Normed Dual Algebras

Abstract

This article is devoted to the investigation of dual and annihilator normed algebras. Their structure is studied in the paper. Extensions of algebras and fields are considered and by using them, core radicals and radicals are investigated. Moreover, for this purpose ∗-algebras and finely regular algebras are also studied. Relations with operator theory and realizations of these algebras by operator algebras are outlined.

1. Introduction

Algebras and operator algebras over the real field $\mathbf{R}$ and the complex field $\mathbf{C}$ were intensively studied. They have found many-sided applications. For these a lot of results already were obtained (see, for example, [1,2,3,4] and references therein). Among them, dual algebras and annihilator algebras play very important roles. However, for such algebras over ultranormed fields, comparatively little is known because of their specific features and additional difficulties arising from structure of fields [5,6,7,8,9,10,11,12].

Many results in the classical case use the fact that the real field $\mathbf{R}$ has the linear ordering compatible with its additive and multiplicative structure and that the complex field $\mathbf{C}$ is algebraically closed, norm complete and locally compact and is the quadratic extension of $\mathbf{R}$, also that there are not any other commutative fields with archimedean multiplicative norms and complete relative to their norms besides these two fields.

For comparison, in the non-archimedean case the algebraic closure of the field ${\mathbf{Q}}_p$ of p-adic numbers is not locally compact. Each ultranormed field can be embedded into a larger ultranormed field. There is not any ordering of an infinite ultranormed field such as ${\mathbf{Q}}_{\mathbf{p}}$, ${\mathbf{C}}_{\mathbf{p}}$ or ${\mathbf{F}}_{\mathbf{p}}\left(t\right)$ compatible with its algebraic structure.

In their turn, non-archimedean analysis, functional analysis and representations theory of groups over non-archimedean fields developed fast in recent years [11,13,14,15,16,17,18]. This is motivated not only by the needs of mathematics, but also their applications in other sciences such as physics, quantum mechanics, quantum field theory, informatics, etc. (see, for example, [19,20,21,22,23,24,25,26] and references therein).

This article is devoted to dual algebras and annihilator algebras over non-archimedean fields. Their structure is studied in the paper. Preliminary results on operator algebras are given in Section 2. Extensions of algebras and fields are considered and using them core radicals and radicals are investigated in Section 3 (see Propositions 1–4 and Theorem 4). It appears to be a specific ultrametric feature to refine to a new notion of a core radical and investigate it. Moreover, for this purpose also ∗-algebras and finely regular algebras are studied. Theorems 5, 7, 8, Propositions 8–11 about ideals, idempotents of algebras and their orthogonality are proven. Division subalgebras related with idempotents are investigated (see Theorem 9). Relations with operator theory and realizations of these algebras by operator algebras are outlined.

In this article a definition or a proposition or a theorem number m in the same section number n is referred to as m, in another section as n.m. A formula number $\left(m\right)$ in a subsection number k is referred to as $\left(m\right)$, in another subsection as $k\left(m\right)$, etc.

All main results of this paper are obtained for the first time. They can be used for further studies of ultranormed algebras and operator algebras on non-archimedean Banach spaces, their cohomologies, spectral theory of operators, the representation theory of groups, algebraic geometry, PDE, applications in the sciences, etc.

2. Normed Algebras and ∗-Algebras

To avoid misunderstandings we first give our definitions and notations (which are used in this paper), because they slightly vary in the literature. Although, a reader familiar with books [8,11,18,27] may skip Section 2.1 and Section 2.2.

2.1. Notation

Let F be an infinite field supplied with a multiplicative non-trivial ultranorm ${|·|}_F$ relative to which it is complete, so that F is non-discrete and ${\mathsf{\Gamma }}_F\subset (0,\infty )=\{r\in \mathbf{R}:0<r<\infty \}$, where ${\mathsf{\Gamma }}_F:={\left\{\right|x|}_F:x\in F\setminus \left\{0\right\}\}$, whilst as usually ${\left|x\right|}_F=0$ if and only of $x=0$ in F, also ${|x+y|}_F\le {max\left(\right|x|}_F{,\left|y\right|}_F)$ and ${\left|xy\right|}_F={\left|x\right|}_F{\left|y\right|}_F$ for each x and y in F. We consider fields with multiplicative ultranorms if something other will not be specified.

If F is such a field, we denote by $E_n\left(F\right)$ the class containing F and all ultranormed field extensions G of F so that these G are norm complete and ${|·|}_G{|}_F={|·|}_F$. By $E_n$ we denote the class of all infinite non-trivially ultranormed fields F which are norm complete.

Henceforward, the terminology is adopted that a commutative field is called shortly a field, while a noncommutative field is called a skew field or a division algebra.

2.2. Definitions

By $c_0(\alpha ,F)$ is denoted a Banach space consisting of all vectors $x=(x_j:\forall j\in \alpha x_j\in F)$ satisfying the condition

$card\{j\in \alpha :|x_j|>ϵ\}<{\aleph }_0$ for each $ϵ>0$

and furnished with the norm

$\left(1\right)$ $\left|x\right|={sup}_{j\in \alpha }\left|x_j\right|$,

where $\alpha$ is a set. For locally convex spaces X and Y over F the family of all linear continuous operators $A:X\to Y$ we denote by $L(X,Y)$. For normed spaces X and Y the linear space $L(X,Y)$ is supplied with the operator norm

$\left(2\right)$ $\left|A\right|:={sup}_{x\in X\setminus \left\{0\right\}}\left|Ax\right|/\left|x\right|$.

For locally convex spaces X and Y over F the space $L(X,Y)$ is furnished with a topology induced by a family of semi-norms

$\left(3\right)$ ${\left|A\right|}_{p,q}:={sup}_{x\in X,p\left(x\right)>0}q\left(Ax\right)/p\left(x\right)$

for all continuous semi-norms p on X and q on Y.

Speaking about Banach spaces and Banach algebras, we stress that a field over which it is defined is ultranorm complete.

If $X=c_0(\alpha ,F)$, then to each $A\in L(X,X)$ an infinite matrix $(A_{i,j}:i\in \alpha ,j\in \alpha )$ corresponds in the standard basis $\{e_j:j\in \alpha \}$ of X, where

$\left(4\right)$ $x={\sum }_j{x}_j{e}_j$

for each $x\in X=c_0(\alpha ,F)$.

For a subalgebra V of $L(X,X)$ an operation $B\mapsto B^t$ from V into $L(X,X)$ will be called a transposition operation if it is induced by that of its infinite matrix such that ${(aA+bB)}^t=a{A}^t+b{B}^t$ and ${\left(AB\right)}^t=B^t{A}^t$ and ${\left(A^t\right)}^t=A$ for every A and B in V and a and b in F, that is ${\left(A^t\right)}_{i,j}=A_{j,i}$ for each i and j in $\alpha$. Then $V^t:=\{A:A=B^t,B\in V\}$.

An operator A in $L(X,X)$ is called symmetric if $A^t=A$.

By $L_0(X,X)$ is denoted the family of all continuous linear operators $U:X\to X$ matrices $(U_{i,j}:i\in \alpha ,j\in \alpha )$ of which fulfill the conditions

$\left(5\right)$$\forall i\exists {lim}_j{U}_{j,i}=0$ and $\forall j\exists {lim}_i{U}_{j,i}=0$.

For an algebra A over F, $F\in E_n$, it is supposed that an ultranorm ${|·|}_A$ on A satisfies the conditions:

${\left|a\right|}_A\in ({\mathsf{\Gamma }}_F\cup \left\{0\right\})$ for each $a\in A$, also

${\left|a\right|}_A=0$ if and only if $a=0$ in A,

${\left|ta\right|}_A={\left|t\right|}_F{\left|a\right|}_A$ for each $a\in A$ and $t\in F$,

${|a+b|}_A\le {max\left(\right|a|}_A{,\left|b\right|}_A)$ and

${\left|ab\right|}_A\le {\left|a\right|}_A{\left|b\right|}_A$ for each a and b in A.

For short it also will be written $|·|$ instead of ${|·|}_F$ or ${|·|}_A$.

A subset V of a Banach space X over F is called a compactoid, if for each $ϵ>0$ there exists a finite subset Y in X such that $V\subset B(X,0,ϵ)+\overline{Co}\left(Y\right)$, where $B(X,w,r):=\{x\in X:|x-w|\le r\}$ denotes a closed ball in X containing a point $w\in X$ and of a radius $r>0$, $\overline{Co}\left(U\right)=\{x\in X:x=a_1{u}_1+\dots +a_j{u}_j+\dots ,\forall j{a}_j\in F,|a_j|\le 1,u_j\in U,{lim}_j{a}_j{u}_j=0\}$, where $j\in \mathbf{N}$.

Theorem 1.

Let V be a subalgebra in $L(X,X)$ such that $V^t=V$. Then J is a left or right ideal in V if and only if $J^t$ is a right or left respectively ideal in V.

Proof. 

For each A and B in V we get ${\left(A{B}^t\right)}^t=B{A}^t$ and $B^t\in V$ and $A^t\in V$, since $V^t=V$. Therefore, for a right ideal J we deduce that $\forall A\in J\forall B\in V(A{B}^t\in J)\iff (B{A}^t\in J^t)$. Moreover, $\forall B\in V\exists U\in V,U^t=B$. The similar proof is for a left ideal J. □

Theorem 2.

Let $X=c_0(\alpha ,F)$, where $F\in E_n$. Then the class $L_c(X,X)$ of all compact operators $T:X\to X$ is a closed ideal in $L(X,X)$, also $L_{t,c}(X,X):=\{A:A\in L_c(X,X)\&A^t\in L_c(X,X)\}$ is a closed ideal in $L_0(X,X)$.

Proof. 

By the definition of a compact operator $T\in L_c(X,X)$ if and only if for the closed unit ball B (of radius 1 and with $0\in B$) in X its image $TB$ is a compactoid in X (see Ch. 4 in [11]). Therefore, if $A\in L(X,X)$, then $AB$ is bounded and convex in X, consequently, $TA\in L_c(X,X)$. On the other hand, if C is a compactoid in X, then $AC$ is a compactoid in X, hence $AT\in L_c(X,X)$. Thus $L_c(X,X)$ is the ideal in $L(X,X)$.

Suppose that $T_n$ is a fundamental sequence in $L_c(X,X)$ relative to the operator norm topology. Then its limit $T={lim}_n{T}_n$ exists in $L(X,X)$, since $L(X,X)$ is complete relative to the operator norm topology. Let $ϵ>0$. There exists $m\in \mathbf{N}$ such that $|T-T_n|<ϵ$ for each $n>m$. Since $T_n$ is the compact operator, there exists a finite set $a_1,\dots ,a_l$ in X such that $\left(T_{n}B\right)\subseteq B(X,0,ϵ)+\overline{Co}(a_1,\dots ,a_l)$, where

$Co(a_1,\dots ,a_l)=\{x\in X:x=t_1{a}_1+\dots +t_l{a}_l,t_1\in B(F,0,1,),$

$\dots ,t_l\in B(F,0,1)\}$ and $B(X,y,r):=\{z\in X:|z-y|\le r\}$,

$0<r$, $\overline{U}$ denotes the closure of a set U in a topological space. Therefore, if $x\in TB$, then there exists $y\in T_{n}B$ such that $|x-y|<ϵ$, consequently, $x\in B(X,0,ϵ)+\overline{Co}(a_1,\dots ,a_l)$ due to the ultrametric inequality, hence

$TB\subseteq B(X,0,ϵ)+\overline{Co}(a_1,\dots ,a_l)$.

This means that the operator T is compact. Thus $L_c(X,X)$ is closed in $L(X,X)$.

The mapping $U\mapsto U^t$ is continuous from $L(X,X)$ into $L(X,X)$, since $\left|U\right|=|U^t|={sup}_{i\in \alpha ,j\in \alpha }\left|U_{i,j}\right|$ for each $U\in L(X,X)$. In view of Theorem 4.39 in [11] for each $A\in L_{t,c}(X,X)$ and $ϵ>0$ operators S and R in $L(X,X)$ exist such that $SX$ and $RX$ are finite dimensional spaces over F and $|A-S|<ϵ$ and $|A^t-R|<ϵ$. Therefore, $L_{t,c}(X,X)\subset L_0(X,X)$ and $L_{t,c}(X,X)$ is the ideal in $L_0(X,X)$. On the other hand, $L_0(X,X)$ is closed in $L(X,X)$, consequently, $L_{t,c}(X,X)$ is closed in $L_0(X,X)$. □

Definition 1.

Suppose that F is an infinite field with a nontrivial non-archimedean norm such that F is norm complete, $F\in E_n$ and of the characteristic $char\left(F\right)\ne 2$ and $B_2=B_2\left(F\right)$ is the commutative associative algebra with one generator $i_1$ such that $i_1^2=-1$ and with the involution ${\left(v{i}_1\right)}^{*}=-v{i}_1$ for each $v\in F$. Let A be a subalgebra in $L(X,X)$ such that A is also a two-sided $B_2$-module, where $X=c_0(\alpha ,F)$ is the Banach space over F, α is a set. We say that A is a ∗-algebra if there is

(1)

a continuous bijective surjective F-linear operator $\mathcal{I}:A\to A$ such that

(2)

$\mathcal{I}\left(ab\right)=\left(\mathcal{I}b\right)\left(\mathcal{I}a\right)$ and

(3)

$\mathcal{I}\left(ga\right)=\left(\mathcal{I}a\right)g^{*}$ and $\mathcal{I}\left(ag\right)=g^{*}\left(\mathcal{I}a\right)$

(4)

$\mathcal{I}\mathcal{I}a=a$

(5)

$\left(\theta \right(y\left)\right)\left(ax\right)=\left(\theta \right(\left(\mathcal{I}a\right)y\left)\right)\left(x\right)$

for every a and b in A and $g\in B_2$ and x and y in X, where $\theta :X\hookrightarrow X^{\prime }$ is the canonical embedding of X into the topological dual space $X^{\prime }$ so that $\theta \left(y\right)x={\sum }_{j\in \alpha }{y}_j{x}_j$. For short we can write $a^{*}$ instead of $\mathcal{I}a$. The mapping $\mathcal{I}$ we call the involution. An element $a\in A$ we call self-adjoint if $a=a^{*}$.

Lemma 1.

Let A be a subalgebra of $L(X,X)$ with transposition and $A^t=A$, where $X=c_0(\alpha ,F)$, $F\in E_n$, $char\left(F\right)\ne 2$. Then the minimal ∗-algebra K generated by A and $B_2$ has an embedding ψ into $L(U,U)$ such that $\psi \left(B_2\right)$ is contained in the center $Z\left(K\right)$ of K, where $U=X\oplus X$.

Proof. 

We put $\psi \left(a\right):=\left(\genfrac{}{}{0pt}{}{a0}{0a}\right)$ and $\psi \left(a{i}_1\right):=\left(\genfrac{}{}{0pt}{}{0a}{-a0}\right)$ and ${\left(\psi \left(a\right)\right)}^{*}:=\left(\genfrac{}{}{0pt}{}{a^{t}0}{0a^t}\right)$ and ${\left(\psi \left(a{i}_1\right)\right)}^{*}:=\left(\genfrac{}{}{0pt}{}{0-a^t}{a^{t}0}\right)$ for each $a\in A$, since $a^t\in A$. Therefore, the minimal algebra containing $\psi \left(A\right)$ and $\psi \left(A{i}_1\right)$ is the ∗-subalgebra in $L(U,U)$. Then ${\left(\psi \left(i_1\right)\right)}^2=-I_U$ and $\psi \left(i_1\right)=\left(\genfrac{}{}{0pt}{}{0I_X}{-I_{X}0}\right)$, where $I_X$ is the unit operator on X. Thus $\psi \left(a{i}_1\right)=\psi \left(a\right)\psi \left(i_1\right)=\psi \left(i_1\right)\psi \left(a\right)=\psi \left(i_{1}a\right)$ for each $a\in A$ and hence $\psi \left(B_2\right)\subset Z\left(K\right)$, where $Z\left(K\right)$ denotes the center of the algebra K. □

Lemma 2.

Let A be a ∗-algebra over F (see Definition 1), then each element $a\in A$ has the decomposition $a=a_0+a_1{i}_1$ with $a_0^{*}=a_0$ and $a_1^{*}=a_1$ in A.

Proof. 

Put $a_0=(a+a^{*})/2$, $a_1=(a{i}_1^{*}+i_1{a}^{*})/2$, since $char\left(F\right)\ne 2$. Then $a_0$ and $a_1$ are in A, since A is the two-sided $B_2$-module and $a^{*}\in A$ and $1\in B_2$ and $i_1\in B_2$ and $i_1^{*}=-i_1$. The algebra A is associative. Therefore, $a_j^{*}=a_j$ and ${\left(i_1{a}_j\right)}^{*}{i}_1=a_j^{*}{i}_1^{*}{i}_1=a_j=i_1{\left(a_j{i}_1\right)}^{*}$ for $j=0$ and $j=1$.

Consider the particular case:

if $a=a^{*}$, then $a_0=a$ and ${\left(a_1{i}_1\right)}^{*}={(a+i_{1}a{i}_1)}^{*}/2=a_1{i}_1$.

The latter together with $a_1^{*}=a_1$ implies that $-i_1{a}_1=a_1{i}_1$ if $a=a^{*}$. On the other hand, $a=2a_1{i}_1-i_{1}a{i}_1$ and $a^{*}=-2a_1{i}_1-i_{1}a{i}_1$ if $a=a^{*}$. Thus $4a_1{i}_1=0$ and hence $a_1=0$, that is, $a{i}_1=i_{1}a$ if $a=a^{*}$, since $a_1=a_1{i}_1{i}_1^{*}$ and $char\left(F\right)\ne 2$. This implies that $a_1{i}_1=i_1{a}_1$ for each $a\in A$, consequently, the decomposition is valid $a=a_0+a_1{i}_1$ with the self-adjoint elements $a_0^{*}=a_0$ and $a_1^{*}=a_1$ in A. □

3. Dual and Annihilator Normed Algebras

At first, necessary definitions are given in this section and then propositions and theorems are proven. Definitions 2, 3, 4 and 5 follow main lines of that of in [4,8,11,28,29,30], but they are added in order to avoid any misunderstanding and because they contain some specific ultranormed features.

Definition 2.

Let A be a topological algebra over a field F and let S be a subset of A. The left annihilator is defined by $\mathsf{L}(A,S):=\{x\in A:xS=0\}$ and the right annihilator is $\mathsf{R}(A,S):=\{x\in A:Sx=0\}$, shortly they also will be denoted by $A_l\left(S\right):=\mathsf{L}(A,S)$ and $A_r\left(S\right):=\mathsf{R}(A,S)$.

Definition 3.

An algebra A is called an annihilator algebra if conditions $(1-3)$ are fulfilled:

(1) $A_l\left(A\right)=A_r\left(A\right)=0$ and

(2) $A_l\left(J_r\right)\ne 0$ and

(3) $A_r\left(J_l\right)\ne 0$

for all closed right $J_r$ and left $J_l$ ideals in A.

If for all closed (proper or improper) left $J_l$ and right $J_r$ ideals in A

(4) $A_l\left(A_r\left(J_l\right)\right)=J_l$ and

(5) $A_r\left(A_l\left(J_r\right)\right)=J_r$

then A is called a dual algebra.

If A is a ∗-algebra (see Definition 1) and for each $x\in A$ elements $a\in A$ and $a_1\in A$ exist such that an ultranorm on A for these elements satisfies the following conditions

(6) $|ax{x}^{*}{a}_1^{*}{|=|x|}^2$ and $\left|a\right||a_1^{*}|\le 1$,

then the algebra A is called finely regular.

Theorem 3.

If A is an ultranormed annihilator finely regular Banach algebra, then A is dual.

Proof. 

Consider arbitrary $x\in A$ and take elements $a\in A$ and $a_1\in A$ fulfilling conditions $\left(6\right)$ of Definition 3, then ${\left|x\right|}^2=|ax{x}^{*}{a}_1^{*}|\le |a\left|\right|x\left|\right|x^{*}\left|\right|a_1^{*}|\le |x\left|\right|x^{*}|$, hence $\left|x\right|\le |x^{*}|$. Substituting x by $x^{*}$ we deduce analogously that $|x^{*}|\le |x|$, consequently, $\left|x\right|=|x^{*}|$.

For a closed left ideal $J_l$ in A if $x\in J_l\cap {\left(A_r\left(J_l\right)\right)}^{*}$, then $x{x}^{*}=0$, consequently, $x=0$ by Formula $\left(6\right)$ in Definition 3 and hence $J_l\cap {\left(A_r\left(J_l\right)\right)}^{*}=0$. Then $V_l:=J_l\oplus {\left(A_r\left(J_l\right)\right)}^{*}$ is a left ideal in A, since $A_r\left(J_l\right)$ is the closed right ideal in A and ${\left(A_r\left(J_l\right)\right)}^{*}$ is the closed left ideal in A.

For an arbitrary $x\in V_l$ there exist elements $y\in J_l$ and $z\in {\left(A_r\left(J_l\right)\right)}^{*}$ such that $x=y+z$. Therefore, $x{z}^{*}=z{z}^{*}$ and $x{y}^{*}=y{y}^{*}$. Using conditions $\left(6\right)$ of Definition 3 we choose elements $a\in A$, $a_1\in A$, $b\in A$ and $b_1\in A$ with $\left|a\right||a_1^{*}|\le 1$ and $\left|b\right||b_1^{*}|\le 1$ such that $|az{z}^{*}{a}_1^{*}{|=|z|}^2$ and $|by{y}^{*}{b}_1^{*}{|=|y|}^2$ and hence $\left|x\right||z^{*}|\ge |a\left|\right|x\left|\right|z^{*}\left|\right|a_1^{*}|\ge |ax{z}^{*}{a}_1^{*}|=|az{z}^{*}{a}_1^{*}{|=|z|}^2$ and $\left|x\right||y^{*}|\ge |b\left|\right|x\left|\right|y^{*}\left|\right|b_1^{*}|\ge |bx{y}^{*}{b}_1^{*}|=|by{y}^{*}{b}_1^{*}{|=|y|}^2$. Therefore, $\left|x\right|\ge \left|z\right|$ and $\left|x\right|\ge \left|y\right|$. Thus $V_l$ is the closed left ideal in A.

From Condition $\left(3\right)$ of Definition 3 it follows that a nonzero element $a\in A$ exists such that $V_{l}a=\left(0\right)$, consequently, $J_{l}a=\left(0\right)$ and ${\left(A_r\left(J_l\right)\right)}^{*}a=\left(0\right)$. Then from the inclusion $a\in A_r\left(J_l\right)$ and hence $a^{*}\in {\left(A_r\left(J_l\right)\right)}^{*}$ it follows that $a^{*}a=0$. The latter contradicts the supposition that the algebra A is completely regular. Thus $V_l=A$ and analogously for each closed right ideal $J_r$ in A the equality $A=V_r$ is valid, where $V_r=J_r\oplus {\left(A_l\left(J_r\right)\right)}^{*}$.

Particularly, for $J_r=A_r\left(J_l\right)$ it implies that $A=A_r\left(J_l\right)\oplus {\left(A_l\left(A_r\left(J_l\right)\right)\right)}^{*}$. The involution of both sides of the latter equality gives $A={\left(A_r\left(J_l\right)\right)}^{*}\oplus A_l\left(A_r\left(J_l\right)\right)$, since $J_l\subseteq A_l\left(A_r\left(J_l\right)\right)$. Thus $J_l=A_l\left(A_r\left(J_l\right)\right)$ for each closed left ideal $J_l$ in A and the involution leads to the equality $J_r=A_r\left(A_l\left(J_r\right)\right)$ for each closed right ideal $J_r$ in A. Thus, conditions $(4,5)$ of Definition 3 are fulfilled. □

Definition 4.

If idempotents $w_1$ and $w_2$ of an algebra A satisfy the conditions $w_1{w}_2=0$ and $w_2{w}_1=0$, then it is said that they are orthogonal. A family $\{w_j:j\}$ of idempotents is said to be orthogonal, if and only if every two distinct of them are orthogonal. An idempotent p is called irreducible, if it cannot be written as the sum of two mutually orthogonal idempotents.

Definition 5.

For two Banach algebras A and B over an ultranormed field F, $F\in E_n$, we consider the completion $A{\widehat{\otimes }}_{F}B$ relative to the projective tensor product topology (see [11,31]) of the tensor product $A{\otimes }_{F}B$ over the field F such that $A{\widehat{\otimes }}_{F}B$ is also a Banach algebra into which A and B have natural F-linear embeddings ${\pi }_1$ and ${\pi }_2$ correspondingly.

For a Banach algebra B over an ultranormed field F, $F\in E_n$, and an element $x\in B$ we say that x has a left core quasi-inverse y if for each $H\in E_n\left(F\right)$ an element $y\in B_H$ exists satisfying the equality $x+y+yx=0$, where $B_H=B{\widehat{\otimes }}_{F}H$. A right core quasi-inverse is defined similarly. Particularly, if only $H=F$ is considered they are shortly called a left quasi-inverse and a right quasi-inverse correspondingly.

For a unital Banach algebra A over F, where $F\in E_n$, if an element $x\in A$ has the property: for each field extension $G\in E_n\left(F\right)$ the left inverse ${(1+yx)}_l^{-1}$ exists in $A_G$ for each $y\in A_G$, then we call x a generalized core nil-degree element. The family of all generalized core nil-degree elements of A we call a core radical and denote it by $R_c\left(A\right)$.

Proposition 1.

Let A be a unital Banach algebra over F, where $F\in E_n$. Then

$R_c\left(A\right)=\bigcap \{A\cap J_l:G\in E_n\left(F\right)\&J_{l}isapropermaximalleftidealin{A}_G\}$.

Proof. 

Consider an element $x\in A$ such that for each $G\in E_n\left(F\right)$ (see Notation 2.1) and each maximal left ideal $J_l$ in $A_G$ the inclusion $x\in J_l$ is valid. If an element $y\in A_G$ is such that ${(1+yx)}_l^{-1}$ does not exist, then an element $z=1+yx$ belongs to some left ideal J in $A_G$. Since $A_G$ is the unital algebra, then z belongs to some proper maximal left ideal M such that $J\subset M$. But $yx$ also belongs to M, since x belongs to each maximal left ideal, consequently, $1=z-yx\in M$. The latter is impossible, since M is the proper left ideal in $A_G$. This means that the left inverse ${(1+yx)}_l^{-1}$ exists for every $G\in E_n\left(F\right)$ and $y\in A_G$. Thus x belongs to the core radical.

Vice versa. Let now $x\in R_c\left(A\right)$. Suppose the contrary that a field extension $G\in E_n\left(F\right)$ and a proper maximal left ideal $J_l$ in $A_G$ exist such that $x\notin J_l$. Consider the set V of all elements $z=b-yx$ with $b\in J_l$ and $y\in A_G$. Evidently V is the left ideal in $A_G$ containing $J_l$, but $J_l$ is maximal, consequently, $V=A_G$. This implies that $1=b-yx$ for some $b\in J_l$ and $y\in A_G$. Therefore, the element $b=1+yx$ has not a left inverse. However, this contradicts the supposition made above. □

Proposition 2.

Suppose that A is a unital Banach algebra over F, where $F\in E_n$. Then

$(x\in R_c\left(A\right))\iff (\forall G\in E_n\left(F\right)\forall y\in A_G\exists {(1+yx)}^{-1}\in A_G)$.

Proof. 

If $\forall G\in E_n\left(F\right)\forall y\in A_G\exists {(1+yx)}^{-1}\in A_G$, then $\forall G\in E_n\left(F\right)\forall y\in A_G\exists {(1+yx)}_l^{-1}\in A_G$, consequently, $x\in R_c\left(A\right)$, where as usually ${(1+yx)}^{-1}$ notates the inverse of $1+yx$.

Vice versa. Let $x\in R_c\left(A\right)$. Then by the definition of the core radical $\forall G\in E_n\left(F\right)\forall y\in A_G\exists {(1+yx)}_l^{-1}\in A_G$. For $G\in E_n\left(F\right)$ denote by $1+b$ a left inverse of $1+yx$ in $A_G$, that is $(1+b)(1+yx)=1$. This implies that $1+yx$ is the right inverse of $1+b$ in $A_G$ and $b=-byx-yx$. From $x\in R_c\left(A\right)$ it follows that $b\in R_c\left(A_G\right)$, since $x\in J_l$ and hence $y\in J_l$ for each proper maximal left ideal $J_l$ in $A_H$ and each $H\in E_n\left(G\right)$. This means that for every $H\in E_n\left(G\right)$ and $z\in A_H$ a left inverse ${(1+zb)}_l^{-1}$ exists in $A_H$, particularly, for $z=1$ also. On the other hand, the right inverse is ${(1+zb)}_r^{-1}=1+yx$ as it was already proved above. Therefore, the inverse (i.e., left and right simultaneously) ${(1+b)}^{-1}=1+yx$ exists. Thus $1+b$ is the inverse of $1+yx$ in $A_G$. □

Proposition 3.

Let A be a unital Banach algebra over F, where $F\in E_n$. Then

$R_c\left(A\right)=\bigcap \{A\cap J_r:G\in E_n\left(F\right)\&J_{r}isapropermaximalrightidealin{A}_G\}$. Moreover, $R_c\left(A\right)$ is the two-sided ideal in A.

Proof. 

Consider the class $Q_e\left(A\right)$ of all elements $x\in A$ such that for each field extension $G\in E_n\left(F\right)$ the right inverse ${(1+xy)}_r^{-1}$ exists in $A_G$ for each $y\in A_G$. Analogously to the proof of Proposition 1 we infer that

$Q_e\left(A\right)=\bigcap \{J_r:G\in E_n\left(F\right)\&J_{r}isapropermaximalrightidealin{A}_G\}$.

Similarly to the proof of Proposition 2 we deduce that

$(x\in Q_e\left(A\right))\iff (\forall G\in E_n\left(F\right)\forall y\in A_G\exists {(1+xy)}^{-1}\in A_G)$.

Suppose that $G\in E_n\left(F\right)$, $x\in A$, $y\in A_G$ and the inverse element exists ${(1+yx)}^{-1}=1+b$ in $A_G$. Then $(1+xy)(1-xy-xby)-1=-x\left(\right(1+yx\left)\right(1+b)-1)y=0$ and $(1-xy-xby)(1+xy)-1=-x\left(\right(1+b\left)\right(1+yx)-1)y=0$, consequently, $1-xy-xby={(1+xy)}^{-1}$. Analogously if the inverse element ${(1+yx)}^{-1}$ exists, then ${(1+xy)}^{-1}$ also exists. This implies that $Q_e\left(A\right)=R_c\left(A\right)$ and hence the core radical is the two-sided ideal in A. □

Proposition 4.

Suppose that A is a unital Banach algebra over F, where $F\in E_n$. Then an extension field $H=H_F\in E_n\left(F\right)$ exists such that $R_c\left(A\right)=A\cap R\left(A_H\right)$, where $R\left(A_H\right)$ denotes the radical of the algebra $A_H$ over H. Moreover, H can be chosen algebraically closed and spherically complete.

Proof. 

Consider an arbitrary element $x\in A\setminus R_c\left(A\right)$. This means that a field extension $G=G_x\in E_n\left(F\right)$ and an element $y\in A_G$ exist such that the element $(1+yx)$ has not the left inverse in $A_G$. For the family $\mathcal{G}:=\{G_x:x\in A\setminus R_c\left(A\right),G_x\in E_n\left(F\right)\}$ a field $H=H_F\in E_n\left(F\right)$ exists such that $G_x\subseteq H$ for each $x\in A\setminus R_c\left(A\right)$ due to Proposition V.3.2.2 [29] and since the multiplicative ultranorm ${|·|}_F$ can be extended to a multiplicative ultranorm ${|·|}_H$ on H (see Proposition 5 in Section VI.3.3 [30], Krull’s existence theorem 14.1 and Theorem 14.2 in [18] or 3.19 in [11], Lemma 1 and Proposition 1 in [32])).

If $H_F$ is not either algebraically closed or spherically complete, one can take the spherical completion of its algebraic closure ${\overline{H}}_F$ (see Corollary 3.25, Theorem 4.48 and Corollary 4.51 in [11]). Then also ${\overline{H}}_F\in E_n\left(F\right)$. Denote shortly ${\overline{H}}_F$ by H.

Therefore, if $G\in \mathcal{G}$, then from $y\in A_G$ it follows that $y\in A_H$. For each $x\in A\setminus R_c\left(A\right)$ an element $y\in A_G$ exists such that $A_G(1+yx)$ is a left proper ideal in $A_G$, consequently, $A_G(1+yx){\widehat{\otimes }}_{G}H=A_H(1+yx)$ is a left proper ideal in $A_H$, since $H\subset Z\left(A_H\right)$. Therefore, $(1+yx)$ has not a left inverse in $A_H$.

Thus for each $x\in A\setminus R_c\left(A\right)$ and $G=G_x\in \mathcal{G}$ and element $y\in A_H$ exists such that $(1+yx)$ has not a left inverse in $A_H$. Therefore, $A\cap R\left(A_H\right)\subset R_c\left(A\right)$. On the other hand, if $x\in R_c\left(A\right)$, then $x\in R\left(A_H\right)$ according to the definition of $R_c\left(A\right)$ in Definition 5. Thus $R_c\left(A\right)=A\cap R\left(A_H\right)$ for the fields H constructed above. □

Theorem 4.

Let A be a unital Banach algebra over F, where $F\in E_n$. Then an extension field $K=K_F\in E_n\left(F\right)$ exists such that

$\left(1\right)$ $R_c\left(A_K\right)=R\left(A_K\right)$.

Moreover, K can be chosen algebraically closed and spherically complete.

Proof. 

Put $K_1=H$, where $H=H_F$ is given by Proposition 4. Then by induction take $K_{n+1}=H_{K_n}$ for each natural number $n=1,2,3,\dots$. There are isometric embeddings $K_n\hookrightarrow K_{n+1}$ for each n. Let K be the norm completion of $K_{\infty }:={\bigcup }_{n=1}^{\infty }{K}_n$, hence $K\in E_n\left(F\right)$. In addition, each field $K_l$ can be chosen algebraically closed and spherically complete due to Proposition 4. Moreover, it is possible to take as K the spherical completion of the algebraic closure of $K_{\infty }$ (see Corollary 3.25, Theorem 4.48 and Corollary 4.51 in [11]).

In view of Proposition 4 $R_c\left(A_{K_l}\right)=A_{K_l}\cap R\left(A_{K_{l+1}}\right)$ for each natural number l. Let $x\in R_c\left(A_K\right)$, that is for each $G\in E_n\left(K\right)$ and $y\in A_K$ a left inverse ${(1+yx)}_l^{-1}$ exists in $A_K$. The algebra $A{\otimes }_F{K}_{\infty }$ over the field $K_{\infty }$ is everywhere dense in $A_K=A{\widehat{\otimes }}_{F}K$. Therefore, there exist sequences $x_n$ and $y_n$ in $A_K$ such that $x_n\in A_{K_n}$ and $y_n\in A_{K_n}$ for each n and ${lim}_n{x}_n=x$ and ${lim}_n{y}_n=y$. Since $(1+z)$ is invertible in $A_K$ for each $z\in A_K$ with $\left|z\right|<1$, then a natural number m exits such that a left inverse ${(1+y_n{x}_n)}_l^{-1}$ exists for each $n>m$.

From $G\in E_n\left(K\right)$ and $K_l\in E_n\left(F\right)$, $K_l\subseteq K$ it follows that $G\in E_n\left(K_l\right)$ for each $l=1,2,3,\dots$. On the other hand, an element $y\in A_K$ can be any marked element in particularly belonging to $A_{K_l}$. Thus ${\bigcup }_l{R}_c\left(A_{K_l}\right)$ is dense in $R_c\left(A_K\right)$. Similarly, considering $G=K$ one gets that ${\bigcup }_{l}R\left(A_{K_l}\right)$ is dense in $R\left(A_K\right)$. Mentioning that ${\bigcup }_l{A}_{K_l}$ is dense in $A{\otimes }_F{K}_{\infty }$ one gets that ${\bigcup }_l{A}_{K_l}$ is dense in $A_K$. Therefore, we infer that

$$R_c\left(A_K\right)=c{l}_{A_K}(\bigcup _l{R}_c\left(A_{K_l}\right))=c{l}_{A_K}(\bigcup _l(A_{K_l}\cap R\left(A_{K_{l+1}}\right)))$$

$$=c{l}_{A_K}(\bigcup _{l}R\left(A_{K_{l+1}}\right))=R\left(A_K\right),$$

where $c{l}_{A_K}B$ denotes the closure of a subset B, $B\subset A_K$, in $A_K$. □

Proposition 5.

Let A be a Banach algebra over F, $F\in E_n$, also let a field K fulfill Condition $\left(1\right)$ of Theorem 4 for $A_1$, where $A_1=A$ if $1\in A$, while $A_1=A\oplus 1F$ if $1\notin A$. Then an element $x\in A_K$ is not core left quasi-invertible if and only if $J_{l,G}:=\{z+zx:z\in A_G\}$ is a proper left ideal in $A_G$ for each $G\in E_n\left(K\right)$. If so $J_{l,G}$ is a proper regular left ideal in $A_G$ such that $x\notin J_{l,G}$.

Proof. 

By virtue of Theorem 4 $R_c\left(A_{1,K}\right)=R\left(A_{1,K}\right)$. Hence for each $G\in E_n\left(K\right)$ an element $x\in A_K$ is not core left quasi-invertible in $A_G$ if and only if it does not belong to $R\left(A_{1,K}\right)$. If $u=y+yx$ and $v=z+zx$ belong to $J_{l,G}$, b and c are in G, where y and z belong to $A_G$, then $bu+cv=(by+cz)+(by+cz)x$, consequently, $cu+bv\in J_{l,G}$. That is $A_G{J}_{l,G}\subseteq J_{l,G}$. If $J_{l,G}$ is not a proper left ideal, then $J_{l,G}=A_G$. This implies that an element $z\in A_G$ exists such that $x+zx=-x$. The latter is equivalent to the equality $x+zx+x=0$. Thus z is a left quasi-inverse of x.

Vise versa if x has a left quasi-inverse in $A_G$, then $x\in J_{l,G}$, hence $-zx\in J_{l,G}$. Therefore, $z=(z+zx)-zx\in J_{l,G}$ for each $z\in A_G$, consequently, $A_G=J_{l,G}$. Thus if $J_{l,G}$ is a proper left ideal, then $x\notin J_{l,G}$. Mention that the element $w=-x$ is unital modulo the proper left ideal $J_{l,K}$, consequently, this ideal is regular. □

Proposition 6.

Suppose that A is a Banach algebra over F, $F\in E_n$, also a field K satisfies Condition $\left(1\right)$ of Theorem 4 for $A_1$. Then the following conditions are equivalent:

$\left(1\right)$ an element $x\in A_K$ possesses a left quasi-inverse in $A_G$ for each $G\in E_n\left(K\right)$;

$\left(2\right)$ for every $G\in E_n\left(K\right)$ and a maximal regular proper left ideal $M_{l,G}$ in $A_G$ an element $y\in A_G$ exists such that $x+y+yx\in M_{l,G}$.

Proof. 

If an element $x\in A_K$ possesses a left quasi-inverse $y_G$ in $A_G$ for each $G\in E_n\left(K\right)$, then $x+y_G+y_{G}x\in M_{l,G}$ for each maximal regular proper left ideal $M_{l,G}$ in $A_G$ due to Theorem 4.

Vise versa suppose that Condition $\left(2\right)$ is fulfilled, but x is not left quasi-invertible in $A_G$ for some $G\in E_n\left(K\right)$. Then $J_{l,G}$ is a regular proper left ideal in $A_G$ according to Proposition 5. Therefore, a maximal regular proper left ideal $M_{l,G}$ in $A_G$ exists containing $J_{l,G}$. Thus an element $y\in A_G$ exists such that $x+y+yx\in M_{l,G}$. On the other hand, the inclusion $y+yx\in J_{l,G}$ is accomplished, consequently, $y\in M_{l,G}$ and hence $-zx\in M_{l,G}$ for each $z\in A_G$. This implies that $z\in M_{l,G}$ for each $z\in A_G$, since $z=-zx+(z+zx)$. However, this leads to the contradiction $A_G=M_{l,G}$. Thus $\left(2\right)\Rightarrow \left(1\right)$. □

Proposition 7.

Suppose that A is a Banach annihilator algebra over an ultranormed field F, $F\in E_n$. Then a field extension K, $K\in E_n\left(F\right)$, exists such that if an element $-p\in A_K$ is not core left quasi-invertible, then a nonzero element $x\in A_K\setminus \left\{0\right\}$ exist satisfying the equation $px=x$.

Proof. 

We take a field K, $K\in E_n\left(F\right)$, given by Theorem 4 for a unital algebra $E=A_1$, where $E=A\oplus 1F$ if $1\notin A$, while $E=A$ if $1\in A$. Therefore, $R_c\left(E_K\right)=R\left(E_K\right)$.

By virtue of Proposition 5 $J_{l,K}:=\{yp-p:y\in A_K\}$ is a regular proper left ideal in $A_K$. Since $E_K$ is the unital Banach algebra over K, then it is with continuous inverse. Hence if A is not unital, then $A_K$ is with the continuous quasi-inverse. Mention that an element v is a left quasi-inverse of q in $A_K$ if and only if $1+v$ is a left inverse of $1+q$ in $E_K$.

Therefore, if $1\notin A$, then a bijective correspondence exists: Q is a left (maximal) ideal of $E_K$ which is not contained entirely in A if and only if $Q\cap A_K$ is a regular (maximal respectively) left ideal of $A_K$. If $1\in A$, then each left ideal in $A_K$ is regular.

Recall that a ring B satisfying the identities

$\left(1\right)$ $\mathsf{L}(B,B)=\left(0\right)$ and $\mathsf{R}(B,B)=\left(0\right)$ is called annihilator, where

$\left(2\right)$ $\mathsf{L}(B,S)=\{x\in B:xS=(0\left)\right\}$ and $\mathsf{R}(B,S)=\{x\in B:Sx=(0\left)\right\}$

denote a left annihilator and a right annihilator correspondingly of a subset S in B. Thus

$\left(3\right)$ $\mathsf{L}(A_K,A_K)=\left(0\right)$ and $\mathsf{R}(A_K,A_K)=\left(0\right)$,

since $A_K=A{\widehat{\otimes }}_{F}K$, since by the conditions of this proposition $\mathsf{L}(A,A)=\left(0\right)$ and $\mathsf{R}(A,A)=\left(0\right)$, also A and $A_K$ are Banach algebras. Next we take the closure $c{l}_{A_K}\left(J_{l,K}\right)$ of $J_{l,K}$ in $A_K$. Therefore, $\mathsf{R}(A_K,c{l}_{A_K}\left(J_{l,K}\right))$ is not nil, $\mathsf{R}(A_K,c{l}_{A_K}\left(J_{l,K}\right))\ne \left(0\right)$.

Suppose that x is a nonzero element in $\mathsf{R}(A_K,c{l}_{A_K}\left(J_{l,K}\right))$, consequently, $x\in \mathsf{R}(A_K,J_{l,K})$.

If $z\in \mathsf{R}(A_K,J_{l,K})$, then $y(pz-z)=(yp-y)z=0$ for each $y\in A_K$. From $\mathsf{L}(A_K,A_K)=\left(0\right)$ and $\mathsf{R}(A_K,A_K)=\left(0\right)$ it follows that $pz-z=0$. Vice versa, if $pz-z=0$ for some $z\in A_K$, then $(yp-y)z=y(pz-z)=0$ and hence $z\in \mathsf{R}(A_K,J_{l,K})$. Therefore,

$\left(4\right)$ $\mathsf{R}(A_K,J_{l,K})=\{z\in A_K:pz=z\}$.

Thus $px=x$. □

Theorem 5.

Suppose that A is a Banach annihilator algebra over a field $F\in E_n$ such that $R_c\left(A\right)=R\left(A\right)$ and $M_r$ is a proper maximal closed right ideal in A satisfying the condition $\mathsf{L}(A,M_r)\cap R\left(A\right)=\left(0\right)$. Then $\mathsf{L}(A,M_r)$ contains an idempotent p and

$\left(1\right)$ $\mathsf{L}(A,M_r)=Ap$ and

$\left(2\right)$ $M_r=\{z-pz:z\in A\}$.

Proof. 

A nonzero element b in $\mathsf{L}(A,M_r)$ exists, since $\mathsf{L}(A,M_r)\ne \left(0\right)$, since $M_r$ is a proper right ideal in A. Therefore, $M_r\subset \mathsf{R}(A,\left\{b\right\})\ne A$ and consequently,

$\left(3\right)$ $\mathsf{R}(A,\left\{b\right\})=M_r$,

since the right ideal $M_r$ is maximal. The element b does not belong to $R\left(A\right)$, since $\mathsf{L}(A,M_r)\cap R\left(A\right)=\left(0\right)$ by the conditions of this theorem.

In view of Theorem 4 and Propositions 5 and 6 a scalar $t\in F$ and an element $y\in A$ exist such that the element $-p=tb+yb$ has not a left quasi-inverse in $A_G$ for each $G\in E_n\left(F\right)$. Thus $p\ne 0$ and $p\in \mathsf{L}(A,M_r)$. By virtue of Proposition 7 a nonzero element $x\in A\setminus \left(0\right)$ exists such that $px=x$, consequently, $(p^2-p)x=0$.

Suppose that $p^2-p$ is not nil, $p^2-p\ne 0$. We have $p^2-p\in \mathsf{L}(A,M_r)$. Taking $b=p^2-p$ in $\left(3\right)$ one gets $\mathsf{R}(A,p^2-p)=M_r$, consequently, $(p^2-p)x\in M_r$ and inevitably $x=px=0$. This leads to the contradiction. Thus $p^2=p$.

On the other hand, $p\in \mathsf{L}(A,M_r)$ and p is not nil. Taking $b=p$ in $\left(3\right)$ provides $M_r=\mathsf{R}(A,\left\{p\right\})$ and $\mathsf{R}(A,\{p\left\}\right)=\{z-pz:z\in A\}$, since $p(y-py)=py-p^{2}y=0$, also if $pz=0$, then $z=z-pz$. Therefore, $\mathsf{L}(A,M_r)=Ap$ due to formula $\left(4\right)$ of Proposition 7 and since p is the idempotent. □

Corollary 1.

If conditions of Theorem 5 are fulfilled, then $M_r$ is a maximal right ideal and $\mathsf{L}(A,M_r)$ is a minimal left ideal, also $pA$ is a minimal right ideal and $\mathsf{L}(A,pA)$ is a maximal left ideal.

Theorem 6.

Let A be a Banach annihilator algebra over a field $F\in E_n$ such that $R_c\left(A\right)=R\left(A\right)$, let also $J_l$ be a minimal left (may be closed) ideal which is not contained in $R\left(A\right)$, $J_l\setminus R\left(A\right)\ne \varnothing$. Then $J_l$ contains an idempotent p for which $J_l=Ap$ and $\mathsf{R}(A,Ap)=\{x-px:x\in A\}$.

Proof. 

Take $x\in J_l\setminus R\left(A\right)$. From Propositions 5 and 6 it follows that $b\in F$ and $y\in A$ exist such that the element $-p=bx+yx$ has not a left quasi-inverse, consequently, $p\ne 0$.

In view of Proposition 7 an element $v\in A$ exists having the property $pv=v$. Therefore, $Y_l:=\{z\in J_l:xv=0\}$ is a left ideal such that it is contained in $J_l$ and $J_l\ne Y_l$, since $p\in J_l\setminus Y_l$. This ideal $Y_l$ is closed, if $J_l$ is closed. The ideal $J_l$ is minimal, hence $Y_l=\left(0\right)$. This implies that $zv\ne 0$ if $z\in J_l\setminus \left\{0\right\}$. On the other hand, $p^2-p\in J_l$ and $(p^2-p)v=0$, hence $p^2-p=0$. Thus p is the idempotent.

For each $z\in Ap$ the condition $z=zp$ is valid, consequently, $Ap$ is a closed left ideal contained in $J_l$ and hence $Ap=J_l$, since the left ideal $J_l$ is minimal. Therefore, $\mathsf{R}(A,J_l)=\{x-px:x\in A\}$. □

Lemma 3.

If A is a Banach annihilator semi-simple algebra over a field $F\in E_n$ with $R_c\left(A\right)=R\left(A\right)$ and J is a left (or right, or two-sided) ideal in A such that $J^2=\left(0\right)$, then $J=\left(0\right)$.

Proof. 

Suppose that J is a left ideal in A with $J^2=\left(0\right)$. Therefore, ${(tx+yx)}^2=0$ for every $t\in F$, $x\in J$ and $y\in A$, since $tx+yx\in J$. In this case the element $z=tx+yx$ has the left quasi-inverse $-z$. By virtue of Propositions 5 and 6 $x\in R\left(A\right)$, since $R_c\left(A\right)=R\left(A\right)$ by the conditions of this lemma. The algebra A is semi-simple, consequently, $J=\left(0\right)$. □

For a right ideal or a two-sided ideal the proof is analogous.

Lemma 4.

If A is a Banach annihilator semi-simple algebra over a field $F\in E_n$ with $R_c\left(A\right)=R\left(A\right)$ and $J_r$ is a right minimal ideal in A, then a closed two-sided ideal $Y=Y\left(J_r\right)$ generated by $J_r$ is minimal and closed in A.

Proof. 

If X is a closed two-sided ideal contained in Y, then $J_r\cap X$ is a right ideal contained in $J_r$, consequently, either $J_r\cap X=J_r$ or $J_r\cap X=\left(0\right)$, since $J_r$ is minimal. If $J_r\cap X=J_r$, then $Y\subset X$, hence $Y=X$.

If $J_r\cap X=\left(0\right)$, then $J_{r}X\subset J_r\cap X=\left(0\right)$, consequently, $J_r\subset \mathsf{L}(A,X)$. Then $\mathsf{L}(A,X)$ is the closed two-sided ideal, consequently, $Y\subset \mathsf{L}(A,X)$. Therefore, $X\subset \mathsf{L}(A,X)$ and consequently, $X^2=\left(0\right)$. Applying Lemma 3 we get that $X=\left(0\right)$.

Thus Y is minimal. □

Theorem 7.

Let A be a Banach annihilator semi-simple algebra over a field $F\in E_n$ with $R_c\left(A\right)=R\left(A\right)$. Then the sum of all left (or right) ideals of A is dense in A.

Proof. 

Suppose that U is a sum of all minimal right ideals and $\overline{U}$ is its closure in A. If $\overline{U}\ne A$, then $\overline{U}$ is the closed right ideal in A, consequently, a nonzero element y in A exists such that $y\overline{U}=\left(0\right)$. This implies that y belongs to all left annihilators of all minimal right ideals and hence it belongs to the intersection V of all maximal left regular ideals. In view of Proposition 3 one gets that this intersection is $R_c\left(A\right)$. By the conditions of this theorem $R_c\left(A\right)=R\left(A\right)$, hence V is zero, since A is semi-simple. Thus $y=0$ providing the contradiction. Thus $\overline{U}=A$. □

Proposition 8.

Let conditions of Theorem 7 be fulfilled and let J be a right ideal in A. Then J contains a minimal right ideal and an irreducible idempotent s.

Proof. 

Suppose that J does not contain a minimal right ideal and $sA$ is a minimal right ideal for some irreducible idempotent s in A. This implies that $J\cap \left(sA\right)=\left(0\right)$. Hence for each $a\in A$ either $asA=\left(0\right)$ or $asA$ is also a minimal right ideal, consequently, $\left(asA\right)\cap J=\left(0\right)$ for all $a\in A$ and hence $\left(as\right)\cap J=\left(ass\right)\cap J\subset \left(asA\right)\cap J=\left(0\right)$ for all $a\in A$. Thus $\left(aS\right)\cap J=\left(0\right)$. Therefore $JAs=\left(0\right)$, since $JAs\subset \left(As\right)\cap J$. This means that $JAs=\left(0\right)$ for all minimal left ideals $As$. In view of Theorem 7 $JA=\left(0\right)$, consequently, $J=\left(0\right)$. □

Proposition 9.

If conditions of Theorem 7 are satisfied and s is an irreducible idempotent in A, then $sA$ and $As$ are minimal right and left ideals correspondingly.

Proof. 

Suppose that $sA$ is not minimal. By virtue of Proposition 8 it contains a minimal right ideal $rA$ such that $rA\ne sA$, $rA\subset sA$. Then an element $a\in A$ exists such that $r=sa$, consequently, $rs=sas\in rA$. This implies that t is a nonzero idempotent contained in $rA$ such that the element $t=rs$ satisfies the equalities $st=ts=t$ and $s-t$ is also a nonzero idempotent providing the contradiction, since $s=t+(s-t)$ and $t(s-t)=(s-t)t=0$, but s is irreducible by the conditions of this proposition. Thus $sA$ is minimal. □

Proposition 10.

If conditions of Theorem 7 are satisfied and J is a closed two-sided ideal in A, then $\mathsf{L}(A,J)=\mathsf{R}(A,J)$ and $J+\mathsf{R}(A,J)$ is dense in A.

Proof. 

In view of Lemma 3 $J\cap \mathsf{R}(A,J)=\left(0\right)$, since $J\cap \mathsf{R}(A,J)=:V$ is the right ideal possessing the property $V^2=V$. Therefore, $\mathsf{R}(A,J)J=\left(0\right)$ and hence $\mathsf{R}(A,J)\subset \mathsf{L}(A,J)$. Similarly $\mathsf{L}(A,J)\subset \mathsf{R}(A,J)$, consequently, $\mathsf{L}(A,J)=\mathsf{R}(A,J)$.

If $J+\mathsf{R}(A,J)$ would be not dense in A, then its closure should be a proper ideal in A, consequently, a nonzero element x in A exists such that $(J+\mathsf{R}(A,J\left)\right)x=\left(0\right)$. Therefore $J(\alpha x+xy)=\left(0\right)$ and $\mathsf{R}(A,J)(\alpha x+xy)=\left(0\right)$ for each $y\in A$ and $\alpha \in F$, hence $(\alpha x+xy)\in \mathsf{R}(A,J)$ and consequently, ${(\alpha x+xy)}^2=0$ for each $y\in A$ and $\alpha \in F$. However, in the semi-simple algebra A with $R_c\left(A\right)=R\left(A\right)$ this is impossible for $x\ne 0$. □

Proposition 11.

If conditions of Theorem 7 are met and J is a minimal closed two-sided ideal in A, then J is an annihilator algebra with $R_c\left(J\right)=R\left(J\right)$. If in addition A is dual, then J is also dual.

Proof. 

If $x\in J$ and $Jx=\left(0\right)$, then $x=0$, since $J\cap \mathsf{R}(A,J)=\left(0\right)$ due to Proposition 10. Analogously if $xJ=\left(0\right)$ and $x\in J$, then $x=0$. Thus $\mathsf{L}(A,J)=\mathsf{R}(A,J)=\left(0\right)$.

If $V_l$ is a closed left ideal in J, then $(J+\mathsf{L}(A,J))V_l=J{V}_l\subset V_l$, hence $A{V}_l\subset V_l$, since $J+\mathsf{L}(A,J)$ is dense in A by Proposition 10. Thus $V_l$ is the closed left ideal in A.

Put $H_l=V_l+\mathsf{R}(A,J)$. Then either $H_l$ is dense in A or $\mathsf{R}(A,H_l)\ne \left(0\right)$. From Lemmas 3, 4 and Proposition 8 one gets $J\cap \mathsf{R}(A,H_l)\ne \left(0\right)$ and hence $J\cap \mathsf{R}(A,V_l)\ne \left(0\right)$. Analogously $J\cap \mathsf{L}(A,V_r)\ne \left(0\right)$ for a closed right ideal $V_r$ in J.

Suppose now that the algebra A is dual. In view of Lemma 3 and Proposition 10 if $x\in J$ and $[\mathsf{L}(A,V_r)\cap J]x=\left(0\right)$, then $x\in \mathsf{R}(A,\mathsf{L}(A,V_r)\cap J)=c{l}_A(\mathsf{R}\left(\mathsf{L}(A,V_r)\right)+\mathsf{R}(A,J))=c{l}_A(V_r+\mathsf{R}(A,J))=c{l}_A(V_r+\mathsf{L}(A,J))$. Then $(V_r+\mathsf{L}(A,J))J=V_{r}J\subset V_r$, since $V_r$ is a right ideal in J, consequently, $c{l}_A(V_r+\mathsf{L}(A,J))J\subset V_r$, hence $xJ\subset V_r$ and consequently, $\mathsf{L}(A,V_r)xJ\subset \mathsf{L}(A,V_r)V_r=\left(0\right)$. On the other hand, $\mathsf{L}(A,V_r)x\mathsf{R}(A,J)=\left(0\right)$, since $x\in J$, consequently, $\mathsf{L}(A,V_r)x(J+\mathsf{R}(A,J))=\left(0\right)$. We have that $J+\mathsf{R}(A,J)$ is dense in A due to Proposition 10, hence $\mathsf{L}(A,V_r)xA=\left(0\right)$ and consequently, $\mathsf{L}(A,V_r)x\subset \mathsf{L}(A,A)=\left(0\right)$. From the duality of A it follows that $x\in V_r$. Therefore, $\mathsf{R}(J,\mathsf{L}(J,V_r))=V_r$ and similarly $\mathsf{L}(J,\mathsf{R}(J,V_l))=V_l$. Thus J is also dual. □

Theorem 8.

Let A be a Banach semi-simple annihilator algebra over $F\in E_n$ with $R_c\left(A\right)=R\left(A\right)$. Then A is the completion of the direct sum of all its minimal closed two-sided ideals $H_k$, each of which is a simple annihilator algebra over F. Moreover, if A is dual, then each $H_k$ is simple and dual.

Proof. 

By virtue of Proposition 8 each closed minimal two-sided ideal J in A contains a minimal right ideal $V_r$, hence $J=V_r$ according to Lemma 4. Then the closure $c{l}_A{V}_r$ is a closed minimal two-sided ideal for each minimal right ideal $V_r$ due to the same lemma. According to Proposition 11 $c{l}_A{V}_r$ is the annihilator algebra, which is also dual if A is dual. If H is a closed two-sided ideal in $c{l}_A{V}_r$, then it is such in A also. However, $c{l}_A{V}_r$ is minimal, hence the algebra $c{l}_A{V}_r$ is simple.

By virtue of Theorem 7 the sum of all minimal right ideals $V_r$ is dense in A. Let K and M be two minimal closed two-sided ideals which are different, $K\ne M$. Therefore $KM\subset K\cap M=\left(0\right)$, since $K\cap M$ is the closed two-sided ideal contained in minimal closed two-sided ideals K and in M and different from them. If $x+y=0$ for some $x\in K$ and $y\in M$, then $Kx=\left(0\right)$ and $My=\left(0\right)$, consequently, ${\left(xA\right)}^2\subset K\left(xA\right)=\left(0\right)$ and analogously ${\left(yA\right)}^2=\left(0\right)$. Therefore $xA=\left(0\right)$ and $yA=\left(0\right)$, since A is semi-simple, consequently, $x=0$ and $y=0$. Thus the considered sum is direct. □

Theorem 9.

If A is a Banach simple annihilator algebra over a field $F\in E_n$ with $R_c\left(A\right)=R\left(A\right)$, if also p is an irreducible idempotent, then $pAp=:H$ is an ultranormed division algebra over F. Moreover, if A and F are ultranormed and A is commutative, then a multiplicative ultranorm ${|·|}_H$ on H exists extending that of F such that it induces a topology on H not stronger than the topology inherited from A.

Proof. 

From the conditions of this proposition it follows that $pH=Hp=H$, since $p^2=p$ and the algebra A is associative. Evidently, H is the algebra over F, since A is the algebra over F. The restriction of p to H is the identity on H, since $ps=p^{2}s=p\left(ps\right)$ for each $s\in A$ and hence $pr=r$ for each $r\in H$, similarly $rp=r$ for each $r\in H$ and hence $pr=rp=r=prp$. For each nonzero element r in H the set $Ar$ is a left ideal in A and $Ar\ne \left(0\right)$ due to Condition $\left(1\right)$ in Definition 3. In view of Propositions 8 and 9 $Ar\subset Ap$ and $Ap$ is a minimal left ideal, since p is the irreducible idempotent. Thus $Ar=Ap$ and hence an element $y\in A$ exists such that $yr=p^2=p$, consequently, $pyr=py\left(pr\right)=\left(pyp\right)r$. Therefore, $\left(pyp\right)r=\left(pyp\right)\left(prp\right)=pyprp=pyr=pp=p$, consequently, $pyp$ is a left inverse of r in H. Similarly r has a right inverse in H. Thus H is the division algebra such that F is isomorphic with $Fp$ and $Fp\subset H$. From the continuity of the algebraic operations on A it follows that they are continuous on H. The norm on A induces a norm on H, since $H\subset A$. Since H is the topological ring with the continuous quasi-inverse and H possesses the unit, then H is with the continuous inverse.

If A and F are ultranormed and A is commutative, then the ultranorm ${|·|}_A$ on A induces the ultranorm on H and H is also commutative. Therefore, ${\left|p\right|}_A=|p^2{|}_A\le {\left|p\right|}_A^2$ and hence $1\le {\left|p\right|}_A$. On the other hand, on H as the field extension of F there exists a multiplicative ultranorm ${|·|}_H$ extending ${|·|}_F$ that of the field F (see Proposition 5 in Section VI.3.3 [30], Krull’s existence theorem 14.1 and Theorem 14.2 in [18] or 3.19 in [11]). We have that ${\left|1\right|}_F=1$, $1\le {\left|p\right|}_A$, also p plays the role of the unit in H, while ${\left|bx\right|}_A={\left|b\right|}_F{\left|x\right|}_A$ for each $b\in F$ and $x\in A$.

If A is not unital, we consider the algebra $A_1$ obtained from it by adjoining the unit. The norms on A and F induce the norm on $A_1=A\oplus F$. Therefore, it is sufficient to consider the case of the unital algebra A. Mention that ${(1-p)}^2=1-p$ and $A(1-p)$ is the ideal in A such that $A=Ap+A(1-p)$ with $Ap\cap A(1-p)=\left(0\right)$. Moreover, $Ap=pAp=A{p}^2$, since A is commutative. This implies that H is isomorphic with the quotient algebra $J=A/\left(A\right(1-p\left)\right)$. Then the ultranorm on A induces the quotient ultranorm on J such that ${\left|xy\right|}_J\le {\left|x\right|}_J{\left|y\right|}_J$ and ${\left|xyp\right|}_J\le {\left|xp\right|}_J{\left|yp\right|}_J$ for each x and y in J, since $pxp=xp$ and $xpyp=xyp$ for each of elements x and y in the commutative algebra A. At the same time, ${\left|xyp\right|}_H={\left|pxppyp\right|}_H={\left|xp\right|}_H{\left|yp\right|}_H$ for each x and y in A.

The ultranorm ${|·|}_A$ on $Fp$ induced from A is equivalent with the multiplicative ultranorm ${|·|}_F$ on F, since $Fp$ is isomorphic with F and consequently, ${\left|xpypz\right|}_A={\left|xp\right|}_A{\left|yp\right|}_A{\left|z\right|}_A$ for every $xp\in Fp$, $yp\in Fp$ and z in A, since ${\left|xp\right|}_F={\left|xp\right|}_A$. Then ${\left|xyp\right|}_J={\left|xp\right|}_J{\left|yp\right|}_J$ if $xp\in Fp$ and $yp\in Fp$, where x and y are in A. The inequality ${\left|p\right|}_J^{-1}{\left|xp\right|}_J\le {\left|x\right|}_J$ is also fulfilled for each $x\in A$. Therefore, H can be supplied with a multiplicative norm ${|·|}_H$ extending that of F and satisfying the inequality ${\left|x\right|}_H\le {\left|x\right|}_J$ for each $x\in H$ according to Theorems 1.15 and 1.16 [7]. □

4. Conclusions

It is worth mentioning the following. Several applications were mentioned in the introduction. Besides these applications, the results of this article can be used for further studies in different areas of fundamental mathematics. Among them are topological algebra, non-archimedean functional analysis, representation theory of totally disconnected topological groups, operator theory in ultranormed spaces. It is also interesting to mention possible applications in mathematical coding theory and its technical applications [33,34,35,36,37,38,39,40,41], because frequently codes are based on binary systems and algebras over non-archimedean fields.

For example, the obtained results (see Theorems 4, 5, 7 and 9 in Section 3) on annihilator algebras can be used for subsequent investigations of invariant subspaces of operator algebras in ultranormed spaces. Moreover, they can also be used for studies of decompositions of totally disconnected topological group representations by operators in ultranormed spaces into irreducible representations (see also Propositions 5, 6, 8–10 in Section 3).

On the other hand, it can also serve for advances in non-archimedean quantum field theory and quantum mechanics. This is natural, because they are based on algebras and operator algebras over ultranormed fields.