On strongly $\oplus$-supplemented modules
In this work, strongly $\oplus$-supplemented and strongly cofinitely $\oplus$-supplemented modules are defined and some properties of strongly $\oplus$-supplemented and strongly cofinitely $\oplus$-supplemented modules are investigated. Let $R$ be a ring. Then every $R$-module is strongly $\oplus$-s...
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| author | Nebiyev, C. Pancar, A. Небієв, С. Пансар, А. |
| author_facet | Nebiyev, C. Pancar, A. Небієв, С. Пансар, А. |
| author_sort | Nebiyev, C. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-03-18T19:35:13Z |
| description | In this work, strongly $\oplus$-supplemented and strongly cofinitely $\oplus$-supplemented modules are defined and some properties of strongly $\oplus$-supplemented and strongly cofinitely $\oplus$-supplemented modules are investigated. Let $R$ be a ring. Then every $R$-module is strongly $\oplus$-supplemented if and only if R is perfect. Finite direct sum of $\oplus$-supplemented modules is $\oplus$-supplemented. But this is not true for strongly $\oplus$-supplemented modules. Any direct sum of cofinitely $\oplus$-supplemented modules is cofinitely $\oplus$-supplemented but this is not true for strongly cofinitely $\oplus$-supplemented modules. We also prove that a supplemented module is strongly $\oplus$-supplemented if and only if every supplement submodule lies above a direct summand. |
| first_indexed | 2026-03-24T02:29:34Z |
| format | Article |
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UDC 512.5
C. Nebiyev, A. Pancar (Ondokuz Mayis Univ., Turkey)
ON STRONGLY
⊕
-SUPPLEMENTED MODULES
ПРО СИЛЬНО
⊕
-ДОПОВНЕНI МОДУЛI
In this work, strongly ⊕-supplemented and strongly cofinitely ⊕-supplemented modules are defined and some
properties of strongly ⊕-supplemented and strongly cofinitely ⊕-supplemented modules are investigated. Let
R be a ring. Then every R-module is strongly ⊕-supplemented if and only if R is perfect. Finite direct sum of
⊕-supplemented modules is ⊕-supplemented. But this is not true for strongly ⊕-supplemented modules. Any
direct sum of cofinitely ⊕-supplemented modules is cofinitely ⊕-supplemented but this is not true for strongly
cofinitely ⊕-supplemented modules. We also prove that a supplemented module is strongly ⊕-supplemented
if and only if every supplement submodule lies above a direct summand.
Визначено сильно ⊕-доповненi та сильно кофiнiтно ⊕-доповненi модулi i дослiджено деякi властивостi
сильно ⊕-доповнених та сильно кофiнiтно ⊕-доповнених модулiв. Припустимо, що R — кiльце. У цьому
випадку кожен R-модуль є сильно ⊕-доповненим тодi i тiльки тодi, коли R є досконалим. Скiнченна
пряма сума ⊕-доповнених модулiв є ⊕-доповненою. Але це не справджується для сильно ⊕-доповнених
модулiв. Будь-яка пряма сума кофiнiтно ⊕-доповнених модулiв є кофiнiтно ⊕-доповненою, але це не
справджується для сильно кофiнiтно ⊕-доповнених модулiв. Доведено також, що доповнений модуль є
сильно ⊕-доповненим модулем тодi i тiльки тодi, коли кожен пiдмодуль-доповнення розташований над
прямим доданком.
1. Introduction. In this work R will denote an arbitrary ring with unity and M will
state for an unitary left R-module. Let M be an R-module. N ≤ M will mean N is
a submodule of M. Let K ≤ M. If L = M for every submodule L of M such that
K + L = M then K is called a small submodule of M and written by K � M . Let
U ≤ M and V ≤ M. If V is minimal with respect to M = U + V then V is called a
supplement of U in M. This equivalent to M = U + V and U ∩ V � V. M is called
supplemented if every submodule of M has a supplement in M. M is called finitely
supplemented if every finitely generated submodule of M has a supplement in M. M
is called ⊕-supplemented if every submodule of M has a supplement that is a direct
summand of M. M is called completely ⊕-supplemented if every direct summand
of M is ⊕-supplemented. A submodule U of M is called cofinite if M/U is finitely
generated. M is called cofinitely supplemented if every cofinite submodule of M has a
supplement in M. We say a submodule U of the R-module M has ample supplements
in M if for every V ≤ M with U + V = M, there exists a supplement V ′ of U with
V ′ ≤ V. If every submodule of M has ample supplements in M, then we call M is
amply supplemented.
M is called a projective cover of N, if M is a projective module and there exists
an epimorphism f : M → N such that Ke f � M. A module M is called semiperfect
if every factor module of M has a projective cover. M is called π-projective module
if there exists an endomorphism f of M such that lm f ≤ U, Im(1− f) ≤ V for every
submodules U, V of M such that M = U + V.
Let V ≤M. V is called lies above a direct summand of M if there exist submod-
ules M1 and M2 of M such that M =M1 ⊕M2,M1 ≤ V, V ∩M2 �M2.
In this work JacR will denote intersection of all maximal left ideals of R.
Let M be an R-module. We consider the following conditions.
(D1) Every submodule of M lies above a direct summand of M.
(D3) If M1 and M2 are direct summands of M with M =M1+M2, then M1∩M2
is also a direct summand of M.
c© C. NEBIYEV, A. PANCAR, 2011
662 ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 5
ON STRONGLY
⊕
-SUPPLEMENTED MODULES 663
Lemma 1.1 (Modular law). Let M be an R-module, K, N and H are submodules
of M and H ≤ N. Then N ∩ (H +K) = H +N ∩K (see [1]).
Lemma 1.2. Let V be a supplement of U in M, K and T be submodules of V.
Then T is a supplement of K in V if and only if T is a supplement of U +K in M.
Proof. (⇒) Let T be a supplement of K in V. Let U+K+L =M for a submodule
L ≤ T. In this case K + L ≤ V and because V is a supplement of U, K + L = V.
Since L ≤ T and T is a supplement of K in V, L = T and then T is a supplement of
U +K in M.
(⇐) Let T be a supplement of U + K in M. This can be found that because of
U + K + T = M and K + T ≤ V, then we can have K + T = V. Since K ∩ T ≤
≤ (U +K) ∩ T � T, K ∩ T � T and then T is a supplement of K in V.
2. Strongly ⊕-supplemented modules.
Definition 2.1. Let M be a supplemented module. If every supplement submodule
of M is a direct summand of M then M is called a strongly ⊕-supplemented module.
Corollary 2.1. Strongly ⊕-supplemented modules are ⊕-supplemented.
Lemma 2.1. Let M be supplemented and π-projective module. Then M is a
strongly ⊕-supplemented module.
Proof. See [21].
Lemma 2.2. Let M be a strongly ⊕-supplemented module. Then every direct
summand of M is strongly ⊕-supplemented.
Proof. Let L be a direct summand of M and M = L⊕ T. Let K be a supplement
of U in L. By Lemma 1.2 K is a supplement of U ⊕ T in M. Because M is strongly
⊕-supplemented, K is a direct summand of M. Let M = K ⊕ P. By Modular law
L = L ∩M = L ∩ (K ⊕ P ) = K ⊕ (L ∩ P ). Thus K is a direct summand of L and L
is strongly ⊕-supplemented.
Corollary 2.2. Strongly ⊕-supplemented modules are completely ⊕-supplemented.
Theorem 2.1. Every (D1) module is strongly ⊕-supplemented.
Proof. See [21].
Theorem 2.2. Let R be a Prüfer ring. Then every finitely generated torsion free
supplemented R-module is strongly ⊕-supplemented.
Proof. Because R is a Prüfer ring, then every finitely generated torsion free R-
module is projective (see [21]). Because every projective module is π-projective, by
Lemma 2.1 every finitely generated torsion free supplemented R-module is strongly
⊕-supplemented.
Theorem 2.3. Let Mi, 1 ≤ i ≤ n, are projective modules. Then ⊕n
i=1 Mi is
strongly ⊕-supplemented if and only if every Mi is strongly ⊕-supplemented.
Proof. (⇒) Because every Mi is direct summand of ⊕n
i=1Mi, by Lemma 2.2 every
Mi is strongly ⊕-supplemented.
⇐ Because every Mi is supplemented by [21], ⊕n
i=1Mi is supplemented. Be-
cause every Mi is projective modules by [21], ⊕n
i=1Mi is projective module. Thus
by Lemma 2.1 ⊕n
i=1Mi is strongly ⊕-supplemented.
Lemma 2.3. Let M be a projective module. Then the followings are equivalent.
(i) M is semiperfect.
(ii) M is supplemented.
(iii) M is ⊕-supplemented.
(iv) M is strongly ⊕-supplemented.
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 5
664 C. NEBIYEV, A. PANCAR
Proof. (i) ⇔ (ii) ⇔ (iii) are proved in [10]. (ii) ⇒ (iv) Because M is a pro-
jective module, M is a π-projective module. Thus by Lemma 2.1 M is strongly ⊕-
supplemented.
(iv) ⇔ (ii) Clear.
Theorem 2.4. For every ring R, the following statements are equivalent.
(i) R is semiperfect.
(ii) Every finitely generated free R-module is ⊕-supplemented.
(iii) Every finitely generated free R-module is strongly ⊕-supplemented.
(iv) RR is ⊕-supplemented.
(v) RR is strongly ⊕-supplemented.
(vi) For every left ideal A of R, there exists an idempotent e ∈ R\A such that
A ∩ eR ⊂ JacR.
Proof. (i)⇔ (ii)⇔ (iv)⇔ (vi) are proved in [11].
Because RR is a projective module, (ii)⇔ (iii)⇔ (iv)⇔ (v) are hold.
Theorem 2.5. A commutative ring R is semiperfect if and only if every π-projective
cyclic R-module is strongly ⊕-supplemented.
Proof. (⇒) Let R be semiperfect. By [11] every cyclic R-module is⊕-supplemented.
Thus by Lemma 2.1 every π-projective cyclic R-module is strongly ⊕-supplemented.
(⇐) Since RR is cyclic and π-projective, by hypothesis RR is strongly⊕-supplemen-
ted. By Lemma 2.3 RR is semiperfect.
Theorem 2.6. Let M be a finitely generated strongly ⊕-supplemented R-module.
Then M is direct sum of cyclic submodules.
Proof. Since M is a strongly ⊕-supplemented module, by Corollary 2.2 M is com-
pletely ⊕-supplemented. In case by [11] M is direct sum of cyclic submodules.
Theorem 2.7. For any ring R, the following statements are equivalent.
(i) R is perfect.
(ii) R(N) is ⊕-supplemented.
(iii) R(N) is strongly ⊕-supplemented.
(iv) Every countable generated free R-module is ⊕-supplemented.
(v) Every countable generated free R-module is strongly ⊕-supplemented.
(vi) Every free R-module is ⊕-supplemented.
(vii) Every free R-module is strongly ⊕-supplemented.
Proof. (i)⇔ (ii)⇔ (iv)⇔ (vi) are proved in [11].
Since RR is a projective module, every free R-module is projective. Thus every free
R-module is π-projective. By Lemma 2.1 (iv)⇔ (v)⇔ (vi)⇔ (vii) are hold.
Theorem 2.8. For a supplemented module M, the following statements are equi-
valent.
(i) M is strongly ⊕-supplemented.
(ii) Every supplement submodule of M lies above a direct summand.
(iii) (a) Every non zero supplement submodule of M contains a non zero direct
summand of M.
(b) Every supplement submodule of M contains a maximal direct summand of M.
Proof. (i)⇒ (ii) Clear from definitions.
(ii) ⇒ (i) Let V be any supplement submodule of M. Let V is a supplement of U
in M. By hypothesis there exist M1 ≤ M and M2 ≤ M such that M = M1 ⊕M2,
M1 ≤ V and V ∩M2 � M2. In this case V = V ∩M = M1 ⊕ V ∩M2 and by
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 5
ON STRONGLY
⊕
-SUPPLEMENTED MODULES 665
V ∩M2 �M, M = U + V = U + V ∩M2 +M1 = U +M1. Since V is supplement
of U, V = M1. Thus M = V ⊕M2 and V is a direct summand of M. That is M is
strongly ⊕-supplemented.
(i)⇒ (iii) Clear from definitions.
(iii) ⇒ (i) Let V be a supplement of U in M and assume X to be a maximal direct
summand of M with X ≤ V and M = X ⊕ Y. Then V = X ⊕ V ∩ Y and by Lemma
1.2 V ∩ Y is a supplement of U +X in M. If V ∩ Y is not zero then by (iii, a) there
exists a non zero direct summand N of M such that N ≤ V ∩ Y. In this case X ⊕ N
is a direct summand of M and X ⊕ N ≤ V. This contradicts the choice of X. Thus
V ∩ Y = 0 and V = X. In this case V is direct summand of M and M is a strongly
⊕-supplemented module.
Let M be an R-module. If rM = M for every r ∈ R which not zero divisor, then
M is called a divisible R-moduIe. Let R be a domain. If every submodule of left
R-module RR is projective, then R is called a Dedekind domain. Let R be a principal
ideal domain. If R has the unique prime element (up to unit), then R is called a discrete
valuation ring.
Remark 2.1. Let R be a discrete valuation ring and p be the unique prime element
of R. Then every ideal of R is of the form Rpk which k ∈ Z. If we take these ideals to
be neighborhoods of 0 in R, we define a topology in R, making R a topological ring. If
R is complete in this topology, we call it a complete discrete valuation ring.
Example 2.1. Let R be a discrete valuation ring which not complete and K be a
quotient field of R. Then M = K2 is strongly ⊕-supplemented but not amply supple-
mented.
Proof. By [21] Theorem 2.2, M is supplemented but not amply supplemented. Let
V be a supplement submodule in M. Assume that V is a supplement of U in M. Since
M is divisible, then M = rM = rU + rV = U + rV for every r ∈ R which r 6= 0.
Since V is a supplement of U in M, V = rV and then V is divisible. Since R is a
Dedekind domain, V is injective (see [19], 40.5) and a direct summand of M. Thus M
is strongly ⊕-supplemented.
Example 2.2. Let R be a discrete valuation ring with quotient field K, let p be
the unique prime element and let N = Rp. Then M = K/R ⊕ R/N is completely
⊕-supplemented but is not strongly ⊕-supplemented.
Proof. By [10] Example 2.1, M is completely ⊕-supplemented but not (D1). More-
over M satisfies (D3). Let L = R(p−2 + R, 1 + N) ≤ M. Then we can prove
K/R + L = M. Let x ∈ (K/R) ∩ L. Then x = (rp−2 + R, r + N) for some
r ∈ R. Since (rp−2 + R, r + N) ∈ K/R, r + N = 0 and then there exists r′ ∈ R
with r = r′p. Then x = (r′pp−2 + R, 0) = (r′p−1 + R, 0) ∈ R(p−1 + R, 0). Since
R(p−1 +R, 0) ≤ (K/R)∩L, K/R∩L = R(p−1 +R, 0). Let R(p−1 +R, 0)+T = L
with T ≤ L. Then there exists s ∈ R such that s(p−2 +R, 1+N) ∈ T and s+N 6= 0.
Since s+N 6= 0, s 6∈ N. Since p is the unique prime element of R, s is invertible in R,
i.e., there exists s′ ∈ R with s′s = 1. Then (p−2+R, 1+N) = s′s(p−2+R, 1+N) ∈ T
and then L = R(p−2 + R, 1 + N) ≤ T. Thus T = L, R(p−1 + R, 0) � L and L is
a supplement of K/R in M. If L is a direct summand of M, by M = K/R + L and
M satisfying (D3), (K/R) ∩ L = R(p−1 + R, 0) is also direct summand of M. This
contradicts R(p−1 +R, 0)�M. Hence L is not a direct summand of M and M is not
strongly ⊕-supplemented.
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 5
666 C. NEBIYEV, A. PANCAR
Remark 2.2. In Example 2.2 K/R is hollow and strongly ⊕-supplemented. Since
R/N is simple, it is strongly ⊕-supplemented. But the direct sum of K/R and R/N
is not strongly ⊕-supplemented. Zöschinger has proved that if R is a Dedekind domain
then an R-module M is supplemented if and only if M is ⊕-supplemented. But this not
true for strongly ⊕-supplemented by Example 2.2.
Definition 2.2. Let M be an R-module. If M is cofinitely supplemented and every
supplement of cofinite submodules of M is a direct summand of M then M is called a
strongly cofinitely ⊕-supplemented module.
Corollary 2.3. Every strongly cofinitely ⊕-supplemented module is cofinitely sup-
plemented.
Theorem 2.9. Let M be a strongly cofinitely ⊕-supplemented module. Then every
direct summand of M is strongly cofinitely ⊕-supplemented.
Proof. Let N be a direct summand of M and let M = N⊕T. Since M is cofinitely
supplemented, N ∼= M/T is also cofinitely supplemented. Let U be a cofinite submod-
ule of N and V be a supplement of U in N. Then by Lemma 1.2 V is a supplement of
U ⊕ T in M. Since U ⊕ T is a cofinite submodule of M and M is strongly cofinitely
⊕-supplemented, V is a direct summand of M. Let M = V ⊕ X. Then by Modular
law N = V ⊕ (N ∩ X) and then V is a direct summand of N. Hence N is strongly
cofinitely ⊕-supplemented.
Theorem 2.10. Let M be a π-projective and finitely supplemented R-moduIe. If
M is cofinitely supplemented then M is strongly cofinitely ⊕-supplemented.
Proof. Let U be a cofinite submodule of M and V be a supplement of U in M. Then
V is finitely generated. Since M is finitely supplemented, V has a supplement X in M.
Since M is π-projective, there exists f ∈ End(M) such the Im f ≤ U, Im(1− f) ≤ V.
Then we can prove (1− f)(U) ≤ U and f(V ) ≤ V. Then M = f(M)+ (1− f)(M) =
= f(V )+f(X)+V = V +f(X). Let ν ∈ V ∩f(X). Then there exists x ∈ X with ν =
= f(x). Since x−ν = x−f(x) = (1−f)(x) ∈ V, x ∈ V. Hence ν = f(x) ∈ f(V ∩X).
Since V ∩X � X, f(V ∩X) � f(X) and V ∩ f(X) ≤ f(V ∩X) � f(X). Hence
f(X) is a supplement of V in M. Since f(X) ≤ U, then V is a supplement of f(X) in
M. Hence V and f(X) are mutual supplements in M. Since M is π-projective, then by
[19] M = V ⊕ f(X) and V is a direct summand of M. Thus M is strongly cofinitely
⊕-supplemented.
Theorem 2.11. If M is cofinitely supplemented, then M/Rad (M) is strongly
cofinitely ⊕-supplemented.
Proof. Since M is cofinitely supplemented, M/Rad(M) is also cofinitely supple-
mented. Let U/Rad(M) be a cofinite submodule of M/Rad(M) and V/Rad(M) be
a supplement of U/Rad(M) in M/Rad(M). Since
U/Rad(M) ∩ V/Rad(M)�M/Rad(M),
U/Rad(M) ∩ V/Rad(M) ≤ Rad(M/Rad(M)) = 0
and then M/Rad(M) = U/Rad(M) ⊕ V/Rad(M). Hence V/Rad(M) is a direct
summand of M/Rad(M) and M/Rad(M) is strongly cofinitely ⊕-supplemented.
Example 2.3. Let M be a direct sum of an infinite number of copies of the
Prüferp-group Zp∞ . Then M is strongly cofinitely ⊕-supplemented but not strongly
⊕-supplemented.
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 5
ON STRONGLY
⊕
-SUPPLEMENTED MODULES 667
Proof. By [10] M is not supplemented, i.e., not strongly ⊕-supplemented. By [2]
M is cofinitely supplemented. Let L be a supplement submodule of M and L be a
supplement of K in M. We can prove that M is a divisible Z-module. Let n ∈ Z. Since
nM = M, M = nM = nK + nL = K + nL. Since L is a supplement of K in M,
nL = L and L is divisible. Since Z is a Dedekind domain, L is injective ([19], 40.5)
and a direct summand of M. Hence M is strongly cofinitely ⊕-supplemented.
Remark 2.3. In Example 2.2 M = K/R ⊕ R/N is cofinitely supplemented but
not strongly cofinitely ⊕-supplemented. Also in Example 2.2 K/R and R/N is strongly
cofinitely ⊕-supplemented but the direct sum of K/R and R/N is not strongly cofinitely
⊕-supplemented.
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Received 30.11.10
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| resource_txt_mv | umjimathkievua/0a/9740c41a1674e5bc28ed4b8bf8cd740a.pdf |
| spelling | umjimathkievua-article-27512020-03-18T19:35:13Z On strongly $\oplus$-supplemented modules Про сильно $\oplus$-доповненi модулi Nebiyev, C. Pancar, A. Небієв, С. Пансар, А. In this work, strongly $\oplus$-supplemented and strongly cofinitely $\oplus$-supplemented modules are defined and some properties of strongly $\oplus$-supplemented and strongly cofinitely $\oplus$-supplemented modules are investigated. Let $R$ be a ring. Then every $R$-module is strongly $\oplus$-supplemented if and only if R is perfect. Finite direct sum of $\oplus$-supplemented modules is $\oplus$-supplemented. But this is not true for strongly $\oplus$-supplemented modules. Any direct sum of cofinitely $\oplus$-supplemented modules is cofinitely $\oplus$-supplemented but this is not true for strongly cofinitely $\oplus$-supplemented modules. We also prove that a supplemented module is strongly $\oplus$-supplemented if and only if every supplement submodule lies above a direct summand. Визначено сильно $\oplus$-доповненi та сильно кофiнiтно $\oplus$-доповненi модулi i дослiджено деякi властивостi сильно $\oplus$-доповнених та сильно кофiнiтно $\oplus$-доповнених модулiв. Припустимо, що $R$ — кiльце. У цьому випадку кожен $R$-модуль є сильно $\oplus$-доповненим тодi i тiльки тодi, коли $R$ є досконалим. Скiнченна пряма сума $\oplus$-доповнених модулiв є $\oplus$-доповненою. Але це не справджується для сильно $\oplus$-доповнених модулiв. Будь-яка пряма сума кофiнiтно $\oplus$-доповнених модулiв є кофiнiтно $\oplus$-доповненою, але це не справджується для сильно кофiнiтно $\oplus$-доповнених модулiв. Доведено також, що доповнений модуль є сильно $\oplus$-доповненим модулем тодi i тiльки тодi, коли кожен пiдмодуль-доповнення розташований над прямим доданком. Institute of Mathematics, NAS of Ukraine 2011-05-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2751 Ukrains’kyi Matematychnyi Zhurnal; Vol. 63 No. 5 (2011); 662-667 Український математичний журнал; Том 63 № 5 (2011); 662-667 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2751/2258 https://umj.imath.kiev.ua/index.php/umj/article/view/2751/2259 Copyright (c) 2011 Nebiyev C.; Pancar A. |
| spellingShingle | Nebiyev, C. Pancar, A. Небієв, С. Пансар, А. On strongly $\oplus$-supplemented modules |
| title | On strongly $\oplus$-supplemented modules |
| title_alt | Про сильно $\oplus$-доповненi модулi |
| title_full | On strongly $\oplus$-supplemented modules |
| title_fullStr | On strongly $\oplus$-supplemented modules |
| title_full_unstemmed | On strongly $\oplus$-supplemented modules |
| title_short | On strongly $\oplus$-supplemented modules |
| title_sort | on strongly $\oplus$-supplemented modules |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2751 |
| work_keys_str_mv | AT nebiyevc onstronglyoplussupplementedmodules AT pancara onstronglyoplussupplementedmodules AT nebíêvs onstronglyoplussupplementedmodules AT pansara onstronglyoplussupplementedmodules AT nebiyevc prosilʹnooplusdopovnenimoduli AT pancara prosilʹnooplusdopovnenimoduli AT nebíêvs prosilʹnooplusdopovnenimoduli AT pansara prosilʹnooplusdopovnenimoduli |