The Stacks project

Lemma 87.6.5. Let $\varphi : A \to B$ be a continuous map of linearly topologized rings. If $\varphi $ is taut, $A$ is pre-adic and has a finitely generated ideal of definition, and $B$ is complete, then $B$ is adic and has a finitely generated ideal of definition and the ring map $\varphi $ is adic.

Proof. Choose a finitely generated ideal of definition $I$ of $A$. Let $J_ n$ be the closure of $\varphi (I^ n)B$ in $B$. Since $B$ is complete we have $B = \mathop{\mathrm{lim}}\nolimits B/J_ n$. Let $B' = \mathop{\mathrm{lim}}\nolimits B/I^ nB$ be the $I$-adic completion of $B$. By Algebra, Lemma 10.96.3, the $I$-adic topology on $B'$ is complete and $B'/I^ nB' = B/I^ nB$. Thus the ring map $B' \to B$ is continuous and has dense image as $B' \to B/I^ nB \to B/J_ n$ is surjective for all $n$. Finally, the map $B' \to B$ is taut because $(I^ nB')B = I^ nB$ and $A \to B$ is taut. By Lemma 87.5.10 we see that $B' \to B$ is open and surjective. Thus the topology on $B$ is the $I$-adic topology and the proof is complete. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0APU. Beware of the difference between the letter 'O' and the digit '0'.