# Notes (part 1): S. Bosch, U Güntzer and R. Remmert, Non-Archimedean Analysis

## Semi-normed and normed groups

$$\newcommand{\mmm}{\mathfrak{m}} \newcommand{\qqq}{\mathfrak{q}} \newcommand{\rr}{\mathbb{R}} \newcommand{\rnneg}{\rr_{\geq 0}} \newcommand{\rpos}{\rr_{> 0}} \newcommand{\zz}{\mathbb{Z}} \newcommand{\znneg}{\zz_{\geq 0}}$$All groups are going to abelian with the group operation denoted by “+”. A filtration, a generalization of valuation, is a function $$\nu$$ from a group $$G$$ to $$\rr \cup \{\infty\}$$ such that

• $$\nu(0) = \infty$$
• $$\nu(x-y) \geq \min\{\nu(x), \nu(y)\}$$

Filtrations are in one-to-one correspondence with ultrametric functions or semi-norms, i.e. functions $$|\cdot|: G \to \rnneg$$ such that

• $$|0| = 0$$
• $$|x-y| \leq \max\{|x|, |y|\}$$

A semi-norm $$|\ |$$ is called a norm if $$\ker(|\ |) = 0.$$ The correspondence between filtrations and semi-norms is given by

• $$|\cdot| = \alpha^{-\nu(\cdot)}$$
• $$\nu(\cdot) = -\log_\alpha|\cdot|$$

where $$\alpha$$ is any fixed real number greater than $$1.$$ There are natural subgroups \begin{align*} G^0 & := \{g: \nu(g) \geq 0\} = \{g: |g| \leq 1\} \\ G^{\vee} &:= \{g: \nu(g) > 0\} = \{g: |g| < 1\} \end{align*}

Elementary properties of a semi-norm $$|\cdot|$$:

• $$|-y| = |y|$$
• $$|x+y| \leq \max\{|x|, |y|\}$$
• $$|x+y| = \max\{|x|, |y|\}$$ if $$|x| \neq |y|$$

A semi-norm $$|\ |$$ defines a pseudometric topology on $$G$$ via the distance function
$d(x,y) := |x-y|$
This makes $$G$$ into a topological group, i.e. the group operation $$(x,y) \mapsto x – y$$ is continuous.

$$G$$ is Hausdorff if and only if $$|\ |$$ is a norm.

Proposition 1. Every open subgroup of a topological group is also closed.

Proof. For any $$g \in G$$, the map $$x \mapsto x + g$$ is continuous and has a continuous inverse, so that it is a homeomorphism. Now if $$H$$ is a subgroup of $$G$$ and $$g \not\in H,$$ then $$g + H \subseteq G \setminus H.$$ If $$H$$ is open, then it follows that $$g + H$$ is an open neighborhood of $$g$$ completely contained in the complement of $$H,$$ so that $$H$$ is also closed.

Given a point $$a$$ on a semi-normed group $$G$$ and $$r > 0,$$ let
\begin{align*} B^0(a,r) & := \{x \in G: |x – a| \leq r\} \\ B^{\vee} (a,r) & := \{x \in G: |x – a| < r\} \end{align*}
respectively be the closed and the open ball of radius $$r$$ centered at $$a.$$ Note that

• $$B^\vee(a,r)$$ is open and $$B^0(a,r)$$ is closed in $$G$$ by definition of the topology on $$G,$$
• $$B^0(a,r) = a + G^0(r)$$ and $$B^{\vee}(a,r) = a + G^{\vee}(r)$$ where
\begin{align*} G^0(r) & := \{x \in G: |x| \leq r\} \\ G^{\vee}(r) & := \{x \in G: |x| < r\} \end{align*}
• The ultrametric inequality implies that
• $$x + G^\vee(r) = G^\vee(r)$$ for each $$x \in G^\vee(r)$$
• $$x + G^\vee(r) \subseteq x + G^0(r) = G^0(r)$$ for each $$x \in G^0(r)$$

These observations immediately implies the following results:

Proposition 2. Every ball $$B^0(a,r)$$ and $$B^\vee(a,r)$$ of positive radius $$r$$ is both open and closed in $$G.$$ Every sphere $$S(a,r) := \{x \in G: |x-a| = r\}$$ of positive radius $$r$$ is both open and closed in $$G.$$

Corollary 3. Every normed group $$G$$ is totally disconnected (i.e. the path components are precisely the singletons) in the ultrametric topology.

Given a subgroup $$H$$ of a semi-normed group $$G$$ and $$a \in G,$$ the distance from $$a$$ to $$H$$ is by definition:
$|a, H| := \inf_{y \in H} |a + y|$
The following observation seems to be important:

Proposition 3 (Section 1.1.4, Proposition 2). Let $$G$$ be a normed group and $$H$$ be a subgroup of $$G$$ which is “$$\epsilon$$-dense” for some nonnegative real number $$\epsilon < 1$$ in the sense that for all $$g \in G$$ there is $$h \in H$$ such that $$|g+h| \leq \epsilon|g|.$$ Then $$H$$ is dense in $$G.$$

Proof. Since $$G$$ is normed, it suffices to show that $$|x,H| = 0$$ for all $$x \in H.$$ If $$\epsilon = 0$$ it is obvious. So assume $$\epsilon > 0$$ and, in order to proceed by contradiction, that $$|x,H| > 0.$$ Since $$\epsilon < 1,$$ there is $$h \in H$$ such that $$|x+h| < \epsilon^{-1}|x,H|.$$ Then by the $$\epsilon$$-density, there is $$h’ \in H$$ such that
$|x+h+h’| \leq \epsilon|x+h| < |x,H|$
which gives the required contradiction and proves the proposition.

A subgroup $$H$$ of a normed group $$G$$ is called strictly closed if for each $$a \in G$$ there is $$y \in H$$ such that $$|a,H| = |a + y|.$$ Since in a normed space the a point is in the closure of a subset if and only if its distance from the subset is zero, it follows that a strictly closed subgroup is also closed. The ultrametric inequality implies that the ball groups $$G^0(r)$$ and $$G^\vee(r)$$ are strictly closed: we can take $$y = a$$ if $$a \in H$$ and $$y = 0$$ if $$a \not\in H$$ (where $$H$$ is either $$G^0(r)$$ or $$G^\vee(r)$$.

Proposition 4 (Section 1.1.5, Proposition 4). Let $$G$$ be a normed group and $$H$$ be a closed subgroup of $$G$$ such that $$|H \setminus \{0\}|$$ is discrete in $$\rpos.$$ Then $$H$$ is strictly closed.

Proof. Given $$x \in G,$$ we need to produce $$y \in H$$ such that $$|x+y| = |x,H|.$$ We may assume $$x \not \in H.$$ Since $$H$$ is closed, it follows that $$\epsilon := |x,H| > 0.$$ Pick $$\delta > 0$$. It suffices to show that $$V := \{|x+y’|: y’ \in H,\ \epsilon <|x+y’| < \epsilon + \delta\}$$ is finite. Indeed, if $$y’$$ is as above, then since $$|x+y’| > |x,H|,$$ there is $$y” \in H$$ such that $$|x+y’| > |x+y”|.$$ But then $$|y’-y”| = |x+y’|,$$ so that $$V \subseteq |H \setminus \{0\}|.$$ The discreteness assumption on $$|H \setminus \{0\}|$$ then implies that $$V$$ is finite, as required.

The functor $$G \to G^\sim$$: $$G^\sim := G^0/G^\vee.$$

Convergence: in a complete normed group a sum $$\sum_{k \geq 0} a_k$$ turns out to be convergent if and only if $$|a_k| \to 0$$ as $$k \to \infty.$$ It then turns out that if $$\sum_k a_k$$ converges, then $$\sum_k a_{\pi_k}$$ converges to the same limit for every bijection $$\pi: \znneg \to \znneg.$$

## Semi-normed, normed and valued rings

A pair $$(A, |\ |)$$, where $$A$$ is a commutative ring with identity, is called a semi-normed (respectively, normed) ring if

1. $$|\ |$$ is a semi-norm (respectively, norm) on the additive group $$A^+$$ of $$A$$. (Note: $$A^+$$ is simply $$A$$ considered an abelian group with respect to $$+$$.)
2. $$|xy| \leq |x||y|$$ for each $$x, y \in A$$
3. $$|1| \leq 1$$

$$|\ |$$ is called a valuation and $$A$$ is called a valued ring if

• $$|\ |$$ is a norm on $$A,$$ and
• property 2 above is always satisfied with an equality.

Note that for a valuation, property 3 above is redundant: it follows from the second property with equality.

## Strictly convergent power series

Let $$(A, |\ |)$$ be a semi-normed ring. A formal power series $$\sum_{k \geq 0} a_kX^k \in A[[X]]$$ is strictly convergent if $$\lim_{k \to \infty} |a_k| = 0.$$ The set of all strictly convergent power series over $$A$$ is denoted by $$A\langle X \rangle.$$ The Gauss norm on $$A\langle X \rangle,$$ which we also denote as $$|\ |,$$ is defined as:
$|\sum_{k \geq 0} a_k X^k| := \max_k |a_k|$

## Normed and faithfully normed modules

Let $$(A, |\ |)$$ be a normed ring. A pair $$(M, |\ |)$$ is normed $$A$$-module if

1. $$(M, |\ |)$$ is a normed group (with respect to addition in $$M$$), and
2. $$|ax| \leq |a||x|$$ for all $$a \in A,$$ $$x \in M.$$

If in addition $$A$$ is a valued ring and the second property above always holds with equality, then we say that $$(M, |\ |)$$ is a faithfully normed $$A$$-module.

Remark. If $$M \neq 0,$$ then the condition that “$$A$$ is a valued ring” in the definition of a “faithful norm” is unnecessary: indeed picking $$x \neq 0$$ in $$M,$$ so that $$|x| > 0.$$ Now for all $$a_1, a_2 \in A,$$ the chain of equalities
$|a_1a_2||x| = |a_1a_2x| = |a_1||a_2x| = |a_1||a_2||x|$
implies that $|a_1a_2 = |a_1||a_2|$

## Examples

1. Non-closed ideals.
1. A (non-complete) Noetherian example. Let $$A := k[x]$$ where $$k$$ is a field equipped with a nontrivial valuation. Consider the topology on $$A$$ induced by the Gauss norm.
Lemma. For each $$a \in K,$$ the ideal $$\mmm_a := \langle x – a \rangle$$ is closed in $$A$$ if and only if $$|a| \leq 1.$$
Proof. Since $$1 – x^n/a^n \in \mmm_a$$ for $$a \in K \setminus 0,$$ it follows that $$|1, \mmm_a| \leq |a|^{-n}.$$ Therefore if $$|a| > 1,$$ then $$|1, \mmm_a| = 0,$$ so that $$\mmm_a$$ is not closed. On the other hand, if $$|a| \leq 1,$$ then we claim that
$|1 + (x-a)f| \geq 1$
for all $$f.$$ Indeed, write $$f = \sum_{j=0}^n c_jx^j$$ so that
$|1 + (x-a)f| = \max\{|1 – ac_0|, |c_0 – ac_1|, \ldots, |c_{n-1} – ac_n|, |c_n|\}$
If $$|1-ac_0| < |1| = 1,$$ then one must have $$|ac_0| = |1| = 1,$$ so that $$|c_0| = 1/|a| \geq 1.$$ Continuing with $$|c_0 – ac_1|$$ and so on, it follows that if each of the terms $$|c_0 – ac_1|, \ldots, |c_{n-1} – ac_n|$$ is smaller than $$1$$, then $$|c_n| = |a|^{-n}$$ so that $$|1 + (x-a)f| = |c_n| = |a|^{-n} \geq 1.$$ It follows that $$1$$ is not in the closure $$\bar \mmm_a$$ of $$\mmm_a.$$ Since $$\bar \mmm_a$$ is an ideal of $$A$$ and $$\mmm_a$$ is a maximal ideal of $$A,$$ this implies that $$\bar \mmm_a = \mmm_a,$$ as required.
2. Contrast the above with the following:
1. If $$A$$ is a complete normed ring, then every maximal ideal is closed. This follows since if a maximal ideal $$\mmm$$ is not closed, then its closure, being an ideal, must contain $$1,$$ i.e. there are elements in $$\mmm$$ arbitrarily close to $$1.$$ However, since $$A$$ is complete, any element of the form $$1 + f$$ with $$|f| < 1$$ is invertible.
2. If $$A$$ is complete Noetherian normed ring, then every ideal is closed, at least if $$A$$ contains a field $$k$$ such that the restriction of the norm on $$k$$ is nontrivial. Indeed, then $$A$$ is a $$k$$-Banach algebra. Let $$\hat \qqq$$ be the completion of an ideal $$\qqq.$$ By Noetherianity there is a surjective $$A$$-module map $$\pi: A^n \to \hat \qqq$$ which is open due to the Open Mapping Theorem. Then $$\pi((A^\vee)^n)$$ is open in $$\hat \qqq.$$ Since $$\qqq$$ is dense in $$\hat \qqq,$$ it follows that $$\hat \qqq = \qqq + \pi((A^\vee)^n).$$ Then $$\hat \qqq = \qqq$$ by Nakayama’s lemma [BGR 1.2.4/6]. Thus $$\qqq$$ is complete, and therefore closed.
3. A quasi-Noetherian but complete example. Let $$A := k[[x_i: i \geq 1]]$$ be the ring of power series ring in infinitely many variables over a field $$k.$$ Then $$A$$ is a local ring with maximal ideal $$m$$ consisting of power series with zero constant term. Let $$m_0$$ be the ideal of $$A$$ generated by all $$x_i,$$ $$i \geq 1.$$ Then $$m_0 \subsetneq m,$$ since e.g. the element
$x_1 + x_2^2 + x_3^3 + \cdots$
can not be expressed as an $$A$$-linear combination of finitely many $$x_i.$$ However, the closure of $$m_0$$ is $$m,$$ e.g. with respect to the topology induced by valuation corresponding to the order of power series. This remains true if the topology on $$A$$ is induced by a weighted order such the weight of $$x_i$$ goes to $$\infty$$ as $$i \to \infty,$$ even though in that case $$A$$ is quasi-Noetherian, i.e. each ideal $$I$$ of $$A$$ has a (possibly infinite) sequence of elements $$a_i,$$ $$i \geq 1,$$ such that $$\lim_{i \to \infty}|a_i| = 0,$$ and in addition, every $$a \in I$$ can be expressed as a possibly infinite $$A$$-linear combination of the $$a_i.$$
2. “Unexpected” behaviour of residues: equivalent norm with transcendental residue. Take a normed ring $$A$$ such that there is real number $$\rho > 1$$ such that $$\rho^{-1} \not\in |A|.$$ (E.g. one can take $$A$$ to be a discrete valuation ring.) Take $$B :=A\langle X \rangle.$$ Let $$|\ |_1$$ be the Gauss norm on $$B$$ and define $$|\ |_2$$ on $$B$$ as:
$|\sum a_k X^k| := \max\{|a_0|, \rho|a_k|: k \geq 1\}$
Then $$|\ |_2$$ is also a norm (one needs $$\rho \geq 1$$ for the ultametric inequality to hold) and
$|\cdot|_1 \leq |\cdot|_2 \leq \rho|\cdot|_1$
so that $$|\ |_1$$ and $$|\ |_2$$ induce the same topology on $$B.$$ However,
$B^\sim_1 \cong A^\sim[X],\ \text{whereas}\ B^\sim_2 \cong A^\sim$
so that $$B^\sim_1$$ is transcendental over $$B^\sim_2.$$
3. “Unexpected” behaviour of residues: trivial residue. Take a cyclic/principal faithfully normed $$A$$-module $$M := Ax$$ such that $$|x|^{-1} \not\in |A|.$$ Then $$M^\vee = M^0,$$ so that $$M^\sim = 0.$$
4. A discrete valuation ring $$A$$ which is not Japanese, i.e. the field of fractions of $$A$$ has a finite algebraic extension $$K$$ such that the integral closure of $$A$$ in $$K$$ is not a finite $$A$$-module [F. K. Schmidt, Math. Z. 41, 443-450, 1936].
Take a prime $$p.$$ Set $$k := (\zz/p\zz)(t_0, t_1, t_2, \ldots)$$ where all $$t_i$$ are indeterminates. Take a new indeterminate $$T,$$ define
$f(T) := t_0 + t_1T + t_2T^2 + \cdots \in k[[T]]$
The lemma below implies that $$f$$ is not algebraic over $$k[T]$$, so that the homomorphism $$k[X,Y] \to k[[T]]$$ defined by $$x \mapsto T,$$ $$y \mapsto f(T)$$ is injective. Let $$K := k(X,Y)$$ and $$|\ |_f$$ be the valuation on $$K$$ induced by the standard valuation on $$k[[T]].$$ Let $$Q := k(X, Y^p) \subseteq K$$ and $$A$$ be the (discrete) valuation ring of the restriction of $$|\ |_f$$ to $$Q,$$ i.e.
$A := Q^0 := \{g \in Q: |g|_f \leq 1\}$
The following claim shows that the integral closure $$\bar A$$ of $$A$$ in $$K$$ is not a finite $$A$$-module, as required. The lemma used above is stated and proved after the claim.
Claim. $$\bar A = K^0 := \{g \in K: |g|_f \leq 1\}.$$ Moreover, $$K^0$$ is not a finite $$A$$-module.
Proof. Since $$char(K) = p > 0,$$ every $$g \in K^0$$ is a $$p$$-th root of some element in $$A.$$ On the other hand, if $$g \in K$$ satisfies an integral equation
$g^n + a_1g^{n-1} + \cdots + a_n = 0$
with $$a_i \in A$$ then one must have $$|g|_f \leq 1.$$ It follows that $$\bar A = K^0.$$ To prove the second assertion proceed by contradiction and assume $$e_1, \ldots, e_n \in K^0$$ generate $$K^0$$ as an $$A$$-module. Since $$1, Y, \ldots, Y^{p-1}$$ generate $$K$$ as a vector space over $$Q$$, there are expressions
$e_i = \sum_{j=0}^{p-1}c_{ij}Y^j,\ c_{ij} \in Q$
Since $$|X|_f < 1,$$ there is $$s \geq 1$$ such that $$|c_{ij}X^s|_f \leq 1$$ for each $$i,j.$$ Then each $$X^se_i$$ is in the $$A$$-module $$B$$ generated by $$1, Y, \ldots, Y^{p-1}.$$ It follows that
$X^s K^0 \subseteq B$
Let
$Z_s := X^{-(s+1)}(Y – t_0 – t_1X – \cdots – t_sX^s) \in K.$
Then $$Z_s \in K^0$$ and therefore
$X^sZ_s = X^{-1}(Y-t_0) – (t_1 + t_2X + \cdots + t_sX^{s-1}) \in B$
Since $$t_1 + t_2X + \cdots + t_sX^{s-1} \in A \subseteq B,$$ it follows that
$X^{-1}(Y – t_0) \in B = A + AY + \cdots + AY^{p-1}$
But this is impossible since $$X^{-1} \not\in A,$$ and $$1, Y, \ldots, Y^{p-1}$$ is a vector space basis of $$K$$ over $$Q.$$ This contradiction finishes the proof of the claim.
Lemma. Let $$k$$ be a field and $$k_0$$ be the prime field of $$k.$$ Let $$f := \sum_i c_i T^i \in k[[T]],$$ where $$T$$ is an indeterminate, be such that $$k_0(c_0, c_1, c_2, \ldots)$$ has infinite transcendence degree over $$k_0.$$ Then $$f$$ is not algebraic over $$k[T].$$
Proof. Assume to the contrary that there is a nonzero polynomial $$q(T,W) \in K[T,W]$$ such that $$q(T, f(T)) = 0.$$ Let $$k’$$ be the subfield of $$k$$ generated by all coefficients of $$q(T,W).$$ We claim that each coefficient $$c_n$$ of $$f$$ is algebraic over $$k’.$$ We proceed by induction on $$n$$. Write $$q = T^sq_0$$ where $$T$$ does not divide $$q_0(T,W).$$ Then $$q_0(T, f(T)) = 0,$$ and in addition, $$q_0(0,W)$$ is a nonzero polynomial over $$k’.$$ Since $$q(0,f(0)) = q(0,c_0) = 0,$$ it follows that $$c_0$$ is algebraic over $$k’,$$ so that the claim holds for $$n = 0.$$ For $$n \geq 1,$$ define
$q'(T,W) := q(T, c_0 + c_1T + \cdots + c_{n-1}T^{n-1} + WT^n) \in k'(c_0, \ldots, c_{n-1})[T,W]$
Since $$\deg_W(q’) \geq 1$$ it is straightforward to check by looking at the highest degree term in $$W$$ that
$\deg_W(q’) = \deg_W(q) > 0.$
In particular, $$q’$$ is a nonzero element in $$k'(c_0, \ldots, c_{n-1})[T,W]$$ such that
$q'(T, c_n + c_{n+1}T + \cdots) = q(T, f(T)) = 0$
The $$n = 0$$ case of the claim then implies that $$c_n$$ is algebraic over $$k'(c_0, \ldots, c_{n-1}),$$ so that $$c_n$$ is algebraic over $$k’$$ due to the induction hypothesis. This proves the claim and consequently, the lemma.
5. A bounded (in particular, continuous) bijective $$A$$-module map may not have a continuous inverse. Pick a field $$K$$ with a non-complete valuation. Pick the completion $$\hat K$$ of $$K$$ and take $$x \in \hat K \setminus K.$$ The map $$\phi: K^2 \to V := K + Kx$$ given by $$(a,b) \mapsto a + bx$$ is a continuous (in fact, bounded) and bijective $$K$$-vector space map. However, $$K$$ is not closed in $$V$$ so that $$\phi$$ is not an homeomorphism; in particular
1. $$\phi^{-1}:V \to K^2$$ is a $$K$$-vector space isomorphism which is not continuous (let alone, bounded)! Indeed, choose $$a_s \in K$$ such that $$a_s \to x$$ as $$s \to \infty.$$ Then $$a_s – x \to 0$$ but $$|\phi^{-1}(a_s -x)| = |(a_s, -1)| \geq 1.$$
2. Contrast the above with the following:
1. Every $$K$$-linear map from $$K^n$$ to a $$K$$-vector space is continuous.
3. Also note that $$\phi^{-1}$$ is not continuous, even though $$\ker(\phi^{-1}) = \{0\}$$ is closed in $$V.$$ In particular, the following fact does not always hold if $$K$$ is replaced by $$K^n$$ for $$n \geq 2$$: given a $$K$$-linear map $$\lambda: V \to K,$$ where $$V$$ is a normed $$K$$-vector space, if $$\ker(\lambda)$$ is closed, then $$\lambda$$ is bounded. (Indeed, if $$x \in V \setminus \ker(\lambda),$$ then $$\alpha := |x, \ker(\lambda)| > 0,$$ and the $$K$$-faithfulness of the norm implies that $$|ax, \ker(\lambda)| = |a|\alpha$$ for all $$a \in K.$$ Then for all $$y \in V \setminus \ker(\lambda),$$ writing $$y = ax + z$$ with $$z \in \ker(\lambda)$$ one has that $$|y| \geq |a|\alpha$$ so that $$|\lambda(y)| = |a||x| \leq (|x|/\alpha)|y|.$$)
4. In this example, define $$\lambda: V \to K$$ as
$\lambda(a + bx) := b$
Then $$\ker(\lambda) = K$$ is not closed in $$V.$$ Does it imply that $$V \to V/\ker(\lambda)$$ is not continuous? Well, the question is more basic. What is the topology to put on $$V/\ker(\lambda)$$? Since $$\ker(\lambda) = K$$ is dense in $$V,$$ it follows that in the residue topology every point has zero norm! So in this topology the open sets are only the whole space and the empty set. So the (surjective) map $$V/\ker(\lambda) \to K$$ is not continuous with respect to that semi-norm.
6. For every ring $$A$$ equipped with a degenerate valuation. there are $$A$$-module maps which are homeomorphisms (i.e. in particular, continuous), but not bounded. Let $$M_i, := Ae_i$$ $$i = 1, 2, \ldots,$$ be a sequence of free cyclic $$A$$-modules. Set $$M := \bigoplus M_i.$$ For each $$m \in \zz,$$ define an $$A$$-module norm $$|\ |_m$$ on $$M$$ by
$| \sum_{i=1}^n a_ie_i|_m := \max\{i^{-m}|a_i|\}$
Since $$|\ |_m \leq |\ |_{m-1},$$ the identity map $$(M, |\ |_{m-1}) \to (M, |\ |_m)$$ is bounded and continuous. The inverse map
$\iota_m: (M, |\ |_{m}) \to (M, |\ |_{m-1})$
is not bounded (look at the norms of $$e_i$$ for $$i \gg 1$$). However, if $$A$$ is a valued ring with a degenerate valuation, then there is $$m$$ such that $$\iota_m$$ is continuous. Indeed, in that case there are two possibilities:
1. If $$|\ | \geq 1$$ on $$A \setminus \{0\}$$ then whenever $$m \leq 0,$$ one has $$|\ |_m \geq 1$$ on $$M \setminus \{0\},$$ so that $$|\ |_m$$ induces the discrete topology on $$M$$ and every map from $$(M, |\ |_m)$$ is continuous.
2. If $$|\ | \leq 1$$ on $$A,$$ then $$|a_i| \leq |a_i|^{1/m}$$ for $$m \geq 1.$$ So that for $$m \geq 2$$
$i^{-m}|a_i| < \epsilon^m \Rightarrow |a_i|^{1/m} < i\epsilon \Rightarrow |a_i| < i^{m-1} \epsilon \Rightarrow |a_ie_i|_{m-1} < \epsilon$
which implies that $$\iota_m$$ is continuous, as claimed.
7. Contrast the preceding example with the following fact [BGR, Proposition 2.1.8/2]: if $$A$$ is equipped with a non-degenerate valuation, then for $$A$$-linear maps between faithfully normed $$A$$-modules, continuity is equivalent to boundedness. The key observation is [BGR, Proposition 2.1.8/1] which says that if the valuation of $$A$$ is non-degenerate, then for each faithfully normed $$A$$-module $$M$$ there is a fixed $$\rho > 1$$ such that for each $$x \in M\setminus \{0\},$$ there is $$c \in A$$ such that $$1 \leq |cx| < \rho.$$