, n)(X).$ The upshot is that the following are equivalent:

• we have $\pi^{*}\phi \in \mathrm{Gr}_{S}(r, n)(X)_{I}$;
• $\pi(X) \subset V_{I}.$

This immediately gives an explicit bijection $$\mathrm{Gr}_{S}(r, n)_{I}(X) \times_{\mathrm{Gr}_{S}(r, n)(X)} h_{Y}(X) \simeq h_{V_{I}}(X) = \mathrm{Hom}_{S}(V_{I}, X),$$ which does the job. $\Box$

Our $V_{I}$'s form an open cover of $Y$. For each $y \in U,$ we have a surjective $\kappa(y)$-linear map $\mathscr{O}_{Y}^{n}|_{y} \twoheadrightarrow \mathscr{Q}|_{y},$ so there exists at least one $I \subset \{1, \dots, n\}$ with $|I| = r$ such that $y \in D_{U}(f_{I}).$ This implies that we have an open cover $$U = \bigcup_{I \subset \{1, \dots, n\}}D_{U}(f_{I}).$$ Denote by $V_{I}$ the union of all $D_{U}(f_{I}) \subset Y$ where $U$ varies among all the trivializing open subsets of $Y$ for $\mathscr{Q}.$ This gives an open cover $$Y = \bigcup_{I \subset [n] \text{ with } |I| = r} V_{I},$$ where we started to write $[n] = \{1, \dots, n\}$ since the notations are getting messy.

Each $\mathrm{Gr}_{S}(n, k)_{I}$ is represented by $\mathbb{A}^{r(n-r)}_{S}$. Let $Y$ be an $S$-scheme. Given $[\phi] \in \mathrm{Gr}_{S}(r, n),$ the map $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ can be described by $n$ sections $\phi_{j} : \mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q}.$ That is, we have $\phi = \phi_{1} \oplus \cdots \oplus \phi_{n}.$ 

Fix $I = \{i_{1}, \dots, i_{r}\}$ with $1 \leq i_{1} < \cdots < i_{r} \leq n.$ If $[\phi] \in \mathrm{Gr}_{S}(r, n)_{I},$ then $\phi_{i_{1}} \oplus \cdots \oplus \phi_{i_{r}} : \mathscr{O}_{Y}^{\oplus r} \rightarrow \mathscr{Q}$ is an isomorphism. 

Proof. To see this, it is enough to show that its localization at a point $y \in Y$ is an isomorphism. Surjectivity is immediate from Nakayama. To show injectivity, consider the map $\mathscr{O}_{Y,y}^{\oplus r} \rightarrow \mathscr{Q}_{y} \simeq \mathscr{O}_{Y,y}^{\oplus r}$ in question as an $r \times r$ matrix whose columns are given by $$\phi_{i_{1},y}(1), \dots, \phi_{i_{r},y}(1) \in \mathscr{Q}_{y} \simeq \mathscr{O}_{Y,y}^{\oplus r}.$$ The determinant of this matrix is not in the maximal ideal $\mathfrak{m}_{Y,y}$ of $\mathscr{O}_{Y,y},$ so it must be invertible in the local ring. By Cramer's rule, this implies that the matrix is invertible over $\mathscr{O}_{Y,y},$ which implies the injectivity we wanted. $\Box$

Hence, we now may write $$\mathscr{Q}(Y) = \mathscr{O}_{Y}(Y)\phi_{i_{1},Y}(1) \oplus \cdots \oplus \mathscr{O}_{Y}(Y)\phi_{i_{r},Y}(1).$$ This implies that given $j \in [n] \setminus I,$ we may write $$\phi_{j,Y}(1) = a_{1,j}(\phi)\phi_{i_{1},Y}(1) + \cdots +a_{r,j}(\phi)\phi_{i_{r},Y}(1) \in \mathscr{Q}(Y),$$ where $a_{i,j}(\phi) \in \mathscr{O}_{Y}(Y) = \Gamma(Y, \mathscr{O}_{Y}),$ which can be viewed as an $S$-scheme map $a_{i,j}(\phi) : Y \rightarrow \mathbb{A}^{1}_{S},$ by gluing $R$-algebra maps $R[t] \rightarrow \Gamma(Y, \mathscr{O}_{Y})$ given by $t \mapsto a_{ij}(\phi)$ for affine open $\mathrm{Spec}(R) \subset S.$

Thus, we have $$(a_{i,j}(\phi))_{1 \leq i \leq r, j \in [n] \setminus I} : Y \rightarrow \mathbb{A}^{r(n-r)}_{S}.$$ This defines a map $$\mathrm{Gr}_{S}(r, n)(Y)_{I} \rightarrow \mathrm{Hom}_{S}(Y, \mathbb{A}^{r(n-r)}_{S}),$$ as we may note that $a_{i,j}(\phi)$ only depends on the isomorphism class of $\phi.$

Given a map $\pi : X \rightarrow Y$ of $S$-schemes, we have $$(\pi^{*}\mathscr{Q})(X) = \mathscr{O}_{X}(X)(\pi^{*}\phi_{i_{1},Y})(1) \oplus \cdots \oplus \mathscr{O}_{X}(X)(\pi^{*}\phi_{i_{r},Y})(1),$$ so we may see that $a_{i,j}(\pi^{*}\phi) = \pi^{*}a_{i,j}(\phi) \in \mathscr{O}_{X}(X)$ for $j \in [n] \setminus I.$ Since $\pi : X \rightarrow Y$ comes with the map $\Gamma(Y, \mathscr{O}_{Y}) \rightarrow \Gamma(X, \mathscr{O}_{X})$ such that $a_{i,j}(\phi) \mapsto \pi^{*}a_{i,j}(\phi),$ we see that $$(a_{i,j}(\phi))_{1 \leq i \leq r, j \in [n] \setminus I} \circ \pi = (\pi^{*}a_{i,j}(\phi))_{1 \leq i \leq r, j \in [n] \setminus I} = (a_{i,j}(\pi^{*}\phi))_{1 \leq i \leq r, j \in [n] \setminus I}.$$ This shows that our map is actually a map $$\mathrm{Gr}_{S}(r, n)(-)_{I} \rightarrow \mathrm{Hom}_{S}(-, \mathbb{A}^{r(n-r)}_{S})$$ of functors.

Conversely, given $$(a_{i,j})_{1 \leq i \leq r, j \in [n] \setminus I} : Y \rightarrow \mathbb{A}^{r(n-r)}_{S},$$ which we interpret as $a_{ij} \in \Gamma(Y, \mathscr{O}_{Y}) = \mathscr{O}_{Y}(Y).$ We define $\phi : \mathscr{O}_{Y}^{n} \rightarrow \mathscr{O}_{Y}^{\oplus r}$ as follows. We declare $\phi_{i_{1},Y}(1) := e_{1}, \dots, \phi_{i_{r},Y}(1) := e_{r} \in \mathscr{O}_{Y}(Y)^{\oplus r}$ so that $$\phi_{i_{1},Y} \oplus \cdots \oplus \phi_{i_{r},Y} : \mathscr{O}_{Y}^{\oplus r} \rightarrow \mathscr{O}_{Y}^{\oplus r}$$ is the identity map. For $j \in [n] \setminus I,$ we may define $$\phi_{j,Y}(1) := a_{1,j}\phi_{i_{1},Y}(1) + \cdots + a_{r,j}\phi_{i_{r},Y}(1) \in \mathscr{O}_{Y}(Y)^{\oplus r}.$$ One may check this gives a desired inverse so that we establish $$\mathrm{Gr}_{S}(r, n)(-)_{I} \simeq \mathrm{Hom}_{S}(-, \mathbb{A}^{r(n-r)}_{S}).$$ That is, the $S$-scheme $\mathbb{A}^{r(n-r)}_{S}$ respresents the functor $\mathrm{Gr}_{S}(r, n)(-)_{I}.$

Upshot. Since these are open subfunctors that cover $\mathrm{Gr}_{S}(r, n),$ we see that $\mathrm{Gr}_{S}(r, n)$ is representable by an $S$-scheme.

Grassmannians - Part 1

In this posting, we follow Vakil's notes to start discussing about Grassmannians. Let us fix the base scheme $S$ once and for all. Given an $S$-scheme $Y$ and integers $0 \leq r \leq n,$ we denote by $\mathrm{Gr}_{S}(r, n)(Y)$ the set of isomorphism classes exact sequences of the form $$\mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q} \rightarrow 0,$$ where $\mathscr{Q}$ is a locally free sheaf (or more precisely $\mathscr{O}_{X}$-module) of rank $r$ on $Y.$

Unimportant remark. The latter $\mathscr{Q}$ is supposed to be "Q", for "quotient".

Given any epimorphism $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ of $S$-schemes where $\mathscr{Q}$ is a locally free sheaf of rank $r$ on $Y,$ its kernel $\ker(\phi)$ is locally free of rank $n - r.$ (See Vakil 13.5.B.(a) for details.) Thus, we see that $\mathrm{Gr}_{S}(r, n)(Y)$ can be described as the set of isomorphism classes of short exact sequences of the form $$0 \rightarrow \mathscr{S} \rightarrow \mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q} \rightarrow 0,$$ where $\mathscr{S}$ has rank $n - r$ and $\mathscr{Q}$ has rank $r.$

Remark. Given an exact sequence $$0 \rightarrow \mathscr{S} \rightarrow \mathscr{O}_{Y}^{\oplus n}$$ with a locally free sheaf $\mathscr{S}$ of rank $r,$ its cokernel is not necessarily locally free. An example (which we learn from Vakil 13.4.1) is as follows: take $$Y = \mathbb{A}^{1}_{k} = \mathrm{Spec}(k[t]),$$ where $k$ is a field. If we take $n = 1$ and $\mathscr{S}$ the locally free sheaf of rank $r = 1$ associated to the $k[t]$-submodule $k[t]t$ of $k[t]$ generated by $t.$ Then the quotient $\mathscr{O}_{Y}/\mathscr{S}$ is the coherent sheaf associated to $k$ by giving $t$-action as multiplication by $0$, which is not locally free.

We want to make $\mathrm{Gr}_{S}(r, n)$ into a functor $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set},$ so let's describe where a morphism $\pi : X \rightarrow Y$ of $S$-schemes is mapped to. Given an exact sequence $$\mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q} \rightarrow 0$$ of $\mathscr{O}_{Y}$-modules, since $\pi^{*}\mathscr{O}_{Y} = \mathscr{O}_{X}$ and $\pi^{*}$ is right-exact, we have an exact sequence $$\mathscr{O}_{X}^{\oplus n} = \pi^{*}(\mathscr{O}_{Y}^{\oplus n}) \rightarrow \pi^{*}\mathscr{Q} \rightarrow 0$$ of $\mathscr{O}_{X}$-modules. Since $\mathscr{Q}$ is a locally free $\mathscr{O}_{Y}$-module of rank $r,$ it follows that $\pi^{*}\mathscr{Q}$ is a locally free $\mathscr{O}_{X}$-module of rank $r$ (e.g., Vakil 16.3.7.(3)). If there is another map $\rho : Y \rightarrow Z$ of $S$-schemes, applying $\rho^{*}$ to the last sequence gives $$\mathscr{O}_{Z} = (\pi \circ \rho)^{*}(\mathscr{O}_{Y}^{\oplus n}) = \rho^{*}\pi^{*}(\mathscr{O}_{Y}^{\oplus n}) \twoheadrightarrow \rho^{*}\pi^{*}\mathscr{Q} = (\pi \circ \rho)^{*}\mathscr{Q},$$ which is the same as applying $(\pi \circ \rho)^{*}$ to $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ we began with. It is not hard to see that our discussion is compatible with isomorphisms of sequences, so we have constructed a functor $$\mathrm{Gr}_{S}(r, n) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set},$$ which we call the Grassmanian of $r$-subspaces in $n$-spaces over $S.$

Special case: rank $1$. Consider the case $r = 1.$ Locally free sheaves of rank $1$ are line bundles by definition. Given $1 \leq i \leq n,$ denote by $$e_{i} = (0, \dots, 0, 1, 0, \dots, 0) \in \mathscr{O}_{Y}(Y)^{\oplus n},$$ where $1$ only appears in the $i$-th spot. For any open subset $U \subset Y,$ this element restricts to an element $\mathscr{O}_{Y}(U)^{\oplus n}$ with the same description. Fix a line bundle $\mathscr{L}.$ Then a map $\mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{L}$ of sheaves is an epimorphism if and only if its localization $\mathscr{O}_{Y, y}^{\oplus n} \rightarrow \mathscr{L}_{y}$ is surjective for all $y \in Y.$ If we denote by $s_{i} \in \mathscr{L}(Y)$ the image of $e_{i}$ under the map $\mathscr{O}_{Y}(Y)^{\oplus n} \rightarrow \mathscr{L}(Y),$ the global section of the given sheaf map, then the last statement is equivalent to saying that one of $s_{1}, \dots, s_{n}$ must have a nonzero image in $\mathscr{L}_{y}$ for each $y \in Y$ (because $\mathscr{L}_{y} \simeq \mathscr{O}_{Y,y},$ so any nonzero element would generate the whole module), or in other words, the global sections $s_{1}, \dots, s_{n}$ do not share a common zero in $Y.$

This gives rise to an $S$-scheme map $[s_{1} : \cdots : s_{n}] : Y \rightarrow \mathbb{P}^{n-1}_{S},$ which can be painlessly obtained by first working over $\mathrm{Spec}(\mathbb{Z})$ and then base change over $S.$ Given another line bundle $\mathscr{L}'$ on $Y,$ an isomorphism of two maps $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{L}$ and $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{L}'$ is an isomorphism $\mathscr{L} \rightarrow \mathscr{L}'$ of $\mathscr{O}_{Y}$-modules such that $s_{i} \mapsto s'_{i}$ for $1 \leq i \leq n,$ where $s'_{i} \in \mathscr{L}'(Y)$ is the image of $e_{i} \in \mathscr{O}_{Y}(Y)^{\oplus n}.$ This implies that for each $y \in Y,$ there is an affine open $U \ni y$ in $Y$ such that $\mathscr{L}|_{U} \simeq \mathscr{O}_{Y}|_{U}$ and for $1 \leq i \leq n,$ the image of $s'_{i}$ is the product of the image of $s_{i}$ and an invertible element in $\mathscr{O}_{Y}(U).$ This implies that $[s_{1} : \cdots : s_{n}] = [s'_{1} : \cdots : s'_{n}].$

On the other hand, any $S$-scheme map $\pi : Y \rightarrow \mathbb{P}^{n-1}_{S}$ can be obtained as $\pi = [\pi^{*}x_{0} : \cdots : \pi^{*}x_{n-1}],$ where $x_{0}, \dots, x_{n-1}$ are given by the global sections of $\mathscr{O}_{\mathbb{P}^{n-1}_{S}}(1).$ Explicitly, if $\mathrm{Spec}(R) \subset S$ is an affine open subset, then $$x_{0}, \dots, x_{n-1} \in R[x_{0}, \dots, x_{n-1}].$$ Let's review the meaning of the pullback of a section of a line bundle:

What do we mean by pullback of a global section of a line bundle? Given a global section $s \in \mathscr{L}(Y)$ of a line bundle $\mathscr{L}$ on $Y,$ we can consider the map $\mathscr{O}_{Y}(Y) \rightarrow \mathscr{L}(Y)$ given by $1 \mapsto s.$ This extends to a unique $\mathscr{O}_{Y}$-module map $\mathscr{O}_{Y} \rightarrow \mathscr{L}$ per each $s.$ This way, the section $s$ can be viewed as a map $\mathscr{O}_{Y} \rightarrow \mathscr{L}.$ Now, if $\pi : X \rightarrow Y$ is a map of $S$-schemes, then we may consider the pullback $\pi^{*}s : \mathscr{O}_{X} = \pi^{*}\mathscr{O}_{Y} \rightarrow \pi^{*}\mathscr{L}.$ The image of $1 \in \mathscr{O}_{X}(X)$ under the map $$\mathscr{O}_{X}(X) = (\pi^{*}\mathscr{O}_{Y})(X) \rightarrow (\pi^{*}\mathscr{L})(X)$$ given by $\pi^{*}s$ is also denoted as $\pi^{*}s.$

Going back to the discussion, open sets of the form $\pi^{-1}(U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R))$ cover $Y,$ where $U_{i} = D_{\mathbb{P}^{n-1}}(x_{i})$ and $\mathrm{Spec}(R) \subset S$ is affine open. The restriction $$\pi^{-1}(U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R)) \rightarrow U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R) = \mathrm{Spec}(R[x_{0}/x_{i}, \dots, x_{n-1}/x_{i}])$$ of $\pi$ corresponds to the ring map $$R[x_{0}/x_{i}, \dots, x_{n-1}/x_{i}] \rightarrow \Gamma(\pi^{-1}(U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R)), \mathscr{O}_{Y})$$ such that $x_{j}/x_{i} \mapsto \pi^{*}x_{j}/\pi^{*}x_{i}.$ Some more details can be found in (the proof of) Vakil 16.4.1. Hence, we have established an explicit bijection $$\mathrm{Gr}_{S}(1, n)(Y) \leftrightarrow \mathrm{Hom}_{S}(Y, \mathbb{P}^{n-1}_{S}),$$ which we can briefly write as $(s_{1}, \dots, s_{n}) \mapsto [s_{1} : \cdots : s_{n}].$ Given any map $\rho : X \rightarrow Y,$ the induced map $\mathrm{Gr}_{S}(1, n)(Y) \rightarrow \mathrm{Gr}_{S}(1, n)(X)$ can be described as $(s_{1}, \dots, s_{n}) \mapsto (\rho^{*}s_{1}, \dots, \rho^{*}s_{n}),$ while $\mathrm{Hom}_{S}(Y, \mathbb{P}^{n-1}_{S}) \rightarrow \mathrm{Hom}_{S}(X, \mathbb{P}^{n-1}_{S})$ is given by $\phi \mapsto \rho \circ \phi.$ We can write $$\phi = [s_{1} : \cdots : s_{n}] = [\phi^{*}x_{1} : \cdots : \phi^{*}x_{n}]$$ and $$\begin{align*}\rho \circ [s_{1} : \cdots : s_{n}] &= \phi \circ \rho \\ &= [(\phi \circ \rho)^{*}x_{1} : \cdots : (\phi \circ \rho)^{*}x_{n}] \\ &= [\rho^{*}\phi^{*}x_{1} : \cdots : \rho^{*}\phi^{*}x_{n}] \\ &= [\rho^{*}s_{n} : \cdots : \rho^{*}s_{n}].\end{align*}.$$ This means that our bijection is now checked to be a natural trasformation so that it is an isomorphism of functors: $$\mathrm{Gr}_{S}(1, n)(-) \simeq \mathrm{Hom}_{S}(-, \mathbb{P}^{n-1}_{S}).$$ This, by definition, means that the functor $\mathrm{Gr}_{S}(1, n)(-)$ is representable by the projective space $\mathbb{P}^{n-1}_{S}$ over $S.$

Example. Take $S = \mathrm{Spec}(k),$ where $k$ is a field. Then we have $$\mathrm{Gr}(1, n)(k) \simeq \mathbb{P}^{n-1}(k),$$ given by $$(a_{1}, \dots, a_{n}) \mapsto [a_{1} : \cdots : a_{n}],$$ where $a_{i} \in k,$ at least one of which is nonzero. We understand the right-hand side quite well, so let's consider the left-hand side. An element is given by (the isomorphism class of) a map $\mathscr{O}_{\mathrm{Spec}(k)}^{\oplus n} \twoheadrightarrow \mathscr{L}.$ This is nothing more than a surjective $k$-linear map $k^{n} \twoheadrightarrow k,$ so the image of each $e_{i}$ is $a_{i} \in k$ we are considering.

Next time. We describe what happens to $\mathrm{Gr}_{S}(r, n),$ when $r$ may be larger than $1.$

$l$-adic cohomology: Lecture 4

References. The following are the references I use for writing this posting:

• Freitag and Kiehl,
• Milne,
• Hochster,
• Vakil,
• Stacks Project (which will be referred as "SP").

Of course, I may cite more references as I go.

Goal. The goal is to follow the first chapter of the first reference.

Presheaves/sheaves on a site (valued in abelian groups). Let $\mathcal{C}$ be any category. A presheaf (valued in the category $\textbf{Ab}$ of abelian groups) on $\mathcal{C}$ is a functor $\mathscr{F} : \mathcal{C}^{\mathrm{op}} \rightarrow \textbf{Ab}.$

Now, suppose that $\mathcal{C}$ is equipped with a Grothendieck topology (i.e., $\mathcal{C}$ is a site). A presheaf $\mathscr{F}$ on $\mathcal{C}$ is called a sheaf if for any object $U$ of $\mathcal{C}$ and a cover $\{U_{i} \rightarrow U\}_{i \in I}$ of $U,$ the induced diagram $$\mathscr{F}(U) \rightarrow \prod_{i \in I}\mathscr{F}(U_{i}) \rightrightarrows \prod_{(i, j) \in I^{2}} \mathscr{F}(U_{i} \times_{U} U_{j})$$ is an equilizer diagram in $\textbf{Ab}.$

How is the above diagram defined? The first map is obtined by applying $\mathscr{F}$ to maps $U_{i} \rightarrow U$ and then using the universal property of the product. There are two maps from the second term to the third. The first one comes from the maps of the form $U_{i} \times_{U} U_{j} \rightarrow U_{i},$ and the second one comes from the maps of the form $U_{i} \times_{U} U_{j} \rightarrow U_{j}.$

Notation. If $\mathcal{C}$ is a site, for any map $U \rightarrow V$ in $\mathcal{C},$ we describe the induced map $\mathscr{F}(V) \rightarrow \mathscr{F}(U)$ as $s \mapsto s|_{U}.$ This notation does not mean that there is a unique map from $U$ to $V,$ but we will use it as long as it does not create any confusion.

Remark. Let $\mathcal{C}$ be a site. Note that a presheaf $\mathscr{F}$ is a sheaf if and only if for every cover $\mathscr{U} = \{U_{i} \rightarrow U\}_{i \in I}$ of each object $U$ in $\mathcal{C},$ the following conditions are satisfied :

• for each $s, t \in \mathscr{F}(U),$ if $s|_{U_{i}} = t|_{U_{i}}$ for every $U_{i} \rightarrow U$ in $\mathscr{U},$ then $s = t$.
• given any $(s_{i})_{i \in I} \in \prod_{i \in I}\mathscr{F}(U_{i}),$ if $s_{i}|_{U_{i} \times_{U} U_{j}} = s_{i}|_{U_{i} \times_{U} U_{j}}$ for every $U_{i} \rightarrow U$ and $U_{j} \rightarrow U$ in $\mathscr{U}$ (including the case $i = j$), then there is $s \in \mathscr{F}(U)$ such that $s|_{U_{i}} = s_{i}$ for every $U_{i} \rightarrow U$ in $\mathscr{U}.$

We want to define cohomology of a sheaf on the étale site $X_{ét}$ of a scheme $X.$ The following facts let us do this:

Fact 1 (SP03NT). The category of sheaves $\textbf{Sh}(X_{ét})$ is an abelian category that has enough injectives.

A part of showing the above fact also yields the following:

Theorem. The global section functor functor $\Gamma : \textbf{Sh}(X_{ét}) \rightarrow \textbf{Ab}$ given by $\mathscr{F} \mapsto \Gamma(X, \mathscr{F}) := \mathscr{F}(X)$ (with morphisms are mapped to their restrictions to the global sections) is left-exact.

Proof. (cf. SP03CN). Let $$0 \rightarrow \mathscr{F} \xrightarrow{\phi} \mathscr{G} \xrightarrow{\psi} \mathscr{H}$$ be an exact sequence in $\textbf{Sh}(X_{ét}).$ We want to show that the induced sequence $$0 \rightarrow \mathscr{F}(X) \xrightarrow{\phi} \mathscr{G}(X) \xrightarrow{\psi} \mathscr{H}(X)$$ is exact.

To show the exactness at $\mathscr{F}(X),$ it is enough to show the following:

Lemma 1. Let $\mathscr{F} \xrightarrow{\phi} \mathscr{G}$ be any morphism of sheaves in $\textbf{Sh}(X_{ét}).$ Then $\ker(\phi_{U}) = \ker(\phi)(U)$ for every $U$ in $\textbf{Sh}(X_{ét}).$

Proof of Lemma 1. We construct a presheaf $\mathscr{K}$ on $X_{ét}$ by defining $$\mathscr{K}(U) := \ker(\phi_{U}) \subset \mathscr{F}(U).$$ Given any map $U \rightarrow V$ of $X_{ét},$ the restriction map $\mathscr{K}(V) \rightarrow \mathscr{K}(U)$ is given by restricting the map $\mathscr{F}(V) \rightarrow \mathscr{F}(U).$ This defines a functor $\mathscr{K} : X_{ét}^{\mathrm{op}} \rightarrow \textbf{Ab},$ because $\mathscr{F}$ is a functor.

We first show that $\mathscr{K}$ is a sheaf. Let $s \in \mathscr{K}(U)$ and say $\{U_{i} \rightarrow U\}_{i \in I}$ is a cover of $U.$ If $s|_{U_{i}} = 0 \in \mathscr{K}(U) \subset \mathscr{F}(U)$ for each $i,$ then $s = 0$ in $\mathscr{F}(U),$ so $s = 0$ in $\mathscr{K}(U).$ Next, suppose that we have $s_{i} \in \mathscr{K}(U_{i})$ for each $i \in I$ such that $$s_{i}|_{U_{i} \times_{U} U_{j}} = s_{j}|_{U_{i} \times_{U} U_{j}} \in \mathscr{K}(U_{i} \times_{U} U_{j}) \subset \mathscr{F}(U_{i} \times_{U} U_{j})$$ for all $i, j \in I.$ Since $\mathscr{F}$ is a sheaf, we may find $s \in \mathscr{F}(U)$ such that $$s|_{U_{i}} = s_{i} \in \mathscr{F}(U_{i}).$$ For all $i,$ we have $$\phi_{U}(s)|_{U_{i}} = \phi_{U_{i}}(s_{i}) = 0$$ because $s_{i} \in \mathscr{K}(U_{i}) = \ker(\phi_{U_{i}}).$ This implies that $\phi_{U}(s) = 0,$ so $s \in \mathscr{K}(U).$ This establishes the fact that $\mathscr{K}$ is a sheaf.

We now claim that $\mathscr{K} = \ker(\phi).$ The inclusions $\mathscr{K}(U) \hookrightarrow \mathscr{F}(U)$ build into a map $\iota : \mathscr{K} \rightarrow \mathscr{F}$ of presheaves. It is immediate that the composition $\mathscr{K} \xrightarrow{\iota} \mathscr{F} \xrightarrow{\phi} \mathscr{G}$ is the zero map (i.e., $\phi \circ \iota = 0$), so to finish the check, we fix any other sheaf map $j : \mathscr{K}' \rightarrow \mathscr{F}$ such that $\phi \circ j = 0.$ We want to show that there exists a unique map $\eta : \mathscr{K}' \rightarrow \mathscr{K}$ such that $j = \iota \circ \eta.$ This requirement forces $j_{U} = \iota_{U} \circ \eta_{U},$ so given any object $U$ of $X_{ét},$ the map $\eta_{U} : \mathscr{K}'(U) \rightarrow \mathscr{K}(U)$ has to be given by the usual property of kernels in $\textbf{Ab}.$ More explicitly, since the image of $j_{U}$ lies in $\ker(\phi_{U}) = \mathscr{K}(U),$ it factors as $$\mathscr{K}'(U) \rightarrow \mathscr{K}(U) \hookrightarrow \mathscr{F}(U).$$ It is immediate that these map build a map $\eta : \mathscr{K}' \rightarrow \mathscr{K}$ of sheaves. Hence, we have proved that $\mathscr{K} = \ker(\phi).$ This finishes the proof (of Lemma 1). $\Box$

Going back to the proof of Theorem, it remains to show the exactness at $\mathscr{G}(X).$ That is, we need to show that $\mathrm{im}(\phi_{X}) = \ker(\psi_{X}).$ We have $$\ker(\mathrm{coker}(\phi)) = \mathrm{im}(\phi) = \ker(\psi),$$ the second of which uses the exactness at $\mathscr{G}.$ Thus, by Lemma 1, we have $$\ker(\mathrm{coker}(\phi)(X)) = \ker(\psi_{X}).$$ The inclusion $$\mathrm{im}(\phi_{X}) = \ker(\mathrm{coker}(\phi_{X})) \subset \ker(\mathrm{coker}(\phi)(X)) = \ker(\psi_{X})$$ follows immediately from diagram chasing.

To show the reverse inclusion, we actually need to understand what $\mathrm{coker}(\phi)$ is. It is the "sheafification" of the presheaf on $X_{ét},$ which we briefly discuss now:

Sheafification. Given a presheaf $\mathscr{F}$ on $X_{ét},$ a sheafification $\tilde{\mathscr{F}}$ of $\mathscr{F}$ is a sheaf on $X_{ét}$ together with a presheaf map $\mathscr{F} \rightarrow \tilde{\mathscr{F}}$ satisfying the following property: for any sheaf $\mathscr{G}$ on $X_{ét}$ and a presheaf map $\phi : \mathscr{F} \rightarrow \mathscr{G},$ there is a unique sheaf map $\tilde{\mathscr{F}} \rightarrow \mathscr{G}$ that $\phi$ factors through the sheafification.

Remark. Note that if a sheafification of $\mathscr{F}$ exists, it is unique up to a unique isomorphism. It is also immediate that taking sheafification defines a functor $\textbf{Psh}(X_{ét}) \rightarrow \textbf{Sh}(X_{ét})$ from the category of presheaves over $X_{ét}$ to that of sheaves.

The sheafification always exists, but we will not discuss the proof of it but cite a reference instead. At this point, if we denote by $\mathrm{coker}(\phi_{-})$ the presheaf given by $U \mapsto \mathrm{coker}(\phi_{U}),$ one can check that the cokernel $\mathscr{G} \rightarrow \mathrm{coker}(\phi)$ of the sheaf map $\phi : \mathscr{F} \rightarrow \mathscr{G}$ can be constructed as the following composition: $$\mathscr{G} \rightarrow \mathrm{coker}(\phi_{-}) \rightarrow \widetilde{\mathrm{coker}(\phi_{-})} =: \mathrm{coker}(\phi).$$ Note that the usual proof for the small Zariski site of $X$ (Vakil 2.6.1) works here without any change.

We need a little bit more to finish our proof:

Fact 2 (Vistoli Theorem 2.64). The desired sheafification can be constructed so that it satisfies the following two properties:

1. for any $s \in \mathscr{F}(U),$ if its image in $\tilde{\mathscr{F}}(U)$ is zero, then there is a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $s|_{U_{i}} = 0$ for each $i.$
2. for any $t \in \tilde{\mathscr{F}}(U),$ there is a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $t|_{U_{i}}$ is in the image of $\mathscr{F}(U_{i}) \rightarrow \tilde{\mathscr{F}}(U_{i})$ for each $i.$

Remark. Note that if we have any sheaf $\tilde{\mathscr{F}}$ and a presheaf map $\mathscr{F} \rightarrow \tilde{\mathscr{F}}$ satisfying the above two properties, then $\mathscr{F} \rightarrow \tilde{\mathscr{F}}$ is necessarily a sheafification.

We are now ready to finish our proof of Theorem. Denote by $\eta : \mathscr{G} \rightarrow \mathrm{coker}(\phi)$ the map comes with $\mathrm{coker}(\phi).$ Consider $\eta_{X} : \mathscr{G}(X) \rightarrow \mathrm{coker}(\phi)(X).$ We know $\ker(\psi_{X}) = \ker(\eta_{X}),$ so what remains for us to show is $\ker(\eta_{X}) \subset \mathrm{im}(\phi_{X}).$ 

Fix any $s \in \ker(\eta_{X}) \subset \mathscr{G}(X).$ We want to show that $s \in \mathrm{im}(\phi_{X}).$ Since $\eta_{X}(s) = 0 \in \mathrm{coker}(\phi)(X),$ we have a map $\mathrm{coker}(\phi_{X}) \rightarrow \mathrm{coker}(\phi)(X)$ given by the universal property of the cokernel (source) such that $\bar{s} \mapsto \eta_{X}(s) = 0.$ Note that this is exactly the sheafification map, so by Fact 2-1, we can find a cover $\{U_{i} \rightarrow U\}_{i \in I}$ such that $\overline{s|_{U_{i}}} = 0$ in $\mathrm{coker}(\phi_{U_{i}})$ (i.e., $s|_{U_{i}} \in \mathrm{im}(\phi_{U_{i}})$) for each $i.$ Hence, we can write $s|_{U_{i}} = \phi_{U_{i}}(t_{i})$ for some (necessarily unique) $t_{i} \in \mathscr{F}(U_{i})$ for each $i.$ Since $\phi_{U_{i} \times_{X} U_{j}} : \mathscr{F}(U_{i} \times_{X} U_{j}) \rightarrow \mathscr{G}(U_{i} \times_{X} U_{j})$ is injective, we have $$t_{i} |_{U_{i} \times_{X} U_{j}} = t_{j} |_{U_{i} \times_{X} U_{j}}$$ for all $i, j \in I.$ As $\mathscr{F}$ is a sheaf, this implies that we have $t \in \mathscr{F}(X)$ such that $t|_{U_{i}} = t_{i}$ for all $i.$ Since $\mathscr{G}$ is a sheaf, it follows that $\phi_{X}(t) = s,$ so $s \in \mathrm{im}(\phi_{X}),$ as desired. This finishes the proof (of Theorem). $\Box$

Definition of étale cohomology group. Using the facts above, by looking at any sheaf $\mathscr{F}$ on $X_{ét}$ as an object of $\textbf{Sh}(X_{ét}),$ one writes $H^{i}(X_{ét}, -)$ to be the $i$-th right derived functor of the global section functor. Given any sheaf $\mathscr{F}$ on $X_{ét},$ the abelian group $H^{i}(X_{ét}, \mathscr{F})$ is called the $i$-th étale cohomology group of the sheaf $\mathscr{F}.$

Remark. It is important to note that our discussion works even if we used the category $\textbf{Mod}_{R}$ of $R$-modules instead of $\textbf{Ab}$ for any given ring $R.$

$l$-adic cohomology: Lecture 3

References. The following are the references I use for writing this posting:

• Freitag and Kiehl,
• Milne,
• Hochster,
• Vakil.

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

1. it is locally of finitely presentation, and
2. 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:

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:

• 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.

1. Any base change of an étale map in the category of schemes is étale.
2. A composition of any two étale maps is étale.
3. (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.
4. 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 $X.$

Proposition (big étale site). The étale coverings over a scheme $X$ give a Grothendieck topology on $\mathrm{Sch}_{X}.$ In other words, we have: 

1. any isomorphism into $U$ is in $\mathrm{Cov}(U)$ as a singleton;
2. 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 $\mathrm{Cov}(V)$;
3. 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.

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 $X_{ét}.$

$l$-adic cohomology: Lecture 2

References. The following are the references I use for writing this posting:

• Freitag and Kiehl,
• Milne,
• Hochster,
• Vakil.

Of course, I may cite more references as I go.

Goal. The goal is to follow the first chapter of the first reference.

Faithfully flatness. A module $M$ over a ring $R$ is said to be faithfully flat if for every sequence $Q_{1} \rightarrow Q_{2} \rightarrow Q_{3}$ of $R$-modules, the sequence $$0 \rightarrow Q_{1} \rightarrow Q_{2} \rightarrow Q_{3} \rightarrow 0$$ is exact if and only if its induced sequence $$0 \rightarrow Q_{1} \otimes_{R} M \rightarrow Q_{2} \otimes_{R} M \rightarrow Q_{3} \otimes_{R} M \rightarrow 0$$ is exact.

Remark. In particular, if $M$ is faithfully flat, then $M$ is flat.

The following statement is from Stacks Project 00H9:

Theorem. Given any flat $R$-module $M$, the following are equivalent:

1. $M$ is faithfully flat;
2. for any nonzero $R$-module $Q,$ the module $Q \otimes_{R} M$ is nonzero;
3. for any prime $\mathfrak{p}$ of $R,$ we have $M/\mathfrak{p}M \neq 0$;
4. for any maximal ideal $\mathfrak{m}$ of $R,$ we have $M/\mathfrak{m}M \neq 0.$

Proof. The only nontrivial implication is to show that the fourth statement implies the first. We will show that the fourth implies the second and then show that the second implies the first. Assume the fourth statement and fix any $R$-module $P$ such that $P \otimes_{R} M = 0.$ Our goal is to show that $P = 0.$

To show that $P \neq 0,$ we can show that every element $x \in P$ is zero. We will make use of $R/\mathrm{Ann}(x) \hookrightarrow P,$ where $$\mathrm{Ann}(x) = \{r \in R : rx = 0\}.$$ Since $M$ is flat, we get $$M/\mathrm{Ann}(x)M \simeq R/\mathrm{Ann}(x) \otimes_{R} M \hookrightarrow P \otimes_{R} M = 0,$$ which guarantees that $M/\mathrm{Ann}(x)M = 0.$

If $x \neq 0,$ then $$\mathrm{Ann}(x) \subsetneq R,$$ so there must be a maximal ideal $\mathfrak{m}$ of $R$ such that $\mathrm{Ann}(x) \subset \mathfrak{m}.$ This gives us $0 = M/\mathrm{Ann}(x)M \twoheadrightarrow M/\mathfrak{m}M,$ so $M/\mathfrak{m}M = 0,$ contradicting the fourth condition. Thus, it must have been that $x = 0,$ as desired. Hence, this shows that $P = 0,$ so the second statement is established, assuming the fourth.

It remains to show that the second statement implies the first. Assume the second statement and fix a sequence $Q_{1} \rightarrow Q_{2} \rightarrow Q_{3}$ of $R$-modules such that $$0 \rightarrow Q_{1} \otimes_{R} M \rightarrow Q_{2} \otimes_{R} M \rightarrow Q_{3} \otimes_{R} M \rightarrow 0$$ is exact. We want to show that $$0 \rightarrow Q_{1} \rightarrow Q_{2} \rightarrow Q_{3} \rightarrow 0$$ is exact. Showing at $Q_{1}$ and $Q_{3}$ are immediate using the second statement, so it remains to show the exactness at $Q_{2}.$ To do so, we write $\phi : Q_{1} \rightarrow Q_{2}$ and $\psi : Q_{2} \rightarrow Q_{3}$ for given maps. Our goal is to show that $\mathrm{im}(\phi) = \ker(\psi).$

To show $\mathrm{im}(\phi) \subset \ker(\psi),$ it is enough to show $\psi \circ \phi = 0.$ For this, it is enough to show that $\ker(\psi \circ \phi) = Q_{1},$ or equivalently, we may show $Q_{1}/\ker(\psi \circ \phi) = 0.$ Using the second statement, this reduces to showing that $$(Q_{1}/\ker(\psi \circ \phi)) \otimes_{R} M = 0,$$ and since $$(Q_{1}/\ker(\psi \circ \phi)) \otimes_{R} M \simeq (Q_{1} \otimes_{R} M)/(\ker(\psi \circ \phi) \otimes_{R} M),$$ our goal reduces to showing that $$\ker(\psi \circ \phi) \otimes_{R} M = Q_{1} \otimes_{R} M,$$ but this already follows from our assumption since $M$ is assumed to be flat.

It now remains to show that $\ker(\psi)/\mathrm{im}(\phi) = 0,$ but this follows easily from the second statement, using the assumption on the exactness of the sequence we have after applying $( - ) \otimes_{R} M$ and flatness of $M.$ This finishes the proof. $\Box$

Remark. We will use the fourth condition soon, with the observation that $$M/\mathfrak{m} \simeq (R/\mathfrak{m}) \otimes_{R} M.$$

Geometric interpretation. We study what is the geometric meaning of an étale ring map $B \rightarrow C$ being faithfully flat. Recall that étaleness includes flatness.

Lemma. Let $\phi : B \rightarrow C$ be a flat ring map. Then $\phi$ is faithfully flat if and only if its induced map $\mathrm{Spec}(C) \rightarrow \mathrm{Spec}(B)$ is surjective.

Proof. First, assume that $\phi : B \rightarrow C$ is faithfully flat. Fix any prime $\mathfrak{q}$ of $B.$ We want to show that the fiber at $\mathfrak{q}$ is nonempty, which is equivalent to showing that $(B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}) \otimes_{B} C$ is nonzero. Since $B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$ is nonzero (as it is a field, which must have at least two elements), the assumption that $B \rightarrow C$ is faithfully flat guarantees this.

Conversely, assume that $\mathrm{Spec}(C) \rightarrow \mathrm{Spec}(B)$ is surjective. Then for every any maximal ideal $\mathfrak{m}$ of $B,$ the ring $$(B/\mathfrak{m}) \otimes_{B} C \simeq C/\mathfrak{m}C$$ is nonzero, so $C$ must be faithfully flat over $B.$ This finishes the proof. $\Box$

Remark. In particular, if $\phi : B \rightarrow C$ is an étale ring map, saying that $\phi$ is faithfully flat is equivalent to saying that the induced map $\mathrm{Spec}(C) \rightarrow \mathrm{Spec}(B)$ is surjective.

Grothendieck topology on a category. Given a category $\mathcal{C},$ a Grothendieck topology on $\mathcal{C}$ is an assignment $\mathrm{Ob}(\mathcal{C}) \rightarrow \textbf{Set}$ given by $U \mapsto \mathrm{Cov}(U),$ where $\mathrm{Cov}(X)$ is a set whose elements are collections $\{U_{i} \rightarrow X\}_{i \in I}$ of morphisms satisfying the following properties.

1. Any isomorphism into $U$ is in $\mathrm{Cov}(U)$ as a singleton.
2. Given any $\{U_{i} \rightarrow U\}_{i \in I}$ in $\mathrm{Cov}(U),$ for any $V \rightarrow U,$ the fiber product $V \times_{U} U_{i}$ exists in $\mathcal{C}$ for each $i \in I$ and the induced collection $\{V \times_{U} U_{i} \rightarrow V\}_{i \in I}$ is an element of $\mathrm{Cov}(V).$
3. 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 $\mathrm{Cov}(U).$

An element of $\mathrm{Cov}(U)$ is called a covering (or a cover) of $U.$ If $\mathcal{C}$ is equipped with a Grothendieck topology, we call it altogether a site. Note that due to the first condition $\mathrm{Cov}(U)$ is never empty because it has at least one element $\{\mathrm{id}_{U}\}.$

Example (small Zariski). Given any topological space $X,$ denote by $\mathrm{O}(X)$ the category whose objects are open subsets of $X$ and morphisms are inclusions. For each object $U$ of $\mathrm{O}(X),$ define $\mathrm{Cov}(U)$ to be the set of all open coverings $\{U_{i} \hookrightarrow U\}_{i \in I}$ of $U.$ We claim that this defines a Grothendieck topology on $\mathrm{O}(X).$

Proof. Indeed, the only isomorphism into $U$ is the identity, and since $U$ is an open subset itself, the first condition is satisfied. For the second condition, let $\{U_{i} \hookrightarrow U\}_{i \in I}$ be an open cover of $U.$ For any open inclusion $V \hookrightarrow U,$ one may check that $V \cap U_{i} = V \times_{U} U_{i}$ in $\mathrm{O}(X).$ The induced collection is $\{V \cap U_{i} \hookrightarrow V\}_{i \in I},$ which is an open cover of $V.$ This checks the second condition. For the third condition, given any open cover $\{U_{i} \hookrightarrow U\}_{i \in I}$ of $U,$ fix any open cover $\{V_{ij} \hookrightarrow U_{i}\}_{j \in J_{i}}$ for each $i \in I.$ Then each composition $V_{ij} \hookrightarrow U_{i} \hookrightarrow U$ is noting more than the inclusion $V_{ij} \hookrightarrow U$ and $\{V_{ij} \hookrightarrow U\}_{i \in I, j \in J_{i}}$ gives an open cover of $U,$ which checks the third condition. $\Box$

We call the site structure of $\mathrm{O}(X)$ the small site with respect to the topology given on $X.$ When $X$ is a scheme, we call $\mathrm{O}(X)$ the small Zariski site of $X$ (i.e., the small site with respect to the Zariski topology on $X$). We write $X_{\mathrm{Zar}} := \mathrm{O}(X)$ to mean the site structure.

Example (big Zariski). Given a scheme $X,$ consider the category $\textbf{Sch}_{X}$ of schemes over $X.$ Given any object $U$ of $\textbf{Sch}_{X}$ (i.e., a scheme map $U \rightarrow X$), define $\mathrm{Cov}(U)$ the set whose elements are given by collections $\{U_{i} \hookrightarrow U\}_{i \in I}$ of open embeddings over $X$ such that $U = \bigcup_{i \in I}U_{i},$ where the $U_{i}$ in the union should be considered as the image of the open embedding $U_{i} \hookrightarrow U.$ We claim that this gives a Grothendieck topology on $\textbf{Sch}_{X}.$

Proof. The previous proof goes through here word by word. $\Box$

We call this site structure of $\textbf{Sch}_{X}$ the big Zariski site of $X.$

Remark. Note that any open embedding $U_{i} \hookrightarrow U$ is an open embedding over $X,$ as we can get the structure map of $U_{i}$ by the composition $U_{i} \hookrightarrow U \rightarrow X.$ e write $(\mathrm{Sch}/X)_{\mathrm{Zar}} := \mathrm{Sch}_{X}$ to mean the site structure.

Plus, the only difference to the small Zariski site is that the structure maps into $X$ before were required to be open inclusions, whereas now they can be any scheme maps. This explains the word "big" versus "small".

Next time. We shall define étale maps between two schemes and define big/small étale sites.

$l$-adic cohomology: Lecture 1

Goal. I have decided to follow the first chapter of Freitag and Kiehl, as their exposition seems to be relatively readable in my perspective. I could be wrong, but I plan to read it as much as I can. As a learning trick, I will pretend as if I am writing my own lecture notes based on this reading; hence the title of the posting. Most likely, I will cite other sources (e.g., lecture notes of Hochster and those of Vakil) as I go. Most arguments I use are from the sources, and I do not claim any originality for them.

Convention. Throughout this series of postings, any ring is assumed to be unital and commutative (unless mentioned otherwise).

Unramified ring maps. We say that a ring map $A \rightarrow B$ is unramified if

1. it is finitely presented, and
2. for any $\mathfrak{q} \in \mathrm{Spec}(B),$ writing $\mathfrak{p} \in \mathrm{Spec}(A)$ for its contraction, the induced map $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is a finite separable field extension.

Remark. Note that requiring that $B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is a field guarantees that $\mathfrak{p}B_{\mathfrak{q}}$ is a maximal ideal of $B_{\mathfrak{q}},$ which necessarily means that $\mathfrak{p}B_{\mathfrak{q}} = \mathfrak{q}B_{\mathfrak{q}}.$

Remark. It is often the case that $A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}$ is not of finite type in this scenario. However, note that that $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is the base change of the map $A \rightarrow B$ under $A \rightarrow A_{\mathfrak{p}} \rightarrow A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}},$ so (without any of the two conditions above)

• if $A \rightarrow B$ is of finite type, then so is $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$;
• if $A \rightarrow B$ is finitely presented, then so is $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}.$

Theorem. A finitely presented ring map $A \rightarrow B$ is unramified if and only if $\Omega^{1}_{B/A} = 0.$

Proof. Let $A \rightarrow B$ be unramified. Fix any $\mathfrak{q} \in \mathrm{Spec}(B)$ and write $\mathfrak{p} \in \mathrm{Spec}(A)$ to mean its contraction. We have $$B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{q}} \otimes_{B_{\mathfrak{q}}} \Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}} \simeq \Omega^{1}_{(B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}})/(A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}})} = 0.$$ The vanishing occurred because $(B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}})/(A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}})$ is a finite separable field extension. The isomorphism is established because $$B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}} \simeq A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \otimes_{A_{\mathfrak{p}}} B_{\mathfrak{q}}.$$ Since $$(\Omega^{1}_{B/A})_{\mathfrak{q}} \simeq \Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}}$$ and $A \rightarrow B$ is of finite type, we see that $\Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}}$ is a finitely generated $B_{\mathfrak{q}}$-module. Thus, noting that $\mathfrak{p}B_{\mathfrak{q}} = \mathfrak{q}B_{\mathfrak{q}},$ by Nakayama's lemma, we must have $$(\Omega^{1}_{B/A})_{\mathfrak{q}} \simeq \Omega^{1}_{B_{\mathfrak{q}}/A_{\mathfrak{p}}} = 0,$$ and since $\mathfrak{q}$ was arbitrary, this implies that $\Omega^{1}_{B/A} = 0.$

Conversely, suppose that $\Omega^{1}_{B/A} = 0.$ Fix any $\mathfrak{q} \in \mathrm{Spec}(B)$ and denote again by $\mathfrak{p} \in \mathrm{Spec}(A)$ its contraction. Having $\Omega^{1}_{B/A} = 0$ implies that $\Omega^{1}_{(B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}})/(A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}})} = 0.$ As we have observed before, the map $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is of finite type because $A \rightarrow B$ is. Since $B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is a local ring, our desired conclusion follows from the following lemma:

Lemma. Let $k$ be a field and $R$ a finite type $k$-algebra. Then the following are equivalent:

• $\Omega^{1}_{R/k} = 0$;
• $R$ is a finite product of fields that are finite separable over $k.$

Proof of Lemma. It is immediate that the second implies the first applying the localization at each prime of $R,$ so we focus on proving the converse. Assume $\Omega^{1}_{R/k} = 0.$ We first claim that $\dim(R) = 0.$ We have $\dim(\bar{k} \otimes_{k} R) = \dim(R)$ (e.g., Vakil 11.1.G) so we may instead show that $\dim(\bar{k} \otimes_{k} R) = 0.$ For this, it is important to observe that $$\Omega^{1}_{(\bar{k} \otimes_{k} R)/\bar{k}} \simeq \bar{k} \otimes_{k} \Omega^{1}_{R/k} = 0.$$ Write $A := \bar{k} \otimes_{k} R$ for convenience, and note that $A$ is of finite type over $\bar{k}.$ Fix any maximal ideal $\mathfrak{m} \subset A.$ Since $\bar{k}$ is algebraically closed, we have $A/\mathfrak{m} \simeq \bar{k}.$ Therefore, we have $$\dim(A) = \dim(A_{\mathfrak{m}}) \leq \dim_{\bar{k}}(\mathfrak{m}/\mathfrak{m}^{2}) = \dim_{\bar{k}}(\Omega^{1}_{A/\bar{k}}) = 0.$$ This shows that $\dim(A) = 0$ as we claimed, which implies that $\dim(R) = 0,$ but it also shows that $\mathrm{Spec}(A)$ is regular at every closed point. Since $A$ is Noetherian, it follows that $A$ is isomorphic to a finite product of $\bar{k}$ as a $\bar{k}$-algebra. Since $$R \hookrightarrow \bar{k} \otimes_{k} R = A,$$ it follows that $R$ is a finite product of $0$-dimensional reduced Noetherian local rings. Each such local ring is regular because their base change over $\bar{k}$ is (applying Vakil 12.2.O), so it must be a field. Fix one of them and write $K$ to mean it. Since $K$ is finitely generated over $k,$ by Nullstellensatz (e.g., Vakil 3.2.5), the field $K$ must be a finite extension over $k.$ It remains to show that $K$ is separable over $k.$ To do so, fix any $\alpha \in K$ and denote by $f(x) \in k[x]$ the minimal polynomial of $\alpha$ over $k.$ We have $$\frac{\bar{k}[x]}{(f(x))} \simeq \bar{k} \otimes_{k} \frac{k[x]}{(f(x))} \simeq \bar{k} \otimes_{k} k[\alpha] \hookrightarrow \bar{k} \otimes_{k} R = A,$$ but since $A$ is reduced, the polynomial $f(x)$ must be square-free. This shows that $\alpha$ is separable over $K,$ as desired. $\Box$ 

Remark. Note that in the above proof, we have accidentally showed that for a finite type algebra $R$ over a field $k,$ the following are equivalent:

• $R$ is unramified over $k$;
• $R$ is a finite prodcut finite separable field extension of $k$;
• $\bar{k} \otimes_{k} R$ is a finite product of $\bar{k}.$

A ring map $A \rightarrow B$ is called unramified if it is of finite type and for every $\mathfrak{q} \in \mathrm{Spec}(B),$ writing $\mathfrak{p} \in \mathrm{Spec}(A)$ to mean its contraction, the induced local ring map $A_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}}$ is unramified.

Remark. The proof for Theorem works if we replace "finitely presented" to "finite type". Some people (e.g., Vakil 21.6) use this more general definition for the definition of unramified maps. In (the first chapter of) Freitag and Kiehl, this is moot, because rings are assumed to be Noetherian.

Remark. The criterion for unramified ring maps using the module of differential $1$-forms is quite useful. For instance, let's show the following:

Lemma. The composition of two unramified ring maps is unramified.

Proof. Let $A \rightarrow B \rightarrow C$ be two unramified ring maps. Then consider the relative cotangent exact sequence (e.g., Vakil 21.2.9) $$C \otimes_{B} \Omega^{1}_{B/A} \rightarrow \Omega^{1}_{C/A} \rightarrow \Omega^{1}_{C/B} \rightarrow 0.$$ Our hypothesis guarantees that $\Omega^{1}_{B/A} = 0$ and $\Omega^{1}_{C/B} = 0,$ so it follows that $\Omega^{1}_{C/A} = 0.$ Since being finite type is preserved under taking compositions, the composition in question is unramified. $\Box$

Lemma. Let $A \rightarrow B$ be an unramified ring map. Given any ring map $A \rightarrow C,$ the induced map $C \rightarrow C \otimes_{A} B$ is unramified.

Proof. It is evident that $C \rightarrow C \otimes_{A} B$ is of finite type when $A \rightarrow B$ is. Since $$\Omega^{1}_{(C \otimes_{A} B)/C} \simeq (C \otimes_{A} B) \otimes_{B} \Omega^{1}_{B/A} = 0,$$ as $\Omega^{1}_{B/A} = 0$ by hypothesis, we are done. $\Box$

Lemma (unramified cancellation). Let $A \rightarrow B$ and $B \rightarrow C$ be finite type ring maps. If $A \rightarrow B \rightarrow C$ is unramified, then so is $B \rightarrow C.$

Proof. This immediately follows from the exact sequence $$C \otimes_{B} \Omega^{1}_{B/A} \rightarrow \Omega^{1}_{C/A} \rightarrow \Omega^{1}_{C/B} \rightarrow 0,$$ although we do not need the first term. $\Box$

We say a ring map $A \rightarrow B$ is étale if it satisfies the following two properties:

1. it is flat;
2. it is unramified.

It is immediate that the composition of two étale ring maps is étale.

Lemma. Let $A \rightarrow B$ be an étale ring map. Given any ring map $A \rightarrow C,$ the induced map $C \rightarrow C \otimes_{A} B$ is étale.

This follows from the following:

Lemma. Let $A \rightarrow B$ be an flat ring map. Given any ring map $A \rightarrow C,$ the induced map $C \rightarrow C \otimes_{A} B$ is flat.

Proof. This immediately follows from $M \otimes_{C} C \otimes_{A} B \simeq M \otimes_{A} B,$ for any $C$-module $M.$ $\Box$

Theorem (étale cancellation). Let $A \rightarrow B$ and $B \rightarrow C$ be finite type ring maps. If $A \rightarrow B$ and $A \rightarrow B \rightarrow C$ are étale, then so is $B \rightarrow C.$

Proof. This follows from the unramified cancellation and an analogous statement about flat ring maps. $\Box$

Remark. The statement about flat ring maps can be done painlessly with even greater generality by citing Vakil 10.1.19 and Vakil 21.2.F, the latter of which we will discuss in the third lecture.

Étale algebras over a field. Let's classify all étale algebras $A$ over a field $k$ (i.e., all étale ring maps $k \rightarrow A$). For any $k$-algebra $A,$ we can see $A$ as a vector space over a field $k,$ which implies that it is free over $k,$ so it must be flat over $k.$ Hence, we immediately see that $A$ is unramified over $k$ if and only if $A$ is étale over $k.$ Thus, we have already shown the following above:

Theorem. Let $A$ be any algebra over a field $k.$ The following are equivalent:

• $A$ is étale over $k$;
• $A$ is unramified over $k$;
• $A$ is a finite product of finite separable field extensions of $k$;
• $\bar{k} \otimes_{k} A$ is a finite product of $\bar{k}.$