For our further purposes, we often use inverse limit to define topology on a commutative ring $A$.
Definition. Let $A$ be a topological group (ring), we say $A$ is linearly topologized group (ring, resp.) if $0$ has a fundamental system of open neighborhoods $\mathscr{N}$ consisting of subgroups (ideals, resp.) of $A$. Such a fundamental system is called linearly topologized fundamental system. And we denote $(A, \mathscr{N})$ for the topological group (ring) $A$ with linearly topologized fundamental system $\mathscr{N}$.
The following lemma is quite useful to detect when are two linearly topoligized fundamental system generate the same topology.
Lemma 2.1. If $A$ is an abelian group (commutative ring) and $\mathscr{N}, \mathscr{M}$ are two sets containing subgroups of $A$ satisfying conditions of Proposition 1.3 (Proposition 1.6, respectively), then they induces the same topology on $A$ iff for any $I\in \mathscr{N}$, there exists some $J\in \mathscr{M}$ such that $J\subset I$, and vice versa.
Proof. It follows directly from our construction in Proposition 1.3, since they generate the same base for the topology on $A$. (Q.E.D)
By the following lemma, it makes sense to give the definition of the completion of a linearly topologized ring $A$.
Lemma 2.2. Let $A$ be an abelian group (commutative ring), $\mathscr{M}, \mathscr{N}$ are sets of subgroups of $A$, such that for every $G\in \mathscr{M}$, there exists $H_G\in \mathscr{N}$, such that $G\subset H_G$, and for every $H\in \mathscr{N}$, there exists $G_H\in \mathscr{M}$ such that $G_H\subset H$. Then there is a canonical isomorphism
$$\varprojlim_{G\in \mathscr{M}}A/G\cong \varprojlim_{H\in \mathscr{N}}A/H$$
Proof. For any $G\subset H$: subgroups of $A$, we denote $\psi^G_H$ the canonical map $A/G\to A/H$. Let $(a_G\mod G)_{G\in \mathscr{M}}\in \varprojlim_{G\in \mathscr{M}} A/G$, for $a_G\in A$, due to the assumption, for any $H\in \mathscr{N}$, there is $G_H\in \mathscr{M}$ such that $G_H\subset H$, and we can define a map
$$\psi: \varprojlim_{G\in \mathscr{M}}A/G\to \varprojlim_{H\in \mathscr{N}}A/H$$
by $\psi((a_G\mod G)_G)=(\psi^{G_H}_H(a_{G_H}))_H=(a_{G_H}\mod H)_H$. It is well-defined, since $(\psi^{G_H}_H)$ is compatible with the inverse system $\varprojlim_{H\in \mathscr{M}}A/H$.
Conversely, for any $G\in \mathscr{N}$, there exists $H_G\in \mathscr{M}$, such that $H_G\subset G$, and one can also define a map
$$\theta: \varprojlim_{H\in \mathscr{N}}A/H\to \varprojlim_{G\in \mathscr{M}}A/G$$
as follows $\theta(b_H\mod H)_H=(b_{H_G}\mod G)_{G\in \mathscr{N}}$, and we have the composition $\theta\circ \psi$ would be
$$(a_G\mod G)_G\mapsto (a_{G_H}\mod H)_H\mapsto (a_{G_{H_G}}\mod G)_G$$
We have $G_{H_G}\subset H_G\subset G$, and because we are in inverse systems, we have
$$a_{G_{H_G}}\mod G_{H_G}=a_G\mod G_{H_G}$$
And that means $a_{G_{H_G}}-a_G\in G_{H_G}\subset G$. So, $a_{G_{H_G}}\mod G=a_G\mod G$, and $\theta\circ\psi$ is just the identity map. It is similar to show that $\psi\circ\theta$ is also the identity map. And this yields the desired isomorphism. (Q.E.D)
Via Lemma 2.1 and Lemma 2.2 we have this
Definition. Let $(A, \mathscr{N})$ be a topological group (ring), we say that $A$ is complete if the canonical map $A\to \lim_{I\in \mathscr{N}}A/I$ is surjective.
We can also obtain an algebraic characterization of Hausdorff on a linearly topologized ring $A$.
Proposition 2.3. Let $(A, \mathscr{N})$ be a topological group (ring), then $A$ is Hausdorff iff the canonical map
$$\phi: A\to \varprojlim_{I\in \mathscr{N}} A/I$$
is injective. Also, $A$ is Hausdorff iff $\cap_{I\in \mathscr{N}} I=0$.
Proof. Assume $\phi$ is injective, but $A$ is not Hausdorff, i.e. there exists $0\ne a\in A$, such that all open neighborhoods at $0$ contain $a$. But then, it is a contradiction, since $\phi(a)=0$, and $\phi$ is not injective.
Conversely, assume that $A$ is Hausdorff, and $\phi$ is not injective, that means, there exists $0\ne a\in A$ and $a\in I(\forall I\in \mathscr{N})$. Due to the definition of a fundamental system, if we take any open neighborhood $U$ of 0, then there exists $I\in \mathscr{N}$, and $I\subset U$, and $I$ contains both $0$ and $a$, and hence, so does $U$. That means, $A$ is not Hausdorff, a contradiction.
For the second statement, assume that
$$\phi: A\to \varprojlim_{I\in \mathscr{N}}A/I$$
is injective, and $0\ne a\in \cap_{I\in \mathscr{N}}I$, then $\phi(a)=0$, so $\phi$ is not injective, a contradiction. Conversely, assume that $\cap _{I\in \mathscr{N}}I=0$, and $\phi$ is not injective, i.e. there exists $0\ne a\in A$ such that $\phi(a)=0$. That means $a\in I(\forall I\in \mathscr{N})$, a contradiction. We hence obtain the statements. (Q.E.D)
Remark 2.4. This follows directly from the definition that if $(A, \mathscr{N})$ is Hausdorff then if any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$, then the convergence is unique.
Proof. Assume that $A$ is Hausdorff, and $(x_n)_n$ is a Cauchy sequence converges in $A$ to $x, x'$, then it is clear that for all $I\in \mathscr{N}$, $x-x'\in I$. But then, since $\cap_{I\in \mathscr{N}}I=0$, we have $x=x'$. (Q.E.D)
For the converse of the remark above, we need to have a more special linearly topologized fundamental system. Let $A$ be an abelian group (commutative ring), and $I_1 \supset I_2\supset ...$ is a chain of ideals in $A$. We can see that $\mathscr{N}:=\{I_1,I_2,...\}$ satisfies conditions Proposition 1.6. Hence, there exists a unique topology on $A$ such that $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$. We say such $\mathscr{N}$ a (countable) filtered fundamental system.
Remark 2.5. The converse of Remark 2.4 holds for a topological group (ring) $(A, \mathscr{N})$, where $\mathscr{N}$ is a filtered fundamental system. That means, in this case, $(A, \mathscr{N})$ is Hausdorff if and only if for any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$, then the convergent value is unique.
Proof. For the converse, assume that $A$ is not Hausdorff, i.e. there exists $0\ne a\in \cap_{I\in \mathscr{N}}I$, then we can choose a sequence $(x_n)_n$ with $x_n\in I_n$, where $I_n\in \mathscr{N}$. This sequence is Cauchy, since we always have $x_{n+1}-x_n\in I_n$, for all $n$. This follows that both $a\ne -a$ are convergent values of $(x_n)_n$, a contradiction. (Q.E.D)
Lemma 2.5. Let $(A, \mathscr{N})$ be a topological group (ring), where $\mathscr{N}:=\{(I_n)_n\}$ is a filtered fundamental system, then the canonical map
$$\phi: A\to \varprojlim_n A/I_n$$
is surjective iff $A$ is complete with respect to $\mathscr{N}$.
Proof. Assume that $\phi: A\to \varprojlim_n A/I_n$ is surjective, and $(x_n)_n$ is a Cauchy sequence. Then for any $k\ge 1$, there exists $n_k$ such that $n_{k+1}> n_{k}$ and for all $m,n\ge n_k$, we have $x_m-x_n\in I_k$. We can define $y_k:=x_{n_k}$, then $(y_k)_k$ is an element in $\varprojlim_k A/I_k$, because $y_{k+1}-y_k=x_{n_{k+1}}-x_{n_k}\in I_k$. By the surjectivity of $\phi$, there exists $y\in A$ such that $\phi(y)=(y_k)_k$, i.e. $y\mod I_k=y_k, \forall k$ and we can see easily that $y$ is a convergent value of $(y_k)_k$. This also follows that $y$ is a convergent value of $(x_n)_n$ because for any $k$, and for all $n\ge n_k$, $y-x_n=y-x_{n_k}+x_{n_k}-x_n\in I_k$.
Conversely, let $(x_n)_n\in \varprojlim_n A/I_n$ be any element, we can see that $x_{n+1}-x_n\in I_n$, and $(x_n)_n$ is a Cauchy sequence in $A$, and there exists $x\in A$ is a convergent value of $(x_n)_n$. Due to the definition, for every $n$, and there exists some $m_n$ such that for all $m\ge m_n$, $x-x_m\in I_n$. We can choose $m\ge \max\{n,m_n\}$, and $x-x_n=x-x_m+x_m-x_n\in I_n$. And this yields, for all $n$, $x-x_n\in I_n$. Hence, $\phi(x)=(x_n)_n$, and $\phi$ is surjective. (Q.E.D)
Via this lemma, for the case $A$ has a filtered fundamental system, we obtain an equivalence between the notions of sequential complete and complete. We conclude this section by a
Proposition 2.7. Let $A$ be an topological group (ring), and $I_1\supset I_2\supset ...$ a linearly topologized fundamental system around $0$. Then $A$ is Hausdorff and complete iff the canonical map
$$\phi: A\to \varprojlim_n A/I_n$$
is bijective.
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