Strongly radical supplemented modules
Zoschinger studied modules whose radicals have supplements and called these modules radical supplemented. Motivated by this, we call a module strongly radical supplemented (briefly srs) if every submodule containing the radical has a supplement. We prove that every (finitely generated) left module...
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2011
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508763173158912 |
|---|---|
| author | Büyükaşık, Е. Türkmen, E. Бюкасік, Є. Тюркмен, Є. |
| author_facet | Büyükaşık, Е. Türkmen, E. Бюкасік, Є. Тюркмен, Є. |
| author_sort | Büyükaşık, Е. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-03-18T19:36:38Z |
| description | Zoschinger studied modules whose radicals have supplements and called these modules radical supplemented.
Motivated by this, we call a module strongly radical supplemented (briefly srs) if every submodule containing
the radical has a supplement. We prove that every (finitely generated) left module is an srs-module if and
only if the ring is left (semi)perfect. Over a local Dedekind domain, srs-modules and radical supplemented
modules coincide. Over a no-local Dedekind domain, an srs-module is the sum of its torsion submodule and
the radical submodule. |
| first_indexed | 2026-03-24T02:30:22Z |
| format | Article |
| fulltext |
К О Р О Т К I П О В I Д О М Л Е Н Н Я
UDC 512.5
E. Büyükaşık (Izmir Inst. Technol., Turkey),
E. Türkmen (Amasya Univ., Turkey)
STRONGLY RADICAL SUPPLEMENTED MODULES
СИЛЬНО РАДИКАЛЬНО ДОПОВНЕНI МОДУЛI
Zöschinger studied modules whose radicals have supplements and called these modules radical supplemented.
Motivated by this, we call a module strongly radical supplemented (briefly srs) if every submodule containing
the radical has a supplement. We prove that every (finitely generated) left module is an srs-module if and
only if the ring is left (semi)perfect. Over a local Dedekind domain, srs-modules and radical supplemented
modules coincide. Over a no-local Dedekind domain, an srs-module is the sum of its torsion submodule and
the radical submodule.
Зошiнгер вивчав модулi, радикали яких мають доповнення, i назвав цi модулi радикально-доповненими.
Мотивуючись цим, будемо називати модуль сильно радикально доповненим (або, скорочено, srs-модулем)
якщо кожен пiдмодуль, що мiстить радикал, має доповнення. Доведено, що кожен (скiнченнопородже-
ний) лiвий модуль є srs-модулем тодi i тiльки тодi, коли кiльце є лiвим (напiв)досконалим. Над локаль-
ною дедекiндовою областю srs-модулi та радикально доповненi модулi збiгаються. Над нелокальною
дедекiндовою областю srs-модуль є сумою свого пiдмодуля скруту i радикального пiдмодуля.
1. Introduction. Throughout, R is an associative ring with identity and all modules
are unital left R-modules. Let M be an R-module. By N ⊆ M , we mean that N is a
submodule of M . A submodule L ⊆M is said to be essential in M , denoted as L�M ,
if L ∩ N 6= 0 for every nonzero submodule N ⊆ M . A submodule S of M is called
small (in M), denoted as S � M , if M 6= S + L for every proper submodule L of
M . By RadM we denote the sum of all small submodules of M or, equivalently the
intersection of all maximal submodules of M . A module M is called supplemented (see
[1]), if every submodule N of M has a supplement, i.e., a submodule K minimal with
respect to N +K = M . K is a supplement of N in M if and only if N +K = M and
N ∩K � K (see [1]). An R-module M is said to be radical supplemented if RadM
has a supplement in M . Radical supplemented modules are studied by Zöschinger in [2]
and [3]. Motivated by this definition, we call a module strongly radical supplemented
if every submodule containing the radical has a supplement. srs-modules lies between
radical supplemented modules and supplemented modules. Some examples are provided
to show that these inclusions are proper.
In this paper, among other results, we prove that srs-modules are closed under factor
modules and finite sums. Every left R-module is an srs-module if and only if R is
left perfect. For modules with small radical the notions of supplemented and being srs-
module coincide. This gives us, every finitely generated R-module is an srs-module
if and only if R is semiperfect. Over a commutative non-local domain, we prove that
every reduced srs-module M is of the form M = T (M) +RadM , where T (M) is the
torsion submodule of M . A commutative domain is h-local if and only if every finitely
generated torsion module is an srs-module. Over a local Dedekind domain (i.e., over
c© E. BÜYÜKAŞıK, E. TÜRKMEN, 2011
1140 ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 8
STRONGLY RADICAL SUPPLEMENTED MODULES 1141
a DVR), a module is an srs-module if and only if it is radical supplemented. Over a
non-local Dedekind domain an srs-module M is of the from M = T (M) + RadM .
2. Strongly radical supplemented modules. Firstly we show some properties of
srs-modules.
Proposition 2.1. Every homomorphic image of an srs-module is an srs-module.
Proof. Let L ⊆ N ⊆ M and Rad(M/L) ⊆ N/L. Since (RadM + L)/L ⊆
⊆ Rad(M/L), we have RadM ⊆ N. By assumption N has a supplement, say K, in
M. Then by [1] (41.1(7)), (K +L)/L is a supplement of N/L in M/L. Hence M/L is
an srs-module.
Proposition 2.2. If M is an srs-module, then M/RadM is semisimple.
Proof. By Proposition 2.1, M/RadM is an srs-module. Rad(M/RadM) = 0,
therefore M/RadM is supplemented. By [1] (41.2(3)), M/RadM is semisimple.
To prove that the finite sum of srs-modules is an srs-module, we use the following
standard lemma (see [1] (41.2)).
Lemma 2.1. Let M be an R-module and M1, N be submodules of M with
RadM ⊆ N. If M1 is an srs-module and M1 +N has a supplement in M, then N has
a supplement in M.
Proof. Let L be a supplement of M1 + N in M. Since RadM1 ⊆ RadM ⊆ N ,
we have RadM1 ⊆ (L+N) ∩M1. Then (L+N) ∩M1 has a supplement, say K, in
M1 because M1 is an srs-module. So
M = M1 +N + L = K + [(L+N) ∩M1] +N + L = (K +N) + L.
Since N + K ⊆ N + M1, L is also a supplement of N + K in M. Then by [4]
(Lemma 1.3a), K + L is a supplement of N in M.
Proposition 2.3. Let M = M1 +M2, where M1 and M2 are srs-modules, then
M is an srs-module.
Proof. Suppose that N ⊆ M with RadM ⊆ N. Clearly M1 + M2 + N has the
trivial supplement 0 in M , so by Lemma 2.1, M1+N has a supplement in M. Applying
the Lemma once more, we obtain a supplement for N in M.
Corollary 2.1. Every finite sum of srs-modules is an srs-module.
Lemma 2.2. Let M be a module with RadM = M . Then M is an srs-module.
Proof. Clearly M has the trivial supplement 0 in M. Since M = RadM is the
unique submodule containing the radical, M is an srs-module.
Let M be an R-module. By P (M) we denote the sum of all submodules V of M
such that RadV = V .
Corollary 2.2. Let M be an R-module. Then P (M) is an srs-module.
Proof. For any module M, RadP (M) = P (M). Then by Lemma 2.2, P (M) is an
srs-module.
The following example shows that srs-modules need not be supplemented.
Example 2.1. Consider the Z-module M =Z Q. Then M is an srs-module, because
RadQ = Q. On the other hand, M is not supplemented by [4] (Theorem 3.1).
Proposition 2.4. Let M be anR-module with RadM �M. Then M is supplemen-
ted if and only if M is an srs-module.
Proof. One direction is clear. Suppose that M is an srs-module. Let N be a
submodule of M. Then N+RadM has a supplement, say L, in M. So N+RadM+L =
= M and (N +RadM)∩L� L. Since RadM �M , we have N +L = M and also
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 8
1142 E. BÜYÜKAŞıK, E. TÜRKMEN
N ∩ L ⊆ (N + RadM) ∩ L � L, i.e., N ∩ L � L. Hence N has a supplement L in
M. Thus M is supplemented.
In [6], a ring R is called left max if every non-zero R-module has a maximal sub-
module. It is well known that R is a left max ring if and only if RadM �M for every
non-zero left R-module M. By using Proposition 2.4, we obtain the following corollary.
Corollary 2.3. Every srs-module over a left max ring is supplemented.
Proposition 2.5. Let M be an R-module. Suppose that RadM is supplemented
and M is an srs-module. Then M is supplemented.
Proof. Let N be a submodule of M. By the hypothesis, RadM+N has a supplement
in M. Since RadM is supplemented, N has a supplement in M by [1] (41.2). Hence
M is supplemented.
A submodule U ⊆ M is said to be cofinite if M/U is finitely generated. In [5], M
is called cofinitely supplemented if every cofinite submodule of M has a supplement
in M. It is also shown that M is cofinitely supplemented if and only if every maximal
submodule of M has a supplement in M (see [5], Theorem 2.8). Since RadM is contai-
ned in every maximal submodule of M, every srs-module is cofinitely supplemented.
But the converse need not be true in general, as it is shown in the following example.
Firstly, we need the following lemma.
Lemma 2.3. Let M be an R-module and U, V ⊆ M. If V is a supplement of U
in M and RadV ⊆ U, then RadV � V.
Proof. Suppose that RadV + T = V for some T ⊆ V. Then
M = U + V = U +RadV + T = U + T.
Since V is a supplement and T ⊆ V, we have T = V. Hence RadV � V.
Example 2.2. Let Z be the ring of integers and p be a prime in Z. Consider
the Z-module, M =
⊕
n>1 Zpn , where Zpn = Z/pnZ. Then M is a torsion module
and it is cofinitely supplemented by [5] (Corollary 4.7). To see that M is not an srs-
module, consider the submodule pM of M. Since M/pM is a semisimple module,
RadM ⊆ pM. We shall prove that pM has not a supplement in M. Suppose pM
has a supplement, say N in M. Then RadN � N by Lemma 2.3. Now since every
element of M is annihilated by some power of p, the module M can be considered as
a module over the local ring Z(p). Then N is a bounded module by [5] (Lemma 2.1).
Therefore pnN = 0 for some n > 1. On the other hand, since N is a supplement of
pM, we have M = pM +N, and so pnM = pn+1M + pnN = pn+1M. So that pnM
is divisible module by [5] (Lemma 4.4). But M has no nonzero divisible submodule.
Hence pnM = 0, a contradiction. Therefore pM has not a supplement in M , i.e., M is
not an srs-module.
Proposition 2.6. Let R be any ring and M be an R-module. Suppose that
M/RadM is finitely generated. Then M is cofinitely supplemented if and only if it
is an srs-module.
Proof. Let M be an R-module and N be a submodule of M with RadM ⊆ N.
Note that
[M/RadM ]/[N/RadM ] ∼= M/N
is finitely generated and thus N is a cofinite submodule of M . Since M is cofinitely
supplemented, N has a supplement in M. Therefore M is an srs-module. The converse
is clear.
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STRONGLY RADICAL SUPPLEMENTED MODULES 1143
Now, we have the following implications on modules:
supplemented =⇒ srs-module =⇒ cofinitely supplemented.
Proposition 2.7. Let M be an R-module and RadM ⊆ U ⊆ M. If V is a
supplement of U in M, then RadV � V.
Proof. Since RadM ⊆ U , we have RadV ⊆ U . Then RadV � V by Lemma 2.3.
Recall from [6] that a submodule L of a module M is called a Rad-supplement of a
submodule N of M in M if N+L = M and N ∩L ⊆ RadL. Clearly every supplement
submodule is a Rad-supplement.
Corollary 2.4. Let M be an R-module and N ⊆ M such that RadM ⊆ N .
Suppose that N + L = M for some L ⊆M . Then L is a supplement of N in M if and
only if L is a Rad-supplement of N and RadL� L.
In the following proposition, we characterize supplements of the radical of a module
over semilocal rings.
Proposition 2.8. Let R be a semilocal ring and M be an R-module. A submodule
N ⊆ M is a supplement of RadM in M if and only if N is coatomic, M/N has no
maximal submodules and RadN = N ∩ RadM.
Proof. (⇒) Let N be a supplement of RadM in M . Then by [1] (41.1(5)), RadN =
= N ∩RadM. If N = M, then clearly RadM �M. Since R is semilocal, M/RadM
is semisimple. Therefore every proper submodule of M is contained in a maximal sub-
module, i.e., M is coatomic. Suppose that N is a proper submodule of M . If K is a
maximal submodule of M with N ⊆ K, then M = RadM +N ⊆ K, a contradiction.
So that N is not contained in any maximal submodule of M, i.e., M/N has no maximal
submodules. By Proposition 2.7, we have RadN � N. Since N/RadN is semisimple,
N is coatomic.
(⇐) Suppose that N + RadM 6= M. Then (N + RadM)/RadM $ M/RadM.
Since R is semilocal, M/RadM is semisimple and so there exists a maximal submodule
K/RadM of M/RadM such that (N + RadM)/RadM ⊆ K/RadM. So N +
+RadM ⊆ K, this implies N ⊆ K. Therefore K/N is a maximal submodule of M/N ,
a contradiction. So N +RadM = M. By the hypothesis, N ∩RadM = RadN � N.
Hence N is a supplement of RadM in M.
Now, we shall characterize the rings over which all (finitely generated) modules are
srs-modules.
Corollary 2.5. For a ring R, the following statements are equivalent.
(1) R is semiperfect.
(2) RR is an srs-module.
(3) Every finitely generated left R-module is an srs-module.
Proof. For every finitely generated module M , we have RadM �M . On the other
hand, by [1] (42.6), R is semiperfect if and only if every finitely generated R-module is
supplemented. From this fact and Proposition 2.4, the implications (1)⇔ (2)⇔ (3) are
clear.
Corollary 2.6. For a ring R, the following statements are equivalent.
(1) R is left perfect.
(2) The left R-module R(N) is an srs-module.
(3) Every left R-module is an srs-module.
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 8
1144 E. BÜYÜKAŞıK, E. TÜRKMEN
Proof. (1)⇒ (3) and (3)⇒ (2) are clear.
(2) ⇒ (1) By Proposition 2.1, RR is an srs-module. So R is semilocal by Proposi-
tion 2.2. Since R(N) is an srs-module, RadR(N) has a (weak) supplement in R(N).
Therefore R is left perfect by [7] (Theorem 1).
The following is a slight modification of [4] (Lemma 1.3 (Folgerung)).
Proposition 2.9. Let M be an R-module and K be a submodule of M. If K and
M/K are srs-modules and K has a supplement L in P for every submodule P with
K ⊆ P ⊆M, then M is an srs-module.
Proof. Let N be a submodule of M with RadM ⊆ N. It follows from [4] (Lem-
ma 1.1(d)) that we can write Rad(M/K) = (RadM +K)/K ⊆ (N +K)/K. Since
M/K is an srs-module, (N +K)/K has a supplement in M/K. That is, there exists a
submodule V/K of M/K such that (N +K)/K + V/K = M/K and [(N +K)/K]∩
∩ [V/K]� V/K. Since K ⊆ V, K has a supplement in V. Therefore V = K + L and
K ∩ L� L for some L ⊆ V . Now
M = N + V = N + (K + L) = (N +K) + L.
Suppose that M = (N +K)+L′ for some L′ ⊆ L. Then M/K = (N +K)/K+(L′+
+K)/K. But V/K is a supplement of (N +K)/K in M/K and (L′+K)/K ⊆ V/K.
By minimality of V/K, we obtain (L′ +K)/K = V/K. It follows that V = L′ +K.
Since L is a supplement of K in V, we have L′ = L. So L is a supplement of N +K
in M. By Lemma 2.1, N has a supplement in M. Hence M is an srs-module.
The following corollary is a direct consequence of Proposition 2.9.
Corollary 2.7. Let M be an R-module which contains an artinian submodule K.
Then M is an srs-module if and only if M/K is an srs-module.
Proof. One direction follows from Proposition 2.1. Conversely, suppose that M/K
is an srs-module. By assumption, K is supplemented and so it is an srs-module. It
follows from [3] that K has a supplement in every P with K ⊆ P ⊆M . Therefore M
is an srs-module by Proposition 2.9.
3. srs-Modules over Dedekind domains. Throughout this section, unless otherwise
stated, we shall consider commutative rings. The following result is due to Zöschinger.
Lemma 3.1 [3] (Satz 3.1). For a module over a discrete valuation ring (DVR), the
following statements are equivalent.
(1) M is radical supplemented,
(2) M = T (M) ⊕X , where the reduced part of T (M) is bounded and X/RadX
is finitely generated,
Now we shall prove that radical supplemented modules and srs-modules coincide
over discrete valuation rings. Firstly we need the following lemma.
Lemma 3.2. Let R be a local ring and M be an R-module. If M/RadM is
finitely generated, then M is an srs-module.
Proof. Let N be a submodule of M such that RadM ⊆ N . Then M/N is finitely
generated, and so M = N + L for some finitely generated submodule L of M . Since
RR is supplemented, L is also supplemented as it is finitely generated. So N has a
supplement in M by Lemma 2.1.
Proposition 3.1. Let R be a DVR and M be an R-module. Then M is an srs-
module if and only if M is radical supplemented.
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STRONGLY RADICAL SUPPLEMENTED MODULES 1145
Proof. One direction is clear. Suppose that M is radical supplemented. Then M =
= T (M) ⊕ X as in Lemma 3.1. Since T (M) is bounded, it is supplemented by [4]
(Theorem 2.4). By Lemma 3.2, X is an srs-module. Therefore M is an srs-module by
Corollary 2.1.
Note that, by Example 2.2, Proposition 3.1 is not true in general for modules over
Dedekind domains which are not DVR.
Proposition 3.2. Let R be a non-local domain and M be a reduced R-module. If
M is an srs-module, then M = T (M) + RadM .
Proof. Suppose that T (M) + RadM 6= M. Since RadM ⊆ T (M) + RadM ,
T (M) + RadM has a supplement, say L, in M. Then L has a maximal submodule K,
because M is reduced. Let K ′ = T (M) + RadM + K. It is easy to see that K ′ is a
maximal submodule of M. Then K ′ has a supplement V in M . By [1] (41.1(3)), V is
local, and so V ∼= R/I for some nonzero I ⊆ R. Therefore V is torsion, and so V ⊆
⊆ T (M). We get M = K ′+V = T (M)+RadM+K+V = T (M)+RadM+K = K ′,
a contradiction. Hence M = T (M) + RadM.
Now we shall prove that, the converse of Proposition 3.2 is true, under a certain
condition.
Proposition 3.3. Let R be a domain and M be an R-module. Suppose that M =
T (M) + RadM and T (M) is supplemented. Then M is an srs-module.
Proof. Let N be a submodule of M such that RadM ⊆ N. Then N = N ∩
∩ T (M) +RadM = T (N) +RadM. Let L be a supplement of T (N) in T (M). Then
T (N) + L = T (M) and T (N) ∩ L � L. It follows that M = T (M) + RadM =
= T (N) + L + RadM ⊆ N + L and so M = N + L. Since L is torsion, N ∩ L =
= T (N) ∩ L. Therefore L is a supplement of N in M.
Let R be a Dedekind domain and M be an R-module. Since R is a dedekind domain,
P (M) is the divisible part of M . By [5] (Lemma 4.4), P (M) is (divisible) injective and
so there exists a submodule N of M such that M = P (M)⊕N. Here N is called the
reduced part of M . Note that P (M) ⊆ RadM. By Corollary 2.2, we know that P (M)
is an srs-module. Using these facts, we have the following result.
Proposition 3.4. Let R be a Dedekind domain and M be an R-module. Then M
is an srs-module if and only if the reduced part N of M is an srs-module.
Proof. N is an srs-module as a homomorphic image of M by Proposition 2.1. The
converse is by Proposition 2.3.
Proposition 3.5. Let R be a non-local Dedekind domain and M be an srs-module.
Then M = T (M) + RadM.
Proof. Let M = P (M)⊕N with N reduced. Then N is an srs-module as a direct
summand of M . By Proposition 3.2, N = T (N) + RadN . So that
M = P (M)⊕N = P (M) + T (N) + RadN ⊆ T (M) + RadM.
Hence M = T (M) + RadM.
Recall from [5] that a commutative domain R is called h-local if every non-zero
non-unit of R belongs to only finitely many maximal ideals and R/P is a local ring for
every prime ideal P of R. It is also proved that a commutative domain R is h-local if and
only if R/I is a semiperfect ring for every non-zero ideal I of R (see [5], Lemma 4.5).
In [5], it is proved that, R is h-local if and only if every finitely generated torsion
R-module is supplemented. Since for finitely generated modules supplemented modules
and srs-modules coincide, we obtain the following .
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 8
1146 E. BÜYÜKAŞıK, E. TÜRKMEN
Proposition 3.6. Let R be a commutative domain. Then R is h-local if and only
if every finitely generated torsion R-module is an srs-module.
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Received 04.02.10,
after revision — 20.03.11
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 8
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| resource_txt_mv | umjimathkievua/77/41b46a6decebced711cbac8d0c97e277.pdf |
| spelling | umjimathkievua-article-27922020-03-18T19:36:38Z Strongly radical supplemented modules Сильно радикально доповненi модулi Büyükaşık, Е. Türkmen, E. Бюкасік, Є. Тюркмен, Є. Zoschinger studied modules whose radicals have supplements and called these modules radical supplemented. Motivated by this, we call a module strongly radical supplemented (briefly srs) if every submodule containing the radical has a supplement. We prove that every (finitely generated) left module is an srs-module if and only if the ring is left (semi)perfect. Over a local Dedekind domain, srs-modules and radical supplemented modules coincide. Over a no-local Dedekind domain, an srs-module is the sum of its torsion submodule and the radical submodule. Зошiнгер вивчав модулi, радикали яких мають доповнення, i назвав цi модулi радикально-доповненими. Мотивуючись цим, будемо називати модуль сильно радикально доповненим (або, скорочено, srs-модулем) якщо кожен пiдмодуль, що мiстить радикал, має доповнення. Доведено, що кожен (скiнченнопороджений) лiвий модуль є srs-модулем тодi i тiльки тодi, коли кiльце є лiвим (напiв)досконалим. Над локальною дедекiндовою областю srs-модулi та радикально доповненi модулi збiгаються. Над нелокальною дедекiндовою областю srs-модуль є сумою свого пiдмодуля скруту i радикального пiдмодуля. Institute of Mathematics, NAS of Ukraine 2011-08-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2792 Ukrains’kyi Matematychnyi Zhurnal; Vol. 63 No. 8 (2011); 1140-1146 Український математичний журнал; Том 63 № 8 (2011); 1140-1146 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2792/2337 https://umj.imath.kiev.ua/index.php/umj/article/view/2792/2338 Copyright (c) 2011 Büyükaşık Е.; Türkmen E. |
| spellingShingle | Büyükaşık, Е. Türkmen, E. Бюкасік, Є. Тюркмен, Є. Strongly radical supplemented modules |
| title | Strongly radical supplemented modules |
| title_alt | Сильно радикально доповненi модулi |
| title_full | Strongly radical supplemented modules |
| title_fullStr | Strongly radical supplemented modules |
| title_full_unstemmed | Strongly radical supplemented modules |
| title_short | Strongly radical supplemented modules |
| title_sort | strongly radical supplemented modules |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2792 |
| work_keys_str_mv | AT buyukasıke stronglyradicalsupplementedmodules AT turkmene stronglyradicalsupplementedmodules AT bûkasíkê stronglyradicalsupplementedmodules AT tûrkmenê stronglyradicalsupplementedmodules AT buyukasıke silʹnoradikalʹnodopovnenimoduli AT turkmene silʹnoradikalʹnodopovnenimoduli AT bûkasíkê silʹnoradikalʹnodopovnenimoduli AT tûrkmenê silʹnoradikalʹnodopovnenimoduli |