# Completed stalks of the pushforward of the structure sheaf

Let $$\pi:X\to S$$ be a proper morphism of Noetherian schemes. Is it possible that $$\pi_*\mathcal{O}_X\neq \mathcal{O}_S$$ but the natural map $$\mathcal{O}_s^\wedge\to (\pi_*\mathcal{O}_X)_s^\wedge$$ is an isomorphism for all points $$s\in S$$? The completion is with respect to $$\mathfrak{m}_s$$ on both sides.

## 1 Answer

This has little to do with morphisms, and follows immediately from the following commutative algebra lemma:

Lemma. Let $$R$$ be a Noetherian ring, and $$f \colon M \to N$$ a morphism of finite $$R$$-modules. Then $$f$$ is an isomorphism if and only if $$f^\wedge_{\mathfrak m} \colon M^\wedge_{\mathfrak m} \to N^\wedge_{\mathfrak m}$$ is an isomorphism for all maximal ideals $$\mathfrak m \subseteq R$$.

Proof. There are many ways to prove this. For example, you can use that $$R_\mathfrak m \to R^\wedge_\mathfrak m$$ is faithfully flat (Tag 00MC), and for finite $$R_{\mathfrak m}$$-modules $$P$$ we have $$P^\wedge _{\mathfrak m} = P \otimes_{R_{\mathfrak m}} R^\wedge_{\mathfrak m}$$ (Tag 00MA). Thus a finite $$R_\mathfrak m$$-module $$P$$ is zero if and only if $$P ^\wedge_{\mathfrak m} = 0$$ (this also follows from Nakayama's lemma), and $$f^\wedge_{\mathfrak m}$$ is an isomorphism if and only if $$f_{\mathfrak m} \colon M_{\mathfrak m} \to N_{\mathfrak m}$$ is. Now the result follows from Tag 00HN. $$\square$$