This section is a continuation of More on Algebra, Section 15.36. 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 87.4.1. Let $R$ be a linearly topologized ring and let $M$ be a linearly topologized $R$-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 87.4.2. 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 $0 \to N \to M \to Q \to 0$ be 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 87.4.3. 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 87.4.4. 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 system $(M_ n + N)/N$ 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 10.86.4. As $N$ is closed we see that the kernel of $M \to \mathop{\mathrm{lim}}\nolimits M/(M_ n + N)$ is $N$ (see Lemma 87.4.2). Finally, $M \to M/N$ is open by definition of the quotient topology.
$\square$

reference
Lemma 87.4.5. 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 87.4.3. 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 87.4.4. 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 87.4.6. 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 10.96.3. 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 87.4.5 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 87.4.7. 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 87.2.3.

Definition 87.4.8. 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 87.4.9. 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 87.4.10. Let $B$ be a linearly topologized ring. The set of topologically nilpotent elements of $B$ is a closed, radical ideal of $B$. 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.**
Let $\mathfrak b \subset B$ be the set of topologically nilpotent elements. We omit the proof of the fact that $\mathfrak b$ is a radical ideal (good exercise in the definitions). Let $g$ be an element of the closure of $\mathfrak b$. Our goal is to show that $g$ is topologically nilpotent. Let $J \subset B$ be an open ideal. We have to show $g^ e \in J$ for some $e \geq 1$. We have $g \in \mathfrak b + J$ by Lemma 87.4.2. Hence $g = f + h$ for some $f \in \mathfrak b$ and $h \in J$. Pick $m \geq 1$ such that $f^ m \in J$. Then $g^{m + 1} \in J$ as desired.

Let $\varphi : A \to B$ be as in the statement of the lemma. Assertion (1) is clear and assertion (2) follows from this and the fact that $\mathfrak b$ is a closed ideal.
$\square$

Lemma 87.4.11. 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 87.4.2. 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 87.4.1. Hence the limit can be written as $\mathop{\mathrm{lim}}\nolimits B/(J_\lambda + IB)$ and the result is clear.
$\square$

Lemma 87.4.12. Let $B \to A$ and $B \to C$ be continuous homomorphisms of linearly topologized rings.

If $A$ and $C$ are weakly pre-admissible, then $A \widehat{\otimes }_ B C$ is weakly admissible.

If $A$ and $C$ are pre-admissible, then $A \widehat{\otimes }_ B C$ is admissible.

If $A$ and $C$ have a countable fundamental system of open ideals, then $A \widehat{\otimes }_ B C$ has a countable fundamental system of open ideals.

If $A$ and $C$ are pre-adic and have finitely generated ideals of definition, then $A \widehat{\otimes }_ B C$ is adic and has a finitely generated ideal of definition.

If $A$ and $C$ are pre-adic Noetherian rings and $B/\mathfrak b \to A/\mathfrak a$ is of finite type where $\mathfrak a \subset A$ and $\mathfrak b \subset B$ are the ideals of topologically nilpotent elements, then $A \widehat{\otimes }_ B C$ is adic Noetherian.

**Proof.**
Let $I_\lambda \subset A$, $\lambda \in \Lambda $ and $J_\mu \subset C$, $\mu \in M$ be fundamental systems of open ideals, then by definition

\[ A \widehat{\otimes }_ B C = \mathop{\mathrm{lim}}\nolimits _{\lambda , \mu } A/I_\lambda \otimes _ B C/J_\mu \]

with the limit topology. Thus a fundamental system of open ideals is given by the kernels $K_{\lambda , \mu }$ of the maps $A \widehat{\otimes }_ B C \to A/I_\lambda \otimes _ B C/J_\mu $. Note that $K_{\lambda , \mu }$ is the closure of the ideal $I_\lambda (A \widehat{\otimes }_ B C) + J_\mu (A \widehat{\otimes }_ B C)$. Finally, we have a ring homomorphism $\tau : A \otimes _ B C \to A \widehat{\otimes }_ B C$ with dense image.

Proof of (1). If $I_\lambda $ and $J_\mu $ consist of topologically nilpotent elements, then so does $K_{\lambda , \mu }$ by Lemma 87.4.10. Hence $A \widehat{\otimes }_ B C$ is weakly admissible by definition.

Proof of (2). Assume for some $\lambda _0$ and $\mu _0$ the ideals $I = I_{\lambda _0} \subset A$ and $J_{\mu _0} \subset C$ are ideals of definition. Thus for every $\lambda $ there exists an $n$ such that $I^ n \subset I_\lambda $. For every $\mu $ there exists an $m$ such that $J^ m \subset J_\mu $. Then

\[ \left(I(A \widehat{\otimes }_ B C) + J(A \widehat{\otimes }_ B C)\right)^{n + m} \subset I_\lambda (A \widehat{\otimes }_ B C) + J_\mu (A \widehat{\otimes }_ B C) \]

It follows that the open ideal $K = K_{\lambda _0, \mu _0}$ satisfies $K^{n + m} \subset K_{\lambda , \mu }$. Hence $K$ is an ideal of definition of $A \widehat{\otimes }_ B C$ and $A \widehat{\otimes }_ B C$ is admissible by definition.

Proof of (3). If $\Lambda $ and $M$ are countable, so is $\Lambda \times M$.

Proof of (4). Assume $\Lambda = \mathbf{N}$ and $M = \mathbf{N}$ and we have finitely generated ideals $I \subset A$ and $J \subset C$ such that $I_ n = I^ n$ and $J_ n = J^ n$. Then

\[ I(A \widehat{\otimes }_ B C) + J(A \widehat{\otimes }_ B C) \]

is a finitely generated ideal and it is easily seen that $A \widehat{\otimes }_ B C$ is the completion of $A \otimes _ B C$ with respect to this ideal. Hence (4) follows from Algebra, Lemma 10.96.3.

Proof of (5). Let $\mathfrak c \subset C$ be the ideal of topologically nilpotent elements. Since $A$ and $C$ are adic Noetherian, we see that $\mathfrak a$ and $\mathfrak c$ are ideals of definition (details omitted). From part (4) we already know that $A \widehat{\otimes }_ B C$ is adic and that $\mathfrak a(A \widehat{\otimes }_ B C) + \mathfrak c(A \widehat{\otimes }_ B C)$ is a finitely generated ideal of definition. Since

\[ A \widehat{\otimes }_ B C / \left(\mathfrak a(A \widehat{\otimes }_ B C) + \mathfrak c(A \widehat{\otimes }_ B C)\right) = A/\mathfrak a \otimes _{B/\mathfrak b} C/\mathfrak c \]

is Noetherian as a finite type algebra over the Noetherian ring $C/\mathfrak c$ we conclude by Algebra, Lemma 10.97.5.
$\square$

## Comments (2)

Comment #5983 by Dario Weißmann on

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