Lemma 87.5.10 (0APT)—The Stacks project

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Table of contents
Part 5: Topics in Geometry
Chapter 87: Formal Algebraic Spaces
Section 87.5: Taut ring maps
Lemma 87.5.10 (cite)

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Lemma 87.5.10. Let $\varphi : A \to B$ be a continuous homomorphism of linearly topologized rings. Assume

$\varphi$ is taut and has dense image,
$A$ is complete and has a countable fundamental system of open ideals, and
$B$ is separated.

Then $\varphi$ is surjective and open, $B$ is complete, and $B = A/K$ for some closed ideal $K \subset A$ .

Proof. By the open mapping lemma (More on Algebra, Lemma 15.36.5) combined with tautness of $\varphi$ , we see the map $\varphi$ is open. Since the image of $\varphi$ is dense, we see that $\varphi$ is surjective. The kernel $K$ of $\varphi$ is closed as $\varphi$ is continuous. It follows that $B = A/K$ is complete, see for example Lemma 87.4.4. $\square$

Comments (2)

Comment #1727 by Matthew Emerton on December 14, 2015 at 22:12

In line four of the proof, I think it should read " $J_{n}$ is the closure", rather than $J_{n+1}$ .

Comment #1766 by Johan on December 15, 2015 at 19:10

Thanks, fixed here.

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