Effective ring
In this paper we will investigate commutative Bezout domains whose finite homomorphic images are semipotent rings. Among such commutative Bezout rings we consider a new class of rings and call them an effective rings. Furthermore we prove that effective rings are elementary divisor rings.
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irk-123456789-1533522019-06-15T01:28:33Z Effective ring Zabavsky, B.V. Kuznitska, B.M. In this paper we will investigate commutative Bezout domains whose finite homomorphic images are semipotent rings. Among such commutative Bezout rings we consider a new class of rings and call them an effective rings. Furthermore we prove that effective rings are elementary divisor rings. 2014 Article Effective ring / B.V. Zabavsky, B.M. Kuznitska // Algebra and Discrete Mathematics. — 2014. — Vol. 18, № 1. — С. 149–156. — Бібліогр.: 7 назв. — англ. 1726-3255 2010 MSC:13F99. http://dspace.nbuv.gov.ua/handle/123456789/153352 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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In this paper we will investigate commutative Bezout domains whose finite homomorphic images are semipotent rings. Among such commutative Bezout rings we consider a new class of rings and call them an effective rings. Furthermore we prove that effective rings are elementary divisor rings. |
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Effective ring |
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Effective ring |
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Effective ring |
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Effective ring |
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Effective ring |
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Інститут прикладної математики і механіки НАН України |
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Effective ring / B.V. Zabavsky, B.M. Kuznitska // Algebra and Discrete Mathematics. — 2014. — Vol. 18, № 1. — С. 149–156. — Бібліогр.: 7 назв. — англ. |
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Algebra and Discrete Mathematics |
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2025-07-14T04:34:39Z |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 18 (2014). Number 1, pp. 149 – 156
© Journal “Algebra and Discrete Mathematics”
Effective ring
B. V. Zabavsky, B. M. Kuznitska
Communicated by M. Ya. Komarnytskyj
Abstract. In this paper we will investigate commutative
Bezout domains whose finite homomorphic images are semipotent
rings. Among such commutative Bezout rings we consider a new
class of rings and call them an effective rings. Furthermore we prove
that effective rings are elementary divisor rings.
Throughout this paper all rings will be commutative with nonzero
identity. We say that R is a Bezout ring if every finitely generated ideal
of R is principal. A nonzero element a in R is said to be adequate to the
element b ∈ R (aAb denote this fact), if we can find such two elements
r, s ∈ R that the decomposition a = rs satisfies the following properties:
1) rR + bR = R,
2) s′R + bR 6= R for any noninvertible divisor s′ of element s.
If for any element b ∈ R we have aAb, then we say that element a is
adequate. If any nonzero element of ring R is an adequate element then
R is called an adequate ring [2]. In addition we notice one simple fact: for
any nonzero a ∈ R we have aAa. The most obvious examples of adequate
elements are units, square free elements, and factorial elements [7].
Following [4] a commutative ring R is called an exchange ring if for
any element a ∈ R one can find such idempotent e ∈ R that e ∈ aR and
(1 − e) ∈ (1 − a)R. We say that an element a ∈ R is clean element if
a = u + e for some idempotent e = e2 and unit u. If all elements of a
2010 MSC: 13F99.
Key words and phrases: Bezout ring, exchange ring, clean ring, effective ring,
elementary divisor ring, idempotent of stable range 1, neat ring.
150 Effective ring
ring R are clean then R is called a clean ring. It is proved in [4] that for
commutative rings class of exchange rings coincide with class of clean
rings.
We say that commutative ring R is semipotent if for any element
a ∈ R that is not contained in Jacobson radical there is some a proper
idempotent e ∈ R that e ∈ aR [4].
Throughout this paper J(R) will denote a Jacobson radical of ring
R, mspec(a) - set of all maximal ideals that contain an element a and
J(aR) =
⋂
M∈mspec(a) M .
Let’s start our research with few simple useful properties of adequacy.
Proposition 1. Let R be a commutative Bezout domain, and a ∈ R is
some nonzero element. Then aAb iff aAb+at for any t ∈ R.
Proof. Let a be adequate to element b in a Bezout ring R. Then by
definition there are some elements r, s ∈ R that a = rs and rR + bR = R
as well as s′R + bR 6= R for any element s′ ∈ R that sR ⊂ s′R 6= R.
Taking arbitrary t ∈ R we consider ideal rR + (b + at)R. Since R is
a Bezout ring then rR + (b + at)R = hR for some element h ∈ R. As
aR ⊂ rR ⊂ hR and (b + at)R ⊂ hR then bR ⊂ hR, that is possible iff h
is a unit because rR + bR = R.
Since s′R + bR 6= R, for any element s′ ∈ R that sR ⊂ s′R 6= R, then
s′R + (b + at)R 6= R for any t ∈ R.
Thus we have obtained that aAb+at for any t ∈ R, so necessity is
proved.
For the converse statement suppose that aAb+at, that means that one
can find some elements r, s ∈ R that a = rs, rR + (b + at)R = R, and
s′R + (b + at)R 6= R for any element s′ ∈ R that sR ⊂ s′R 6= R. If
rR + bR = hR 6= R for some element h ∈ R, then as aR ⊂ rR ⊂ hR
and bR ⊂ hR we have (b + at)R ⊂ hR that is impossible since rR ⊂ hR.
Next, assume that there is some element s′ ∈ R that sR ⊂ s′R 6= R and
s′R + bR = R. Let we have s′R + (b + at)R = hR 6= R. Then bR ⊂ hR
and this contradicts assumption s′R + bR = R. Proposition is proved.
As a consequence this result motivates us to consider cosets b = b+aR
in quotient ring R = R/aR. The following statement determines the
correspondence between property aAb of a commutative Bezout ring R
and structure of homomorphic image of element b in R = R/aR.
Proposition 2. Let R be a commutative Bezout domain and aAb. Then
element b = b + aR is a clean element in R = R/aR.
B. V. Zabavsky, B. M. Kuznitska 151
Proof. We start with obvious equality (−1)R + bR + aR = R and hence
−1R + bR = R. Since a is adequate to element b in a Bezout ring R then
there are such elements r, s ∈ R that a = rs and rR + bR = R as well as
s′R + bR 6= R for any element s′ ∈ R that sR ⊂ s′R 6= R. Going down to
quotient ring R we have that rR + bR = R and s′R + bR 6= R. Let t be
some noninvertible in R divisor of element s. Then there is some element
k ∈ R that (s + ak)R ⊂ tR. Let’s show that sR + tR 6= R. Suppose the
contrary, that is sR + tR = R. Since (s + ak)R ⊂ tR then s + ak = tβ for
some element β ∈ R. Equality s + rsk = tβ implies s(1 + rk) = tβ. As
sR + tR = R then (1 + rk)R ⊂ tR, so tR + rR = R. Since tR + sR = R
and tR+rR = R then tR+rsR = R, thus tR+aR = R and then tR = R
that contradicts assumption about noninvertibility of t. Thus we have
proved that sR + tR = uR 6= R. This also means that uR + bR 6= R.
Since u is a divisor of an element t then tR + bR 6= R. So, we obtained
that 0 = rs is an adequate element to an element b.
Moreover, we notice that rR + sR = R. In fact, if rR + sR = hR 6= R,
then according adequacy of an element 0 to an element b ∈ R we know that
hR + bR = R (as h is a divisor of r) and on the other hand hR + sR 6= R
(as h is noninvertible divisor of s). But this is impossible. So, rR+sR = R
and there are such elements u, v ∈ R that ru + sv = 1.
Additionally, we are going to prove that elements ru and sv are
idempotents in ring R. This can be shown by next expression: (ru)2 =
ru(1 − sv) = ru − rusv = ru. Similarly we have (sv)2 = sv.
Let’s denote e = ru. Now we want to prove that b − e is a unit in
R. Suppose that (b − e)R = hR 6= R. Consider an ideal hR + rR = tR.
If t is nonunit in R then (b − e)R ⊂ hR ⊂ tR, hence bR ⊂ tR that is
impossible since rR ⊂ tR, bR ⊂ tR, and bR + rR = R. So, hR + rR = R.
Now we’ll prove that hR + sR = R. Assume that sR + hR = tR 6= R.
Since t is nonunit and it divides s, then by defining property of an element
s we have tR + bR = kR 6= R. On the other hand (b − e)R = hR and
eR + sR = R, so we have that eR + tR = R implies eR ⊂ kR. Last
inclusion is impossible since bR ⊂ kR and bR + (−1)R = R. At last we
have that sR + hR = R. Since rR + hR = R then rsR + hR = R. As we
know that rs = 0 then we’ll obtain that h is a unit of ring R.
So, we have proved that b − e = u is a unit in a ring R, and hence
b = u + e is a clean element. Proposition is proved completely.
Definition 1. Element a ∈ R is said to be an exchange element if there
exists idempotent e ∈ R such that e ∈ aR and (1 − e) ∈ (1 − a)R [4]. We
say that an element a ∈ R is element of idempotent stable range 1 if for
152 Effective ring
every element b ∈ R such that aR + bR = R exists idempotent e ∈ R
such that a + be is invertible element of R.
By [3, 4, 5] and proposition 2 we have
Theorem 1. Let R be a commutative Bezout domain and aAb. Then:
1) b = b + aR is a clean element of R/aR.
2) b = b + aR is an exchange element of R/aR.
3) b = b + aR is element of idempotent stable range 1 of R/aR.
Note that entry aAb implies, in particular, that a 6= 0
Theorem 2. Let R be a commutative Bezout domain and let a ∈ R \ {0}.
If aAb and b /∈ J(aR) then an ideal bR contains a proper idempotent.
Proof. Since a is adequate to b then a = rs, and rR + bR = R as well
as s′R + bR 6= R, for any element s′ ∈ R that sR ⊂ s′R 6= R. Since
b /∈ J(aR) then by simple observation element r is not a unit in a ring R.
Then equality rR + bR = R implies that ru + bv = 1 for some elements
u, v ∈ R. Then rsu + bsv = s. Since a = rs then rs = 0 and we have
sR ⊂ bR and s̄R̄ 6= (0). Moreover, since rR + sR = R then rx + sy = 1
for some elements x, y ∈ R. As a consequence we have s2v = s that means
that element s is a regular element. As R is commutative then there is
such idempotent e that sR = eR. Last fact means that we have found a
proper idempotent in an ideal bR as was desired. Theorem is proved.
It is natural to ask about the answer to the converse statement.
Theorem 3. Let R be a commutative Bezout domain and let a ∈ R \ {0}.
If quotient ring R = R/aR is semipotent then for any element b ∈
R \ J(aR) there is some element u ∈ R that aAbu.
Proof. Let R = R/aR be a semipotent ring and b ∈ R \ J(aR). By
semipotency of R for a coset b = b + aR there is some nonzero idempotent
e ∈ bR. Hence exist some elements u, t ∈ R that e − bu = at. Moreover,
since e = e2 then e(1 − e) = as, for some element s ∈ R. Let’s show that
aAe. In fact, as R is a commutative Bezout ring then eR + aR = dR, for
some element d ∈ R. By [6] we have e = e0d,a = a0d and e0p + a0q = 1,
for some elements e0, a0, p, q ∈ R. Furthermore, equality e(1 − e) = as
implies e0(1 − e) = a0s. Since e0p + a0q = 1 and e0(1 − e) = a0s then one
can deduce that a0k = 1 − e for some element k ∈ R. Hence e + a0k = 1.
B. V. Zabavsky, B. M. Kuznitska 153
Taking r = a0 and s = d we obtain a decomposition a = rs, where
rR + eR = R as well as s′R + eR 6= R for any element s′ ∈ R that
sR ⊂ s′R 6= R. Thus aAe and since bu = e + at we conclude aAbu by
Proposition 1. Theorem is proved.
As a corollary of two previous theorems we have next result.
Theorem 4. Let R be a commutative Bezout domain and let a ∈ R \ {0}.
Then quotient ring R = R/aR is semipotent iff for any element b ∈
R \ J(aR) there is some element u ∈ R that aAbu, bu /∈ aR.
All mentioned results allows us to define a new subclass of commutative
Bezout ring that are so called effective ring, that are also elementary
divisor rings.
Recall that a commutative ring R is said to be an elementary divisor
ring if for every matrix A over R one can find such invertible matrices
P and Q of appropriate sizes that PAQ = diag(ε1, ..., εr, 0, ..., 0), where
εi+1R ⊂ εiR, for any i ∈ 1, r − 1 [2].
Definition 2. A commutative Bezout domain R is said to be effective if
for any elements a, b, c ∈ R that aR + bR + cR = R and aR + bR 6= R
there exists such element p ∈ R that cApa and pR + bR + cR = R.
An obvious example of effective ring is any adequate ring. Henriksen’s
example [1], that is commutative Bezout domain R = {z0 + a1x + a2x2 +
...|z0 ∈ Z, ai ∈ Q}, is also an effective ring that is not adequate. Notice
that in this ring aR + bR + cR = R implies that at least one of these
elements is an adequate element of R.
Theorem 5. Effective ring R is an elementary divisor ring.
Proof. Due to [7] it is sufficient to prove that for any coprime triple
of elements aR + bR + cR = R there are some elements p, q ∈ R that
(pa + qb)R + qcR = R. If aR + bR = R then obviously, exist element
p, g ∈ R such that (pa + gb)R + qcR = R [1,2,7]. By definition of an
effective ring there is some element p ∈ R that cAap and pR+bR+cR = R.
As aR + bR + cR = R and pR + bR + cR = R then apR + bR + cR = R.
Since cAap then c = qs, where qR + apR = R and s′R + apR 6= R, for
any s′ ∈ R that sR ⊂ s′R 6= R. Let’s show that (ap + bq)R + cqR = R.
Suppose the contrary, that is (ap + bq)R + cqR = hR 6= R. Since cq = sq2,
then let qR + hR = dR 6= R. As qR ⊂ dR and hR ⊂ dR then apR ⊂ dR
that is impossible since qR + apR = R. Thus sR ⊂ hR. By definition of
154 Effective ring
element s we have that hR + apR = kR 6= R. Let kR + aR = xR 6= R,
then bqR ⊂ xR. Since cR ⊂ xR, aR ⊂ xR and aR + bR + cR = R then
xR + bR = R. So qR ⊂ xR, that is impossible as qR + sR = R. As a
result we have pR ⊂ kR. Then bqR ⊂ kR. If bR + kR = αR 6= R then
cR ⊂ αR, bR ⊂ αR and pR ⊂ αR. Using fact that pR + bR + cR = R
we obtain a contradiction, so α must be a unit. So qR ⊂ kR, but this
is impossible as qR + sR = R and sR ⊂ kR. So, we have proved that
(ap + bq)R + cqR = R that was desired. Theorem is proved.
Neat rings are studied in [3] as rings whose homomorphic images are
exchange rings. We’ll show that in case of commutative Bezout ring neat
rings are effective rings. To complete this purpose we need the following.
Proposition 3. A commutative ring R is an exchange ring iff for any
pair of elements a, b ∈ R that aR + bR = R there is an idempotent e ∈ R
that e ∈ aR and (1 − e) ∈ bR.
Proof. By [4, Prop.1.1.1] in every exchange ring equality aR + bR = R
implies that there are orthogonal idempotents e and 1 − e that e ∈ aR
and (1−e) ∈ bR, so necessity is proved. If aR+bR = R implies that there
is an idempotent e ∈ aR that e ∈ aR and (1 − e) ∈ bR then considering
equality a + (1 − a) = 1 we obtain that e ∈ R and (1 − e) ∈ (1 − a)R, so
R is an exchange ring by definition in [4]. Proposition is proved.
Theorem 6. Let R be a commutative Bezout domain whose finite homo-
morphic image R/cR is an exchange ring for any c ∈ R \ {0}. Then R is
an effective ring.
Proof. Let R = R/cR be an exchange ring for any c ∈ R \ {0}. By [3] R
is a neat ring. Then by Proposition 3 an equality aR + bR = R implies
that we can find an idempotent e ∈ R, that e ∈ aR and 1 − e ∈ bR. Let’s
notice that condition aR + bR = R implies aR + bR + cR = R. Since
e ∈ aR then there is some element p ∈ R that e − ap = cs for some
element s ∈ R. Similarly 1 − e − bα = cβ for some elements α, β ∈ R. By
substitution e = cs + ap in 1 − e − bα = cβ we’ll get ap + cw + bα = 1,
that means pR + cR + bR = R. Let’s prove that cAap. Since e = e2 then
e(1 − e) = ct for some element t ∈ R. We consider an ideal eR + cR = dR.
By [6], we have e = de0 and c = dc0 for some elements e0, c0 ∈ R that
e0R + c0R = R, then e0(1 − e) = c0t, and then e + c0γ = 1 for some
element γ ∈ R. Taking r = c0, s = d we obtain a decomposition c = rs,
where rR + eR = R and sR ⊂ eR. So, we have cAe. Since e = ap + cs
B. V. Zabavsky, B. M. Kuznitska 155
then by Proposition 1 and we conclude cAap, as was desired. Theorem is
proved.
Proposition 4. Let R be a commutative Bezout domain, which for any
elements a, b, c ∈ R that aR + bR + cR = R there exists such element
p ∈ R that cApa. Let c = rs where rR + apR = R and s′R + apR 6= R for
any noninvertible divisor s′ of element s and then pR + bR + cR = R iff
sR + bR = R.
Proof. Let c = rs where rR + apR = R and s′R + apR 6= R for any
noninvertible divisor s′ of elements s. If aR + bR + cR = R and pR +
bR + cR = R then apR + bR + cR = R. If sR + bR = δR 6= R, we have
δR + apR = hR 6= R. It is impossible, since apR + bR + cR = R.
Let sR+bR = R. We prove that apR+bR+cR = R. If pR+bR+cR =
hR 6= R, then pR ⊂ hR, cR ⊂ hR and bR ⊂ hR. Since rR + apR = R,
then h is noninvertible divisor of s. By bR ⊂ hR and sR ⊂ hR, we have
sR + bR ⊂ hR 6= R. It is impossible, since sR + bR = R. Proposition is
proved.
Theorem 7. Let R be a commutative Bezout domain in which for any
elements a, b, c ∈ R that aR + bR + cR = R there exists such element
p ∈ R that cAap and pR + bR + cR = R. Then R/cR is an exchange ring
for every c ∈ R \ {0}.
Proof. Let R̄ = R/cR and āR̄ + b̄R̄ = R̄, where ā = a + cR, b̄ = b + cR.
Then aR+ bR+cR = R and exists such element p ∈ R that c = rs, where
rR + apR = R and s′R + apR 6= R for each s′R such that sR ⊂ s′R 6= R
and pR + bR + cR = R. Obviously, we have ru + sv = 1. Since r̄ū + s̄v̄ = 1
and r̄s̄ = 0̄, we have r̄2ū = r̄ and s̄2v̄ = s̄.
Notice ē = s̄v̄. Obviously 1̄ − ē = r̄ū and s̄R̄ = ēR̄, (1̄ − ē)R̄ = r̄R̄.
Since, by proposition 4, we have sR+bR = R then sα+bp = 1, α, β ∈ R.
Since rsα + rbp = r, we have r̄b̄p̄ = r̄ and r̄R̄ ⊂ b̄R̄. Since rR + apR = R,
t, k ∈ R. Since rsp+apsk = s we have s̄R̄ ⊂ āR̄. We proved if āR̄+b̄R̄ = R̄
then exists idempotents ē ∈ āR̄ and 1̄ − ē ∈ b̄R̄. By proposition 3, R/cR
is an exchange ring. Theorem is proved completely.
References
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156 Effective ring
[3] McGovern W. Wm. Neat ring, J. of Pure and Appl. Algebra, 2006, 205, pp. 243 –
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[4] Nicholson W. K. Lifting idempotents and exchange rings, Trans. Amer. Math. Soc.,
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[5] Zabavsky B., Bilavsky S. Every zero adequate ring is an exchange ring, Fund i
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[6] Zabavsky B. V. Reduction of matrices over Bezout rings of stable range not higher
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Contact information
B. V. Zabavsky,
B. M. Kuznitska
Department of Mechanics and Mathematics,
Ivan Franko National Univ., Lviv, Ukraine
E-Mail: zabavskii@gmail.com,
kuznitska@ukr.net
Received by the editors: 04.01.2014
and in final form 26.07.2014.
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