Lemma 85.4.15. Let $\varphi : A \to B$ be a continuous map of linearly topologized rings. Let $I \subset A$ be an ideal. Assume

$I$ is finitely generated,

$A$ has the $I$-adic topology,

$B$ is complete, and

$\varphi $ is taut.

Then the topology on $B$ is the $I$-adic topology.

**Proof.**
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 10.95.3, 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 85.4.14 we see that $B' \to B$ is open and surjective which implies the lemma.
$\square$

## Comments (0)