# Survey on the Kakutani problem in p-adic analysis II

Alain Escassut

► To cite this version:

Alain Escassut. Survey on the Kakutani problem in p-adic analysis II. Sarajevo Journals of Mathematics, 2020, 16 (1), pp.55-70. hal-03017652

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Submitted on 21 Nov 2020

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# Survey on the Kakutani problem in p-adic analysis II

by Alain Escassut

## Abstract

Let $\mathbb{K}$ be a complete ultrametric algebraically closed field and let $A$ be the Banach $\mathbb{K}$-algebra of bounded analytic functions in the "open" unit disk $D$ of $\mathbb{K}$ provided with the Gauss norm. Let $\mathrm{Mult}(A, \| \cdot \|)$ be the set of continuous multiplicative semi-norms of $A$ provided with the topology of pointwise convergence, let $\mathrm{Mult}_m(A, \| \cdot \|)$ be the subset of the $\phi \in \mathrm{Mult}(A, \| \cdot \|)$ whose kernel is a maximal ideal and let $\mathrm{Mult}_1(A, \| \cdot \|)$ be the subset of the $\phi \in \mathrm{Mult}(A, \| \cdot \|)$ whose kernel is a maximal ideal of the form $(x-a)A$ with $a \in D$. By analogy with the Archimedean context, one usually calls *ultrametric Corona problem, or ultrametric Kakutani problem* the question whether $\mathrm{Mult}_1(A, \| \cdot \|)$ is dense in $\mathrm{Mult}_m(A, \| \cdot \|)$. In a previous paper, we have recalled the characterization of a large set of continuous multiplicative semi-norms and why the multibijectivity of the algebra $A$ would solve the Corona problem. Here we are going to prove that multibijectivity in the general case which will prove that $\mathrm{Mult}_1(A, \| \cdot \|)$ is dense in $\mathrm{Mult}_m(A, \| \cdot \|)$, beginning by the case when $\mathbb{K}$ is spherically complete and generalizing next.

2000 Mathematics subject classification: Primary 12J25 Secondary 46S10

Keywords: p-adic analytic functions, corona problem, multiplicative spectrum

## III. Special prime closed ideals of $A$.

**Notations:** Let $\mathbb{K}$ be an algebraically closed field complete with respect to an ultrametric absolute value $| \cdot |$. Given $a \in \mathbb{K}$ and $r > 0$, we denote by $d(a, r)$ the disk $\{x \in \mathbb{K} \mid |x-a| \le r\}$, by $d(a, r^{-})$ the disk $\{x \in \mathbb{K} \mid |x-a| < r\}$, by $C(a, r)$ the circle $\{x \in \mathbb{K} \mid |x-a| = r\}$ and set $D = d(0, 1^{-})$. Let $a \in D$. Given $r, s \in [0, 1]$ such that $0 < r < s$ we set $\Gamma(a, r, s) = \{x \in \mathbb{K} \mid r < |x-a| < s\}$.

Let $A$ be the $\mathbb{K}$-algebra of bounded power series converging in $D$ which is complete with respect to the Gauss norm defined as $\| \sum_{n=1}^{\infty} a_n x^n \| = \sup_{n \in \mathbb{N}} |a_n|$: we know that this norm actually is the norm of uniform convergence on $D$ [5].

We will denote by $B(\mathbb{N}, \mathbb{K})$ the $\mathbb{K}$-algebra of bounded sequences of $\mathbb{K}$. Let $S = (a_n)_{n \in \mathbb{N}}$ be a sequence in $D$ thinner than $W$. We will denote by $T_S$ the mapping from $A$ into $B(\mathbb{N}, \mathbb{K})$ which associates to each $f(x) = \sum_{n=0}^{\infty} a_n x^n$ the sequence $(f(a_n)_{n \in \mathbb{N}})$, we will denote by $\Sigma(S)$ the set of ultrafilters thinner than $S$ and by $\mathcal{I}(S)$ the ideal of the $f \in A$ such that $f(a_n) = 0 \forall n \in \mathbb{N}$.

Given $a \in \mathbb{K}$ and $r > 0$, we denote by $\Phi(a, r)$ the set of circular filters secant with $d(a, r)$ i.e. the circular filters of center $b \in d(a, r)$ and radius $s \in [0, r]$.

We denote by $\mathcal{W}$ the circular filter on $D$ of center 0 and diameter 1 and by $\mathcal{Y}$ the filter admitting for basis the family of sets of the form $\Gamma(0, r, 1) \setminus (\bigcup_{n=0}^{\infty} d(a_n, r_n^{-}))$ with $a_n \in D$, $r_n \le |a_n|$ and $\lim_{n \to \infty} |a_n| = 1$.

An ultrafilter $\mathcal{U}$ on $D$ will be called *coroner ultrafilter* if it is thinner than $\mathcal{W}$. Similarly, a sequence $(a_n)$ on $D$ will be called *a coroner sequence* if its filter is a coroner filter, i.e. if $\lim_{n \to +\infty} |a_n| = 1$.
Two coroner ultrafilters $\mathcal{F}$, $\mathcal{G}$ are said to be contiguous if for every subsets $F \in \mathcal{F}$, $G \in \mathcal{G}$ of $D$ the distance from $F$ to $G$ is null.

Let $\psi \in \mathrm{Mult}(A, \| \cdot \|)$ be different from $\| \cdot \|$. Then $\psi$ will be said to be coroner if its restriction to $\mathbb{K}[x]$ is equal to $\| \cdot \|$.

In [8] regular ultrafilters were defined: Let $(a_n)_{n \in \mathbb{N}}$ be a coroner sequence in $D$. The sequence is called a regular sequence if $\inf_{j \in \mathbb{N}} \prod_{n \in \mathbb{N}, n \neq j} |a_n - a_j| > 0$.

An ultrafilter $\mathcal{U}$ is said to be regular if it is thinner than a regular sequence. Thus, by definition, a regular ultrafilter is a coroner ultrafilter.

By Corollary (4.6) in [19], we have Theorem III.1:

**Theorem III.1:** Let $S$ be a sequence in $D$ thinner than $W$. Then $T_S$ is surjective on $B(\mathbb{N}, \mathbb{K})$ if and only if the sequence $S$ is regular.

**Notations:** Let $S$ be a regular sequence. Since $T_S$ is surjective, there exists a $\mathbb{K}$-algebra isomorphism $\Lambda_S$ from $A/Ker(T_S)$ onto $B(\mathbb{N}, \mathbb{K})$, where $Ker(T_S) = \mathcal{I}(S)$.

For every ultrafilter $\mathcal{G}$ on $\mathbb{N}$ we will denote by $\Theta(\mathcal{G})$ the ideal of $B(\mathbb{N}, \mathbb{K})$ consisting of sequences $(a_n)_{n \in \mathbb{N}}$ such that $\lim_{\mathcal{G}} a_n = 0$.

The following Theorem III.2 is classical:

**Theorem III.2:** $\Theta$ is a bijection from the set of ultrafilters on $\mathbb{N}$ onto $\mathrm{Max}(B(\mathbb{N}, \mathbb{K}))$. The restriction of $\Theta$ to the subset of non-principal ultrafilters on $\mathbb{N}$ is a bijection from this set onto the set of non-principal maximal ideals of $B(\mathbb{N}, \mathbb{K})$. Moreover, a maximal ideal of $B(\mathbb{N}, \mathbb{K})$ is principal if and only it is of codimension 1.

By Theorem 23 [11], we have this:

**Theorem III.3:** Let $S$ be a regular sequence and let $\mathcal{M}$ be a maximal ideal of $A$. The following two statements are equivalent:

(i) $\mathcal{I}(S) \subset \mathcal{M}$,

(ii) There exists an ultrafilter $\mathcal{U}$ thinner than $S$ such that $\mathcal{M} = \mathcal{J}(\mathcal{U})$.

Moreover, the mapping $\Psi$ which associates to each ultrafilter $\mathcal{U}$ thinner than $S$ the ideal $\mathcal{J}(\mathcal{U})$ is a bijection from $\Sigma(S)$ onto the set of maximal ideals of $A$ containing $\mathcal{I}(S)$.

**Proof:** Obviously, (ii) implies (i). Thus, suppose (i) true. Let $S = (a_n)_{n \in \mathbb{N}}$. By Theorem III.2, the isomorphism $\Lambda_S$ makes a bijection $\Psi$ from the set of maximal ideals of $A$ containing $\mathcal{I}(S)$ to the set of maximal ideals of $B(\mathbb{N}, \mathbb{K})$ and more precisely, it makes a bijection from the set of maximal ideals of $A$ of infinite codimension containing $\mathcal{I}(S)$ to the set of maximal ideals of $B(\mathbb{N}, \mathbb{K})$ of infinite codimension which actually are the non-principal maximal ideals of $B(\mathbb{N}, \mathbb{K})$. Let $\mathcal{N} = \Psi(M)$. By Theorem III.1, there exists an ultrafilter $\mathcal{U}$ on $\mathbb{N}$ such that $\mathcal{N}$ is the ideal of the bounded sequences tending to zero along $\mathcal{U}$. Now, let $\Xi$ be the natural bijection from the set of non-principal ultrafilters of $\mathbb{N}$ onto the set of ultrafilters thinner than $S$ and let $V = \Xi(\mathcal{U})$. Then $\mathcal{N} = \Lambda_S(M)$ hence $\mathcal{M} = \mathcal{I}(S)$. Moreover, in this way, we can see that $\Psi \circ \Xi^{-1}$ is a bijection from $\Sigma(S)$ onto the set of maximal ideals of $A$ containing $\mathcal{I}(S)$.
**Corollaire III.3.1:** If $\mathcal{U}$ is a regular ultrafilter on $D$, $\mathcal{J}(\mathcal{U})$ is a maximal ideal of $A$.

**Corollary III.3.2:** If $A$ is multbijective, then for every $\phi \in \mathrm{Mult}_m(A, \| \cdot \|)$ there exists a coroner ultrafilter $\mathcal{U}$ such that $\phi = \phi_{\mathcal{U}}$.

**Notation:** Let $F$ be a field, let $R$ be a commutative $F$-algebra with unity and let $\mathcal{D}$ be a derivation on $R$. Let $J$ be an ideal of $R$. We will denote by $\tilde{J}$ the set $\{f \in R \mid \mathcal{D}^{(n)}(f) \in J \ \forall n \in \mathbb{N}\}$.

**Theorem III.4:** There exist regular maximal ideals $\mathcal{M}$ of $A$ and $f \in \mathcal{M}$, having a sequence of zeros of order 1 and no other zeros, such that $f' \notin \mathcal{M}$, and such that $\tilde{M} \neq \{0\}$.

**Proof:** By Theorem II.13 we know that there exist bounded sequences $(a_n)_{n \in \mathbb{N}}$ in $D$ such that the sequence $\left| \frac{a_n}{a_{n+1}} \right|$ is strictly increasing and then the function $f(x) = \sum_{n=0}^{\infty} a_n x^n$ admits a sequence of zeros $(\alpha_n)_{n \in \mathbb{N}^*}$ satisfying $|\alpha_n| = \left| \frac{a_n}{a_{n+1}} \right|$. Thus, particularly, if we set $r_n = \left| \frac{a_n}{a_{n+1}} \right|$ then $f$ admits exactly a unique zero in each circle $C(0, r_n)$, each of order 1, and has no other zero in $D$. Consequently, by Lemma I.8, we can see that $|f'(\alpha_n)| = |f'|(r_n) \ \forall n \in \mathbb{N}^*$. Now, let $\mathcal{U}$ be an ultrafilter thinner than the sequence $(\alpha_n)_{n \in \mathbb{N}^*}$. On the other hand, we can check that the sequence $(\alpha_n)$ is regular, hence $\mathcal{U}$ is a regular ultrafilter. Consequently, by Corollary III.3.1, $\mathcal{J}(\mathcal{U})$ is a maximal ideal of $A$. Thus $f$ belongs to $\mathcal{J}(\mathcal{U})$ but $f'$ doesn't.

Now, for each $n \in \mathbb{N}$, let $u_n = \log(r_n)$. Then, by Lemma I.3, we have $\log(\|f\|) = -\sum_{n=0}^{\infty} u_n$. By Lemma I.2, there obviously exists an increasing sequence $(s_n)_{n \in \mathbb{N}}$ of $\mathbb{N}$ such that $\lim_{n \to \infty} s_n = +\infty$ and such that the series $\sum_{n=0}^{\infty} s_n u_n$ converges.

Now, by Corollary II.8.1, there exists $g \in A$ (not identically zero) such that for each $n \in \mathbb{N}$, $\alpha_n$ is a zero of $g$ of order $z_n \ge s_n$. And since $\lim_{n \to \infty} s_n = +\infty$, for every fixed $k \in \mathbb{N}$, we can see that $f^{(k)}(\alpha_n) = 0$ when $n$ is big enough, therefore $g^{(k)}$ belongs to $\mathcal{M}$. Consequently, $\tilde{\mathcal{M}}$ is not null, which ends the proof.

**Remark:** Theorem III.4 shows that ideals of the form $\tilde{\mathcal{M}}$ are not null and are not equal to $\mathcal{M}$ in general.

The following Theorem III.5 only concerns algebraic considerations but will be useful next:

**Theorem III.5:** Let $F$ be a field, let $R$ be a commutative $F$-algebra with unity and let $\mathcal{D}$ be a derivation on $R$. Let $J$ be an ideal of $R$ and let $J'$ be the set of $f \in J$ such that $\mathcal{D}f \in J$. Then $J'$ is an ideal of $R$. Let $\tilde{J}$ be the set of $f \in J$ such that $\mathcal{D}^n f \in J \ \forall n \in \mathbb{N}^*$. Then $\tilde{J}$ is an ideal of $R$ and $(\tilde{J}) = \tilde{J}$. Moreover, if $F$ is of characteristic 0 and if $J$ is prime, so are $J'$ and $\tilde{J}$.

**Notation:** On $A$ we shall apply this notation to the usual derivation of functions. Let $\psi \in \mathrm{Mult}(A, \| \cdot \|)$. Here we set $\mathrm{Subker}(\psi) = \mathrm{Ker}(\psi)$.

Since $\|f'\| \le \|f\|$ $\forall f \in A$, we can derive Corollary III.5.1:

**Corollary III.5.1:** Suppose $\mathbb{K}$ is of characteristic zero. Let $\mathcal{P}$ be a prime ideal of $A$. Then $P'$ and $\tilde{\mathcal{P}}$ are prime ideals of $A$ such that $(\tilde{\mathcal{P}}) = \tilde{\mathcal{P}}$. Moreover, if $\mathcal{P}$ is closed, so are $P'$ and $\tilde{\mathcal{P}}$.

**Corollary III.5.2:** Suppose $\mathbb{K}$ is of characteristic zero. Let $\psi \in \mathrm{Mult}(A, \| \cdot \|)$. Then $\mathrm{Subker}(\psi)$ is a prime closed ideal.
**Remark:** Suppose $\mathbb{K}$ is of characteristic 0. By Corollary III.5.1, given a prime closed ideal $P$, both $P'$ and $\tilde{P}$ are prime closed ideals. An interesting question is whether they are kernels of elements of $\mathrm{Mult}(A, \| \cdot \|)$. If one of them were the kernel of an element of $\mathrm{Mult}(A, \| \cdot \|)$, this would be a new kind of continuous multiplicative semi-norm. If none of them is the kernel of an element of $\mathrm{Mult}(A, \| \cdot \|)$, then they are new kinds of prime closed ideals which are not the kernel of an element of $\mathrm{Mult}(A, \| \cdot \|)$. Actually, we do not imagine what element of $\mathrm{Mult}(A, \| \cdot \|)$ might admit $P'$ or $\tilde{P}$ for kernel.

In order to prove Theorem III.8 and give a counterexample to Theorem III.5 when $K$ is of characteristic $p \neq 0$, we shall state Theorem III.6:

**Theorem III.6:** There exist regular maximal ideals $\mathcal{M}$ of $A$ and $f \in \mathcal{M}$, having a sequence of zeros of order 1 and no other zeros, such that $f' \notin \mathcal{M}$, and such that $\tilde{M} \neq \{0\}$.

**Proof:** By Theorem II.13 there exist bounded sequences $(a_n)_{n \in \mathbb{N}}$ in $D$ such that $a_0 = 1$ and such that the sequence $\left| \frac{a_n}{a_{n+1}} \right|$ is strictly increasing and then the function $f(x) = \sum_{n=0}^{\infty} a_n x^n$ admits a sequence of zeros $(\alpha_n)_{n \in \mathbb{N}^*}$ satisfying $|\alpha_n| = \left| \frac{a_n}{a_{n+1}} \right|$. Thus, particularly, if we set $r_n = \left| \frac{a_n}{a_{n+1}} \right|$, then by Theorem II.13 $f$ admits exactly a unique zero in each circle $C(0, r_n)$, each of order 1, and has no other zero in $D$. Consequently, by Lemma I.8, we can see that $|f'(\alpha_n)| = |f'(r_n) \forall n \in \mathbb{N}^*$. Now, let $\mathcal{U}$ be an ultrafilter thinner than the sequence $(\alpha_n)_{n \in \mathbb{N}^*}$. We can check that the sequence $(\alpha_n)_{n \in \mathbb{N}}$ is regular, hence $\mathcal{U}$ is a regular ultrafilter. Consequently, by Corollary III.3.1 $\mathcal{J}(\mathcal{U})$ is a maximal ideal of $A$. Now, by construction, $f$ belongs to $\mathcal{J}(\mathcal{U})$. But $\lim_{n \to +\infty} |f'(\alpha_n)| = \lim_{n \to +\infty} |f'(r_n) = \|f'\| \neq 0$, hence $f'$ does not belong to $\mathcal{J}(\mathcal{U})$.

Now, for each $n \in \mathbb{N}$, set $u_n = \log(r_n)$. By Lemma I.3, we have $\log(\|f\|) = - \sum_{n=0}^{\infty} u_n$. By Lemma I.2, there exists an increasing sequence $(s_n)_{n \in \mathbb{N}}$ of $\mathbb{N}$ such that $\lim_{n \to +\infty} s_n = +\infty$ and such that the series $\sum_{n=0}^{\infty} s_n u_n$ converges.

Now, by Theorem II.14, there exists $g \in A$ (not identically zero) such that for each $n \in \mathbb{N}$, $\alpha_n$ is a zero of $g$ of order $z_n \geq s_n$. And since $\lim_{n \to +\infty} s_n = +\infty$, for every fixed $k \in \mathbb{N}$, we can see that $f^{(k)}(\alpha_n) = 0$ when $n$ is big enough, therefore $g^{(k)}$ belongs to $\mathcal{M}$. Consequently, $\tilde{\mathcal{M}}$ is not null, which ends the proof.

**Lemma III.7:** Let $(a_n)_{n \in \mathbb{N}}$ be a regular sequence, let $\delta = \inf_{k \in \mathbb{N}} \prod_{n \neq k, n \in \mathbb{N}} |a_n - a_k|$ and let $\rho = \inf_{k \neq n, k \in \mathbb{N}} |a_n - a_k|$. Let $f \in A$ admit each $a_n$ as a zero of order 1 and have no other zero. Then $|f'(x)| \geq \|f\|_{\rho}^{\delta} \forall x \in \bigcup_{n=0}^{\infty} d(a_n, (\delta\rho)^{-})$.

**Proof:** Let us fix $t \in \mathbb{N}$, let $r = |a_t|$. Set $u = x - a_t$, $g(u) = f(x)$ and consider $|g|(\rho)$. Since $g$ has a unique zero in $d(0, \rho^{-})$ and admits all the the $a_n - a_t$ as zeros, by Lemma I.7 and I.8 we can check that $|g|(\rho) \geq \|g\|\delta = \|f\|\delta$. Inside $d(0, \rho^{-})$, $g(u)$ is of the form $b_1 u + \sum_{n=2}^{\infty} b_n u^n$ with $|b_1|\rho \geq |b_n|\rho^n \forall n \geq 2$. Consequently, $|g'(u)|\rho \geq \|f\|\delta$ Now $|g'(u)| = |b_1| = \frac{|g(u)|}{\rho} \forall u \in d(0, \rho^{-})$. Now, of course $f'(x) = g'(u)$ hence

$$|f'(x)| = \frac{|f(x)|}{\rho} \geq \|f\|_{\rho}^{\delta} \quad \forall x \in d(a_t, \rho^{-}).$$

That holds for every $t \in \mathbb{N}$, therefore that ends the proof.

**Remarks:** Lemma III.7 is a correction to Lemma 2.13 in [9].
**Theorem III.8:** Suppose $\mathbb{K}$ is spherically complete and let $\mathcal{M}$ be a regular maximal ideal of $A$. There exists $f \in \mathcal{M}$, having a sequence of zeros of order 1 and no other zeros, such that $f' \notin \mathcal{M}$.

**Proof:** Since $\mathcal{M}$ is a regular maximal ideal, there exists a regular sequence $(a_n)_{n \in \mathbb{N}}$ and a regular ultrafilter $\mathcal{U}$ thinner than the sequence $(a_n)$ such that $\mathcal{M} = \mathcal{J}(\mathcal{U})$. Since the sequence is regular, we have $\delta = \inf_{k \in \mathbb{N}} \prod_{n \neq k, n \in \mathbb{N}} |a_n - a_k| > 0$ and $\rho = \inf_{k \neq n, k, n \in \mathbb{N}} |a_n - a_k| > 0$.

Since $\mathbb{K}$ is spherically complete, since $\delta > 0$ we may apply Corollary II.15.1 showing there exists $f \in \mathcal{A}$ admitting each $a_n$ as a zero of order 1 and no other zero. Now by Lemma III.7, we have $|f'(x)| \ge \|f\|_{\rho}^{\delta} \quad \forall x \in \bigcup_{n=0}^{\infty} d(a_n, \delta^{-})$ which shows that $\varphi_{\mathcal{U}}(f') > 0$ because the set $E = \bigcup_{n=0}^{\infty} d(a_n, (\delta\rho)^{-})$ obviously belongs to $\mathcal{U}$. Consequently, $f'$ does not belong to $\mathcal{M}$.

**Remarks:** 1) Now we may notice that when the field is of characteristic 2, it is easy to show that for certain maximal ideals $\mathcal{M}$ of $A$, $\tilde{\mathcal{M}}$ is not prime. Indeed, by Theorem III.6, there exists a regular maximal ideal $\mathcal{M}$ and $f \in \mathcal{M}$ such that $f' \notin \mathcal{M}$. Hence $f$ does not belong to $\tilde{\mathcal{M}}$. Now consider $g = f^2$. Then $g' = 2ff' = 0$ hence $g^{(n)} \in \mathcal{M} \quad \forall n \in \mathbb{N}$. If $\mathbb{K}$ is of characteristic 3, we can also construct a similar but less simple counter-example.

2) In the algebra of bounded complex holomorphic functions in the open unit disk of $\mathbb{C}$, the derivation is not an endomorphism. Consequently, ideals of the form $\tilde{P}$ do not exist.

By Theorem II.14, we may notice the following proposition III.9:

**Proposition III.9:** Let $\mathbb{K}$ be spherically complete. Let $(a_j)_{j \in \mathbb{N}}$ be a coroner sequence such that $\prod_{n=0}^{\infty} |a_n| > 0$. There exists $f \in \mathcal{A}$ admitting each $a_n$ as a zero of order 1 and having no other zeros.

*Proof.* Let $E$ be the divisor on $D(a_n, 1)_{n \in \mathbb{N}}$ By Theorem II.15, there exists $f \in B$ such that $\mathcal{T}(f) \ge E$ and such that $|f|(r) \le 2|E|(r) \quad \forall r \in ]0, 1[$. But since $\prod_{n \in \mathbb{N}} |a_n| > 0$, by Corollary II.8.1 $|E|(r)$ is bounded in $]0, 1[$ and hence $f$ belongs to $A$. Consequently, $\mathcal{I}$ is not null. $\square$

**Notation:** Recall that we denote by $U$ the disk $d(0, 1)$. Considering the ring $U[x]$ of polynomials with coefficients in $U$, we denote by $\mathcal{H}$ be the family of ideals $J$ of $U[x]$ such that $J \cap U \neq \{0\}$ and, given an integer $s \in \mathbb{N}^*$, let $\mathcal{H}_s$ be the set of $J \in \mathcal{H}$ generated by $s$ elements. For every ideal $J \in \mathcal{H}$ we put $t(J) = \sup\{|x| \mid x \in J \cap U\}$ and $\ell(J) = \inf\{\sup_{f \in J} |f(x)| \mid x \in V\}$ and we denote by $u(J)$ the number such that $t(J) = \ell(J)^{u(J)}$. Finally, we put $m(s) = \sup\{u(J) \mid J \in \mathcal{J}_s\}$. Henceforth, given $f_1, \dots, f_s \in \mathcal{H}(U)$ such that $\|f_i\| < 1 \quad \forall i = 1, \dots, s$, we set $w(f_1, \dots, f_s) = \inf\{\max_{1 \le i \le s} |f_i(x)| \mid x \in U\}$. Moreover, given $f_1, \dots, f_s \in A$ we set $\lambda(f_1, \dots, f_s) = \inf\{\max_{1 \le i \le s} |f_i(x)| \mid x \in D\}$.

**Remark:** Characterizing the coroner ultrafilters $\mathcal{U}$ such that $\mathcal{J}(\mathcal{U})$ is a maximal ideal appears very hard. For instance, consider an ultrafilter $\mathcal{U}$ thinner than $\mathcal{Y}$. It is a coroner ultrafilter. But $\mathcal{J}(\mathcal{U}) = \{0\}$. Indeed, suppose a non-identically zero function $f$ lies in $\mathcal{J}(\mathcal{U})$. Let $(a_n)$ be its sequence of zeros, set $r_n = |a_n|$, $n \in \mathbb{N}$, and let $E = D \setminus \bigcup_{n=0}^{\infty} d(a_n, r_n^-)$. Clearly $|f(x)| = |f|(|x|) \quad \forall x \in E$. However, $E$ belongs to $\mathcal{Y}$ and therefore, $\mathcal{U}$ is secant with $E$, a contradiction with the hypothesis $f \in \mathcal{J}(\mathcal{U})$.

On the other hand, the mapping $\mathcal{J}$ from the set of coroner ultrafilters to the set of ideals of $A$ is not injective: as noticed in [8], two contiguous coroner ultrafilters define the same ideal.

Thus, by Theorem II.12 if an element $\psi \in \mathrm{Mult}(A, \| \cdot \|)$ is neither the Gauss norm nor of the form $\varphi_{\mathcal{F}}$ on the whole set $A$, with $\mathcal{F}$ a circular filter on $D$ of diameter $r < 1$, then, its restriction
to $\mathbb{K}[x]$ must be the Gauss norm on $\mathbb{K}[x]$. So its kernel is a prime closed ideal included in a
maximal ideal of the form $\mathcal{J}(\mathcal{U})$, with $\mathcal{U}$ a coroner ultrafilter.

Here we shall first examine the problem of the continuation of $\varphi_W$ to $A$ through multiplicative
norms, what was not done in [8].

**Notation:** Let $F$ be a field, let $R$ be a commutative $F$-algebra with unity and let $\mathcal{D}$ be a
derivation on $R$. Let $J$ be an ideal of $R$. We will denote by $\tilde{J}$ the set $\{f \in R \mid \mathcal{D}^{(n)} \in J \ \forall n \in \mathbb{N}\}$.

On $A$ we shall apply this notation to the usual derivation of functions. Let $\psi \in \text{Mult}(A, \|\cdot\|)$.
Here we set $\text{Subker}(\psi) = \widetilde{\text{Ker}}(\psi)$.

In [11], we asked the question whether there exist prime closed ideals which are neither zero
nor maximal ideals. Theorem II.16 shows that such ideals do exists and can be the kernel of an
element of $\operatorname{Mult}(A, \|\cdot\|)$. Now we will construct another kind of prmie closed ideal which is
neither maximal nor null.

Theorem III.10 comes from classical algebraic results:

**Theorem III.10:** Let $F$ be a field, let $R$ be a commutative $F$-algebra with unity and let $\mathcal{D}$ be a derivation on $R$. Let $J$ be an ideal of $R$. Then $\tilde{J}$ is an ideal of $R$ and $(\tilde{J}) = \tilde{J}$. Moreover, if $F$ is of characteristic $0$ and if $J$ is prime, so is $\tilde{J}$.

Since $\|f'\| \le \|f\| \quad \forall f \in A$, we can derive Corollary III.10.1:

**Corollary III.10.1:** Suppose $\mathbb{K}$ is of characteristic zero. Let $\mathcal{P}$ be a prime ideal of $A$. Then $\tilde{\mathcal{P}}$ is a prime ideal of $A$ such that $(\tilde{\mathcal{P}}) = \tilde{\mathcal{P}}$. Moreover, if $\mathcal{P}$ is closed, so is $\tilde{\mathcal{P}}$.

**Corollary III.10.2:** Suppose $\mathbb{K}$ is of characteristic zero. Let $\psi \in \operatorname{Mult}(A, \|\cdot\|)$. Then $\operatorname{Subker}(\psi)$ is a prime closed ideal.

**Theorem III.11:** Let $\mathcal{U}$ be a regular ultrafilter. Then the ideal $\mathcal{M} = \mathcal{J}(\mathcal{U})$ is a maximal ideal of $A$ and there exists $f \in \mathcal{M}$, having a sequence of zeros of order 1 and no other zeros, such that $f' \notin \mathcal{M}$, and such that $\tilde{M} \ne \{0\}$.

**Proof:** By Theorem II.13 we know that there exist sequences $(a_n)_{n \in \mathbb{N}}$ in $D$ such that the sequence $\left| \frac{a_n}{a_{n+1}} \right|$ is strictly increasing and then the function $f(x) = \sum_{n=0}^{\infty} a_n x^n$ admits a sequence of zeros $(\alpha_n)_{n \in \mathbb{N}^*}$ satisfying $|\alpha_n| = \left| \frac{a_n}{a_{n+1}} \right|$. Thus, particularly, if we set $r_n = \left| \frac{a_n}{a_{n+1}} \right|$ then $f$ admits exactly a unique zero in each circle $C(0, r_n)$, each of order 1, and has no other zero in $D$. Consequently, by Lemma I.8, we can see that $|f'(\alpha_n)| = |f'|(r_n) \quad \forall n \in \mathbb{N}^*$. Now, let $\mathcal{U}$ be an ultrafilter thinner than the sequence $(\alpha_n)_{n \in \mathbb{N}^*}$. We can check that the sequence $(\alpha_n)$ is regular, hence $\mathcal{U}$ is a regular ultrafilter. Consequently, by Corollary III.3.1 the ideal $\mathcal{M} = \mathcal{J}(\mathcal{U})$ is a maximal ideal of $A$. On the other hand, by lemma III.7, $f'$ does not belong to $\mathcal{M}$.

For each $n \in \mathbb{N}$, let $u_n = \log(r_n)$. Then, by Theorem II.8, we have $\log(\|f\|) = -\sum_{n=0}^{\infty} u_n$. By Lemma I.2, there exists an increasing sequence $(s_n)_{n \in \mathbb{N}}$ of $\mathbb{N}$ such that $\lim_{n \to \infty} s_n = +\infty$ and such that the series $\sum_{n=0}^{\infty} s_n u_n$ converges.

Now, by Corollary II.14.1, there exists $g \in A$ (not identically zero) such that for each $n \in \mathbb{N}$, $\alpha_n$ is a zero of $g$ of order $z_n \ge s_n$. And since $\lim_{n \to \infty} s_n = +\infty$, for every fixed $k \in \mathbb{N}$, we can see that $f^{(k)}(\alpha_n) = 0$ when $n$ is big enough, therefore $g^{(k)}$ belongs to $\mathcal{M}$. Consequently, $\tilde{\mathcal{M}}$ is not null, which ends the proof.
By Theorem III.11 and Corollary 7.2, we now have this corollary:

**Corollary III.11.1: A admits maximal ideals of infinite codimension.**

**Remarks:** 1) Now we may notice that when the field is of characteristic 2, it is easy to show that for certain maximal ideals $\mathcal{M}$ of $A$, $\tilde{\mathcal{M}}$ is not prime. Indeed, by Theorem III.4, there exists a regular maximal ideal $\mathcal{M}$ and $f \in \mathcal{M}$ such that $f' \notin \mathcal{M}$. Hence $f$ does not belong to $\tilde{\mathcal{M}}$. Now consider $g = f^2$. Then $g' = 2ff' = 0$ hence $g^{(n)} \in \mathcal{M} \ \forall n \in \mathbb{N}$. If $\mathbb{K}$ is of characteristic 3, we can also construct a similar but less simple counter-example. 2) In the algebra of bounded complex holomorphic functions in the open unit disk of $\mathbb{C}$, the derivation is not an endomorphism. Consequently, ideals of the form $\tilde{P}$ do not exist.

Following Corollary II.12.1, we can now complete the characterization of continuous multiplicative norms on $A$.

**Theorem III.12:** Let $\psi \in \text{Mult}(A, \|\cdot\|)$ be coroner. Then $\text{Subker}(\psi)$ is not null. Moreover, if $\mathbb{K}$ is spherically complete, then, for every $f \in A$ such that $\psi(f) < \|f\|$, there exists $g \in \text{Subker}(\psi)$ admitting no zero but zeros of $f$ and admitting each zero of $f$ as a zero of order superior or equal to its order as a zero of $f$.

**Proof:** The proof takes advantage of the proof of a theorem in [3]. Suppose the claim is wrong. Let $\psi \in \text{Mult}(A, \|\cdot\|)$ be an absolute value on $A$ different from the Gauss norm $\|\cdot\|$ on $A$. So, there exists a circular filter $\mathcal{F}$ on $D$, of diameter $r \le 1$ such that $\psi(P) = \varphi_{\mathcal{F}}(P) \ \forall P \in K[x]$. But by Corollary II.12.1, we know that $r = 1$ and hence, the restriction of $\psi$ to $K[x]$ is the Gauss norm. Now, since $\psi$ is not the Gauss norm on $A$, there exists $f \in A$ such that $\psi(f) < \|f\|$. Actually, without loss of generality, we can choose $f \in A$ such that $\psi(f) < 1 \le \|f\|$. Let $\rho = \psi(f)$. And, up a change of origin, we can also assume that $f(0) \ne 0$. By Proposition II.4, $f$ is not quasi-invertible, hence $f$ has a sequence of zeroes $(a_n)_{n \in \mathbb{N}}$ in $D$, with $|a_n| \le |a_{n+1}|$. For each $n \in \mathbb{N}$, let $q_n$ be the multiplicity order of $a_n$. By Theorem II.8 we know that $\sum_{n=0}^{\infty} -q_n \log |a_n| < +\infty$. Now, there clearly exists a sequence $t_n$ of strictly positive integers satisfying

$$t_n \le t_{n+1}, \ n \in \mathbb{N},$$

$$\lim_{n \to \infty} t_n = +\infty,$$

$$\sum_{n=0}^{\infty} t_n q_n \log(|a_n|) < +\infty.$$

By Theorem II.14 there exists a function $g \in A$ admitting each $a_n$ as a zero of order $s_n \ge t_n q_n$, such that $|g|(|a_n|) \le 2 \left| \prod_{k=0}^{n} \left( \frac{a_n}{a_k} \right)^{t_n q_n} \right| \ \forall n \in \mathbb{N}$ and consequently, $g$ belongs to $A$.

Now, for each $n \in \mathbb{N}$ and for each $k = 0, \dots, n$, let $u_{n,k} = \max(0, t_n q_k - s_k)$ and let $P_n(x) = \prod_{k=0}^{n} (x - a_k)^{u_{n,k}}$. Clearly, all coefficients of $P_n$ lie in $D$ except the leading coefficient that is 1. Consequently, $\|P_n\| = 1 \ \forall n \in \mathbb{N}$ and therefore

$$(1) \quad \|P_n g\| = \|g\|.$$

On the other hand, since the sequence $t_n$ is increasing, we can check that for each fixed $n \in \mathbb{N}$, each zero $a_k$ of $f^{t_n}$ is a zero of $P_n g$ of order $\ge t_k q_k$. Consequently, by Lemma I.5 in the ring $A$ we can write $P_n g$ in the form $f^{t_n} \sigma_n$, with $\sigma_n \in A$.
By (1), we have $\|\sigma_n\|\|f^{t_n}\| = \|g\|$ hence, since $\|f\| \ge 1$, we can see that $\|\sigma_n\| \le \|g\|$. But now, since the restriction of $\psi$ to $K[x]$ is $\| \cdot \|$, we have $\psi(P_n) = 1$, hence $\psi(P_n g) = \psi(P_n)\psi(g) = \psi(g)$ and therefore

$$ (2) \quad \psi(g) = \psi(f^{t_n} \sigma_n) = \psi(f)^{t_n} \psi(\sigma_n) \le \rho^{t_n} \|g\|. $$

Relation (2) holds for every $n \in \mathbb{N}$ hence $\lim_{n \to +\infty} \rho^{t_n} \|g\| = 0$. Consequently, $\psi(g) = 0$, a contradiction. This ends the proof.

**Notation:** We will denote by $\mathrm{Mult}_n(A, \| \cdot \|)$ the set of continuous multiplicative norms of $A$.

**Corollary III.12.1:** Let $\psi \in \mathrm{Mult}(A, \| \cdot \|)$ be coroner. Then $\psi$ is not a norm.

By Theorems II.12 and III.5, we now can also state Corollary III.12.2:

**Corollary III.12.2:** Let $\psi \in \mathrm{Mult}_n(A, \| \cdot \|)$. If $\psi$ is not $\| \cdot \|$, there exists a circular filter $\mathcal{F}$ on $D$, of diameter $r < 1$, such that $\psi = \varphi_{\mathcal{F}}$.

On the other hand, each coroner maximal ideal is the kernel of some coroner continuous multiplicative semi-norm of $A$.

**Corollary III.12.3:** Let $\mathcal{M}$ be a coroner maximal ideal of $A$. Then $\widetilde{\mathcal{M}}$ is not null.

Concerning the Corona Problem, we may notice this:

**Corollary III.12.4:** $\mathrm{Mult}_n(A, \| \cdot \|)$ is included in the closure of $\mathrm{Mult}_1(A, \| \cdot \|)$.

**Theorem III.13:** Let $G$ be the set of circular filters on $D$ of diameter $r \in ]0, 1[$. For every circular filter $\mathcal{F} \in G$, $\varphi_{\mathcal{F}}$ is a multiplicative norm on $A$ and the mapping from $G$ into $\mathrm{Mult}(A, \| \cdot \|)$ that associates to each $\mathcal{F} \in \Phi(0, 1)$ the norm $\varphi_{\mathcal{F}}$ is a bijection from $G$ onto $\mathrm{Mult}_n(A, \| \cdot \|) \setminus \{\| \cdot \|\}$.

**Proof:** By Corollary II.12.1, we know that for each $\mathcal{F} \in G$, if $0 < \mathrm{diam}(\mathcal{F}) < 1$, $\varphi_{\mathcal{F}}$ has extension to a multiplicative norm on $A$ and the mapping $G$ into $\mathrm{Mult}(A, \| \cdot \|)$ that associates to each $\mathcal{F} \in G$ the norm $\varphi_{\mathcal{F}}$ is obviously injective because its restriction to $H(D)$ is already injective.

Now, consider a multiplicative semi-norm $\phi$ on $A$ which is not of the form $\varphi_{\mathcal{F}}$ with $\mathcal{F}$ a circular filter of diameter $r < 1$. Then, if $\phi$ is not $\| \cdot \|$, $\phi$ is coroner, therefore by Corollary III.12.1, it is not a norm. Consequently, the mapping from $G$ into $\mathrm{Mult}(A, \| \cdot \|)$ that associates to each $\mathcal{F} \in G$ the norm $\varphi_{\mathcal{F}}$ is a surjection from $G$ onto the set of multiplicative norms of $A$ other than $\| \cdot \|$, which ends the proof.

Thus, Theorem III.13 lets us characterize the continuous multiplicative norms of $A$, which we can summarize in this way:

**Corollary III.13.1:** Let $\phi \in \mathrm{Mult}(A, \| \cdot \|)$ be different from $\| \cdot \|$. Then $\phi$ is a norm if and only if it is of the form $\varphi_{\mathcal{F}}$ with $\mathcal{F}$ a circular filter of diameter $r \in ]0, 1[$.

# IV. Density of $\mathrm{Mult}_1(A, \| \cdot \|)$ in $\mathrm{Mult}_m(A, \| \cdot \|)$.

Now we will consider the question whether $\mathrm{Mult}_1(A, \| \cdot \|)$ is dense in $\mathrm{Mult}_m(A, \| \cdot \|)$. We have to prove many intermediate results.
The following Proposition IV.2 is also called Corona Statement in dimension 1. However, on a non-archimedean field, it is not at all proven to be equivalent to a property of density for maximal ideals defined by points of D inside the set $Mult(A, \| \cdot \|)$. We will prove Proposition IV.2 in the same way as in [19]. Proposition IV.1 is proven in [19] and is indispensable for further results.

**Notation:** Recall that we denote by $U$ the disk $d(0, 1)$. Let $\mathcal{H}$ be the family of ideals $J$ of $U[x]$ such that $J \cap U \neq \{0\}$ and, given an integer $s \in \mathbb{N}^*$, let $\mathcal{H}_s$ be the set of $J \in \mathcal{H}$ generated by $s$ elements. For every ideal $J \in \mathcal{H}$ we put $t(J) = \sup\{|x| \mid x \in J \cap U\}$ and $\ell(J) = \inf\{\sup_{f \in J} |f(x)| \mid x \in V\}$ and we denote by $u(J)$ the number such that $t(J) = \ell(J)^{u(J)}$. Finally, we put $m(s) = \sup\{u(J) \mid J \in \mathcal{J}_s\}$. Henceforth, given $f_1, \dots, f_s \in H(U)$ such that $\|f_i\| < 1 \ \forall i = 1, \dots, s$, we set $w(f_1, \dots, f_s) = \inf\{\max_{1 \le i \le s} |f_i(x)| \mid x \in U\}$. Moreover, given $f_1, \dots, f_s \in A$ we set $\lambda(f_1, \dots, f_s) = \inf\{\max_{1 \le i \le s} |f_i(x)| \mid x \in D\}$.

**Proposition IV.1:** Let $J$ be a finitely generated ideal of $U[x]$ such that $J \cap U \neq \{0\}$. Then $m(s) = 2 \ \forall s \ge 2$.

**Proposition IV.2:** Let $s \in \mathbb{N}^*$. For any $f_1, \dots, f_s \in A$ satisfying $\|f_i\| < 1 \ (1 \le i \le s)$ and $\lambda(f_1, \dots, f_s) > 0$. There exist $g_1, \dots, g_s \in A$ satisfying $\sum_{i=1}^s f_ig_i = 1$ and $\|g_i\| < \lambda(f_1, \dots, f_s))^{-2}$.

*Proof.* As a first step, we will prove that for any $f_1, \dots, f_s \in H(U)$ satisfying $\|f_i\| < 1 \ \forall i = 1, \dots, s$ and $w(f_1, \dots, f_s) > 0$, there exist $g_1, \dots, g_s \in H(d(0, R))$ satisfying $\sum_{i=1}^s f_ig_i = 1$ and $\|g_i\| < (w(f_1, \dots, f_s))^{-2} \ \forall i = 1, \dots, s$.

Since $\mathbb{K}[x]$ is dense in $H(U)$, we can find polynomials $P_1, \dots, P_s \in \mathbb{K}[x]$ such that $\|P_i - f_i\| \le (w(f_1, \dots, f_s))^2 \ \forall i = 1, \dots, s$. Since $w(f_1, \dots, f_s) > 0$, by Corollary II.12.1, there exists no maximal ideal $\mathcal{M}$ of $H(U)$ containing the ideal generated by $f_1, \dots, f_s$. Consequently, there exist $g_1, \dots, g_s \in H(U)$ such that $\sum_{i=1}^s g_if_i = 1$. Therefore, we can define $h_1, \dots, h_s \in H(U)$ such that $\|h_i\| \le 1 \ \forall i = 1, \dots, s$ and such that $\sum_{i=1}^s h_if_i$ is an element $P_0 \in U$. Let $I$ be the ideal of $U[x]$ generated by $P_0, P_1, \dots, P_s$. Since $I \cap U \neq \{0\}$, we have $0 < t(I) < \ell(I)$. Consequently, by Proposition IV.1 we can find $Q_0, Q_1, \dots, Q_s \in U[x]$ such that $\sum_{i=0}^s Q_i P_i$ be an element $a \in U$ satisfying $|a| \ge (w(f_1, \dots, f_s))^{m(s)} = (w(f_1, \dots, f_s))^2$. Then

$$ \sum_{i=1}^{s} a^{-1}f_i + a^{-1}Q_0 \sum_{i=1}^{s} h_if_i = 1 + \sum_{i=1}^{s} a^{-1}Q_i(f_i - P_i). $$

By construction, we have $\|\sum_{i=1}^s a^{-1}Q_i(f_i - P_i)\| < 1$ and hence $1 + \sum_{i=1}^s a^{-1}Q_i(f_i - P_i)$ is an invertible element $u$ in $H(U)$. So, $\sum_{i=1}^s u^{-1}(a^{-1}(Q_i + Q_0h_i))f_i = 1$ and hence for every $i = 1, \dots, s$, we have $\|u^{-1}a^{-1}(Q_i + Q_0h_i)\| \le (w(f_1, \dots, f_s))^{-2} \ \forall i = 1, \dots, s$. Thus, putting $g_i = u^{-1}(a^{-1}(Q_i +
$Q_0h_i)$, $i = 1, ..., s$ we have proven that for any $f_1, ..., f_s \in H(U)$ satisfying $\|f_i\| < 1 \quad \forall i = 1, ..., s$ and $w(f_1, ..., f_s) > 0$, there exist $g_1, ..., g_s \in H(U)$ satisfying $\sum_{i=1}^{s} f_i g_i = 1$ and $\|g_i\| < (w(f_1, ..., f_s))^{-2} \quad \forall i = 1, ..., s$.

Now, let us prove the conclusion of Proposition IV.2. So, we take $f_1, ..., f_s \in A$. Let $(u_n)_{n \in \mathbb{N}}$ be a sequence in $\mathbb{K}$ such that $0 < |u_n| < |u_{n+1}| < ... < 1$ and $\lim_{n \to +\infty} |u_n|^n = 1$. For each $i = 1, ..., s$, set $f_{i,n}(x) = f_i(u_n x)$. Since each $u_n$ belongs to $D$, each $f_{i,n}$ is a power series of radius $r > 1$ and hence belongs to $H(U)$. Then by the claim we have just proven, for each $n \in \mathbb{N}$, there are $g_{i,n} \in H(U)$ such that $\sum_{i=1}^{s} g_{i,n} f_{i,n} = 1$ and $\|g_{i,n}\| < (w(f_{1,n}, ..., f_{s,n}))^{-2}$. Now, each $g_{i,n}$ is a power series $\sum_{k=0}^{+\infty} g_{i,n,k} x^k$. Let $h_{i,n}$ be the power series $\sum_{k=0}^{2n} g_{i,n,k} (u_n)^{-k} x^k$. Then we have $\|h_{i,n}\| < |u_n|^{-2n} (w(f_1, ..., f_s))^{-2}$ and hence $\sum_{i=1}^{s} h_{i,n} f_i$ is of the form $1 + x^n t_n(x)$ with $t_n \in H(U)$.

We will get to a conclusion thanks to a Banach process. Let $\mathcal{E}$ be the Banach space of bounded sequences in $\mathbb{K}$ provided with the classic norm $\|(a_n)_{n \in \mathbb{N}}\|_0 = \sup_{n \in \mathbb{N}} |a_n|$ and let $\mathcal{C}$ be the closed sub-space of converging sequences. For every sequence $(a_n)_{n \in \mathbb{N}} \in \mathcal{C}$, we put $\mathcal{L}((a_n)) = \lim_{n \to +\infty} a_n$.

Suppose first that $\mathbb{K}$ is spherically complete. There exists a linear map $\tilde{\mathcal{L}}$ of norm 1 from $\mathcal{E}$ to $\mathbb{K}$ expanding $\mathcal{L}$ to $\mathcal{E}$. Set $h_{i,n} = \sum_{k=0}^{+\infty} x^k$ and let $l_{i,k}$ be the sequence $(h_{i,n,k})_{1 \le n}$. For each pair $(i, k)$, we can now define $g_{i,k} = \tilde{\mathcal{L}}(l_{i,k})$ and put $g_i = \sum_{k=0}^{+\infty} g_{i,k} x^k$. Since $\tilde{\mathcal{L}}$ is of norm 1, each sequence $g_{i,k}$ is bounded and hence $g_i$ belongs to $A$. Moreover, by construction, $g_i$ satisfies $\|g_i\| < (\lambda(f_1, ..., f_s))^{-2} \quad \forall i = 1, ..., s$. Now, we have

$$ (1) \qquad \sum_{i=1}^{s} g_i f_i = \sum_{k=0}^{+\infty} \left( \sum_{i=1}^{s} \left( \sum_{j=0}^{k} \tilde{\mathcal{L}}(l_{i,j}) f_{i,k-j} \right) \right) x^k $$

and for each fixed $k \in \mathbb{N}$,

$$ \sum_{i=1}^{s} \left( \sum_{j=0}^{k} \tilde{\mathcal{L}}(l_{i,j}) f_{i,k-j} \right) = \tilde{\mathcal{L}} \left( \sum_{i=1}^{s} \left( \sum_{j=0}^{k} f_{i,j} h_{i,n,k-j} \right) \right). $$

Consequently, since $\sum_{i=1}^{s} h_{i,n} f_i = 1 + x^n t_n(x)$, we can check that $\lim_{n \to +\infty} \left( \sum_{i=1}^{s} \left( \sum_{j=0}^{k} f_{i,j} h_{i,n,k-j} \right) \right) = 1$ whenever $k = 0$ and $\lim_{n \to +\infty} \left( \sum_{i=1}^{s} \left( \sum_{j=0}^{k} f_{i,j} h_{i,n,k-j} \right) \right) = 0$ whenever $k \ne 0$. Consequently, since $\tilde{\mathcal{L}}$ extends to $\mathcal{E}$ the limit on $\mathcal{C}$, we have $\tilde{\mathcal{L}}\left(\sum_{i=1}^{s} \left(\sum_{j=0}^{k} f_{i,j} h_{i,n,k-j}\right)\right) = 1$ whenever $k = 0$ and

$$ \tilde{\mathcal{L}}\left(\sum_{i=1}^{s} \left(\sum_{j=0}^{k} f_{i,j} h_{i,n,k-j}\right)\right) = 0 \quad \text{whenever } k \ne 0. $$
Therefore by (1) we obtain $\sum_{i=1}^{s} g_i f_i = 1$.

Consider now the general case when $\mathbb{K}$ is no longer supposed to be spherically complete. Consider a spherically complete algebraically closed extension $\hat{\mathbb{K}}$ of $\mathbb{K}$. Let $\hat{D}$ be the disk of $\hat{K}: \{x \in \hat{K} \mid |x| < 1\}$ and let $\hat{A}$ be the algebra of bounded analytic functions from $\hat{D}$ to $\hat{K}$. Set $B = \lambda(f_1, \dots, f_s)^{-2}$. Then by what forgoes, there exist $h_1, \dots, h_s \in \hat{A}$ such that $\sum_{j=1}^{s} h_j f_i = 1$ and $\max_{1 \le j \le s} \|h_j\| < B$.

Inside $\hat{A}$, let $F$ be the closed subspace of the $\mathbb{K}$-Banach space generated by $1$ and all coefficients of all the $h_j$. Let us take $\epsilon > 0$ such that $(1 + \epsilon) \max_{1 \le j \le s} \|h_j\| \le B$. Since $F$ is a $\mathbb{K}$-Banach space of countable type, there exists a $\mathbb{K}$-linear map $\ell$ from $F$ to $\mathbb{K}$ satisfying $\ell(1) = 1$ with a norm $\| \cdot \|^*$ satisfying $\|\ell\|^* \le 1 + \epsilon$. Let $T$ be the closed subspace of $\hat{A}$ consisting of the power series with coefficients in $F$. Then $T$ is a $\mathbb{A}$-module and then we have an extension $\mathcal{L}$ of $\ell$ from $T$ to $\mathbb{A}$ defined as $\mathcal{L}(\sum_{k=0}^{\infty} e_k x^k) = \sum_{k=0}^{\infty} \ell(e_k) x^k$ which is $\mathbb{A}$-linear and its norm satisfies $\|\mathcal{L}\|^* \le 1 + \epsilon$.

Now, putting $g_j = \mathcal{L}(h_j)$, $1 \le j \le s$, we have $\sum_{i=1}^{s} f_i g_i = 1$ and $\max_{1 \le j \le s} \|g_j\| < B$. $\square$

**Notation:** Let $\mathcal{I}$ be an ideal of $\mathbb{A}$. For each $f \in \mathbb{A}$ and for each $\epsilon > 0$, we set $E(f, \epsilon) = \{x \in D \mid |f(x)| \le \epsilon\}$.

**Corollary IV.2.1:** Let $\mathcal{I}$ be a proper ideal of $\mathbb{A}$. The family $(E(f, \epsilon), f \in \mathcal{I}, \epsilon > 0)$ generates a filter $\mathcal{F}$ on $D$ such that $\mathcal{I} \subset \mathcal{J}(\mathcal{F})$.

*Proof.* Let $(E(f_j, \epsilon_j), 1 \le j \le n)$ be such that $\bigcap_{j=1}^n E(f, \epsilon_j) = \emptyset$. Let $\epsilon = \min_{1 \le j \le n}(\epsilon_j)$ and let $F_j = D \setminus E(f_j, \epsilon)$, $1 \le j \le n$. Then $\bigcup_{j=1}^n F_j = D$, hence by Theorem IV.2, $1 \in \mathcal{I}$, a contradiction. $\square$

By Lemma III.7 and Corollary IV.2.1, we can derive Corollary IV.2.2:

**Corollary IV.2.2:** Let $\mathcal{M}$ be a maximal ideal of $\mathbb{A}$. Then there exists an ultrafilter $\mathcal{U}$ on $D$ such that $\mathcal{M} = \mathcal{J}(\mathcal{U})$. Moreover, if $\mathcal{U}$ converges, $\mathcal{M}$ is of codimension 1 and of the form $(x-a)\mathbb{A}$, $a \in D$. If $\mathcal{U}$ does not converge, it is coroner, $\mathcal{M}$ is of infinite codimension and for every $f \in \mathcal{M}$, $f$ is not quasi-invertible.

Let us recall the definition of an increasing pierced filter.

**Definition:** Let $a \in \mathbb{K}$, let $R > 0$ and consider the filter $\mathcal{G}$ admitting for basis the annuli $\Gamma(a, r, R) r \in ]0, R[$. This filter is called increasing filter of center $a$ and diameter $R$. Moreover, if all $F$ is an infraconnected subset of $\mathbb{K}$ and if every annulus $\Gamma(a, r, R)$ contain holes of $F$, then the filter on $F$ induced by $\mathcal{G}$ is said to be pierced. Next, a pierced filter is called a $T$-filter if its holes satisfy a certain relation [5].

By using properties of $T$-filters and particularly idempotent $T$-sequences [17], by Lemma 35.1 and Proposition 37.1 in [5], Proposition 1.6 and [19]), we have the following proposition:
**Proposition IV.3:** Let $(r_n)_{n \in \mathbb{N}}$ be a sequence in $|\mathbb{K}|$ such that $0 < r_n < r_{n+1}$, $\lim_{n \to +\infty} r_n = 1$, let $(q_n)_{n \in \mathbb{N}}$ be a sequence of $\mathbb{N}$ such that $q_n \le q_{n+1}$ and $\lim_{n \to +\infty} \left(\frac{r_n}{r_{n+1}}\right)^{q_n} = 0$. Let $l \in ]0, 1[$ and for each $n \in \mathbb{N}$, let $b_n \in C(0, (r_n)^{q_n})$, let $a_{n,1}, \dots, a_{n,q_n}$ be the $q_n$-th roots of $b_n$ and let $F = D \setminus \left( \bigcup_{n \in \mathbb{N}} \left( \bigcup_{j=1}^{q_n} d(a_{n,j}, l^{-}) \right) \right)$. Set $f_n(x) = \prod_{k=1}^{n} \prod_{j=1}^{q_k} \left( \frac{1}{1 - \frac{x}{a_{k,j}}} \right)$. Then each $f_n$ belongs to $R(F)$ and the sequence $(f_n)_{n \in \mathbb{N}}$ converges in $H(F)$ to an element $f$ strictly vanishing along the pierced increasing filter of center 0 and diameter 1.

**Notation:** Given a unital commutative ultrametric $\mathbb{K}$-algebra $B$ and $f \in B$, we denote by $sp(f)$ the spectrum of $f$ i.e. the set of $x \in \mathbb{K}$ such that $f - x$ is not invertible in $B$.

**Proposition IV.4:** Let $(B, \| \cdot \|)$ be a unital commutative ultrametric Banach $\mathbb{K}$-algebra. Suppose there exist $\ell \in B$, $\phi, \psi \in \mathrm{Mult}(B, \| \cdot \|)$ such that $\psi(\ell) < \phi(\ell)$, $\mathrm{sp}(\ell) \cap \Gamma(0, \psi(\ell), \phi(\ell)) = \emptyset$ and there exists $\epsilon \in ]0, \phi(\ell) - \psi(\ell)[$ satisfying further $\|(\ell - a)^{-1}\| \le M \forall a \in \Gamma(0, \psi(\ell), \phi(\ell) - \epsilon)$. Then there exists $f \in B$ such that $\psi(f) = 1$, $\phi(f) = 0$.

*Proof.* Let $s = \psi(\ell)$, $t = \phi(\ell)$, $Q = \|\ell\|$, $R = t - \epsilon$ and $l = \frac{1}{M}$. Let $r_0 \in ]s, t - \epsilon[$. Consider the sequence $(a_{n,j})_{n \in \mathbb{N}, 1 \le j \le q_n}$ defined in Proposition IV.3 and the set

$$E = d(0, Q^{-}) \setminus \left( \bigcup_{n \in \mathbb{N}} \left( \bigcup_{j=1}^{q_n} d(a_{n,j}, l^{-}) \right) \right). \text{ Then in } H(E) \text{ we have}$$

$$ (1) \qquad \left\| \frac{1}{x-b} \right\|_E \le l \quad \forall b \in \bigcup_{n \in \mathbb{N}} \left( \bigcup_{j=1}^{q_n} d(a_{n,j}, l^{-}) \right). $$

There exists a natural homomorphism $\theta$ from $R(E)$ into $B$ such that $\theta(x) = \ell$. Since $Q = \|\ell\|$ and $\|(\ell-b)^{-1}\| \le M \forall b \in \Gamma(0, s, t)$, by Theorem I.11 and by (1) $\theta$ is clearly continuous with respect to the norms $\| \cdot \|_E$ of $R(E)$ and $\| \cdot \|$ of $B$. Consequently, $\theta$ has continuation to a continuous homomorphism from $H(E)$ to $B$.

Now, let $\psi' = \psi \circ \theta$, $\phi' = \phi \circ \theta$. Then both $\phi'$, $\psi'$ belong to $\mathrm{Mult}(H(E), \| \cdot \|)$ and satisfy $\psi'(x) = s$, $\phi'(x) = t - \epsilon$. So, $\psi'$ is of the form $\varphi_F$ with $F$ a circular filter on $E$ secant with $C(0, s)$ and $\phi'$ is of the form $\varphi_G$ with $G$ a circular filter on $E$ secant with $C(0, t)$.

Consider now the function $f$ constructed in Proposition IV.3 which, by construction, belongs to $H(E)$ and has no zero and no pole in $d(0, s^{-})$. Consequently, $|f(x)| = |f(0)| = 1 \forall x \in d(0, s^{-})$. Moreover, we have $\lim_{g \in G} f(x) = 0$, hence $\phi'(f) = 0$. Let $g = \theta(f)$. Then $\psi(g) = \psi'(f) = 1$ and $\phi(g) = \phi'(f) = 0$, which ends the proof. $\square$

**Proposition IV.5:** Let $\mathcal{M}$ be a non-principal maximal ideal of $A$ and let $\mathcal{U}$ be an ultrafilter thinner than $\mathcal{G}_{\mathcal{M}}$. Then $\varphi_{\mathcal{U}}$ belongs to the closure of $\mathrm{Mult}_1(A, \| \cdot \|)$ in $\mathrm{Mult}_m(A, \| \cdot \|)$.

*Proof.* Let $V$ be neighborhood of $\varphi_{\mathcal{U}}$ in $\mathrm{Mult}(A, \| \cdot \|)$. It contains some set of the form $\mathcal{V}(\varphi_{\mathcal{U}}, f_1, \dots, f_q, \epsilon)$, where $f_1, \dots, f_q \in A$ and $\epsilon > 0$, with respect to the topology of pointwise convergence i.e.

$$ \mathcal{V}(\varphi_r, f_1, \dots, f_q, \epsilon) = \{ \phi \in \mathrm{Mult}(A, \| \cdot \|) \mid \left| \varphi_r(f_j) - \phi f_j \right|_{\infty} \le \epsilon, j = 1, \dots, q, q \in \mathbb{N}^* \}. \text{ For each} \\ j = 1, \dots, q, \text{ there exists } E_j \in \mathcal{U} \text{ such that } \left| f_j(x) - \varphi_{\mathcal{U}}(f_j) \right|_{\infty} \le \epsilon \quad \forall x \in E_j. \text{ Let } E = \bigcap_{j=1}^{q} E_j. $$
Then

$$ \left| |f_j(x)| - \varphi_{\mathcal{U}}(f_j) \right|_{\infty} \le \epsilon \quad \forall x \in E, \forall j = 1, \dots, q. $$

Consequently, $\varphi_a$ belongs to $\mathcal{V}(\varphi_{\mathcal{U}}, f_1, \dots, f_q, \epsilon)$ for all $a \in E$.

<figure><img src="image_2.png" /></figure>

**Corollary IV.5.1:** Let $\mathcal{M}$ be a univalent non-principal maximal ideal of $A$ and let $\phi \in \mathrm{Mult}_m(A, \|\cdot\|)$ satisfy $\mathrm{Ker}(\phi) = \mathcal{M}$. Then $\phi$ is of the form $\phi(f) = \lim_{\mathcal{U}} |f(x)|$ with $\mathcal{U}$ a coroner ultrafilter and such that $\mathcal{J}(\mathcal{U}) = \mathcal{M}$. Moreover, $\phi$ belongs to the closure of $\mathrm{Mult}_1(A, \|\cdot\|)$ in $\mathrm{Mult}_m(A, \|\cdot\|)$.

**Proposition IV.6:** Let $\mathcal{U}$ be a coroner ultrafilter on $D$, let $f \in A \setminus \mathcal{J}(\mathcal{U})$ be non-invertible in $A$, such that $\|f\| \le 1$ and let $g \in A$, $h \in \mathcal{J}(\mathcal{U})$ be such that $fg = 1 + h$. Let $\tau = \varphi_{\mathcal{U}}(f)$, let $\epsilon \in ]0, \tau[$ and let $\Lambda = \{x \in D \mid |f(x)g(x)| - 1|_{\infty} < \epsilon, \mid |f(x)| - \tau|_{\infty} < \epsilon\}$.

Suppose that there exist a function $\tilde{h} \in A$ admitting for zeros in $D$ the zeros of $h$ in $D \setminus \Lambda$ and a function $\bar{h} \in A$ admitting for zeros the zeros of $h$ in $\Lambda$, each counting multiplicities, so that $h = \tilde{h}h$. Then $|\tilde{h}(x)|$ has a strictly positive lower bound in $\Lambda$ and $\bar{h}$ belongs to $\mathcal{J}(\mathcal{U})$.

Moreover, there exists $\omega \in ]0, \tau[$ such that $\omega \le \inf\{\max(|f(x)|, |\bar{h}(x)|) \mid x \in D\}$. Further, for every $a \in d(0, (\tau - \epsilon))$, we have $\omega \le \inf\{\max(|f(x) - a|, |\bar{h}(x)|) \mid x \in D\}$.

*Proof.* Let $u \in \Lambda$ and let $s$ be the distance of $u$ from $\mathbb{K} \setminus \Lambda$. So, the disk $d(u, s^-)$ is included in $\Lambda$, hence $fg$ has no zero inside this disk. Consequently, $|f(x)g(x)|$ is a constant $b$ in $d(u, s^-)$. Consider the family $F_u$ of radii of circles $C(u, r)$, containing at least one zero of $fg$. By Theorem I.4, $F_u$ has no cluster point different from 1. Consequently, there exists $\rho \ge s$ such that $fg$ admits at least one zero in $C(u, \rho)$ and admits no zero in $d(u, \rho^-)$. And then $|f(x)g(x)|$ is a constant $c$ in $d(u, \rho^-)$. But then, at $u$ we see that $b = c$ and therefore $d(u, \rho^-)$ is included in $\Lambda$. Hence $\rho = s$ and therefore $fg$ admits at least one zero $\alpha$ in $C(u, s)$. Thus, at $\alpha$ we have $h(\alpha) = -1$. Therefore, in the disk $d(\alpha, s^-)$ we can check that $\varphi_{\alpha,s}(h) \ge 1$. But by Theorem I.9, we have $\varphi_{\alpha,s}(h) = \varphi_{u,s}(h)$, hence $\varphi_{u,s}(h) \ge 1$.

Now,

$$ \frac{\|h\|}{\varphi_{u,s}(h)} = \frac{\|\tilde{h}\|}{\varphi_{u,s}(\tilde{h})} \frac{\|\bar{h}\|}{\varphi_{u,s}(\bar{h})} \ge \frac{\|\tilde{h}\|}{\varphi_{u,s}(\tilde{h})}. $$

Therefore, since $\varphi_{u,s}(h) \ge 1$, we obtain

$$ (1) \qquad \frac{\|\tilde{h}\|}{\varphi_{u,s}(\tilde{h})} \le \|h\|. $$

But since by definition $d(u, s^-)$ is included in $\Lambda$, $\tilde{h}$ has no zero in this disk, hence $|\tilde{h}(x)|$ is constant and equal to $\varphi_{u,s}(\tilde{h})$. Consequently, by (1) we obtain $\frac{\|\tilde{h}\|}{|\tilde{h}(u)|} \le \|h\|$ and therefore

$$ |\tilde{h}(u)| \ge \frac{\|\tilde{h}\|}{\|h\|} \quad \forall u \in \Lambda. $$

This shows that $\tilde{h}$ does not belong to $\mathcal{J}(\mathcal{U})$, hence, $\varphi_{\mathcal{U}}(\tilde{h}) \ne 0$. Consequently, $\varphi_{\mathcal{U}}(\bar{h}) = 0$.
Now, by hypothesis, we have $fg - \tilde{h}\tilde{h} = 1$. Since both $g, \tilde{h}$ belong to $A$ and therefore are bounded in $D$, it is obvious that $\inf\{\max(|f(x)|, |\tilde{h}(x)|) \mid x \in D\} > 0$. So, we may obviously choose $\omega \in ]0, \tau - \epsilon[$ such that

$$ (2) \qquad \omega \le \inf\{\max(|f(x)|, |\tilde{h}(x)|) \mid x \in D\}. $$

Let us now show that for every $a \in d(0, (\tau - \epsilon))$, we have $\omega \le \inf\{\max(|f(x) - a|, |\tilde{h}(x)|) \mid x \in D\}$.

Let $\Lambda' = \{x \in D \mid |f(x)| \ge \tau - \epsilon\}$ and let $a \in d(0, (\tau - \epsilon)^-)$. When $\beta$ lies in $\Lambda'$, we have $|f(\beta)| > |a|$, hence by (2), $\max(|f(\beta) - a|, |\tilde{h}(\beta)|) \ge \omega$ because by (2), either $\omega \le |\tilde{h}(\beta)|$, or $\omega \le |f(\beta)| = |f(\beta) - a|$.

Now, let $\beta$ lie in $D \setminus \Lambda'$ and let $t$ be the distance from $\beta$ to $\Lambda'$. Since $D \setminus \Lambda'$ is open, $t$ is $> 0$. Consider $\varphi_{\beta,t}(f)$. Either there exists $\mu \in \Lambda'$ such that $|\beta - \mu| = t$ and then $\varphi_{\beta,t}(f) \ge |f(\mu)| \ge \tau - \epsilon$, or there exists a sequence $(x_n)_{n \in \mathbb{N}} \in \Lambda'$ such that $\lim_{n \to +\infty} |\beta - x_n| = t$ and $|x_n - \beta| > t$. Suppose that we are in the second case: there exists a sequence $(x_n)_{n \in \mathbb{N}} \in \Lambda'$ such that $\lim_{n \to +\infty} |\beta - x_n| = t$ and $|x_n - \beta| > t$. Then the sequence is thinner than the circular filter of center $\beta$ and diameter $t$, hence

$$ \lim_{n \to +\infty} |f(x_n)| = \varphi_{\beta,t}(f) $$

hence $\varphi_{\beta,t}(f) \ge \tau - \epsilon$ again. If $f$ has no zero in $d(\beta, t^-)$, then $|f(x)|$ is a constant in that disk, hence of course $\varphi_{\beta,t}(f) < \tau - \epsilon$, a contradiction. Consequently, $f$ must have a zero $\gamma$ in $d(\beta, t^-)$. Therefore, due to (2), we have $|\tilde{h}(\gamma)| \ge \omega$. But since by definition, $\Lambda \subset \Lambda'$, the zeros of $\tilde{h}$ belong to $\Lambda'$. And since $d(\beta, t^-) \cap \Lambda' = \emptyset$ actually $\tilde{h}$ has no zero in $d(\beta, t^-)$. Consequently $|\tilde{h}(x)|$ is constant in $d(\beta, t^-)$ and hence $|\tilde{h}(\beta)| \ge \omega$, which completes the proof. $\square$

The following basic Proposition is easily checked and is an application of Proposition 10 in [3]:

**Proposition IV.7:** Let $S$ be a set and let $E$ be a subset. Let $\mathcal{F}$ be an ultrafilter on $E$. Then the filter $\tilde{\mathcal{F}}$ on $S$ with basis $\mathcal{F}$ is an ultrafilter inducing on $E$ the ultrafilter $\mathcal{F}$.

**Corollary IV.7.1:** Let $S$ be a set and let $E$ be subset of $S$. Let $\mathcal{F}$ be an ultrafilter on $E$ and let $\tilde{\mathcal{F}} = \mathcal{G}$ be the ultrafilter on $S$ having $\mathcal{F}$ as a basis of filter. Let $f$ be a function defined on $S$ with values in a compact topological space $T$. Then $\lim_{\mathcal{G}} f(x) = \lim_{\mathcal{F}} f(x)$.

*Proof.* Suppose that $f$ admits distinct limits on $\mathcal{F}$ and $\mathcal{G}$. Then $\mathcal{F}$ is a basis of a filter on $S$ that is not secant with $\mathcal{G}$, a contradiction since $\mathcal{F}$ is the ultrafilter induced by $\mathcal{G}$ on $E$. $\square$

We can now prove Proposition IV.8 in the general context of a field that is not supposed to be spherically complete:

**Proposition IV.8:** Let $\mathcal{M}$ be a non-principal maximal ideal of $A$ and let $\mathcal{U}$ be an ultrafilter on $D$ such that $\mathcal{M} = \mathcal{J}(\mathcal{U})$. Let $f \in A \setminus \mathcal{M}$ satisfy $\|f\| < 1$, let $\tau = \varphi_{\mathcal{U}}(f)$ and let $\epsilon \in ]0, \tau[$. There exists $c > 0$ such that, for every $a \in d(0, \tau - \epsilon)$, there exists $g_a \in A$ satisfying $(f - a)g_a - 1 \in \mathcal{M}$ and $\|g_a\| \le c$.
Proof. Suppose first that $f$ is invertible in $A$. By Theorem I.4, $|f(x)|$ is a constant and hence is equal to $\tau$. Therefore, $|f(x) - a| = \tau \forall a \in d(0, \tau - \epsilon)$. Consequently, $f - a$ is invertible and its inverse $g_a$ satisfies $\|g_a\| = \tau^{-1}$. Thus, we only have to show the claim when $f$ is not invertible. Since $f$ does not belong to $M$, we can find $g \in A$ and $h \in M$ such that $fg = 1 + h$ with $h \in M$.

Let $\widehat{\mathbb{K}}$ be an algebraically closed spherically complete extension of $\mathbb{K}$ and let $\widehat{D}$ be the disk $\{x \in \widehat{\mathbb{K}} \mid |x| < 1\}$. Let $\widehat{A}$ be the algebra of bounded power series converging in $\widehat{D}$ with coefficients in $\widehat{\mathbb{K}}$. $\mathcal{U}$ makes a basis of a filter $\widehat{\mathcal{U}}$ on $\widehat{D}$ and by definition, $\mathcal{U}$ is the filter induced by $\widehat{\mathcal{U}}$ on $D$. By Corollary IV.7.1 $\widehat{\mathcal{U}}$ is an ultrafilter on $\widehat{D}$.

Consider now $f$ as an element of $\widehat{A}$. Then $\widehat{\mathcal{U}}$ defines an element $\psi$ of $\mathrm{Mult}(\widehat{A}, \| \cdot \|)$ as $\psi(\ell) = \lim_{\widehat{\mathcal{U}}} |\ell(x)|, \forall \ell \in \widehat{A}$. Consequently, by Corollary IV.7.1 $\tau$ is equal to $\lim_{\widehat{\mathcal{U}}} |f(x)|$. Let

$$ \Lambda = \{x \in \widehat{D} \mid |f(x)g(x)| - 1|_{\infty} < \epsilon, \ |f(x)| - \tau|_{\infty} < \epsilon\}. $$

Since $\widehat{\mathbb{K}}$ is spherically complete, by Proposition III.9, we can factorize $h$ in the form $\overline{hh}$ where $\overline{h} \in \widehat{A}$ is a function admitting for zeros in $\widehat{D}$ the zeros of $h$ in $\widehat{D} \setminus \Lambda$ and $\overline{h} \in \widehat{A}$ is a function admitting for zeros the zeros of $h$ in $\Lambda$, each counting multiplicities. Moreover, we can choose $\overline{h}$ so that $\|\overline{h}\| < 1$.

Now, in the field $\widehat{\mathbb{K}}$, by Proposition IV.6, there exists $\omega > 0$ such that for every $a \in \widehat{d}(0, (\tau - \epsilon))$, we have $\omega \le \inf\{\max(|f(x) - a|, |\overline{h}(x)|) \mid x \in \widehat{D}\}$. This implies that $\inf\{\max(|f(x) - a|, |\overline{h}(x)|) \mid x \in D\} \ge \omega \forall a \in \widehat{d}(0, \tau - \epsilon)$. We notice that $\|f - a\| < 1$ for every $a \in \widehat{d}(0, \tau - \epsilon)$, so we may apply Proposition IV.2 and obtain a bound $b$ only depending on $f$ and $h$ and functions $\ell_a$, $h_a \in \widehat{A}$ such that $(f - a)\ell_a + \overline{hh}_a = 1$, with

$$ (1) \qquad \|\ell_a\| < b, \quad \|h_a\| < b \ \forall a \in \widehat{d}(0, \tau - \epsilon). $$

By hypothesis we have $\lim_{\widehat{\mathcal{U}}} h(x) = 0$. Hence by Corollary IV.7.1, on $\widehat{D}$ we have $\lim_{\widehat{\mathcal{U}}} h(x) = 0$. Then, by Corollary IV.7.1 we have $\lim_{\widehat{\mathcal{U}}} \overline{h}(x) = 0$ hence, on $D$,

$$ (2) \qquad \lim_{\widehat{\mathcal{U}}} \overline{hh}_a(x) = 0 \ \forall a \in d(0, \tau - \epsilon). $$

Now, let us fix $a \in d(0, \tau - \epsilon)$. Let $G$ be the closed $\mathbb{K}$-vector subspace of $\widehat{\mathbb{K}}$ (considered as a $\mathbb{K}$-Banach space), linearly generated over $\mathbb{K}$ by 1 and all coefficients of $\ell_a$. Take $\eta > 0$ such that $(1 + \eta) \max(\|\ell_a\|, \|h_a\|) \le b$. We notice that $G$ is a $\mathbb{K}$-Banach space of countable type, hence there exists a $\mathbb{K}$-linear mapping $\Xi$ from $G$ to $\mathbb{K}$ of norm $\le 1 + \eta$, such that $\Xi(1) = 1$ [19]. Let $F$ be the closed $\mathbb{K}$-vector subspace of $\widehat{A}$ consisting of all power series with coefficients in $E$. Then $F$ is a $A$-module and $\Xi$ has continuation to a $A$-linear mapping $\widehat{\Xi}$ from $F$ to $A$ defined as $\widehat{\Xi}(\sum_{n=0}^{\infty} b_n x^n) = \sum_{n=0}^{\infty} \Xi(b_n)x^n$. This mapping $\widehat{\Xi}$ has a norm bounded by $1 + \eta$. Set $g_a = \widehat{\Xi}(\ell_a)$. Then by (1) we have

$$ (3) \qquad \|g_a\| \le b(1 + \eta) \ \forall a \in d(0, \tau - \epsilon). $$

On the other hand, by construction, for every $z \in G$, we have $|\widehat{\Xi}(z)| \le |z|(1 + \eta)$: that holds particularly for elements of $G \cap D$. Now, since $(f - a)(l_a) - \overline{hh}_a = 1$, for all $x \in D$, we have $l_a(x) \in G$, $f(x) - a \in K$ and hence $\overline{hh}_a(x)$ belongs to $G$. Therefore the inequality applies and shows that $|\widehat{\Xi}(\overline{hh}_a)(x)| \le |(\overline{hh}_a)(x)|(1 + \eta)$, hence by (2) we can derive $\lim_{\widehat{\mathcal{U}}} \widehat{\Xi}(\overline{hh}_a)(x) = 0 \ \forall a \in d(0, \tau - \epsilon)$.
But since $\hat{\Xi}$ is a $\mathcal{A}$-module linear mapping, we have $\hat{\Xi}((f-a)h_a-1) = (f-a)g_a-1$. Consequently, $\lim_{\mathcal{U}} |(f(x)-a)g_a(x)-1| = 0 \quad \forall a \in d(0, \tau - \epsilon)$ and hence $(f-a)g_a-1$ belongs to $\mathcal{J}(\mathcal{U})$. Putting $c = b(1+\eta)$, by (3) we are done. $\square$

**Theorem IV.9:** *A is multbijective.*

*Proof.* Suppose $A$ is not multbijective and let $\mathcal{M}$ be a maximal ideal which is not univalent. Let $F$ be the quotient field $\frac{A}{\mathcal{M}}$, let $\theta$ be the canonical surjection from $A$ onto $F$ and let $\|\cdot\|_q$ be the $\mathbb{K}$-Banach algebra quotient norm of $F$. By Corollary IV.2.2 there exists an ultrafilter $\mathcal{U}$ on $D$ such that $\mathcal{M} = \mathcal{J}(\mathcal{U})$. Thus, there exists $\psi \in \text{Mult}(A, \|\cdot\|)$ such that $\text{Ker}(\psi) = \mathcal{M}$ and $\psi \neq \varphi_{\mathcal{U}}$. Consequently, there exists $f \in A$ such that $\psi(f) \neq \varphi_{\mathcal{U}}(f)$, with $\psi(f) \neq 0$, $\varphi_{\mathcal{U}}(f) \neq 0$. We shall check that we may also assume $\psi(f) < \varphi_{\mathcal{U}}(f)$. Indeed, suppose $\psi(f) > \varphi_{\mathcal{U}}(f)$. Let $g \in A$ be such that $\theta(g) = \theta(f)^{-1}$. Then we can see that $\psi(g) = \psi(f)^{-1}$, $\varphi_{\mathcal{U}}(g) = (\varphi_{\mathcal{U}}(f))^{-1}$, therefore $\psi(g) < \varphi_{\mathcal{U}}(g)$. Thus, we may assume $\psi(f) < \varphi_{\mathcal{U}}(f)$ without loss of generality. Similarly, we may obviously assume that $\|f\| < 1$.

By construction, $\varphi_{\mathcal{U}}$ factorizes in the form $\phi_1 \circ \theta$ and similarly, $\psi$ factorizes in the form $\phi_2 \circ \theta$ with $\phi_1, \phi_2 \in \text{Mult}(F, \|\cdot\|_q)$. So, on $F$ we have $\phi_1(\theta(f)) > \phi_2(\theta(f))$.

Let $\sigma = \varphi_{\mathcal{U}}(f)$ and let $\epsilon \in ]0, \sigma[$. By Proposition IV.8, there exists $c > 0$ such that, for every $a \in d(0, \sigma - \epsilon)$, there exists $g_a \in A$ satisfying $(f-a)g_a - 1 \in \mathcal{M}$ and $\|g_a\| \le c$. Now, $\theta(g_a) = (\theta(f-a))^{-1}$. Thus, $\|(\theta(f-a))^{-1}\|_q \le c \forall a \in d(0, \sigma-\epsilon)$. Therefore, by applying Proposition IV.4 to the Banach $\mathbb{K}$-algebra $F$, we can see that there exists $y \in F$ such that $\phi_1(y) = 1, \phi_2(y) = 0$. Therefore, taking $g \in A$ such that $\theta(g) = y$, we get $\varphi_{\mathcal{U}}(g) = 0, \psi(g) = 1$, a contradiction to the hypothesis $\text{Ker}(\varphi_{\mathcal{U}}) = \text{Ker}(\psi)$. This finishes showing that $A$ is multbijective. $\square$

**Corollary IV.9.1:** For every $\phi \in \text{Mult}_m(A, \|\cdot\|) \setminus \text{Mult}_1(A, \|\cdot\|)$, there exists a coroner ultrafilter $\mathcal{U}$ such that $\phi(f) = \lim_{\mathcal{U}} |f(x)| \quad \forall f \in A$.

**Corollary IV.9.2:** $\text{Mult}_1(A, \|\cdot\|)$ is dense in $\text{Mult}_m(A, \|\cdot\|)$.

**Conjecture:** $\text{Mult}_1(A, \|\cdot\|)$ is dense in $\text{Mult}(A, \|\cdot\|)$.

**Acknowledgement:** I am very grateful to Bertin Diarra for many advises and to Jesus Araujo for sending me Theorem II.16.

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Alain Escassut
Laboratoire de Mathématiques Blaise Pascal UMR 6620
Université Clermont Auvergne
(Clermont-Ferrand)
Les Cézeaux
24 Avenue des Landais
63178 AUBIERE CEDEX
FRANCE
Alain.Escassut@math.univ-bpclermont.fr