Spectra of locally matrix algebras
We describe spectra of associative (not necessarily unital and not necessarily countable-dimensional) locally matrix algebras. We determine all possible spectra of locally matrix algebras and give a new proof of Dixmier–Baranov Theorem. As an application of our description of spectra, we determine e...
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Інститут прикладної математики і механіки НАН України
2021
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| Назва журналу: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Цитувати: | Spectra of locally matrix algebras / O. Bezushchak // Algebra and Discrete Mathematics. — 2021. — Vol. 31, № 1. — С. 17–36. — Бібліогр.: 11 назв. — англ. |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1859672789476179968 |
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| author | Bezushchak, O. |
| author_facet | Bezushchak, O. |
| citation_txt | Spectra of locally matrix algebras / O. Bezushchak // Algebra and Discrete Mathematics. — 2021. — Vol. 31, № 1. — С. 17–36. — Бібліогр.: 11 назв. — англ. |
| collection | DSpace DC |
| container_title | Algebra and Discrete Mathematics |
| description | We describe spectra of associative (not necessarily unital and not necessarily countable-dimensional) locally matrix algebras. We determine all possible spectra of locally matrix algebras and give a new proof of Dixmier–Baranov Theorem. As an application of our description of spectra, we determine embeddings of locally matrix algebras.
|
| first_indexed | 2025-11-30T14:19:41Z |
| format | Article |
| fulltext |
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© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 31 (2021). Number 1, pp. 17–36
DOI:10.12958/adm1734
Spectra of locally matrix algebras
O. Bezushchak
Communicated by A. P. Petravchuk
Abstract. We describe spectra of associative (not necessar-
ily unital and not necessarily countable-dimensional) locally matrix
algebras. We determine all possible spectra of locally matrix alge-
bras and give a new proof of Dixmier–Baranov Theorem. As an
application of our description of spectra, we determine embeddings
of locally matrix algebras.
Introduction
Let F be a ground field. Recall that an associative F-algebra A is called
a locally matrix algebra (see [10]) if for an arbitrary finite subset of A
there exists a subalgebra B ⊂ A containing this subset and such that B is
isomorphic to a matrix algebra Mn(F) for some n > 1. In what follows we
will sometimes identify B and Mn(F), that is, assume that Mn(F) ⊂ A.
We call a locally matrix algebra unital if it contains 1.
Let A be a countable-dimensional unital locally matrix algebra. In [7],
J.G. Glimm defined the Steinitz number st(A) of the algebra A and
proved that A is uniquely determined by st(A). J. Dixmier [5] showed
that non-unital countable-dimensional locally matrix algebras over the
field of complex numbers can be parameterized by pairs (s, α), where s is
a Steinitz number and α is a nonnegative real number. A.A. Baranov [1]
extended this parametrization to locally matrix algebras over arbitrary
fields.
2020 MSC: 08A05, 16S50.
Key words and phrases: locally matrix algebra, Steinitz number, spectrum,
embedding.
https://doi.org/10.12958/adm1734
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18 Spectra of locally matrix algebras
In [2], we defined the Steinitz number st(A) for a unital locally matrix
algebra A of an arbitrary dimension. We showed that for a unital locally
matrix algebra A of dimension > ℵ0 the Steinitz number st(A) no longer
determines A; see [3, 4]. However, it determines the universal elementary
theory of A [3].
In this paper for an arbitrary (not necessarily unital and not necessarily
countable-dimensional) locally matrix algebra A, we define a subset of
SN that we call the spectrum of A and denote as Spec(A). We determine
all possible spectra of locally matrix algebras and give a new proof of
Dixmier–Baranov Theorem. As an application of our description of spectra,
we determine embeddings of locally matrix algebras.
1. Spectra of locally matrix algebras
Let P be the set of all primes and N be the set of all positive integers.
A Steinitz number (see [11]) is an infinite formal product of the form
∏
p∈P
prp ,
where rp ∈ N ∪ {0,∞} for all p ∈ P.
Denote by SN the set of all Steinitz numbers. Notice, that the set of
all positive integers N is a subset of SN. The numbers SN \ N are called
infinite Steinitz numbers.
Let A be a locally matrix algebra with a unit 1 over a field F and let
D(A) be the set of all positive integers n such that there is a subalgebra
A′, 1 ∈ A′ ⊆ A, A′ ∼=Mn(F ). Then the least common multiple of the set
D(A) is called the Steinitz number of the algebra A and denoted as st(A);
see [2].
Now let A be a (not necessarily unital) locally matrix algebra. For an
arbitrary idempotent 0 6= e ∈ A the subalgebra eAe is a unital locally
matrix algebra. That is why we can talk about its Steinitz number st(eAe).
The subset
Spec(A) =
{
st(eAe) | e ∈ A, e 6= 0, e2 = e
}
,
where e runs through all nonzero idempotents of the algebra A, is called
the spectrum of the algebra A.
For a Steinitz number s let Ω(s) denote the set of all natural numbers
n ∈ N that divide s.
For a Steinitz numbers s1, s2 we say that s1 finitely divides s2 if there
exists b ∈ Ω(s2) such that s1 = s2/b (we denote: s1
∣∣
fin s2).
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O. Bezushchak 19
Steinitz numbers s1, s2 are rationally connected if s2 = q · s1, where q
is some rational number.
We call a subset S ⊂ SN saturated if
1) any two Steinitz numbers from S are rationally connected;
2) if s2 ∈ S and s1
∣∣
fin s2 then s1 ∈ S;
3) if s, ns ∈ S, where n ∈ N, then is ∈ S for any i, 1 6 i 6 n.
Theorem 1. For an arbitrary locally matrix algebra A its spectrum is a
saturated subset of SN.
Let us consider examples of saturated subsets of SN.
Example 1. For an arbitrary natural number n the set {1, 2, . . . , n} is
saturated.
Example 2. Let s be a Steinitz number. The set
S(∞, s) :=
{ a
b
· s
∣∣∣ a ∈ N, b ∈ Ω(s)
}
is saturated. For an arbitrary Steinitz number s′ ∈ S(∞, s) we have
S(∞, s) = S(∞, s′). If s ∈ N then S(∞, s) = N.
Example 3. Let r be a real number, 1 6 r < ∞. Let s be an infinite
Steinitz number. The set
S(r, s) =
{ a
b
s
∣∣∣ a, b ∈ N; b ∈ Ω(s), a 6 rb
}
is saturated.
Example 4. Let s be an infinite Steinitz number and let r = u/v be a
rational number; u, v ∈ N, v ∈ Ω (s). Then the set
S+(r, s) =
{a
b
s
∣∣∣ a, b ∈ N; b ∈ Ω(s), a < rb
}
is saturated.
Theorem 2. Every saturated subset of SN is one of the following sets:
1) {1, 2, . . . , n}, n ∈ N, or N;
2) S(∞, s), s ∈ SN � N;
3) S(r, s), where s ∈ SN � N, r ∈ [1,∞);
4) S+(r, s), where s ∈ SN � N, r = u/v, u ∈ N, v ∈ Ω(s).
Remark 1. The real number r above is the inverse of the density invariant
of Dixmier–Baranov.
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20 Spectra of locally matrix algebras
Theorem 3. (1) For any saturated subset S ⊆ SN there exists
a countable-dimensional locally matrix algebra A such that Spec(A) = S.
(2) If A, B are countable-dimensional locally matrix algebras and
Spec(A) = Spec(B) then A ∼= B.
Remark 2. The part (2) of Theorem 3 is a new proof of Dixmier–Baranov
Theorem.
Which spectra above correspond to unital algebras?
Theorem 4. A locally matrix algebra A is unital if and only if Spec(A) =
{1, 2, . . . , n}, where n ∈ N, or Spec(A) = S(r, s), where s ∈ SN � N,
r = u/v, u, v ∈ N, v ∈ Ω(s).
Proof of Theorem 1. In what follows, we assume that A is a locally matrix
F-algebra. Recall the partial order on the set of all idempotents of A : for
idempotents e, f ∈ A we define e > f if f ∈ eA e.
We claim that for arbitrary idempotents e1, e2 ∈ A there exists an
idempotent e3 ∈ A such that e1 6 e3, e2 6 e3. Indeed, there exists a
subalgebra A′ ⊂ A such that e1, e2 ∈ A′ and A′ ∼=Mn(F), n > 1. Let e3
be the identity element of the subalgebra A′. Then e1 6 e3, e2 6 e3.
Now suppose that the locally matrix algebra A is unital. Let a ∈ A.
Choose a subalgebra A′ ⊂ A such that 1 ∈ A′, a ∈ A′ and A′ ∼= Mn(F),
n > 1. Let r be the range of the matrix a in A′. Let
r(a) =
r
n
, 0 6 r(a) 6 1.
V.M. Kurochkin [9] noticed that the number r(a) does not depend on a
choice of a subalgebra A′. We call r(a) the relative range of the element
a. In [4], we showed that if A is a unital locally matrix algebra and e ∈ A
is an idempotent, then st(eAe) = r(e) · st(A).
Now let A be a not necessarily unital locally matrix algebra. Let e1,
e2 ∈ A be idempotents. Choose an idempotent e3 ∈ A such that e1 6 e3,
e2 6 e3, i.e. e1, e2 ∈ e3Ae3. Let q1, q2 be relative ranges of the idempotents
e1, e2 in the unital locally matrix algebra e3Ae3. Then
st(e1Ae1) = q1 st(e3Ae3), st(e2Ae2) = q2 st(e3Ae3).
This implies that the Steinitz numbers st(e1Ae1), st(e2Ae2) are ratio-
nally connected. We have checked the condition 1) from the definition of
saturated sets.
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O. Bezushchak 21
Let 0 6= e ∈ A be an idempotent. Let s2 = st(eAe), k ∈ Ω(s2) and let
s1 = s2/k. The unital locally matrix algebra eAe contains a subalgebra
e ∈ Mk(F) ⊂ eAe. Consider the matrix unit e11 of the algebra Mk(F).
The relative range of the idempotent e11 in the unital algebra eAe is equal
to 1/k. Hence
st(e11Ae11) =
1
k
st(eA e) = s1, s1 ∈ Spec(A).
We have checked the condition 2).
Now let n > 1. Suppose that Steinitz numbers s and ns lie in Spec(A).
It means that there exist idempotents e1, e2 ∈ A such that s = st(e1Ae1),
ns = st(e2Ae2). There exists a matrix subalgebra Mk(F) ⊂ A that
contains e1 and e2. As above, let e3 be the identity element of the algebra
Mk(F). Let rk(ei) be the range of the idempotent ei in the matrix algebra
Mk(F). We have
s =
rk(e1)
k
· st(e3Ae3), n s =
rk(e2)
k
· st(e3Ae3),
which implies rk(e2) = n·rk(e1). In particular, n·rk(e1) 6 k. Let 1 6 i 6 n.
Consider the idempotent
e = diag
(
1, 1, . . . , 1,︸ ︷︷ ︸
i·rk(e1)
0, 0, . . . , 0
)
in the matrix algebra Mk(F). We have
st(eA e) =
i · rk(e1)
k
· st(e3Ae3) = i · st(e1Ae1) = i s.
We showed that is ∈ Spec(A). Hence Spec(A) is a saturated subset of SN.
It completes the proof of Theorem 1.
2. Classification of saturated subsets of SN
Our aim in this section is to classify all saturated subsets of SN. We
remark that if at least one Steinitz number from a saturated set S is
infinite then by the condition 1) all Steinitz numbers from S are infinite.
Let S be a saturated subset of SN. For a Steinitz number s ∈ S and
for a natural number b ∈ Ω(s) let
rs(b) = max
{
i > 1
∣∣∣ i ·
s
b
∈ S
}
.
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22 Spectra of locally matrix algebras
Lemma 1. If there exists a Steinitz number s0 ∈ S and a natural number
b0 ∈ Ω (s0) such that rs0(b0) = ∞ then for any s ∈ S and any b ∈ Ω (s)
we have rs(b) = ∞.
Proof. Let us show at first that rs0(b) = ∞ for any b ∈ Ω(s0). Indeed,
there exists a natural number c ∈ Ω(s0) such that both b0 and b divide c.
Then for an arbitrary i > 1 we have
i ·
s0
b0
=
(
i ·
c
b0
)
·
s0
c
∈ S.
This implies that rs0(c) = ∞. Hence,
i ·
s0
b
=
(
i ·
c
b
)
·
s0
c
∈ S,
which proves the claim.
Now choose an arbitrary Steinitz number s ∈ S. By the condition 1),
the Steinitz numbers s and s0 are rationally connected, i.e. there exist
a ∈ N, b ∈ Ω(s0) such that s = (a/b) · s0. By the condition 2), s0/b ∈ S.
Choose a natural number c ∈ Ω(s0/b). Then c ∈ Ω(s) and bc ∈ Ω(s0). For
an arbitrary i > 1 we have i · s/c = i · a · s0/(bc) ∈ S since rs0(bc) = ∞.
This implies rs(c) = ∞ and completes the proof of the lemma.
If a saturated set satisfies the assumptions of Lemma 1 then it is
referred to as a set of infinite type. Otherwise, we talk about a saturated
set of finite type.
Lemma 2. 1) For an arbitrary Steinitz number s0 ∈ SN the set
S(∞, s0) :=
{ a
b
· s0
∣∣∣ a ∈ N, b ∈ Ω(s0)
}
is a saturated set of infinite type.
2) If S is a saturated set of infinite type, then for an arbitrary Steinitz
number s ∈ S we have S = S(∞, s).
Proof. We have to show that the set S(∞, s0) satisfies the conditions 1),
2), 3). The condition 1) is obvious. Let s = (a/b) · s0, b ∈ Ω(s0). Without
loss of generality, we assume that a and b are coprime. Let c ∈ Ω(s) and
let d = gcd(c, a) be the greatest common divisor of a and c, a = a ′d,
c = c ′d, the numbers a ′, c ′ are coprime. Then a · s0/(bc) = a ′ · s0/(bc
′),
which implies that dc ′ ∈ Ω(s0). Hence
s
c
=
a
bc
· s0 =
a ′
bc ′
· s0 ∈ S(∞, s0).
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O. Bezushchak 23
We have checked the condition 2).
Let us check the condition 3). Choose s = (a/b) · s0 ∈ S(∞, s0),
b ∈ Ω(s0). Let c ∈ Ω(s). We need to check that for any i > 1
i ·
s
c
=
ia
bc
· s0 ∈ S(∞, s0).
Let a/(bc) = a ′/b ′, where the natural numbers a ′, b ′ are coprime. Since
a
bc
· s0 =
s
c
∈ SN
it follows that b ′ ∈ Ω(s0). Hence, i · (a ′/b ′) · s0 ∈ S(∞, s0), which implies
that S(∞, s0) satisfies the condition 3) and, therefore, is saturated.
Let S be a saturated subset of SN of infinite type. Choose s0 ∈ S. Our
aim is to show that S = S(∞, s0). Since the subset S is of infinite type it
follows that rs(b) = ∞ for any s ∈ S, b ∈ Ω(s). In particular,
S(∞, s0) =
{ a
b
· s0
∣∣∣ s ∈ Ω(s0)
}
⊆ S.
An arbitrary Steinitz number s ∈ S is rationally connected to s0, hence
there exist a, b ∈ N such that s = (a/b) · s0. Without loss of generality,
we assume that a and b are coprime, which implies b ∈ Ω(s0). We proved
that s ∈ S(∞, s0).
Now let S ⊂ SN be a saturated subset of finite type, that is, for any
s ∈ S, d ∈ Ω (s) we have
rs(b) = max
{
i ∈ N
∣∣ i ·
s
b
∈ S
}
< ∞.
By the condition 3),
{
i ∈ N
∣∣ i ·
s
b
∈ S
}
=
[
1 , rs(b)
]
.
Since b · (s/b) ∈ S it follows that b 6 rs(b). Choose a Steinitz number
s ∈ S and two natural numbers b, c ∈ Ω(s) such that b divides c. If
i · (s/b) ∈ S then (ic/b) · (s/c) ∈ S. Hence rs(b) · (c/b) 6 rs(c). In other
words,
rs(b)
b
6
rs(c)
c
. (1)
Let i ∈ N, s/c ∈ S and let k be a maximal nonnegative integer such that
k · (c/b) 6 i. By the condition 3), k · (c/b) · (s/c) ∈ S, hence k · (s/b) ∈ S.
So, k 6 rs(b). We proved that
[ rs(c)
c / b
]
6 rs(b). (2)
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24 Spectra of locally matrix algebras
The inequalities (1), (2) imply
[ rs(c)
c / b
]
6 rs(b) 6
rs(c)
c / b
.
Hence
rs(b) =
[ rs(c)
c / b
]
. (3)
In particular,
rs(c)
c / b
− 1 < rs(b),
rs(c)
c / b
< rs(b) + 1.
Dividing by b, we get
rs(b)
b
6
rs(c)
c
<
rs(b)
b
+
1
b
. (4)
Lemma 3. Let S ⊂ SN be a saturated subset of finite type and let s ∈ S
be an infinite Steinitz number. Then there exists a limit
rS(s) = lim
b ∈ Ω(s)
b→∞
rs(b)
b
, 1 6 rS(s) <∞.
If the set S is fixed then we denote rS(s) = r(s).
Remark 3. The limit r(s) is equal to the inverse of the density invariant
of Dixmier–Baranov [1, 5].
The proof of Lemma 3. The set {rs(b)/b | b ∈ Ω(s)} is bounded from
above. Indeed, choose b0 ∈ Ω(s). For an arbitrary b ∈ Ω(s) there exists
c ∈ Ω(s) that is a common multiple for b0 and b. Then by (1) and (4),
r(b)
b
6
r(c)
c
<
r(b0)
b0
+
1
b0
.
Let
r = r(s) = sup
{ rs(b)
b
∣∣∣ b ∈ Ω(s)
}
.
Clearly, 1 6 r <∞. Choose ε > 0. Let N(ε) = [2r/ε] + 1. We will show
that for any b ∈ Ω (s), b > N(ε), we have r − ε < rs(b)/b.
Indeed, let b ∈ Ω(s), b > N(ε) > 2r/ε. Then 1/b < ε/(2r) 6 ε/2.
There exists a natural number b0 ∈ Ω (s) such that r − ε/2 < rs(b0)/b0.
Let c ∈ Ω(s) be a common multiple of b0 and b. Then (4) implies
r(b)
b
>
r(c)
c
−
1
b
>
r(b0)
b0
−
1
b
> r −
ε
2
−
ε
2
= r − ε.
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O. Bezushchak 25
So,
r = lim
b ∈ Ω(s)
b→∞
rs(b)
b
and this completes the proof of the lemma.
Lemma 4. Let s, s ′ ∈ S be infinite Steinitz numbers, s ′ = (a/b) · s; a,
b ∈ N; b ∈ Ω(s). Then r(s ′) = (a/b) · r(s).
Proof. It is sufficient to show that if s, ms ∈ S, m ∈ N, then m · r(ms) =
r(s).
Suppose that b ∈ Ω(s) and i · (ms/b) ∈ S. Then i ·m · (s/b) ∈ S. Hence
rms(b) ·m 6 rs(b) and, therefore, r(ms) ·m 6 r(s).
On the other hand, if i · (s/b) ∈ S then [i/m] ·m 6 i and, therefore,
[i/m] ·m · (s/b) ∈ S. We showed that
[ rs(b)
m
]
6 rms(b),
rs(b)
m
− 1 < rms(b),
1
m
·
rs(b)
b
−
1
b
<
rms(b)
b
.
Assuming b→ ∞ we get (1/m) · r(s) 6 r(ms), which completes the proof
of the lemma.
In the inequality (4), let c→ ∞. Then
rs(b)
b
6 r(s) 6
rs(b)
b
+
1
b
, rs(b) 6 r(s) b 6 rs(b) + 1.
If the number r(s) is irrational then rs(b) = [r(s)b] for all b ∈ Ω(s).
Now suppose that the number r = rs(b) is rational; r = u/v; u, v
are coprime. If a number b ∈ Ω(s) is not a multiple of v then, as above,
rs(b) = [(u/v) · b]. If b is a multiple of v then
rs(b) =
[
r b or
r b − 1 .
Lemma 5. If at least for one number b0 ∈ Ω(s)
⋂
vN we have rs(b0) = rb0
then for all b ∈ Ω(s)
⋂
vN we have rs(b) = rb.
Proof. Let b, c ∈ Ω(s)
⋂
vN and b divides c. If rs(b) = rb then, by the
inequality (1), we have
r =
rs(b)
b
6
rs(c)
c
,
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26 Spectra of locally matrix algebras
which implies rs(c) = rc. On the other hand, if rs(c) = rc then, by the
inequality (4),
r =
rs(c)
c
<
rs(b)
b
+
1
b
,
which implies rs(b) > rb− 1. Hence rs(b) = rb. We showed that rs(b) = rb
if and only if rs(c) = rc.
Now choose b1, b2 ∈ Ω(s)
⋂
vN and suppose that rs(b1) = rb1. There
exists c ∈ Ω(s)
⋂
vN such that both b1 and b2 divide c. In view of the
above, rs(b1) = rb1 implies rs(c) = rc, which implies rs(b2) = rb2. This
completes the proof of the lemma.
Recall that for an infinite Steinitz number s and a real number r,
1 6 r <∞,
S(r, s) =
{ a
b
s
∣∣∣ a, b ∈ N; b ∈ Ω(s), a 6 rb
}
,
S+(r, s) =
{ a
b
s
∣∣∣ a, b ∈ N; b ∈ Ω(s), a < rb
}
.
If r is an irrational number or r = u/v, the integers u, v are coprime
and v 6∈ Ω (s) then S(r, s) = S+(r, s). If r = u/v, v ∈ Ω (s) then
S+(r, s) $ S(r, s).
Lemma 6. The subsets S(r, s) and S+(r, s) are saturated.
Proof. The condition 1) in the definition of saturated subsets is obviously
satisfied. Let us check the condition 2). Let (a/b) ·s ∈ S(r, s) (respectively,
(a/b) · s ∈ S+(r, s)), where a, b are coprime natural numbers, b ∈ Ω(s).
Then a 6 rb (respectively, a < rb). Suppose that c ∈ Ω(ab s). We need to
show that (a · s)/(b · c) ∈ S(r, s) (respectively, (a · s)/(b · c) ∈ S+(r, s)).
Let d = gcd(a, c), a = da ′, c = dc ′. Then
a s
b c
=
a ′
b c ′
s ∈ SN.
Since the number bc ′ is coprime with a ′ it follows that bc ′ ∈ Ω(s). The
inequality a ′ 6 rbc ′ (respectively, a ′ < rbc ′) is equivalent to the inequality
a 6 rbc (respectively, a < rbc). The latter inequality follows from a 6 rb
(respectively, a < rb). The condition 2) is verified.
Let us check the condition 3). As above, we assume that a, b are
coprime natural numbers, b ∈ Ω(s) and a/b ∈ S(r, s) (respectively, a/b ∈
S+(r, s)). Let c ∈ Ω((a/b) · s), gcd(a, c) = d, a = da ′, c = dc ′. We
have shown above that bc ′ ∈ Ω(s). Let n ∈ N and n · (as/(bc)) ∈ S(r, s)
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O. Bezushchak 27
(respectively, n · (as/(bc)) ∈ S+(r, s)). Then na ′ 6 rbc ′ (respectively,
na ′ < rbc ′). This immediately implies that for any i, 1 6 i 6 n, we
have ia ′ 6 rbc ′ (respectively, ia ′ < rbc ′). Hence, i · (as/b) ∈ S(r, s)
(respectively, i · (as/b) ∈ S+(r, s)).
Lemma 7. Let r = u/v, where u, v are coprime natural numbers. Let s
be an infinite Steinitz number and v ∈ Ω (s). Then the set S+(r, s) is not
equal to any of the sets S(r ′, s ′), r ′ ∈ [1,∞), s ′ ∈ SN.
Proof. Let s2 ∈ S(r, s1) (respectively, s2 ∈ S+(r, s1)). Then s2 = (a/b)·s1,
where a, b ∈ N, b ∈ Ω(s1). By Lemma 4,
S(r, s1) = S
(
r
b
a
, s2
) (
respectively, S+(r, s1) = S+
(
r
b
a
, s2
))
.
We showed that the set S(r, s) (respectively, S+(r, s)) is determined by
any Steinitz number s ′ ∈ S(r, s) (respectively, s ′ ∈ S+(r, s)) with an
appropriate recalibration of r.
Let S = S(r1, s1) = S+(r2, s2). Choosing an arbitrary Steinitz number
s ∈ S we get S(r ′
1, s) = S+(r ′
2, s) for some r ′
1, r
′
2 ∈ [1,∞). The number
r ′
2 = u/v is rational, gcd(u, v) = 1 and v ∈ Ω(s).
The number r is uniquely determined by a saturated subset S and
a choice of s ∈ S. Hence r ′
1 = r ′
2. Now it remains to notice that for a
rational number r = u/v, gcd(u, v) = 1, and an infinite Steinitz number s
such that v ∈ Ω(s) we have S(r, s) 6= S+(r, s). This completes the proof
of the lemma.
Lemma 8. Let S ⊂ SN�N be a saturated subset of finite type, s ∈ S,
r = rS(s) ∈ [1,∞). Then S = S(r, s) or S = S+(r, s).
Proof. Recall that for a natural number b ∈ Ω(s) we defined
rs(b) = max
{
i ∈ N
∣∣∣ i
s
b
∈ S
}
.
We showed that if r is an irrational number or r = u/v; u, v are coprime
and v 6∈ Ω (s), then rs(b) = [rb] for an arbitrary b ∈ Ω(s).
An arbitrary Steinitz number s ′ ∈ S is representable as s ′ = (a/b) · s,
where a, b are coprime natural numbers. Clearly, b ∈ Ω(s) and a 6 rs(b) =
[rb]. That is why in the case when r is irrational or r = u/v, gcd(u, v) = 1,
v 6∈ Ω(s), we have S = S(r, s) = S+(r, s).
Suppose now that r = u/v, gcd(u, v) = 1, v ∈ Ω(s). If b ∈ Ω(s)�vN
then as above rs(b) = [rb]. By Lemma 5, either for all b ∈ Ω(s) ∩ vN we
have rs(b) = rb or for all b ∈ Ω(s) ∩ vN we have rs(b) = rb − 1. In the
first case S = S(r, s), in the second case S = S+(r, s).
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28 Spectra of locally matrix algebras
Lemma 9. Let S ⊆ N be a saturated subset.Then either S = {1, 2, . . . , n}
for some n ∈ N or S = N.
Proof. First, notice that the subsets {1, 2, . . . , n} and N are saturated.
Now let S ⊆ N be a saturated subset. If n ∈ S then n ∈ Ω(n) and
n · (n/n) ∈ S. By the condition 3), all natural numbers i = i · (n/n),
1 6 i 6 n, lie in S. This implies the assertion of the lemma.
Now, Theorem 2 follows from Lemmas 8, 9.
3. Countable-dimensional locally matrix algebras
For an algebra A and an idempotent 0 6= e ∈ A we call the subalgebra
eAe a corner of the algebra A.
Let A1 ⊂ A2 ⊂ · · · be an ascending chain of unital locally matrix
algebras, Ai is a corner of the algebra Ai+1, i > 1,
A =
∞⋃
i=1
Ai.
Clearly, Spec(A1) ⊆ Spec(A2) ⊆ · · · .
Lemma 10.
Spec(A) =
∞⋃
i=1
Spec(Ai).
Proof. For an arbitrary idempotent e ∈ Ai we have eAie = eAe, hence
Spec(Ai) ⊆ Spec(A). On the other hand, an arbitrary idempotent e ∈ A
lies in one of the subalgebras Ai. Hence st(eAe) = st(eAie) ∈ Spec(Ai).
Proof of Theorem 3 (1). To start with we notice that {1, 2, . . . , n} =
Spec(Mn(F)). Let s be a Steinitz number. In [2], we showed that there
exists a unital locally matrix algebra A, dimFA 6 ℵ0, such that st(A) = s.
Consider the algebra M∞(A) of infinite N× N-matrices, having finitely
many nonzero entries. The algebra Mn(A) of n × n-matrices over A is
embedded in M∞(A) as a north-west corner,
M1(A) ⊂ M2(A) ⊂ · · · , M∞(A) =
∞⋃
n=1
Mn(A).
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O. Bezushchak 29
In particular, it implies that M∞(A) is a locally matrix algebra. We will
show that
Spec
(
M∞(A)
)
= S(∞, s). (5)
Indeed, by Lemma 10,
Spec
(
M∞(A)
)
=
∞⋃
n=1
Spec
(
Mn(A)
)
.
We have st(Mn(A)) = ns. In [4], we showed that
Spec
(
Mn(A)
)
=
{ a
b
n s
∣∣∣ b ∈ Ω(ns), a, b ∈ N; 1 6 a 6 b
}
.
This implies Spec(Mn(A)) ⊆ S(∞, s). A Steinitz number (a/b)·s, b ∈ Ω(s),
lies in Spec(Mn(A)) provided that a/b 6 n. This completes the proof of
(5). In particular, Spec(M∞(F)) = N.
Consider now a saturated subset S = S(r, s) or S = S+(r, s), 1 6
r <∞, where s is an infinite Steinitz number. Choose a sequence b1, b2,
. . . ∈ Ω(s) such that bi divides bi+1, i > 1, and s is the least common
multiple of bi, i > 1. There exists a unique (up to isomorphism) unital
countable-dimensional locally matrix algebra As/bi such that st(As/bi) =
s/bi. Let Ai =Mrs(bi)(As/bi). We have
st(As/bi) = st
((
Mbi+1/bi
(
As/bi+1
))
= s/bi+1.
Hence, by Glimm’s Theorem, As/bi
∼=Mbi+1/bi(As/bi+1
) and, therefore,
Ai = Mrs(bi)
(
As/bi
)
∼= M
rs(bi) ·
bi+1
bi
(
As/bi+1
)
.
By the inequality (1), rs(bi) · (bi+1/bi) 6 rs(bi+1). Hence, the algebra Ai
is embeddable in the algebra Ai+1 as a north-west corner. Let
A =
∞⋃
i=1
Ai.
We will show that Spec(A) = S. Let 0 6= e ∈ A be an idempotent. Then
e ∈ Ai for some i > 1. In [4], we showed that
st(eAie) =
a
b
st(Ai),
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30 Spectra of locally matrix algebras
where a, b ∈ N; a, b are coprime natural numbers; b ∈ Ω(st(Ai)), a 6 b.
Furthermore,
st(Ai) = rs(bi)
s
bi
, st(eAie) =
a
b
rs(bi)
s
bi
.
Let d = gcd(b, rs(bi)), b = db ′, rs(bi) = d · rs(bi)
′. So,
st(eAie) =
a · rs(bi)
′
b ′
·
s
bi
∈ SN.
This implies that b ′ ∈ Ω(s/bi). Therefore, b ′bi ∈ Ω(s). To show that
st(eAie) lies in S(r, s) (respectively, S+(r, s)) we need to verify that
a ·rs(bi)
′ 6 r b ′ bi (respectively, a ·rs(bi)
′ < r b ′ bi). Multiplying both sides
of the inequality by d we get a·rs(bi) 6 r b bi (respectively, a·rs(bi) < r b bi).
This inequality holds since a 6 b and rs(bi) 6 r · bi (respectively, a 6 b
and rs(bi) < r · bi). We proved that Spec(A) ⊆ S.
Let us show that S ⊆ Spec(A). Consider a Steinitz number (a/b)·s ∈ S,
where a, b ∈ N; b ∈ Ω(s), a 6 r b in the case S = S(r, s) or a < r b in the
case S = S+(r, s).
There exists a member of our sequence bi such that b divides bi,
bi = k · b, k ∈ N. Then (a/b) · s = (a k/bi) · s.
We will show that a k 6 rs(bi). Indeed, multiplying both sides of the
inequality by b we get abi 6 rs(bi)b. Let S = S(r, s). Then a 6 rb. Since
a ∈ N it implies a 6 [r b]. Furthermore, rs(bi) = [rbi] = [rbk]. So, it is
sufficient to show that [rb]k 6 [rbk]. This inequality holds since [rb]k is
an integer and [rb]k 6 rbk.
Now suppose that S = S+(r, s). Then a < rb,
rs(bi) =
{
[r bi], if r bi 6∈ N,
r bi − 1, if r bi ∈ N.
There are three possibilities:
1) r b ∈ N and, therefore, r bi ∈ N. In this case a 6 rb − 1, rs(bi) =
rbi − 1. We have abi 6 (rb− 1)bi 6 (rbi − 1)b = rs(bi)b;
2) r b 6∈ N, but r bi ∈ N. In this case a 6 [rb], rs(bi) = rbi − 1, we
have abi 6 [rb]bi, rs(bi)b = (rbi − 1)b. Hence, it is sufficient to show
that [rb]k 6 rbi − 1 = rbk − 1. The number [r b] k is an integer and
[rb]k < rbk since [rb] < rb. This implies the claimed inequality;
3) r bi 6∈ N and, therefore, r b 6∈ N. In this case abi 6 [rb]bk, rs(bi)b =
[rbk]b and it remains to notice that [r b] k 6 [r b k].
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O. Bezushchak 31
We showed that both for S = S(r, s) and for S = S+(r, s) there holds the
inequality ak 6 rs(bi).
Recall that Ai = Mrs(bi)(As/bi). Consider the north-east corner
Mak
(
As/bi
)
of the algebra Ai. We have
st
(
Mak
(
As/bi
))
= a k ·
s
bi
=
a
b
s,
and, therefore, S ⊆ Spec(A). This completes the proof of Theorem 3 (1).
For the proof of Theorem 3 (2) we will need several lemmas on exten-
sions of isomorphisms.
Lemma 11. Let A be a locally matrix algebra and let A1 be a subalgebra
of A such that A1
∼=Mn(F). Then every automorphism of the algebra A1
extends to an automorphism of the algebra A.
Proof. Let e be the identity element of the subalgebra A1. Then the corner
eAe is a unital locally matrix algebra. Let C be the centralizer of the
subalgebra A1 in eAe. By Wedderbern’s Theorem (see [6, 8]), we have
eA e = A1 ⊗F C. An arbitrary automorphism ϕ of the subalgebra A1 is
inner, that is, there exists an invertible element x of the subalgebra A1
such that ϕ(a) = x−1ax for all elements a ∈ A1. The conjugation by the
element x⊗ e extends ϕ to an automorphism of the algebra eAe. Consider
the Peirce decomposition
A = eAe + eA(1− e) + (1− e)Ae + (1− e)A(1− e),
and the mapping
ϕ̃ : A ∋ a 7→ x−1ax + x−1a(1− e) + (1− e)ax + (1− e)a(1− e).
The mapping ϕ̃ extends ϕ and ϕ̃ ∈ Aut(A). This completes the proof of
the lemma.
Lemma 12. Let A be a unital locally matrix algebra with an idempotent
e 6= 0. Then an arbitrary automorphism of the corner eAe extends to an
automorphism of the algebra A.
Proof. Suppose at first that an automorphism ϕ of the algebra eAe is
inner, and there exists an element xe ∈ eAe that is invertible in the algebra
eAe such that ϕ(a) = x−1
e axe for an arbitrary element a ∈ eA e. The
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32 Spectra of locally matrix algebras
element x = xe + (1− e) is invertible in the algebra A. So, conjugation by
the element x extends ϕ.
Now let ϕ be an arbitrary automorphism of the corner eAe. Let A1 ⊆ A
be a subalgebra such that 1, e ∈ A1 and A1
∼= Mm(F) for some m > 1.
Consider A2 ⊆ A such that A1 ⊆ A2, ϕ(eA1 e) ⊆ A2 and A2
∼= Mn(F)
for some n > 1. Consider the embedding
ϕ : eA1 e → ϕ (eA1 e) ⊆ eA2 e
that preserves the identity element e. By Skolem–Noether Theorem (see [6]),
there exists an invertible element xe ∈ eA2e such that ϕ(a) = x−1
e axe for
an arbitrary element a ∈ eA1 e.
As noticed above, there exists an automorphism ψ of the algebra A that
extends the automorphism eAe → eAe, a 7→ x−1
e axe. The composition
ψ−1 ◦ ϕ leaves all elements of the algebra eA1e fixed. Since it is sufficient
to prove that the automorphism ψ−1 ◦ ϕ ∈ Aut(eAe) extends to an
automorphism of A we will assume without loss of generality that the
automorphism ϕ ∈ Aut(eAe) fixes all elements of eA1e.
Let C be the centralizer of the subalgebra A1 in A. Then A = A1⊗FC
and eAe = eA1e⊗F C. Since the subalgebra e⊗F C is the centralizer of
eA1e ⊗F C in the algebra eAe it follows that e ⊗F C is invariant with
respect to the automorphism ϕ. Hence, there exists an automorphism
θ ∈ Aut(C) such that ϕ(a⊗ c) = a⊗ θ(c) for all elements a ∈ eAe, c ∈ C.
So, the automorphism ϕ̃(a⊗ c) = a⊗ θ(c), a ∈ A1, c ∈ C, extends ϕ. This
completes the proof of the lemma.
Lemma 13. Let A be a unital locally matrix algebra with nonzero idem-
potents e1, e2. An arbitrary isomorphism ϕ : e1Ae1 → e2Ae2 extends to
an automorphism of the algebra A.
Proof. There exists a subalgebra A1 ⊆ A such that 1, e1, e2 ∈ A1 and
A1
∼=Mn(F) for some n > 1. Let ri be the matrix range of the idempotent
ei in A1, i = 1, 2. In [4], it was shown that
st (e1 A e1) =
r1
n
· st(A), st (e2 A e2) =
r2
n
· st(A).
Since e1Ae1 ∼= e2Ae2 it follows that r1 = r2. In the matrix algebra Mn(F)
any two idempotents of the same range are conjugate via an automorphism.
Hence, the idempotents e1, e2 are conjugate via an automorphism ofA1. By
Lemma 11, an arbitrary automorphism of A1 extends to an automorphism
of the algebra A. Now the assertion of the lemma follows from Lemma 12.
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O. Bezushchak 33
Lemma 14. Let A, B be isomorphic unital locally matrix algebras with
nonzero idempotents e ∈ A, f ∈ B. An arbitrary isomorphism eAe→ fBf
extends to an isomorphism A→ B.
Proof. Let ϕ : A→ B, ψ : eAe→ fBf be isomorphisms. Then
ϕ−1 ◦ ψ : e A e → ϕ−1(f) A ϕ−1(f)
is an isomorphism of two corners of the algebra A. By Lemma 13, ϕ−1 ◦ψ
extends to an automorphism χ of the algebra A, the isomorphism ϕ ◦ χ
extends ψ.
Lemma 15. Let A be a unital locally matrix algebra and let s1, s2 be
Steinitz numbers from Spec(A). Suppose that s2/s1 > 1. Let e1 ∈ A be
an idempotent such that st(e1Ae1) = s1. Then there exists an idempotent
e2 > e1 such that st(e2Ae2) = s2.
Proof. Since s2 ∈ Spec(A) there exists an idempotent e ′ ∈ A such that
st(e ′Ae ′) = s2. Choose a subalgebra A1 ⊆ A such that e1, e
′ ∈ A1 and
A1
∼=Mn(F).
Let r1, r2 be the matrix ranges of e1, e
′ in Mn(F), respectively. In [4],
it was shown that
st(e1 A e1) = s1 =
r1
n
st(A), st(e ′ A e ′) = s2 =
r2
n
st(A).
Hence r2 > r1. Since every idempotent in the algebra Mn(F) is diago-
nalizable there exist automorphisms ϕ, ψ of the algebra A1 such that
ψ(e ′) > ϕ(e1). By Lemma 11, the automorphisms ϕ, ψ extend to auto-
morphisms ϕ̃, ψ̃ of the algebra A, respectively.
Let e2 = ϕ̃−1 (ψ (e ′)) . Then e2 > e1 and st(e2Ae2) = s2, which
completes the proof of the lemma.
Proof of Theorem 3 (2). Let A, B be countable-dimensional locally ma-
trix algebras, Spec(A) = Spec(B). Choose bases a1, a2, . . . and b1, b2, . . .
in the algebras A, B, respectively.
We will construct ascending chains of corners {0} = A0 ⊂ A1 ⊂ A2 ⊂
· · · in the algebra A and {0} = B0 ⊂ B1 ⊂ B2 ⊂ · · · in the algebra B,
such that
∞⋃
i=0
Ai = A,
∞⋃
i=0
Bi = B
and a1, . . . , ai ∈ Ai, b1, . . . , bi ∈ Bi, st(Ai) = st(Bi) for all i > 1.
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34 Spectra of locally matrix algebras
Suppose that corners {0} = A0 ⊂ A1 ⊂ A2 ⊂ · · · ⊂ An, {0} = B0 ⊂
B1 ⊂ B2 ⊂ · · · ⊂ Bn have already been selected, n > 0. There exist
corners A ′ ⊂ A, B ′ ⊂ B in the algebras A, B, respectively, such that
An ⊂ A ′, an+1 ∈ A ′ and Bn ⊂ B ′, bn+1 ∈ B ′. The Steinitz numbers
st(A ′), st(B ′) lie in the same saturated subset of SN, therefore, they are
rationally connected.
Suppose that st(B ′) > st(A ′). Let = e ′ be an idempotent of the
algebra A such that A ′ = e ′Ae ′. The Steinitz number st(B ′) lies in
Spec(A). Hence, by Lemma 15, there exists an idempotent e ∈ A such
that e > e ′ and st(eAe) = st(B ′). Choose An+1 = eAe, Bn+1 = B ′. The
chains {0} = A0 ⊂ A1 ⊂ A2 ⊂ · · · and {0} = B0 ⊂ B1 ⊂ B2 ⊂ · · · have
been constructed.
By Lemma 14, every isomorphism Ai → Bi extends to an isomorphism
Ai+1 → Bi+1. This gives rise to a sequence of isomorphisms ϕi : Ai → Bi,
i > 0, where each ϕi+1 extends ϕi. Taking the union ∪i>0ϕi we get an
isomorphism from the algebra A to the algebra B. This completes the
proof of the theorem.
Proof of Theorem 4. It is easy to see that a locally matrix algebra A is a
unital if and only if the set of idempotents of A has a largest element: an
identity. This is equivalent to Spec(A) containing a largest Steinitz number.
Among saturated sets of Steinitz numbers only {1, 2, . . . , n} and S(r, s),
s ∈ SN�N, r = u/v; u and v are coprime natural numbers, v ∈ Ω(s),
satisfy this assumption.
4. Embeddings of locally matrix algebras
Lemma 16. Let S1, S2 be saturated sets of Steinitz numbers. Then either
S1
⋂
S2 = ∅ or one of the sets S1, S2 contains the other one.
Proof. Let s ∈ S1
⋂
S2. If s ∈ N then, by Lemma 9, each set Si is either
a segment [1, n], n > 1, or the whole N. In this case the assertion of the
lemma is obvious.
Suppose that the number s is infinite. Then by Theorem 2, Si = S(ri, s)
or Si = S+(ri, s), where ri = rSi
(s) ∈ [1,∞) ∪ {∞}, i = 1, 2. Clearly, if
rS1
(s) < rS2
(s) then S1 $ S2. If rS1
(s) = rS2
(s) then
S1, S2 =
S(r, s)
S+(r, s)
S(∞, s)
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O. Bezushchak 35
and S+(r, s) ⊆ S(r, s) ⊂ S(∞, s) for any r ∈ [1,∞). This completes the
proof of the lemma.
Let A be a locally matrix algebra. A subalgebra B ⊆ A is called an
approximative corner of A if B is the union of an increasing chain of
corners. In other words, there exist idempotents e0, e1, e2, . . . such that
e0 A e0 ⊆ e1 A e1 ⊆ e2 A e2 ⊆ · · · , B =
∞⋃
i=0
ei A ei.
It is easy to see that an approximative corner of a locally matrix
algebra is a locally matrix algebra.
Theorem 5. Let A, B be countable-dimensional locally matrix algebras.
Then B is embeddable in A as an approximative corner if and only if
Spec(B) ⊆ Spec(A).
Proof. If B is an approximative corner of A then every corner of B is a
corner of A, hence Spec(B) ⊆ Spec(A).
Suppose now that Spec(B) ⊆ Spec(A). If the algebra B is unital
then it embedds in the algebra A as a corner. Indeed, the embedding
Spec(B) ⊆ Spec(A) implies that there exists an idempotent e ∈ A such
that st(B) = st(eAe). By Glimm’s Theorem [7], we have B ∼= eAe.
Suppose now that the algebra B is not unital. Then there exists a
sequence of idempotents 0 = f0, f1, f2, . . . of algebra B such that
{0} = f0 B f0 $ f1 B f1 $ f2 B f2 $ · · · ,
∞⋃
i=0
fi B fi = B.
We will construct a sequence of idempotents e0, e1, e2, . . . in the algebra
A such that
e0 A e0 $ e1 A e1 $ e2 A e2 $ · · · , st(fi B fi) = st(ei A ei)
for an arbitrary i > 0. Let e0 = 0. Suppose that we have already selected
idempotents e0, e1, . . . , en ∈ A such that e0Ae0 ⊂ e1Ae1 ⊂ · · · ⊂ enAen
and st(eiAei) = st(fiBfi), 0 6 i 6 n. We have
st(fn+1 B fn+1) > st(fn B fn) = st(en A en)
and st(fn+1Bfn+1) ∈ Spec(A). By Lemma 15, there exists an idempo-
tent en+1 ∈ A such that enAen ⊂ en+1Aen+1 and st(en+1Aen+1) =
st(fn+1Bfn+1), which proves existence of a sequence e0, e1, e2, . . . .
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36 Spectra of locally matrix algebras
The union
A ′ =
∞⋃
i=0
ei A ei
is an approximative corner of the algebra A. By Glimm’s Theorem [7],
eiAei ∼= fiBfi, i > 1. By Lemma 10,
Spec(A ′) =
∞⋃
i=1
Spec(ei A ei) and Spec(B) =
∞⋃
i=1
Spec(fiBfi).
Hence Spec(B) = Spec(A ′). By Theorem 3 (2), we have A ′ ∼= B, which
completes the proof of the theorem.
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Contact information
Oksana Bezushchak Faculty of Mechanics and Mathematics, Taras
Shevchenko National University of Kyiv,
Volodymyrska, 60, Kyiv 01033, Ukraine
E-Mail(s): mechmatknubezushchak@gmail.com
Received by the editors: 08.12.2020
and in final form 15.02.2021.
mailto:mechmatknubezushchak@gmail.com
O. Bezushchak
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| id | nasplib_isofts_kiev_ua-123456789-188675 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1726-3255 |
| language | English |
| last_indexed | 2025-11-30T14:19:41Z |
| publishDate | 2021 |
| publisher | Інститут прикладної математики і механіки НАН України |
| record_format | dspace |
| spelling | Bezushchak, O. 2023-03-11T13:10:28Z 2023-03-11T13:10:28Z 2021 Spectra of locally matrix algebras / O. Bezushchak // Algebra and Discrete Mathematics. — 2021. — Vol. 31, № 1. — С. 17–36. — Бібліогр.: 11 назв. — англ. 1726-3255 DOI:10.12958/adm1734 2020 MSC: 08A05, 16S50 https://nasplib.isofts.kiev.ua/handle/123456789/188675 We describe spectra of associative (not necessarily unital and not necessarily countable-dimensional) locally matrix algebras. We determine all possible spectra of locally matrix algebras and give a new proof of Dixmier–Baranov Theorem. As an application of our description of spectra, we determine embeddings of locally matrix algebras. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics Spectra of locally matrix algebras Article published earlier |
| spellingShingle | Spectra of locally matrix algebras Bezushchak, O. |
| title | Spectra of locally matrix algebras |
| title_full | Spectra of locally matrix algebras |
| title_fullStr | Spectra of locally matrix algebras |
| title_full_unstemmed | Spectra of locally matrix algebras |
| title_short | Spectra of locally matrix algebras |
| title_sort | spectra of locally matrix algebras |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/188675 |
| work_keys_str_mv | AT bezushchako spectraoflocallymatrixalgebras |