## 15.35 Topological rings and modules

Let's quickly discuss some properties of topological abelian groups. An abelian group $M$ is a *topological abelian group* if $M$ is endowed with a topology such that addition $M \times M \to M$, $(x, y) \mapsto x + y$ and inverse $M \to M$, $x \mapsto -x$ are continuous. A *homomorphism of topological abelian groups* is just a homomorphism of abelian groups which is continuous. The category of commutative topological groups is additive and has kernels and cokernels, but is not abelian (as the axiom $\mathop{\mathrm{Im}}= \mathop{\mathrm{Coim}}$ doesn't hold). If $N \subset M$ is a subgroup, then we think of $N$ and $M/N$ as topological groups also, namely using the induced topology on $N$ and the quotient topology on $M/N$ (i.e., such that $M \to M/N$ is submersive). Note that if $N \subset M$ is an open subgroup, then the topology on $M/N$ is discrete.

We say the topology on $M$ is *linear* if there exists a fundamental system of neighbourhoods of $0$ consisting of subgroups. If so then these subgroups are also open. An example is the following. Let $I$ be a directed set and let $G_ i$ be an inverse system of (discrete) abelian groups over $I$. Then

\[ G = \mathop{\mathrm{lim}}\nolimits _{i \in I} G_ i \]

with the inverse limit topology is linearly topologized with a fundamental system of neighbourhoods of $0$ given by $\mathop{\mathrm{Ker}}(G \to G_ i)$. Conversely, let $M$ be a linearly topologized abelian group. Choose any fundamental system of open subgroups $U_ i \subset M$, $i \in I$ (i.e., the $U_ i$ form a fundamental system of open neighbourhoods and each $U_ i$ is a subgroup of $M$). Setting $i \geq i' \Leftrightarrow U_ i \subset U_{i'}$ we see that $I$ is a directed set. We obtain a homomorphism of linearly topologized abelian groups

\[ c : M \longrightarrow \mathop{\mathrm{lim}}\nolimits _{i \in I} M/U_ i. \]

It is clear that $M$ is *separated* (as a topological space) if and only if $c$ is injective. We say that $M$ is *complete* if $c$ is an isomorphism^{1}. We leave it to the reader to check that this condition is independent of the choice of fundamental system of open subgroups $\{ U_ i\} _{i \in I}$ chosen above. In fact the topological abelian group $M^\wedge = \mathop{\mathrm{lim}}\nolimits _{i \in I} M/U_ i$ is independent of this choice and is sometimes called the *completion* of $M$. Any $G = \mathop{\mathrm{lim}}\nolimits G_ i$ as above is complete, in particular, the completion $M^\wedge $ is always complete.

reference
Definition 15.35.1 (Topological rings). Let $R$ be a ring and let $M$ be an $R$-module.

We say $R$ is a *topological ring* if $R$ is endowed with a topology such that both addition and multiplication are continuous as maps $R \times R \to R$ where $R \times R$ has the product topology. In this case we say $M$ is a *topological module* if $M$ is endowed with a topology such that addition $M \times M \to M$ and scalar multiplication $R \times M \to M$ are continuous.

A *homomorphism of topological modules* is just a continuous $R$-module map. A *homomorphism of topological rings* is a ring homomorphism which is continuous for the given topologies.

We say $M$ is *linearly topologized* if $0$ has a fundamental system of neighbourhoods consisting of submodules. We say $R$ is *linearly topologized* if $0$ has a fundamental system of neighbourhoods consisting of ideals.

If $R$ is linearly topologized, we say that $I \subset R$ is an *ideal of definition* if $I$ is open and if every neighbourhood of $0$ contains $I^ n$ for some $n$.

If $R$ is linearly topologized, we say that $R$ is *pre-admissible* if $R$ has an ideal of definition.

If $R$ is linearly topologized, we say that $R$ is *admissible* if it is pre-admissible and complete^{2}.

If $R$ is linearly topologized, we say that $R$ is *pre-adic* if there exists an ideal of definition $I$ such that $\{ I^ n\} _{n \geq 0}$ forms a fundamental system of neighbourhoods of $0$.

If $R$ is linearly topologized, we say that $R$ is *adic* if $R$ is pre-adic and complete.

Note that a (pre)adic topological ring is the same thing as a (pre)admissible topological ring which has an ideal of definition $I$ such that $I^ n$ is open for all $n \geq 1$.

Let $R$ be a ring and let $M$ be an $R$-module. Let $I \subset R$ be an ideal. Then we can consider the linear topology on $R$ which has $\{ I^ n\} _{n \geq 0}$ as a fundamental system of neighbourhoods of $0$. This topology is called the $I$-adic topology; $R$ is a pre-adic topological ring in the $I$-adic topology^{3}. Moreover, the linear topology on $M$ which has $\{ I^ nM\} _{n \geq 0}$ as a fundamental system of open neighbourhoods of $0$ turns $M$ into a topological $R$-module. This is called the *$I$-adic topology* on $M$. We see that $M$ is $I$-adically complete (as defined in Algebra, Definition 10.95.2) if and only if $M$ is complete in the $I$-adic topology^{4}. In particular, we see that $R$ is $I$-adically complete if and only if $R$ is an adic topological ring in the $I$-adic topology.

As a special case, note that the discrete topology is the $0$-adic topology and that any ring in the discrete topology is adic.

Lemma 15.35.2. Let $\varphi : R \to S$ be a ring map. Let $I \subset R$ and $J \subset S$ be ideals and endow $R$ with the $I$-adic topology and $S$ with the $J$-adic topology. Then $\varphi $ is a homomorphism of topological rings if and only if $\varphi (I^ n) \subset J$ for some $n \geq 1$.

**Proof.**
Omitted.
$\square$

Lemma 15.35.3 (Baire category theorem). Let $M$ be a topological abelian group. Assume $M$ is linearly topologized, complete, and has a countable fundamental system of neighbourhoods of $0$. If $U_ n \subset M$, $n \geq 1$ are open dense subsets, then $\bigcap _{n \geq 1} U_ n$ is dense.

**Proof.**
Let $U_ n$ be as in the statement of the lemma. After replacing $U_ n$ by $U_1 \cap \ldots \cap U_ n$, we may assume that $U_1 \supset U_2 \supset \ldots $. Let $M_ n$, $n \in \mathbf{N}$ be a fundamental system of neighbourhoods of $0$. We may assume that $M_{n + 1} \subset M_ n$. Pick $x \in M$. We will show that for every $k \geq 1$ there exists a $y \in \bigcap _{n \geq 1} U_ n$ with $x - y \in M_ k$.

To construct $y$ we argue as follows. First, we pick a $y_1 \in U_1$ with $y_1 \in x + M_ k$. This is possible because $U_1$ is dense and $x + M_ k$ is open. Then we pick a $k_1 > k$ such that $y_1 + M_{k_1} \subset U_1$. This is possible because $U_1$ is open. Next, we pick a $y_2 \in U_2$ with $y_2 \in y_1 + M_{k_1}$. This is possible because $U_2$ is dense and $y_2 + M_{k_1}$ is open. Then we pick a $k_2 > k_1$ such that $y_2 + M_{k_2} \subset U_2$. This is possible because $U_2$ is open.

Continuing in this fashion we get a converging sequence $y_ i$ of elements of $M$ with limit $y$. By construction $x - y \in M_ k$. Since

\[ y - y_ i = (y_{i + 1} - y_ i) + (y_{i + 2} - y_{i + 1}) + \ldots \]

is in $M_{k_ i}$ we see that $y \in y_ i + M_{k_ i} \subset U_ i$ for all $i$ as desired.
$\square$

Lemma 15.35.4. With same assumptions as Lemma 15.35.3 if $M = \bigcup _{n \geq 1} N_ n$ for some closed subgroups $N_ n$, then $N_ n$ is open for some $n$.

**Proof.**
If not, then $U_ n = M \setminus N_ n$ is dense for all $n$ and we get a contradiction with Lemma 15.35.3.
$\square$

Lemma 15.35.5 (Open mapping lemma). Let $u : N \to M$ be a continuous map of linearly topologized abelian groups. Assume that $N$ is complete, $M$ separated, and $N$ has a countable fundamental system of neighbourhoods of $0$. Then exactly one of the following holds

$u$ is open, or

for some open subgroup $N' \subset N$ the image $u(N')$ is nowhere dense in $M$.

**Proof.**
Let $N_ n$, $n \in \mathbf{N}$ be a fundamental system of neighbourhoods of $0$. We may assume that $N_{n + 1} \subset N_ n$. If (2) does not hold, then the closure $M_ n$ of $u(N_ n)$ is an open subgroup for $n = 1, 2, 3, \ldots $. Since $u$ is continuous, we see that $M_ n$, $n \in \mathbf{N}$ must be a fundamental system of open neighbourhoods of $0$ in $M$. Also, since $M_ n$ is the closure of $u(N_ n)$ we see that

\[ u(N_ n) + M_{n + 1} = M_ n \]

for all $n \geq 1$. Pick $x_1 \in M_1$. Then we can inductively choose $y_ i \in N_ i$ and $x_{i + 1} \in M_{i + 1}$ such that

\[ u(y_ i) + x_{i + 1} = x_ i \]

The element $y = y_1 + y_2 + y_3 + \ldots $ of $N$ exists because $N$ is complete. Whereupon we see that $x = u(y)$ because $M$ is separated. Thus $M_1 = u(N_1)$. In exactly the same way the reader shows that $M_ i = u(N_ i)$ for all $i \geq 2$ and we see that $u$ is open.
$\square$

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