The Stacks project

36.32 Applications

Mostly applications of cohomology and base change. In the future we may generalize these results to the situation discussed in Lemma 36.30.1.

Lemma 36.32.1. Let $f : X \to S$ be a flat, proper morphism of finite presentation. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module of finite presentation, flat over $S$. For fixed $i \in \mathbf{Z}$ consider the function

\[ \beta _ i : S \to \{ 0, 1, 2, \ldots \} ,\quad s \longmapsto \dim _{\kappa (s)} H^ i(X_ s, \mathcal{F}_ s) \]

Then we have

  1. formation of $\beta _ i$ commutes with arbitrary base change,

  2. the functions $\beta _ i$ are upper semi-continuous, and

  3. the level sets of $\beta _ i$ are locally constructible in $S$.

Proof. By cohomology and base change (more precisely by Lemma 36.30.4) the object $K = Rf_*\mathcal{F}$ is a perfect object of the derived category of $S$ whose formation commutes with arbitrary base change. In particular we have

\[ H^ i(X_ s, \mathcal{F}_ s) = H^ i(K \otimes _{\mathcal{O}_ S}^\mathbf {L} \kappa (s)) \]

Thus the lemma follows from Lemma 36.31.1. $\square$

Lemma 36.32.2. Let $f : X \to S$ be a flat, proper morphism of finite presentation. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module of finite presentation, flat over $S$. The function

\[ s \longmapsto \chi (X_ s, \mathcal{F}_ s) \]

is locally constant on $S$. Formation of this function commutes with base change.

Proof. By cohomology and base change (more precisely by Lemma 36.30.4) the object $K = Rf_*\mathcal{F}$ is a perfect object of the derived category of $S$ whose formation commutes with arbitrary base change. Thus we have to show the map

\[ s \longmapsto \sum (-1)^ i \dim _{\kappa (s)} H^ i(K \otimes ^\mathbf {L}_{\mathcal{O}_ S} \kappa (s)) \]

is locally constant on $S$. This is Lemma 36.31.2. $\square$

Lemma 36.32.3. Let $f : X \to S$ be a proper morphism of finite presentation. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module of finite presentation, flat over $S$. Fix $i, r \in \mathbf{Z}$. Then there exists an open subscheme $U \subset S$ with the following property: A morphism $T \to S$ factors through $U$ if and only if $Rf_{T, *}\mathcal{F}_ T$ is isomorphic to a finite locally free module of rank $r$ placed in degree $i$.

Proof. By cohomology and base change (more precisely by Lemma 36.30.4) the object $K = Rf_*\mathcal{F}$ is a perfect object of the derived category of $S$ whose formation commutes with arbitrary base change. Thus this lemma follows immediately from Lemma 36.31.3. $\square$

Lemma 36.32.4. Let $f : X \to S$ be a morphism of finite presentation. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module of finite presentation, flat over $S$ with support proper over $S$. If $R^ if_*\mathcal{F} = 0$ for $i > 0$, then $f_*\mathcal{F}$ is locally free and its formation commutes with arbitrary base change (see proof for explanation).

Proof. By Lemma 36.30.1 the object $E = Rf_*\mathcal{F}$ of $D(\mathcal{O}_ S)$ is perfect and its formation commutes with arbitrary base change, in the sense that $Rf'_*(g')^*\mathcal{F} = Lg^*E$ for any cartesian diagram

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

of schemes. Since there is never any cohomology in degrees $< 0$, we see that $E$ (locally) has tor-amplitude in $[0, b]$ for some $b$. If $H^ i(E) = R^ if_*\mathcal{F} = 0$ for $i > 0$, then $E$ has tor amplitude in $[0, 0]$. Whence $E = H^0(E)[0]$. We conclude $H^0(E) = f_*\mathcal{F}$ is finite locally free by More on Algebra, Lemma 15.74.2 (and the characterization of finite projective modules in Algebra, Lemma 10.78.2). Commutation with base change means that $g^*f_*\mathcal{F} = f'_*(g')^*\mathcal{F}$ for a diagram as above and it follows from the already established commutation of base change for $E$. $\square$

Lemma 36.32.5. Let $f : X \to S$ be a morphism of schemes. Assume

  1. $f$ is proper, flat, and of finite presentation, and

  2. for all $s \in S$ we have $\kappa (s) = H^0(X_ s, \mathcal{O}_{X_ s})$.

Then we have

  1. $f_*\mathcal{O}_ X = \mathcal{O}_ S$ and this holds after any base change,

  2. locally on $S$ we have

    \[ Rf_*\mathcal{O}_ X = \mathcal{O}_ S \oplus P \]

    in $D(\mathcal{O}_ S)$ where $P$ is perfect of tor amplitude in $[1, \infty )$.

Proof. By cohomology and base change (Lemma 36.30.4) the complex $E = Rf_*\mathcal{O}_ X$ is perfect and its formation commutes with arbitrary base change. This first implies that $E$ has tor amplitude in $[0, \infty )$. Second, it implies that for $s \in S$ we have $H^0(E \otimes ^\mathbf {L} \kappa (s)) = H^0(X_ s, \mathcal{O}_{X_ s}) = \kappa (s)$. It follows that the map $\mathcal{O}_ S \to Rf_*\mathcal{O}_ X = E$ induces an isomorphism $\mathcal{O}_ S \otimes \kappa (s) \to H^0(E \otimes ^\mathbf {L} \kappa (s))$. Hence $H^0(E) \otimes \kappa (s) \to H^0(E \otimes ^\mathbf {L} \kappa (s))$ is surjective and we may apply More on Algebra, Lemma 15.76.2 to see that, after replacing $S$ by an affine open neighbourhood of $s$, we have a decomposition $E = H^0(E) \oplus \tau _{\geq 1}E$ with $\tau _{\geq 1}E$ perfect of tor amplitude in $[1, \infty )$. Since $E$ has tor amplitude in $[0, \infty )$ we find that $H^0(E)$ is a flat $\mathcal{O}_ S$-module. It follows that $H^0(E)$ is a flat, perfect $\mathcal{O}_ S$-module, hence finite locally free, see More on Algebra, Lemma 15.74.2 (and the fact that finite projective modules are finite locally free by Algebra, Lemma 10.78.2). It follows that the map $\mathcal{O}_ S \to H^0(E)$ is an isomorphism as we can check this after tensoring with residue fields (Algebra, Lemma 10.79.4). $\square$

Lemma 36.32.6. Let $f : X \to S$ be a morphism of schemes. Assume

  1. $f$ is proper, flat, and of finite presentation, and

  2. the geometric fibres of $f$ are reduced and connected.

Then $f_*\mathcal{O}_ X = \mathcal{O}_ S$ and this holds after any base change.

Proof. By Lemma 36.32.5 it suffices to show that $\kappa (s) = H^0(X_ s, \mathcal{O}_{X_ s})$ for all $s \in S$. This follows from Varieties, Lemma 33.9.3 and the fact that $X_ s$ is geometrically connected and geometrically reduced. $\square$

Lemma 36.32.7. Let $f : X \to S$ be a proper morphism of schemes. Let $s \in S$ and let $e \in H^0(X_ s, \mathcal{O}_{X_ s})$ be an idempotent. Then $e$ is in the image of the map $(f_*\mathcal{O}_ X)_ s \to H^0(X_ s, \mathcal{O}_{X_ s})$.

Proof. Let $X_ s = T_1 \amalg T_2$ be the disjoint union decomposition with $T_1$ and $T_2$ nonempty and open and closed in $X_ s$ corresponding to $e$, i.e., such that $e$ is identitically $1$ on $T_1$ and identically $0$ on $T_2$.

Assume $S$ is Noetherian. We will use the theorem on formal functions in the form of Cohomology of Schemes, Lemma 30.20.7. It tells us that

\[ (f_*\mathcal{O}_ X)_ s^\wedge = \mathop{\mathrm{lim}}\nolimits _ n H^0(X_ n, \mathcal{O}_{X_ n}) \]

where $X_ n$ is the $n$th infinitesimal neighbourhood of $X_ s$. Since the underlying topological space of $X_ n$ is equal to that of $X_ s$ we obtain for all $n $ a disjoint union decomposition of schemes $X_ n = T_{1, n} \amalg T_{2, n}$ where the underlying topological space of $T_{i, n}$ is $T_ i$ for $i = 1, 2$. This means $H^0(X_ n, \mathcal{O}_{X_ n})$ contains a nontrivial idempotent $e_ n$, namely the function which is identically $1$ on $T_{1, n}$ and identically $0$ on $T_{2, n}$. It is clear that $e_{n + 1}$ restricts to $e_ n$ on $X_ n$. Hence $e_\infty = \mathop{\mathrm{lim}}\nolimits e_ n$ is a nontrivial idempotent of the limit. Thus $e_\infty $ is an element of the completion of $(f_*\mathcal{O}_ X)_ s$ mapping to $e$ in $H^0(X_ s, \mathcal{O}_{X_ s})$. Since the map $(f_*\mathcal{O}_ X)_ s^\wedge \to H^0(X_ s, \mathcal{O}_{X_ s})$ factors through $(f_*\mathcal{O}_ X)^\wedge _ s / \mathfrak m_ s (f_*\mathcal{O}_ X)_ s^\wedge = (f_*\mathcal{O}_ X)_ s / \mathfrak m_ s (f_*\mathcal{O}_ X)_ s$ (Algebra, Lemma 10.96.3) we conclude that $e$ is in the image of the map $(f_*\mathcal{O}_ X)_ s \to H^0(X_ s, \mathcal{O}_{X_ s})$ as desired.

General case: we reduce the general case to the Noetherian case by limit arguments. We urge the reader to skip the proof. We may replace $S$ by an affine open neighbourhood of $s$. Thus we may and do assume that $S$ is affine. By Limits, Lemma 32.13.3 we can write $(f : X \to S) = \mathop{\mathrm{lim}}\nolimits (f_ i : X_ i \to S_ i)$ with $f_ i$ proper and $S_ i$ Noetherian. Denote $s_ i \in S_ i$ the image of $s$. Then $s = \mathop{\mathrm{lim}}\nolimits s_ i$, see Limits, Lemma 32.4.4. Then $X_ s = X \times _ S s = \mathop{\mathrm{lim}}\nolimits X_ i \times _{S_ i} s_ i = \mathop{\mathrm{lim}}\nolimits X_{i, s_ i}$ because limits commute with limits (Categories, Lemma 4.14.10). Hence $e$ is the image of some idempotent $e_ i \in H^0(X_{i, s_ i}, \mathcal{O}_{X_{i, s_ i}})$ by Limits, Lemma 32.4.7. By the Noetherian case there is an element $\tilde e_ i$ in the stalk $(f_{i, *}\mathcal{O}_{X_ i})_{s_ i}$ mapping to $e_ i$. Taking the pullback of $\tilde e_ i$ we get an element $\tilde e$ of $(f_*\mathcal{O}_ X)_ s$ mapping to $e$ and the proof is complete. $\square$

Lemma 36.32.8. Let $f : X \to S$ be a morphism of schemes. Let $s \in S$. Assume

  1. $f$ is proper, flat, and of finite presentation, and

  2. the fibre $X_ s$ is geometrically reduced.

Then, after replacing $S$ by an open neighbourhood of $s$, there exists a direct sum decomposition $Rf_*\mathcal{O}_ X = f_*\mathcal{O}_ X \oplus P$ in $D(\mathcal{O}_ S)$ where $f_*\mathcal{O}_ X$ is a finite étale $\mathcal{O}_ S$-algebra and $P$ is a perfect of tor amplitude in $[1, \infty )$.

Proof. The proof of this lemma is similar to the proof of Lemma 36.32.5 which we suggest the reader read first. By cohomology and base change (Lemma 36.30.4) the complex $E = Rf_*\mathcal{O}_ X$ is perfect and its formation commutes with arbitrary base change. This first implies that $E$ has tor amplitude in $[0, \infty )$.

We claim that after replacing $S$ by an open neighbourhood of $s$ we can find a direct sum decomposition $E = H^0(E) \oplus \tau _{\geq 1}E$ in $D(\mathcal{O}_ S)$ with $\tau _{\geq 1}E$ of tor amplitude in $[1, \infty )$. Assume the claim is true for now and assume we've made the replacement so we have the direct sum decomposition. Since $E$ has tor amplitude in $[0, \infty )$ we find that $H^0(E)$ is a flat $\mathcal{O}_ S$-module. Hence $H^0(E)$ is a flat, perfect $\mathcal{O}_ S$-module, hence finite locally free, see More on Algebra, Lemma 15.74.2 (and the fact that finite projective modules are finite locally free by Algebra, Lemma 10.78.2). Of course $H^0(E) = f_*\mathcal{O}_ X$ is an $\mathcal{O}_ S$-algebra. By cohomology and base change we obtain $H^0(E) \otimes \kappa (s) = H^0(X_ s, \mathcal{O}_{X_ s})$. By Varieties, Lemma 33.9.3 and the assumption that $X_ s$ is geometrically reduced, we see that $\kappa (s) \to H^0(E) \otimes \kappa (s)$ is finite étale. By Morphisms, Lemma 29.36.17 applied to the finite locally free morphism $\underline{\mathop{\mathrm{Spec}}}_ S(H^0(E)) \to S$, we conclude that after shrinking $S$ the $\mathcal{O}_ S$-algebra $H^0(E)$ is finite étale.

It remains to prove the claim. For this it suffices to prove that the map

\[ (f_*\mathcal{O}_ X)_ s \longrightarrow H^0(X_ s, \mathcal{O}_{X_ s}) = H^0(E \otimes ^\mathbf {L} \kappa (s)) \]

is surjective, see More on Algebra, Lemma 15.76.2. Choose a flat local ring homomorphism $\mathcal{O}_{S, s} \to A$ such that the residue field $k$ of $A$ is algebraically closed, see Algebra, Lemma 10.159.1. By flat base change (Cohomology of Schemes, Lemma 30.5.2) we get $H^0(X_ A, \mathcal{O}_{X_ A}) = (f_*\mathcal{O}_ X)_ s \otimes _{\mathcal{O}_{S, s}} A$ and $H^0(X_ k, \mathcal{O}_{X_ k}) = H^0(X_ s, \mathcal{O}_{X_ s}) \otimes _{\kappa (s)} k$. Hence it suffices to prove that $H^0(X_ A, \mathcal{O}_{X_ A}) \to H^0(X_ k, \mathcal{O}_{X_ k})$ is surjective. Since $X_ k$ is a reduced proper scheme over $k$ and since $k$ is algebraically closed, we see that $H^0(X_ k, \mathcal{O}_{X_ k})$ is a finite product of copies of $k$ by the already used Varieties, Lemma 33.9.3. Since by Lemma 36.32.7 the idempotents of this $k$-algebra are in the image of $H^0(X_ A, \mathcal{O}_{X_ A}) \to H^0(X_ k, \mathcal{O}_{X_ k})$ we conclude. $\square$

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