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 flat, 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 15.73.2 (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 aplitude 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 15.75.2 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 15.73.2 (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 10.79.3). $\square$

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 aplitude 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 15.73.2 (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 15.75.2. 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BDM. Beware of the difference between the letter 'O' and the digit '0'.