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.
$\square$

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