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 and the right annihilator is , shortly they also will be denoted by and .
Definition 3. An algebra A is called an annihilator algebra if conditions are fulfilled:
(1) and
(2) and
(3)
for all closed right and left ideals in A.
If for all closed (proper or improper) left and right ideals in A
(4) and
(5)
then A is called a dual algebra.
If A is a ∗-algebra (see Definition 1) and for each elements and exist such that an ultranorm on A for these elements satisfies the following conditions
(6) and ,
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 and take elements and fulfilling conditions of Definition 3, then , hence . Substituting x by we deduce analogously that , consequently, .
For a closed left ideal in A if , then , consequently, by Formula in Definition 3 and hence . Then is a left ideal in A, since is the closed right ideal in A and is the closed left ideal in A.
For an arbitrary there exist elements and such that . Therefore, and . Using conditions of Definition 3 we choose elements , , and with and such that and and hence and . Therefore, and . Thus is the closed left ideal in A.
From Condition of Definition 3 it follows that a nonzero element exists such that , consequently, and . Then from the inclusion and hence it follows that . The latter contradicts the supposition that the algebra A is completely regular. Thus and analogously for each closed right ideal in A the equality is valid, where .
Particularly, for it implies that . The involution of both sides of the latter equality gives , since . Thus for each closed left ideal in A and the involution leads to the equality for each closed right ideal in A. Thus, conditions of Definition 3 are fulfilled. □
Definition 4. If idempotents and of an algebra A satisfy the conditions and , then it is said that they are orthogonal. A family 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, , we consider the completion relative to the projective tensor product topology (see [11,31]) of the tensor product over the field F such that is also a Banach algebra into which A and B have natural F-linear embeddings and correspondingly. For a Banach algebra B over an ultranormed field F, , and an element we say that x has a left core quasi-inverse y if for each an element exists satisfying the equality , where . A right core quasi-inverse is defined similarly. Particularly, if only 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 , if an element has the property: for each field extension the left inverse exists in for each , 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 .
Proposition 1. Let A be a unital Banach algebra over F, where . Then
.
Proof. Consider an element such that for each (see Notation 2.1) and each maximal left ideal in the inclusion is valid. If an element is such that does not exist, then an element belongs to some left ideal J in . Since is the unital algebra, then z belongs to some proper maximal left ideal M such that . But also belongs to M, since x belongs to each maximal left ideal, consequently, . The latter is impossible, since M is the proper left ideal in . This means that the left inverse exists for every and . Thus x belongs to the core radical.
Vice versa. Let now . Suppose the contrary that a field extension and a proper maximal left ideal in exist such that . Consider the set V of all elements with and . Evidently V is the left ideal in containing , but is maximal, consequently, . This implies that for some and . Therefore, the element 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 . Then
.
Proof. If , then , consequently, , where as usually notates the inverse of .
Vice versa. Let . Then by the definition of the core radical . For denote by a left inverse of in , that is . This implies that is the right inverse of in and . From it follows that , since and hence for each proper maximal left ideal in and each . This means that for every and a left inverse exists in , particularly, for also. On the other hand, the right inverse is as it was already proved above. Therefore, the inverse (i.e., left and right simultaneously) exists. Thus is the inverse of in . □
Proposition 3. Let A be a unital Banach algebra over F, where . Then
. Moreover, is the two-sided ideal in A.
Proof. Consider the class of all elements such that for each field extension the right inverse exists in for each . Analogously to the proof of Proposition 1 we infer that
.
Similarly to the proof of Proposition 2 we deduce that
.
Suppose that , , and the inverse element exists in . Then and , consequently, . Analogously if the inverse element exists, then also exists. This implies that 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 . Then an extension field exists such that , where denotes the radical of the algebra over H. Moreover, H can be chosen algebraically closed and spherically complete.
Proof. Consider an arbitrary element
. This means that a field extension
and an element
exist such that the element
has not the left inverse in
. For the family
a field
exists such that
for each
due to Proposition V.3.2.2 [
29] and since the multiplicative ultranorm
can be extended to a multiplicative ultranorm
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
is not either algebraically closed or spherically complete, one can take the spherical completion of its algebraic closure
(see Corollary 3.25, Theorem 4.48 and Corollary 4.51 in [
11]). Then also
. Denote shortly
by
H.
Therefore, if , then from it follows that . For each an element exists such that is a left proper ideal in , consequently, is a left proper ideal in , since . Therefore, has not a left inverse in .
Thus for each and and element exists such that has not a left inverse in . Therefore, . On the other hand, if , then according to the definition of in Definition 5. Thus for the fields H constructed above. □
Theorem 4. Let A be a unital Banach algebra over F, where . Then an extension field exists such that
.
Moreover, K can be chosen algebraically closed and spherically complete.
Proof. Put
, where
is given by Proposition 4. Then by induction take
for each natural number
. There are isometric embeddings
for each
n. Let
K be the norm completion of
, hence
. In addition, each field
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
(see Corollary 3.25, Theorem 4.48 and Corollary 4.51 in [
11]).
In view of Proposition 4 for each natural number l. Let , that is for each and a left inverse exists in . The algebra over the field is everywhere dense in . Therefore, there exist sequences and in such that and for each n and and . Since is invertible in for each with , then a natural number m exits such that a left inverse exists for each .
From
and
,
it follows that
for each
. On the other hand, an element
can be any marked element in particularly belonging to
. Thus
is dense in
. Similarly, considering
one gets that
is dense in
. Mentioning that
is dense in
one gets that
is dense in
. Therefore, we infer that
where
denotes the closure of a subset
B,
, in
. □
Proposition 5. Let A be a Banach algebra over F, , also let a field K fulfill Condition of Theorem 4 for , where if , while if . Then an element is not core left quasi-invertible if and only if is a proper left ideal in for each . If so is a proper regular left ideal in such that .
Proof. By virtue of Theorem 4 . Hence for each an element is not core left quasi-invertible in if and only if it does not belong to . If and belong to , b and c are in G, where y and z belong to , then , consequently, . That is . If is not a proper left ideal, then . This implies that an element exists such that . The latter is equivalent to the equality . Thus z is a left quasi-inverse of x.
Vise versa if x has a left quasi-inverse in , then , hence . Therefore, for each , consequently, . Thus if is a proper left ideal, then . Mention that the element is unital modulo the proper left ideal , consequently, this ideal is regular. □
Proposition 6. Suppose that A is a Banach algebra over F, , also a field K satisfies Condition of Theorem 4 for . Then the following conditions are equivalent:
an element possesses a left quasi-inverse in for each ;
for every and a maximal regular proper left ideal in an element exists such that .
Proof. If an element possesses a left quasi-inverse in for each , then for each maximal regular proper left ideal in due to Theorem 4.
Vise versa suppose that Condition is fulfilled, but x is not left quasi-invertible in for some . Then is a regular proper left ideal in according to Proposition 5. Therefore, a maximal regular proper left ideal in exists containing . Thus an element exists such that . On the other hand, the inclusion is accomplished, consequently, and hence for each . This implies that for each , since . However, this leads to the contradiction . Thus . □
Proposition 7. Suppose that A is a Banach annihilator algebra over an ultranormed field F, . Then a field extension K, , exists such that if an element is not core left quasi-invertible, then a nonzero element exist satisfying the equation .
Proof. We take a field K, , given by Theorem 4 for a unital algebra , where if , while if . Therefore, .
By virtue of Proposition 5 is a regular proper left ideal in . Since is the unital Banach algebra over K, then it is with continuous inverse. Hence if A is not unital, then is with the continuous quasi-inverse. Mention that an element v is a left quasi-inverse of q in if and only if is a left inverse of in .
Therefore, if , then a bijective correspondence exists: Q is a left (maximal) ideal of which is not contained entirely in A if and only if is a regular (maximal respectively) left ideal of . If , then each left ideal in is regular.
Recall that a ring B satisfying the identities
and is called annihilator, where
and
denote a left annihilator and a right annihilator correspondingly of a subset S in B. Thus
and ,
since , since by the conditions of this proposition and , also A and are Banach algebras. Next we take the closure of in . Therefore, is not nil, .
Suppose that x is a nonzero element in , consequently, .
If , then for each . From and it follows that . Vice versa, if for some , then and hence . Therefore,
.
Thus . □
Theorem 5. Suppose that A is a Banach annihilator algebra over a field such that and is a proper maximal closed right ideal in A satisfying the condition . Then contains an idempotent p and
and
.
Proof. A nonzero element b in exists, since , since is a proper right ideal in A. Therefore, and consequently,
,
since the right ideal is maximal. The element b does not belong to , since by the conditions of this theorem.
In view of Theorem 4 and Propositions 5 and 6 a scalar and an element exist such that the element has not a left quasi-inverse in for each . Thus and . By virtue of Proposition 7 a nonzero element exists such that , consequently, .
Suppose that is not nil, . We have . Taking in one gets , consequently, and inevitably . This leads to the contradiction. Thus .
On the other hand, and p is not nil. Taking in provides and , since , also if , then . Therefore, due to formula of Proposition 7 and since p is the idempotent. □
Corollary 1. If conditions of Theorem 5 are fulfilled, then is a maximal right ideal and is a minimal left ideal, also is a minimal right ideal and is a maximal left ideal.
Theorem 6. Let A be a Banach annihilator algebra over a field such that , let also be a minimal left (may be closed) ideal which is not contained in , . Then contains an idempotent p for which and .
Proof. Take . From Propositions 5 and 6 it follows that and exist such that the element has not a left quasi-inverse, consequently, .
In view of Proposition 7 an element exists having the property . Therefore, is a left ideal such that it is contained in and , since . This ideal is closed, if is closed. The ideal is minimal, hence . This implies that if . On the other hand, and , hence . Thus p is the idempotent.
For each the condition is valid, consequently, is a closed left ideal contained in and hence , since the left ideal is minimal. Therefore, . □
Lemma 3. If A is a Banach annihilator semi-simple algebra over a field with and J is a left (or right, or two-sided) ideal in A such that , then .
Proof. Suppose that J is a left ideal in A with . Therefore, for every , and , since . In this case the element has the left quasi-inverse . By virtue of Propositions 5 and 6 , since by the conditions of this lemma. The algebra A is semi-simple, consequently, . □
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 with and is a right minimal ideal in A, then a closed two-sided ideal generated by is minimal and closed in A.
Proof. If X is a closed two-sided ideal contained in Y, then is a right ideal contained in , consequently, either or , since is minimal. If , then , hence .
If , then , consequently, . Then is the closed two-sided ideal, consequently, . Therefore, and consequently, . Applying Lemma 3 we get that .
Thus Y is minimal. □
Theorem 7. Let A be a Banach annihilator semi-simple algebra over a field with . 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 is its closure in A. If , then is the closed right ideal in A, consequently, a nonzero element y in A exists such that . 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 . By the conditions of this theorem , hence V is zero, since A is semi-simple. Thus providing the contradiction. Thus . □
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 is a minimal right ideal for some irreducible idempotent s in A. This implies that . Hence for each either or is also a minimal right ideal, consequently, for all and hence for all . Thus . Therefore , since . This means that for all minimal left ideals . In view of Theorem 7 , consequently, . □
Proposition 9. If conditions of Theorem 7 are satisfied and s is an irreducible idempotent in A, then and are minimal right and left ideals correspondingly.
Proof. Suppose that is not minimal. By virtue of Proposition 8 it contains a minimal right ideal such that , . Then an element exists such that , consequently, . This implies that t is a nonzero idempotent contained in such that the element satisfies the equalities and is also a nonzero idempotent providing the contradiction, since and , but s is irreducible by the conditions of this proposition. Thus is minimal. □
Proposition 10. If conditions of Theorem 7 are satisfied and J is a closed two-sided ideal in A, then and is dense in A.
Proof. In view of Lemma 3 , since is the right ideal possessing the property . Therefore, and hence . Similarly , consequently, .
If 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 . Therefore and for each and , hence and consequently, for each and . However, in the semi-simple algebra A with this is impossible for . □
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 . If in addition A is dual, then J is also dual.
Proof. If and , then , since due to Proposition 10. Analogously if and , then . Thus .
If is a closed left ideal in J, then , hence , since is dense in A by Proposition 10. Thus is the closed left ideal in A.
Put . Then either is dense in A or . From Lemmas 3, 4 and Proposition 8 one gets and hence . Analogously for a closed right ideal in J.
Suppose now that the algebra A is dual. In view of Lemma 3 and Proposition 10 if and , then . Then , since is a right ideal in J, consequently, , hence and consequently, . On the other hand, , since , consequently, . We have that is dense in A due to Proposition 10, hence and consequently, . From the duality of A it follows that . Therefore, and similarly . Thus J is also dual. □
Theorem 8. Let A be a Banach semi-simple annihilator algebra over with . Then A is the completion of the direct sum of all its minimal closed two-sided ideals , each of which is a simple annihilator algebra over F. Moreover, if A is dual, then each is simple and dual.
Proof. By virtue of Proposition 8 each closed minimal two-sided ideal J in A contains a minimal right ideal , hence according to Lemma 4. Then the closure is a closed minimal two-sided ideal for each minimal right ideal due to the same lemma. According to Proposition 11 is the annihilator algebra, which is also dual if A is dual. If H is a closed two-sided ideal in , then it is such in A also. However, is minimal, hence the algebra is simple.
By virtue of Theorem 7 the sum of all minimal right ideals is dense in A. Let K and M be two minimal closed two-sided ideals which are different, . Therefore , since is the closed two-sided ideal contained in minimal closed two-sided ideals K and in M and different from them. If for some and , then and , consequently, and analogously . Therefore and , since A is semi-simple, consequently, and . Thus the considered sum is direct. □
Theorem 9. If A is a Banach simple annihilator algebra over a field with , if also p is an irreducible idempotent, then is an ultranormed division algebra over F. Moreover, if A and F are ultranormed and A is commutative, then a multiplicative ultranorm 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 , since 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 for each and hence for each , similarly for each and hence . For each nonzero element r in H the set is a left ideal in A and due to Condition in Definition 3. In view of Propositions 8 and 9 and is a minimal left ideal, since p is the irreducible idempotent. Thus and hence an element exists such that , consequently, . Therefore, , consequently, 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 and . 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 . 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
on
A induces the ultranorm on
H and
H is also commutative. Therefore,
and hence
. On the other hand, on
H as the field extension of
F there exists a multiplicative ultranorm
extending
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
,
, also
p plays the role of the unit in
H, while
for each
and
.
If A is not unital, we consider the algebra obtained from it by adjoining the unit. The norms on A and F induce the norm on . Therefore, it is sufficient to consider the case of the unital algebra A. Mention that and is the ideal in A such that with . Moreover, , since A is commutative. This implies that H is isomorphic with the quotient algebra . Then the ultranorm on A induces the quotient ultranorm on J such that and for each x and y in J, since and for each of elements x and y in the commutative algebra A. At the same time, for each x and y in A.
The ultranorm
on
induced from
A is equivalent with the multiplicative ultranorm
on
F, since
is isomorphic with
F and consequently,
for every
,
and
z in
A, since
. Then
if
and
, where
x and
y are in
A. The inequality
is also fulfilled for each
. Therefore,
H can be supplied with a multiplicative norm
extending that of
F and satisfying the inequality
for each
according to Theorems 1.15 and 1.16 [
7]. □