Lemma 85.4.14. Let $\varphi : A \to B$ be a continuous homomorphism of linearly topologized rings. If

1. $\varphi$ is taut,

2. $\varphi$ has dense image,

3. $A$ is complete,

4. $B$ is separated, and

5. $A$ has a countable fundamental system of neighbourhoods of $0$.

Then $\varphi$ is surjective and open, $B$ is complete, and $B = A/K$ for some closed ideal $K \subset A$.

Proof. We may choose a sequence of open ideals $A \supset I_1 \supset I_2 \supset I_3 \supset \ldots$ which form a fundamental system of neighbourhoods of $0$. For each $i$ let $J_ i \subset B$ be the closure of $\varphi (I_ i)B$. As $\varphi$ is taut we see that these form a fundamental system of open ideals of $B$. Set $I_0 = A$ and $J_0 = B$. Let $n \geq 0$ and let $y_ n \in J_ n$. Since $J_ n$ is the closure of $\varphi (I_ n)B$ we can write

$y_ n = \sum \nolimits _ t \varphi (f_ t)b_ t + y'_{n + 1}$

for some $f_ t \in I_ n$, $b_ t \in B$, and $y'_{n + 1} \in J_{n + 1}$. Since $\varphi$ has dense image we can choose $a_ t \in A$ with $\varphi (a_ t) = b_ t \bmod J_{n + 1}$. Thus

$y_ n = \varphi (f_ n) + y_{n + 1}$

with $f_ n = \sum f_ ta_ t \in I_ n$ and $y_{n + 1} = y'_{n + 1} + \sum f_ t(b_ t - \varphi (a_ t)) \in J_{n + 1}$. Thus, starting with any $y = y_0 \in B$, we can find by induction a sequence $f_ m \in I_ m$, $m \geq 0$ such that

$y = y_0 = \varphi (f_0 + f_1 + \ldots + f_ n) + y_{n + 1}$

with $y_{n + 1} \in J_{n + 1}$. Since $A$ is complete we see that

$x = x_0 = f_0 + f_1 + f_2 + \ldots$

exists. Since the partial sums approximate $x$ in $A$, since $\varphi$ is continuous, and since $B$ is separated we find that $\varphi (x) = y$ because above we've shown that the images of the partial sums approximate $y$ in $B$. Thus $\varphi$ is surjective. In exactly the same manner we find that $\varphi (I_ n) = J_ n$ for all $n \geq 1$. This proves the lemma. $\square$

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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