This section is a continuation of More on Algebra, Section 15.35. Let $R$ be a topological ring and let $M$ be a linearly topologized $R$-module. When we say “*let $M_\lambda $ be a fundamental system of open submodules*” we will mean that each $M_\lambda $ is an open submodule and that any neighbourhood of $0$ contains one of the $M_\lambda $. In other words, this means that $M_\lambda $ is a fundamental system of neighbourhoods of $0$ in $M$ consisting of submodules. Similarly, if $R$ is a linearly topologized ring, then we say “*let $I_\lambda $ be a fundamental system of open ideals*” to mean that $I_\lambda $ is a fundamental system of neighbourhoods of $0$ in $R$ consisting of ideals.

Example 85.4.1. Let $R$ be a linearly topologized ring and let $M$ be a linearly topologized $A$-module. Let $I_\lambda $ be a fundamental system of open ideals in $R$ and let $M_\mu $ be a fundamental system of open submodules of $M$. The continuity of $+ : M \times M \to M$ is automatic and the continuity of $R \times M \to M$ signifies

\[ \forall f, x, \mu \ \exists \lambda , \nu ,\ (f + I_\lambda )(x + M_\nu ) \subset fx + M_\mu \]

Since $fM_\nu + I_\lambda M_\nu \subset M_\mu $ if $M_\nu \subset M_\mu $ we see that the condition is equivalent to

\[ \forall x, \mu \ \exists \lambda \ I_\lambda x \subset M_\mu \]

However, it need not be the case that given $\mu $ there is a $\lambda $ such that $I_\lambda M \subset M_\mu $. For example, consider $R = k[[t]]$ with the $t$-adic topology and $M = \bigoplus _{n \in \mathbf{N}} R$ with fundamental system of open submodules given by

\[ M_ m = \bigoplus \nolimits _{n \in \mathbf{N}} t^{nm}R \]

Since every $x \in M$ has finitely many nonzero coordinates we see that, given $m$ and $x$ there exists a $k$ such that $t^ k x \in M_ m$. Thus $M$ is a linearly topologized $R$-module, but it isn't true that given $m$ there is a $k$ such that $t^ kM \subset M_ m$. On the other hand, if $R \to S$ is a continuous map of linearly topologized rings, then the corresponding statement does hold, i.e., for every open ideal $J \subset S$ there exists an open ideal $I \subset R$ such that $IS \subset J$ (as the reader can easily deduce from continuity of the map $R \to S$).

Lemma 85.4.2. Let $R$ be a topological ring. Let $M$ be a linearly topologized $R$-module and let $M_\lambda $, $\lambda \in \Lambda $ be a fundamental system of open submodules. Let $N \subset M$ be a submodule. The closure of $N$ is $\bigcap _{\lambda \in \Lambda } (N + M_\lambda )$.

**Proof.**
Since each $N + M_\lambda $ is open, it is also closed. Hence the intersection is closed. If $x \in M$ is not in the closure of $N$, then $(x + M_\lambda ) \cap N = 0$ for some $\lambda $. Hence $x \not\in N + M_\lambda $. This proves the lemma.
$\square$

Unless otherwise mentioned we endow submodules and quotient modules with the induced topology. Let $M$ be a linearly topologized module over a topological ring $R$, and let $0 \to N \to M \to Q \to 0$ is a short exact sequence of $R$-modules. If $M_\lambda $ is a fundamental system of open submodules of $M$, then $N \cap M_\lambda $ is a fundamental system of open submodules of $N$. If $\pi : M \to Q$ is the quotient map, then $\pi (M_\lambda )$ is a fundamental system of open submodules of $Q$. In particular these induced topologies are linear topologies.

Lemma 85.4.3. Let $R$ be a topological ring. Let $M$ be a linearly topologized $R$-module. Let $N \subset M$ be a submodule. Then

$0 \to N^\wedge \to M^\wedge \to (M/N)^\wedge $ is exact, and

$N^\wedge $ is the closure of the image of $N \to M^\wedge $.

**Proof.**
Let $M_\lambda $, $\lambda \in \Lambda $ be a fundamental system of open submodules. Then $N \cap M_\lambda $ is a fundamental system of open submodules of $N$ and $M_\lambda + N/N$ is a fundamental system of open submodules of $M/N$. Thus we see that (1) follows from the exactness of the sequences

\[ 0 \to N/N \cap M_\lambda \to M/M_\lambda \to M/(M_\lambda + N) \to 0 \]

and the fact that taking limits commutes with limits. The second statement follows from this and the fact that $N \to N^\wedge $ has dense image and that the kernel of $M^\wedge \to (M/N)^\wedge $ is closed.
$\square$

Lemma 85.4.4. Let $R$ be a topological ring. Let $M$ be a complete, linearly topologized $R$-module. Let $N \subset M$ be a closed submodule. If $M$ has a countable fundamental system of neighbourhoods of $0$, then $M/N$ is complete and the map $M \to M/N$ is open.

**Proof.**
Let $M_ n$, $n \in \mathbf{N}$ be a fundamental system of open submodules of $M$. We may assume $M_{n + 1} \subset M_ n$ for all $n$. The $(M_ n + N)/N$ is a fundamental system in $M/N$. Hence we have to show that $M/N = \mathop{\mathrm{lim}}\nolimits M/(M_ n + N)$. Consider the short exact sequences

\[ 0 \to N/N \cap M_ n \to M/M_ n \to M/(M_ n + N) \to 0 \]

Since the transition maps of the system $\{ N/N\cap M_ n\} $ are surjective we see that $M = \mathop{\mathrm{lim}}\nolimits M/M_ n$ (by completeness of $M$) surjects onto $\mathop{\mathrm{lim}}\nolimits M/(M_ n + N)$ by Algebra, Lemma 10.85.4. As $N$ is closed we see that the kernel of $M \to \mathop{\mathrm{lim}}\nolimits M/(M_ n + N)$ is $N$ (see Lemma 85.4.2). Finally, $M \to M/N$ is open by definition of the quotient topology.
$\square$

reference
Lemma 85.4.5. Let $R$ be a topological ring. Let $M$ be a linearly topologized $R$-module. Let $N \subset M$ be a submodule. Assume $M$ has a countable fundamental system of neighbourhoods of $0$. Then

$0 \to N^\wedge \to M^\wedge \to (M/N)^\wedge \to 0$ is exact,

$N^\wedge $ is the closure of the image of $N \to M^\wedge $,

$M^\wedge \to (M/N)^\wedge $ is open.

**Proof.**
We have $0 \to N^\wedge \to M^\wedge \to (M/N)^\wedge $ is exact and statement (2) by Lemma 85.4.3. This produces a canonical map $c : M^\wedge /N^\wedge \to (M/N)^\wedge $. The module $M^\wedge /N^\wedge $ is complete and $M^\wedge \to M^\wedge /N^\wedge $ is open by Lemma 85.4.4. By the universal property of completion we obtain a canonical map $b : (M/N)^\wedge \to M^\wedge /N^\wedge $. Then $b$ and $c$ are mutually inverse as they are on a dense subset.
$\square$

Lemma 85.4.6. Let $R$ be a topological ring. Let $M$ be a topological $R$-module. Let $I \subset R$ be a finitely generated ideal. Assume $M$ has an open submodule whose topology is $I$-adic. Then $M^\wedge $ has an open submodule whose topology is $I$-adic and we have $M^\wedge /I^ n M^\wedge = M/I^ nM$ for all $n \geq 1$.

**Proof.**
Let $M' \subset M$ be an open submodule whose topology is $I$-adic. Then $\{ I^ nM'\} _{n \geq 1}$ is a fundamental system of open submodules of $M$. Thus $M^\wedge = \mathop{\mathrm{lim}}\nolimits M/I^ nM'$ contains $(M')^\wedge = \mathop{\mathrm{lim}}\nolimits M'/I^ nM'$ as an open submodule and the topology on $(M')^\wedge $ is $I$-adic by Algebra, Lemma 10.95.3. Since $I$ is finitely generated, $I^ n$ is finitely generated, say by $f_1, \ldots , f_ r$. Observe that the surjection $(f_1, \ldots , f_ r) : M^{\oplus r} \to I^ n M$ is continuous and open by our description of the topology on $M$ above. By Lemma 85.4.5 applied to this surjection and to the short exact sequence $0 \to I^ nM \to M \to M/I^ nM \to 0$ we conclude that

\[ (f_1, \ldots , f_ r) : (M^\wedge )^{\oplus r} \longrightarrow M^\wedge \]

surjects onto the kernel of the surjection $M^\wedge \to M/I^ nM$. Since $f_1, \ldots , f_ r$ generate $I^ n$ we conclude.
$\square$

Definition 85.4.7. Let $R$ be a topological ring. Let $M$ and $N$ be linearly topologized $R$-modules. The *tensor product* of $M$ and $N$ is the (usual) tensor product $M \otimes _ R N$ endowed with the linear topology defined by declaring

\[ \mathop{\mathrm{Im}}(M_\mu \otimes _ R N + M \otimes _ R N_\nu \longrightarrow M \otimes _ R N) \]

to be a fundamental system of open submodules, where $M_\mu \subset M$ and $N_\nu \subset N$ run through fundamental systems of open submodules in $M$ and $N$. The *completed tensor product*

\[ M \widehat{\otimes }_ R N = \mathop{\mathrm{lim}}\nolimits M \otimes _ R N/(M_\mu \otimes _ R N + M \otimes _ R N_\nu ) = \mathop{\mathrm{lim}}\nolimits M/M_\mu \otimes _ R N/N_\nu \]

is the completion of the tensor product.

Observe that the topology on $R$ is immaterial for the construction of the tensor product or the completed tensor product. If $R \to A$ and $R \to B$ are continuous maps of linearly topologized rings, then the construction above gives a tensor product $A \otimes _ R B$ and a completed tensor product $A \widehat{\otimes }_ R B$.

We record here the notions introduced in Remark 85.2.3.

Definition 85.4.8. Let $A$ be a linearly topologized ring.

An element $f \in A$ is called *topologically nilpotent* if $f^ n \to 0$ as $n \to \infty $.

A *weak ideal of definition* for $A$ is an open ideal $I \subset A$ consisting entirely of topologically nilpotent elements.

We say $A$ is *weakly pre-admissible* if $A$ has a weak ideal of definition.

We say $A$ is *weakly admissible* if $A$ is weakly pre-admissible and complete^{1}.

Given a weak ideal of definition $I$ in a linearly topologized ring $A$ and an open ideal $J$ the intersection $I \cap J$ is a weak ideal of definition. Hence if there is one weak ideal of definition, then there is a fundamental system of open ideals consisting of weak ideals of definition. In particular, given a weakly admissible topological ring $A$ then $A = \mathop{\mathrm{lim}}\nolimits A/I_\lambda $ where $\{ I_\lambda \} $ is a fundamental system of weak ideals of definition.

Lemma 85.4.9. Let $A$ be a weakly admissible topological ring. Let $I \subset A$ be a weak ideal of definition. Then $(A, I)$ is a henselian pair.

**Proof.**
Let $A \to A'$ be an étale ring map and let $\sigma : A' \to A/I$ be an $A$-algebra map. By More on Algebra, Lemma 15.11.6 it suffices to lift $\sigma $ to an $A$-algebra map $A' \to A$. To do this, as $A$ is complete, it suffices to find, for every open ideal $J \subset I$, a unique $A$-algebra map $A' \to A/J$ lifting $\sigma $. Since $I$ is a weak ideal of definition, the ideal $I/J$ is locally nilpotent. We conclude by More on Algebra, Lemma 15.11.2.
$\square$

Lemma 85.4.10. Let $\varphi : A \to B$ be a continuous map of linearly topologized rings.

If $f \in A$ is topologically nilpotent, then $\varphi (f)$ is topologically nilpotent.

If $I \subset A$ consists of topologically nilpotent elements, then the closure of $\varphi (I)B$ consists of topologically nilpotent elements.

**Proof.**
Part (1) is clear. Let $g$ be an element of the closure of $\varphi (I)B$. Let $J \subset B$ be an open ideal. We have to show $g^ e \in J$ for some $e$. We have $g \in \varphi (I)B + J$ by Lemma 85.4.2. Hence $g = \sum _{i = 1, \ldots , n} f_ ib_ i + h$ for some $f_ i \in I$, $b_ i \in B$ and $h \in J$. Pick $e_ i$ such that $\varphi (f_ i^{e_ i}) \in J$. Then $g^{e_1 + \ldots + e_ n + 1} \in J$.
$\square$

Definition 85.4.11. Let $\varphi : A \to B$ be a continuous map of linearly topologized rings. We say $\varphi $ is *taut*^{2} if for every open ideal $I \subset A$ the closure of the ideal $\varphi (I)B$ is open and these closures form a fundamental system of open ideals.

If $\varphi : A \to B$ is a continuous map of linearly topologized rings and $I_\lambda $ a fundamental system of open ideals of $A$, then $\varphi $ is taut if and only if the closures of $I_\lambda B$ are open and form a fundamental system of open ideals in $A$.

Lemma 85.4.12. Let $\varphi : A \to B$ be a continuous map of weakly admissible topological rings. The following are equivalent

$\varphi $ is taut,

for every weak ideal of definition $I \subset A$ the closure of $\varphi (I)B$ is a weak ideal of definition of $B$ and these form a fundamental system of weak ideals of definition of $B$.

**Proof.**
It is clear that (2) implies (1). The other implication follows from Lemma 85.4.10.
$\square$

Lemma 85.4.13. Let $A \to B$ be a continuous map of linearly topologized rings. Let $I \subset A$ be an ideal. The closure of $IB$ is the kernel of $B \to B \widehat{\otimes }_ A A/I$.

**Proof.**
Let $J_\mu $ be a fundamental system of open ideals of $B$. The closure of $IB$ is $\bigcap (IB + J_\lambda )$ by Lemma 85.4.2. Let $I_\mu $ be a fundamental system of open ideals in $A$. Then

\[ B \widehat{\otimes }_ A A/I = \mathop{\mathrm{lim}}\nolimits (B/J_\lambda \otimes _ A A/(I_\mu + I)) = \mathop{\mathrm{lim}}\nolimits B/(J_\lambda + I_\mu B + I B) \]

Since $A \to B$ is continuous, for every $\lambda $ there is a $\mu $ such that $I_\mu B \subset J_\lambda $, see discussion in Example 85.4.1. Hence the limit can be written as $\mathop{\mathrm{lim}}\nolimits B/(J_\lambda + IB)$ and the result is clear.
$\square$

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

$\varphi $ is taut,

$\varphi $ has dense image,

$A$ is complete,

$B$ is separated, and

$A$ 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.
$\square$

The next lemma says “$\varphi $ is taut” if and only if “$\varphi $ is adic” for continuous maps $\varphi : A \to B$ between adic rings if $A$ has a finitely generated ideal of definition. In some sense the previously introduced notion of tautness for continuous ring maps supersedes the notion of an adic map between adic rings. See also Section 85.17.

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)