$l$-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.