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

Comment #1766 by Johan on