The Stacks project

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. $\square$

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