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

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. $\square$

## 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.

## Comments (0)