0
$\begingroup$

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.

$\endgroup$
2
+50
$\begingroup$

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. $\square$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy