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

1. $\varphi$ is taut and has dense image,

2. $A$ is complete and has a countable fundamental system of open ideals, and

3. $B$ 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. $\square$

Comment #1727 by Matthew Emerton on

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

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