Diagonalizability theorem for matrices over certain domains
It is proved that R is a commutative adequate domain, then R is the domain of stable range 1 in localization in multiplicative closed set which corresponds s-torsion in the sense of Komarnitskii.
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irk-123456789-1548562019-06-17T01:31:21Z Diagonalizability theorem for matrices over certain domains Zabavsky, B. Domsha, O. It is proved that R is a commutative adequate domain, then R is the domain of stable range 1 in localization in multiplicative closed set which corresponds s-torsion in the sense of Komarnitskii. 2011 Article Diagonalizability theorem for matrices over certain domains / B. Zabavsky, O. Domsha // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 1. — С. 132–139. — Бібліогр.: 9 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:Type AMS subject classificationhere.. http://dspace.nbuv.gov.ua/handle/123456789/154856 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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It is proved that R is a commutative adequate domain, then R is the domain of stable range 1 in localization in multiplicative closed set which corresponds s-torsion in the sense of Komarnitskii. |
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Zabavsky, B. Domsha, O. Diagonalizability theorem for matrices over certain domains Algebra and Discrete Mathematics |
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Zabavsky, B. Domsha, O. |
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Zabavsky, B. |
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Diagonalizability theorem for matrices over certain domains |
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Diagonalizability theorem for matrices over certain domains |
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Diagonalizability theorem for matrices over certain domains |
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Diagonalizability theorem for matrices over certain domains |
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Diagonalizability theorem for matrices over certain domains |
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diagonalizability theorem for matrices over certain domains |
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Інститут прикладної математики і механіки НАН України |
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2011 |
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http://dspace.nbuv.gov.ua/handle/123456789/154856 |
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Diagonalizability theorem for matrices over certain domains / B. Zabavsky, O. Domsha // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 1. — С. 132–139. — Бібліогр.: 9 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT zabavskyb diagonalizabilitytheoremformatricesovercertaindomains AT domshao diagonalizabilitytheoremformatricesovercertaindomains |
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2025-07-14T06:55:42Z |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 12 (2011). Number 1. pp. 132 – 139
c© Journal “Algebra and Discrete Mathematics”
Diagonalizability theorem for matrices
over certain domains
Bogdan Zabavsky, Olga Domsha
Communicated by V. V. Kirichenko
Abstract. It is proved that R is a commutative adequate
domain, then R is the domain of stable range 1 in localization in
multiplicative closed set which corresponds s-torsion in the sense of
Komarnitskii.
Introduction
A question of quasi-reduction of matrices over a commutative domain
with so-called Lϕ condition is considered by J. Szucs [1]. B. Zabavsky [2]
proved that the Lϕ condition for commutative Bezout domain is nothing
else than of stable range condition. More over increase in the reduction in
localization of given ring to reduction in basic one is shown.
In this work we continued this research and more precisely proved
that an adequate domain is the domain of stable range 1 in localization
in multiplicative closed set which corresponds s-torsion in the sense of
Komarnitskii [3].
Let R be a commutative Bezout domain. An element a ∈ R is called
an adequate element if for any b from R the element a can be represented
as a product a = rs, where rR+bR = R and for any non invertible divisor
s′ of s we have obtain s′R+ bR 6= R.
A ring R is called a Bezout ring if every finitely generated ideal is
principal. A commutative Bezout ring in which any nonzero element is
2000 Mathematics Subject Classification: Type AMS subject classification
here...
Key words and phrases: a Bezout domain, a ring of stable range 1, an adequate
domain, a co-adequate element, an element of almost stable range 1, an elementary
divisors ring.
B. Zabavsky, O. Domsha 133
adequate is called an adequate ring [4]. An element a ∈ R is called co-
adequate if every non zero element b ∈ R can be represented as a product
b = rs where rR+ aR = R and for any non invertible divisor s′ of s we
have obtain s′R+ bR 6= R [5].
A ring R is the ring of stable range 1 (in denotation st.r.(R) = 1), if
the condition aR+ bR = R for every elements a, b ∈ R implies that there
exist element t ∈ R such that a+ bt is an invertible element of the ring R
[6].
An element a ∈ R is called the element of almost stable range 1 if
st.r.(R/aR) = 1 [7]. A ring where every non zero and non invertible
element is almost stable range 1 is called the ring of almost stable range 1
[7,8]. The ring, where every finitely presented module is decomposed in
the direct sum of cyclic modules, is called elementary divisors ring [4].
By a|b we denote the fact that an element a of a ring R is a divisor of
an element b of R. Let’s denote as J(R) a radical of Jacobson of the ring
R and U(R) – the group of units of the ring R.
1. Main result
Let R be a commutative Bezout domain, a – nonzero and non invertible
element of domain R. Let’s denote the set by
Sa = {b|b ∈ R, aR+ bR = R.}
Proposition 1. The set Sa is saturated and multiplicative closed.
Proof. Let c, b ∈ Sa. According to the determination, there exist elements
u1, u2, v1, v2 ∈ R such, that
au1 + bv1 = 1,
au2 + bv2 = 1.
Multiplying these equalities, we get
aw1 + cbw2 = 1
for some elements w1, w2 ∈ R. Therefore cb ∈ Sa.
If b = cd ∈ Sa, then
au+ c(dv) = 1
for some elements u, v ∈ R. So c ∈ Sa and Sa – saturated and multiplicative
closed set.
134 Diagonalizability theorem for matrices
Let a is nonzero and non invertible element of R. Let’s denote
Ra = RS−1
a .
Proposition 2. If a – adequate element of domain R, then st.r.(Ra) = 1.
Proof. Let
b
s
Ra +
c
s
Ra = Ra.
Then
b
s
·
u
s1
+
c
s
·
v
s2
= t,
where s1, s2, t ∈ Sa. Hence bu′ + cv′ = ss1s2t ∈ Sa for some u′, v′ ∈ R. So
(bu+ cv)R+ aR = R and therefore
aR+ bR+ cR = R.
Since element a adequate, there exist element r ∈ R, such as aR +
(b+ cr)R = R, that is u = b+ cr ∈ Sa. Another words, (b+ cr)Ra = Ra.
More,
b
s
+
c
s
·
r
1
= su ∈ Ra,
that is
(
b
s
+
c
s
·
r
1
)Ra = Ra.
And that mean the stable range of the ring Ra is equal 1.
Obviously, from the proposition 2 we get next proposition.
Proposition 3. Let R be an adequate domain. Then for any nonzero and
non invertible element a ∈ R the set Ra is a commutative Bezout domain
of the stable range 1.
The next question arose: what this commutative Bezout domain is,
where for any element a localization is a commutative Bezout domain
with the stable range 1?
It is worth to remarks that the stable range of the commutative Bezout
domain doesn’t exceed 2, that’s why the stable range of Ra doesn’t exceed
2 either [9].
Let for any nonzero and non invertible element x ∈ R the stable
range of the ring Rx is equal 1 and b, d – nonzero elements of R such as
dR = aR + bR, moreover d – non invertible element in R. Then elements
u, v, a0, b0 ∈ R such that au+ bv = d, a = da0, b = db0 exist.
B. Zabavsky, O. Domsha 135
According to the restrictions imposed on R, the stable range of the ring
Rd equals 1. Since R is domain, then a0u+b0v = 1, that is a0R+b0R = R.
Than a0Rd + bdR = Rd.
Once again st.r.(Rd)=1, so elements q ∈ R i u, p ∈ Sd such that
a0
1
q
p
+
b0
1
= u
exist. It follows that a0q + b0p = up. According to the proposition 1
up ∈ Sd, that is
(a0p+ b0q)R+ dR = R
and
pR+ dR = R.
We shall notice that a = da0, b = db0. So for any nonzero and co-prime
elements b, d there exist p, q such that
aq + bp = (a, b) = d
and (p, d) = 1.
So, we proofed the next proposition.
Proposition 4. Let R be such commutative Bezout domain for any
nonzero element a ∈ R st.r.(Ra) = 1. Than for any nonzero and co-
prime a, b ∈ R elements p, q ∈ R such that
aR+ bR = (ap+ bq)R
and
qR+ (ap+ bq)R = R
exist.
As obviously corollary from this proposition we got the next result.
Proposition 5. Let R be such commutative Bezout domain for any
nonzero element a ∈ R st.r.(Ra) = 1. Than R is elementary divisors
ring.
Proof. To prove it is sufficient to show that for any a, b, c ∈ R such that
(a, b, c) = 1 the matrix
A =
(
c a
0 b
)
diagonalizes [4].
Let’s consider possible case.
136 Diagonalizability theorem for matrices
1) c = 0, another words the matrix A look as
A =
(
0 a
0 b
)
.
Let aR+bR = dR. Then the elements a0, b0, u, v ∈ R, such as au+bv = d,
a = da0 b = db0 exist. Hence a0u+ b0v = 1 and
(
u v
−b0 a0
)(
0 a
0 b
)
=
(
0 d
0 0
)
;
(
0 d
0 0
)(
0 1
1 0
)
=
(
d 0
0 0
)
.
Obviously, the matrices
(
u v
−b0 a0
)
and
(
0 1
1 0
)
are invertible.
2) c ∈ U(R). Then
(
c a
0 b
)(
c−1 0
0 1
)
=
(
1 a
0 b
)
;
(
1 a
0 b
)(
1 −a
0 1
)
=
(
1 0
0 b
)
.
Obviously, the matrices
(
c−1 0
0 1
)
and
(
1 −a
0 1
)
are invertible.
3) Let c 6= 0, c ∈ U(R). According to the condition and the proposition
4 the elements p, q ∈ R such as
aR+ bR = (ap+ bq)R
and
qR+ (ap+ bq)R = R
exist. It follows that (ap+bq, p) = 1. Since ap+bq = (a, b) and (a, b, c) = 1,
than (c, ap+ bq) = 1 and therefore (cp, ap+ bq) = 1. We shall notice that
(p, q) = 1 and
(p q)
(
c a
0 b
)
= (cp ap+ bq) .
Another words the matrix
(
c a
0 b
)
diagonalizes.
Proposition 6. Let R a commutative Bezout domain, a – any nonzero
element of R. Then J(Ra) 6= 0.
B. Zabavsky, O. Domsha 137
Proof. Let M a maximal ideal of the ring Ra such as a doesn’t belong to
M . Than M + aRa = Ra, that is the elements m ∈ M and r
s
∈ Ra such
as
m+ a
r
s
= 1.
Hence ms+ ar = s.
Let’s consider (m, a) = n. If n doesn’t belong to U(R), than
n(m0s+ a0r) = s,
where m = nm0, a = na0.
From n
s
∈ S follows that (n, a) = 1. It’s impossible, because n doesn’t
belong to U(R) and n | a = 1. So (m, a) = 1. That is m ∈ U(Ra), but it’s
impossible, because m ∈ M ∈ mspecRa.
So a belongs to all maximal ideals of Ra.
Proposition 7. The element a is a co-adequate element of Ra.
Proof. As in Ra only units are co-prime with a elements, any non invertible
element b has the form b = 1 · b, where 1Ra + aRa = Ra. For any element
b′ doesn’t belong to U(Ra) such as b′ | b execute
b′Ra + aRa 6= Ra.
Proposition 8. Let R is such commutative Bezout domain that J(Ra) =
aRa. Than st.r.(Ra) = 1, that is to say R is elementary divisors ring.
Proof. Let R such commutative Bezout domain that J(Ra) = aRa. That
is to say a ∈ J(Ra) 6= 0 and J(Ra) = aRa. So let’s consider the factor
ring Ra/aRa. Obviously, the Jacobson radical of this factor ring Ra/aRa
equals zero and any element a of Ra/aRa is zero divisor or invertible.
As the factor ring Ra/aRa is reduced, it is possible only if Ra/aRa is a
zero-dimensional ring. So st.r.(Ra/aRa) = 1, this implies Ra is a ring of
the almost stable range 1, which has a nonzero Jacobson radical.
Let b, c ∈ Ra such that bRa + cRa = Ra. Let’s consider a ∈ Ra. Then
aRa + bRa + cRa = Ra. Since Ra is a ring of almost stable range 1, there
an element r ∈ Ra such that aRa + (b+ cr)Ra = Ra exist. It follows that
au+ (b+ cr)v = 1
for any u, v ∈ Ra. Another words, (b+ cr)v = 1− au.
From a ∈ J(Ra) follows (b + cr)v ∈ U(Ra), that is Ra is a ring of
stable range 1, and therefore R is an elementary divisor ring.
138 Diagonalizability theorem for matrices
Proposition 9. Let R is such a commutative Bezout domain that for
any nonzero and non invertible element a ∈ R the localization Ra is an
adequate ring. Then R is an elementary divisors ring.
Proof. According to the proposition 6 Ra is adequate domain with non
zero Jacobson radical. In [9] is proofed that st.r.(Ra) = 1. Then according
to the proposition 5, R is elementary divisor ring.
Let us denote K = Ra and consider K = K/rad(aK). Let’s suppose
that there are regular elements in K. Let it be b. Since ba0 = ab0, where
(a, b) = d, a = a0d, b = b0d, so b · a0 = 0.
As b is regular, there exists n ∈ N such that an
0
= at. It is follows
that an−1
0
= dt. As (a0, b0) = 1, then (an−1
0
, b0) = 1. Since d | an−1
0
, that
(d, b0) = 1. Since (a0, b0) = 1, that (a, b0 = 1.) That’s b0 is invertible in
R. So divisors of a is regular elements in K. Another words a = bc.
Let a = bc. What happens to the image b by homomorphism K −→
K/rad(aK)?
Proposition 10. Let R = Ka = { b
c
| (c, a) = 1}. If a is an adequate
element in R , then st.r.(R) = 1.
Proof. Considering J(R) 6= 0, and st.r.(R/aR) = 1
we get st.r.(R) = 1.
At the end we answer the question: is it possible that non adequate
element a in R is adequate in Ra?
Proposition 11. Let R = Ka = { b
c
| (c, a) = 1} and the element a is
non adequate in R. then rad(r/aR) 6= 0.
Proof. First of all, we mast remark if a is non adequate element in R then
there at least one representation a = bc exist, where (b, c) 6= 1. Really, if
all representation look as a = bc and (b, c) = 1, then – adequate.
Let d ∈ R. If (a, d) = 1, all right. But if (a, d) = δ, where δ does not
belong to U(R), then a = a0δ d = d0δ and (a0, b0) = 1. Since (a0, δ) = 1,
representation a = rs where r = a0 s = δ is desired.
So let a = bc, where (b, c) = δ and δ does not belong to U(R). It
follows that b = b0δ, c = c0δ and
(bc0)
2 = b0δδb0c0c0 = δb0δc0(b0c0) = ab0c0 ∈ aR.
Another words bc0 ∈ rad(R/aR). Remarks that bc0 6= 0. Really bc0 =
ac0δt. Hence δt = 1, δ ∈ U(R). It is contradiction. So this presentation is
proofed.
B. Zabavsky, O. Domsha 139
References
[1] J. Szucs, Diagonalization theorem for matrices over certain domains, Acta sci.math.,
36, N.1-2, 1974, pp.193-201.
[2] B. Zabavsky, Reduction of matrices and simultaneously reduction of pair matrices
over rings, Math. studii, V.24 N.1, 2005, pp.3-11.
[3] J. S. Golan, On S-torsion in the sens of Komarnitskii, Univer. Haifa, Israll, 1984,
prerpint.
[4] Larsen M., Levis W., Shores T., Elementary divisor rings and finitely presrnted
modules, Trans. Amer. Math. Soc., N.187, 1974, pp.231-248.
[5] B. Zabavsky, Generalizated adequate rings, Ukr. math. journ., V.48 N.4, 1996,
pp.554-557.
[6] L. N. Vaserstein, Bass’s first stable range condition, J. Pure Appl. Alg., N.34, 1984,
pp.319-330.
[7] S. Biliavs’ka, Elements of stable and almost stable range 1, Visnik LNU, N.71, 2009,
pp.5-12.
[8] Mc Govern W. Bezout rings with almost stable range 1 are elementary divisors ring,
J. Pure Appl. Alg., N.212, 2007, pp.340-348.
[9] S. Biliavs’ka, B. Zabavsky, A stable range of adequate domain, Math. studii, V.48,
N.2, 2008, pp.212-214
Contact information
B. Zabavsky Ivan Franko National University of Lviv, Univer-
sytetska, 1, Lviv, Ukraine, 79000
E-Mail: b_zabava@ukr.net
O. Domsha Ivan Franko National University of Lviv, Univer-
sytetska, 1, Lviv, Ukraine, 79000
E-Mail: olya.domsha@i.ua
Received by the editors: 06.05.2011
and in final form 07.10.2011.
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