A commutative Bezout PM* domain is an elementary divisor ring
We prove that any commutative Bezout PM∗ domain is an elementary divisor ring.
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Інститут прикладної математики і механіки НАН України
2015
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| Назва видання: | Algebra and Discrete Mathematics |
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| Цитувати: | A commutative Bezout PM* domain is an elementary divisor ring / B. Zabavsky, A. Gatalevych // Algebra and Discrete Mathematics. — 2015. — Vol. 19, № 2. — С. 295–301. — Бібліогр.: 12 назв. — англ. |
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irk-123456789-1542472019-06-16T01:26:53Z A commutative Bezout PM* domain is an elementary divisor ring Zabavsky, B. Gatalevych, A. We prove that any commutative Bezout PM∗ domain is an elementary divisor ring. 2015 Article A commutative Bezout PM* domain is an elementary divisor ring / B. Zabavsky, A. Gatalevych // Algebra and Discrete Mathematics. — 2015. — Vol. 19, № 2. — С. 295–301. — Бібліогр.: 12 назв. — англ. 1726-3255 2010 MSC:13F99. http://dspace.nbuv.gov.ua/handle/123456789/154247 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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We prove that any commutative Bezout PM∗ domain is an elementary divisor ring. |
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Article |
| author |
Zabavsky, B. Gatalevych, A. |
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Zabavsky, B. Gatalevych, A. A commutative Bezout PM* domain is an elementary divisor ring Algebra and Discrete Mathematics |
| author_facet |
Zabavsky, B. Gatalevych, A. |
| author_sort |
Zabavsky, B. |
| title |
A commutative Bezout PM* domain is an elementary divisor ring |
| title_short |
A commutative Bezout PM* domain is an elementary divisor ring |
| title_full |
A commutative Bezout PM* domain is an elementary divisor ring |
| title_fullStr |
A commutative Bezout PM* domain is an elementary divisor ring |
| title_full_unstemmed |
A commutative Bezout PM* domain is an elementary divisor ring |
| title_sort |
commutative bezout pm* domain is an elementary divisor ring |
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Інститут прикладної математики і механіки НАН України |
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2015 |
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http://dspace.nbuv.gov.ua/handle/123456789/154247 |
| citation_txt |
A commutative Bezout PM* domain is an elementary divisor ring / B. Zabavsky, A. Gatalevych // Algebra and Discrete Mathematics. — 2015. — Vol. 19, № 2. — С. 295–301. — Бібліогр.: 12 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
| work_keys_str_mv |
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| first_indexed |
2025-07-14T05:54:23Z |
| last_indexed |
2025-07-14T05:54:23Z |
| _version_ |
1837600566760439808 |
| fulltext |
Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 19 (2015). Number 2, pp. 295–301
© Journal “Algebra and Discrete Mathematics”
A commutative Bezout P M
∗ domain
is an elementary divisor ring
B. Zabavsky, A. Gatalevych
Communicated by V. V. Kirichenko
Abstract. We prove that any commutative Bezout PM∗
domain is an elementary divisor ring.
The aim of this paper is to study the question of diagonalizability
for matrices over a ring. It is well-known that any elementary divisor
domain is a Bezout domain and it is a classical open question to determine
whether the converse statement is true?
The notion of an elementary divisor ring was introduced by Kaplansky
in [6]. There are a lot of researches that deal with the matrix diagonaliza-
tion in different cases (the most comprehensive history of these researches
can be found in [10]). It is an open question dating back at least to
Helmer [5] in 1942 to decide, whether a commutative Bezout domain is
always an elementary divisor domain. Helmer showed that not only does
the domain of entire functions is an elementary divisor domain, it also
has a property which he labeled adequate. Henriksen [4] appears to be
the first person to have given an example to show that being adequate
is a stronger property than that of being an elementary divisor ring. In
proving this, Henriksen observed that in an adequate domain each nonzero
prime ideal is contained in a unique maximal ideal [4]. It is a natural
question to ask whether or not the converse holds and this question is
explicitly raised in [7]. The negative answer to this question is given in [1].
Furthermore, it is shown that there exists an elementary divisor ring
2010 MSC: 13F99.
Key words and phrases: Bezout domain, PM-ring, clean element, neat element,
elementary divisor ring, stable range 1, neat range 1.
296 A commutative Bezout PM∗ domain. . .
which is not adequate but which does have the property that each nonzero
prime ideal is contained in a unique maximal ideal. In this paper we show
that a commutative Bezout domain in which each nonzero prime ideal is
contained in a unique maximal ideal is an elementary divisor ring. Note
that these results are responses to open questions work [12, Questions 10,
Problem 6].
We introduce the necessary definitions and facts.
All rings considered will be commutative and have identity. A ring is
a Bezout ring, if every its finitely generated ideal is principal. A ring R is
an elementary divisor ring if every matrix A (not necessarily square one)
over R admits diagonal reduction, that is, there exist invertible square
matrices P and Q such that PAQ is a diagonal matrix, say (dij), for
which dii is a divisor of di+1,i+1 for each i. A ring R to be right Hermite
if every 1 × 2 matrix over R admits diagonal reduction. Any Hermite ring
is a Bezout ring. For domains, the notions of Hermite and Bezout ring are
equivalent. Gillman and Henriksen showed that any commutative ring R
is an Hermite ring if and only if for all a, b ∈ R there exist a1, b1, d ∈ R
such that a = a1d, b = b1d and a1R + b1R = R [10]. Furthermore, they
proved the following result, which we state formally.
Proposition 1. Let R be a commutative Bezout ring. R is an elementary
divisor ring if and only if R is an Hermite ring that satisfies the extra
condition that for all a, b, c ∈ R with aR+bR+cR = R there exist p, q ∈ R
such that paR + (pb + qc)R = R.
Definition 1. Let R be a commutative Bezout domain. A nonzero
element a in R is called an adequate element if for every b ∈ R there exist
r, s ∈ R such that a = rs, rR + bR = R, and if s′ is a non-unit divisor of
s, then s′R + bR 6= R. If every nonzero element of the ring R is adequate,
then R is called an adequate ring [5, 10].
Definition 2. Let R be a commutative ring. An element a ∈ R is called a
clean element if a can be written as the sum of a unit and an idempotent.
If every element of R is clean, then we say that R is a clean ring [8, 9].
Any clean ring is a Gelfand ring. Recall that a ring R is called a
Gelfand ring if for every a, b ∈ R such that a + b = 1 there are r, s ∈ R
such that (1 + ar)(1 + bs) = 0. A ring R is called a PM-ring if each prime
ideal is contained in a unique maximal ideal. It had been asserted that a
commutative ring is a Gelfand ring if and only if it is a PM-ring [2, 3]. A
ring R is called a PM∗-ring if each nonzero prime ideal is contained in a
B. Zabavsky, A. Gatalevych 297
unique maximal ideal [9]. A ring R is said to be a ring of stable range
1, if for any a, b ∈ R such that aR + bR = R there exist t ∈ R such that
(a + bt)R = R.
Definition 3. An element a ∈ R \ {0} of a commutative ring R is called
a PM-element if the factor ring R/aR is a PM-ring.
Proposition 2. For a commutative ring R the following are equivalent:
1) a ∈ R is a PM-element;
2) for each prime ideal P such that a ∈ P there exists a unique
maximal ideal M such that P ⊂ M .
Proof. This is obvious, since P is a prime ideal of R/aR if and only if
there exists a prime ideal P such that aR ⊂ P and P = P/aR.
As a consequence of Proposition 2 we obtain the following result.
Proposition 3. A commutative domain R is a domain in which each
nonzero prime ideal is contained in a unique maximal ideal of R if and
only if every nonzero element of R is a PM-element.
Proposition 4. An element a of a commutative Bezout domain is
a PM-element if and only if, for every elements b, c ∈ R such that
aR + bR + cR = R, an element a can be represented as a = rs, where
rR + bR = R, sR + cR = R.
Proof. Denote R = R/aR, b = b+aR, c = c+aR. Since aR+bR+cR = R,
we see that bR + cR = R. Therefore, if a = rs where rR + bR = R,
sR + cR = R, then bR + cR = R and 0 = rs where rR + bR = R,
sR + cR = R. By [2], R is a PM-ring.
If R is a PM-ring then, by [9], 0 = rs where rR + bR = R, sR + cR =
R for arbitrary b, c ∈ R such that bR + cR = R. Whence we obtain
aR + bR + cR = R. Because 0 = 0 + aR = rs, we have rs ∈ aR, where
r = r + aR, s = s + aR. Let rR + aR = r1R, sR + aR = s1R. From
this r = r1r0, a = r1a0, s = s1s2, a = s1a2, where r0R + a0R = R,
s2R + a2R = R. Since r0R + a0R = R, we obtain r0u + a0v = 1 for
some u, v ∈ R. Since rs ∈ aR, we see that rs = at for some t ∈ R.
Then r1r0s = r1a0t, because R is a domain, and we have a0t = r0s. By
the equality, r0u + a0v = 1 we have sr0u + sa0v = s, a0(tu + a0v) = s.
Therefore a = r1a0, where r1R + bR + r1a0R = R. Then r1R + bR = R.
Since a0(tu + a0v) = s and a0R + cR + aR = R, we obtain a0R + cR = R.
The proposition is proved.
298 A commutative Bezout PM∗ domain. . .
Theorem 1. A commutative Bezout domain in which each nonzero prime
ideal is contained in a unique maximal ideal is an elementary divisor ring.
Proof. Let R be a commutative Bezout domain with the property that
each nonzero prime ideal is contained in a unique maximal ideal. According
to Proposition 4, let a, b, c ∈ R be such that aR+bR+cR = R. According
to the restrictions imposed on R, by Proposition 4, we have b = rs where
rR + aR = R, sR + cR = R. Let p ∈ R be such that sp + ck = 1 for some
k ∈ R. Hence rsp + rck = r and bp + crk = r. Denoting rk = q and we
obtain (br + cq)R + aR = R. Let pR + qR = dR and d = pp1 + qq1 with
p1R + q1R = R. Hence p1R + (p1b + q1c)R = R and, since pR ⊂ p1R, we
obtain p1R + cR = R and p1R + (p1b + q1c)R = R.
Since bp + cq = d(bp1 + cq1), and (bp + cq)R + aR = R we obtain
(bp1 + cq1)R + aR = R. Finally, we have ap1R + (bp1 + cq1)R = R. By
Proposition 1, we obtain that R is an elementary divisor ring. The theorem
is proved.
Remark 1. Note that in order to prove this theorem, it is necessary that
only the element b ∈ R is a PM-element.
Let R be a commutative Bezout domain. We denote by S = S(R)
the set of all PM-elements of R. Since 1 ∈ R, the set S is nonempty.
Furthermore, we obtain the following result.
Proposition 5. The set S(R) of all PM-elements of a commutative
domain R is a saturated multiplicatively closed set.
Proof. Let a, b ∈ S(R). We show that ab ∈ S(R). Suppose the contrary.
Then there exist a prime ideal P and maximal ideals M1, M2 such that
M1 6= M2 and ab ∈ P ⊂ M1 ∩ M2. Since ab ∈ P , we obtain that a ∈ P
or b ∈ R. It is impossible because a ∈ S(R), b ∈ S(R) and P ⊂ M1 ∩ M2.
Therefore S(R) is a multiplicatively closed set.
Let ab ∈ S(R) for some a, b ∈ R. If a /∈ S(R) then there exists a prime
ideal P such that a ∈ P and P ⊂ M1∩M2 for some maximal ideals M1, M2
and M1 6= M2. Therefore, ab ∈ P and P ⊂ M1 ∩ M2, M1 6= M2. It is
impossible because ab ∈ S(R). Hence S(R) is a saturated multiplicatively
closed set. The Proposition is proved.
Let R be a commutative Bezout domain and S(R) be the set of all
PM-elements of R. Since S(R) is a saturated multiplicatively closed set,
we can consider the localization of R with denominators from S(R) i.e.
the ring of fractious RS . We have:
B. Zabavsky, A. Gatalevych 299
Theorem 2. Let R be a commutative elementary divisor domain. Then
a ring RS is an elementary divisor ring.
Proof. Suppose that R is an elementary divisor ring. We need to show
that RS is also an elementary divisor ring. Let as−1, bs−1, cs−1 be any
elements from RS such that
as−1RS + bs−1RS + cs−1RS = RS .
Then aR+bR+cR = dR, for some element d ∈ S(R). Let a = a1d, b = b1d,
c = c1d for some elements a1, b1, c1 ∈ R such that a1R + b1R + c1R = R.
Since R is an elementary divisor ring, there are elements u, v, p, q ∈ R
such that
a1pu + (b1p + c1q)v = 1.
Then
apRS + (bp + cq)RS = RS .
By [6], RS is an elementary divisor ring. Theorem is proved.
Let R be a commutative Bezout domain and S = S(R) be the set of
all PM-elements of R. Since S(R) is a saturated multiplicatively closed
set, we can construct by transfinite induction a natural chain
{Rα|α is an ordinal}
of the saturated multiplicatively closed sets in R as follows. Let R0 = S(R).
Let α be an ordinal greater than zero and assume Rβ has been defined
and is a saturated multiplicatively closed set in R, whenever β < α and
let Kβ = RRβ
. Then Kβ is a commutative Bezout domain (see [10]) and
hence S(Kβ) is a saturated multiplicatively closed set by Proposition 5.
We define Rα by Rα =
⋃
β<α Rβ if α is a limit ordinal and Rα =
S(Kα−1)∩R otherwise. It is obvious that Rα is a saturated multiplicatively
closed set. If α, β are ordinals such that α 6 β then Rα ⊂ Rβ ⊂ R. Also
Rα = Rα+1 for some ordinal α. In case, when Rα 6= Rα+1 for each ordinal
α, then
card(Rα) > card(α).
Choosing β such that card(β) > card(R) we obtain
card(β) > card(R) > card(Rβ),
a contradiction. We let α0 denote the least ordinal such that
Rα0 = Rα0+1
300 A commutative Bezout PM∗ domain. . .
and we call
{Rα|0 6 α 6 α0}
a D-chain in R. In this situation R−1 will denote the group of units of R.
By Theorem 2 and the fact that union of elementary divisor rings are
an elementary divisor ring and using D-chain of a commutative Bezout
domain we can conclude that the problem of being a commutative Bezout
domain an elementary divisor ring is reduced to the case of a commutative
Bezout domain where PM-elements are the only units, when U(R) = S(R).
Definition 4. Let R be a commutative Bezout domain. An element
a ∈ R is called a neat element if R/aR is a clean ring.
Obvious examples of neat elements are units of a ring, and adequate
elements of a ring [11]. If R is a commutative Bezout domain and a is
a neat element of R, then R/aR is a clean ring [9], that is R/aR is a
PM-ring. Hence we obtain the following result.
Proposition 6. Every neat element of a commutative Bezout domain is
a PM-element.
Definition 5. A commutative ring R is said to be of the neat range 1 if
for any a, b ∈ R such that aR + bR = R there exists t ∈ R such that for
the element a + bt = c the ring R/cR is a clean ring [11].
Theorem 3 ([11]). A commutative Bezout domain is an elementary
divisor ring if and only if R is a ring of the neat range 1.
From this we obtain the following result.
Theorem 4. Let R be a commutative Bezout domain and U(R) = S(R).
Then R is an elementary divisor ring if and only if stable range of R is
equal to 1.
Proof. Since every neat element is a PM-element and U(R) = S(R),
then only units in a ring are neat elements. Then by Theorem 3, R is
an elementary divisor ring if and only if R is a ring of stable range 1.
Theorem is proved.
Let R be a commutative Bezout domain and a ∈ R is a neat element
of R. By [9] the stable range of R/aR is equal to 1. Consequently by
Theorem 4, we have a next result.
Theorem 5. Let R be a commutative Bezout domain such that for every
nonzero element a ∈ R stable range of R/aR is not equal 1. Then R is
not an elementary divisor ring.
B. Zabavsky, A. Gatalevych 301
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Contact information
B. V. Zabavsky,
A. Gatalevych
Department of Mechanics and Mathematics,
Ivan Franko National Univ., Lviv, Ukraine
E-Mail(s): zabavskii@gmail.com,
gatalevych@ukr.net
Received by the editors: 07.03.2015
and in final form 13.07.2015.
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