On Supplement Submodules
We investigate some properties of supplement submodules. Some relations between lying-above and weak supplement submodules are also studied. Let V be a supplement of a submodule U in M. Then it is possible to define a bijective map between the maximal submodules of V and the maximal submodules of M...
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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 | 2020-03-18T19:16:28Z |
| description | We investigate some properties of supplement submodules. Some relations between lying-above and weak supplement submodules are also studied. Let V be a supplement of a submodule U in M. Then it is possible to define a bijective map between the maximal submodules of V and the maximal submodules of M that contain U. Let M be an R-module, U ≤ M, let V be a weak supplement of U, and let K ≤ V. In this case, K is a weak supplement of U if and only if V lies above K in M. We prove that an R-module M is amply supplemented if and only if every submodule of M lies above a supplement in M. We also prove that M is semisimple if and only if every submodule of M is a supplement in M. |
| first_indexed | 2026-03-24T02:24:17Z |
| format | Article |
| fulltext |
UDC 512.5
C. Nebiyev, A. Pancar (Ondokuz Mayis Univ., Turkey)
ON SUPPLEMENT SUBMODULES
ПРО ДОПОВНЮЮЧI ПIДМОДУЛI
We investigate some properties of supplement submodules. Some relations between lying-above and weak supplement
submodules are also studied. Let V be a supplement of a submodule U in M. Then it is possible to define a bijective map
between maximal submodules of V and maximal submodules of M that contain U. Let M be an R-module, U ≤ M, V
be a weak supplement of U , and K ≤ V. In this case, K is a weak supplement of U if and only if V lies above K in M.
We prove that an R-module M is amply supplemented if and only if every submodule of M lies above a supplement in
M. We also prove that M is semisimple if and only if every submodule of M is a supplement in M.
Дослiджено деякi властивостi доповнюючих пiдмодулiв. Також вивчено деякi спiввiдношення мiж вищерозмiще-
ними та слабкими доповнюючими пiдмодулями. Нехай V — доповнення пiдмодуля U в M. Тодi можна означити
бiєкцiю мiж максимальними пiдмодулями V та максимальними пiдмодулями M , що мiстять U. Нехай M — R-
модуль, U ≤ M, V — слабке доповнення U i K ≤ V. У цьому випадку K є слабким доповненням U тодi i тiльки
тодi, коли V лежить вище K у M. Доведено, що R-модуль M є достатньо доповненим тодi i тiльки тодi, коли кожен
пiдмодуль модуля M лежить вище доповнення в M. Також доведено, що M є напiвпростим тодi i тiльки тодi, коли
кожен пiдмодуль модуля M є доповненням у M.
1. Introduction. Throughout this paper R will be an arbitrary ring with identity and all modules are
unital left R-modules. Let M be an R-module and V be a submodule of M. If L = M for every
submodule L of M such that V + L = M then V is called a small submodule of M and written by
V << M. In this work Rad(M) will denote the intersection of all maximal submodules of M. If
M has no maximal submodule then we define Rad(M) = M . Let M be an R-module. N ≤M will
mean N is a submodule of M.
Lemma 1.1 (Modular law). Let M be an R-module, K, N and H be submodules of M and
H ≤ N. Then N ∩ (H + K) = H + N ∩K (see [3]).
Let U be a submodule of MR−module. If a submodule V is minimal in the collection of
submodules L of M such that U +L = M then V is called a supplement of U by addition or simply
a supplement of U in M. In this case U + V = M is clear. Let V be a supplement of U in M.
Then K = V for every K ≤ V such that U + K = M. The modules whose every submodules have
supplements are called supplemented modules. If every submodule of the R-module M has at least
one supplement that is a direct summand in M, then M is called ⊕-supplemented. A submodule V
of M is called supplement in M if V is a supplement of a submodule 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 amply supplemented.
2. Properties of supplement submodules.
Lemma 2.1. A submodule V of M is a supplement of a submodule U in M if and only if
U + V = M and U ∩ V << V (see [14]).
Lemma 2.2. Let M = U + V. If a submodule K is a proper submodule of M which contains
U and distinct from U, then K ∩ V is a proper submodule of V.
Proof. Because of U ≤ K, M = U + V and M 6= K, then V 6⊂ K and V ∩ K 6= V. By
K = M ∩K = (U +V )∩K = U +V ∩K and K 6= U, then V ∩K 6= 0. Hence K ∩V is a proper
submodule of V.
c© C. NEBIYEV, A. PANCAR, 2013
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 7 961
962 C. NEBIYEV, A. PANCAR
Lemma 2.3. Let V be a supplement of a submodule U in M. If U is a maximal submodule,
then V is cyclic and U ∩ V is the unique maximal submodule of V. In this case U ∩ V = Rad(V )
(see [14]).
Lemma 2.4. Let M be an R-module, U and V be proper submodules of M. If M = U + V
and V is simple, then U is a maximal submodule of M.
Proof. If K is a submodule which contains U and distinct from M and U, then by Lemma 2.2
K∩V is a proper submodule of V. This contradicts while V is simple. Hence M have no submodules
which contains U and distinct from M and U. Thus U is a maximal submodule of M.
Corollary 2.1. Let V be a supplement of U in M. Then U is a maximal submodule of M if and
only if V or V/U ∩ V is simple.
Lemma 2.5. Let V be a supplement in M and K be a submodule of V. Then K << M if and
only if K << V (see [4]).
The following lemma is in [4] (Exercise 20.39). We prove this lemma as follows.
Lemma 2.6. 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 L ≤ T. Then 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. Hence T is a supplement of U + K in M.
(⇐) Let T be a supplement of U + K in M. Then by Lemma 2.1 U + K + T = M and
(U + K)∩ T << T. Since 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. Then by Lemma 2.1 T is a supplement of K in V.
Corollary 2.2. Let M = U ⊕V, 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.
Corollary 2.3. Let U and V be mutual supplements in M, L be a supplement of S in U and T
be a supplement of K in V. Then L + T is a supplement of K + S in M.
Proof. Since U = S + L and V is a supplement of U then by Lemma 2.6 T is a supplement of
S +L+K in M and then (S +L+K)∩ T << T. Since V = K + T and U is a supplement of V,
then by Lemma 2.6 L is a supplement of S+K+T in M and then (S+K+T )∩L << L. Because
U = S + L, V = K + T and M = U + V, then we have M = S + L + K + T = S + K + L + T.
We can also have (S + K) ∩ (L + T ) ≤ L ∩ (S + K + T ) + T ∩ (S + K + L) << L + T. Hence
L + T is a supplement of K + S in M.
Corollary 2.4. Let M = U ⊕ V, L be a supplement of S in U and T be a supplement of K in
V. Then L + T is a supplement of K + S in M.
Lemma 2.7. Let V be a supplement of U in M and K be a maximal submodule of V. Then
U + K is a maximal submodule of M. In this case K = (U + K) ∩ V.
Proof. Because K is a maximal submodule of V, K 6= V. Since V is a supplement of U,
U + K 6= M. Since U ∩ V << V and K is a maximal submodule of V, we have U ∩ V ≤ K and
K = U∩V +K = (U+K)∩V. Then by M/(U+K) = (U+K+V )/(U+K) ∼= V/V ∩(U+K) =
= V/K, we have M/(U + K) is simple and U + K is a maximal submodule of M.
Lemma 2.8. Let M be an R-module and V be a submodule of M. If K is a maximal submodule
of M and V 6⊂ K, then V ∩K is a maximal submodule of V.
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ON SUPPLEMENT SUBMODULES 963
Proof. Because of V 6⊂ K, V ∩K 6= V. Let v ∈ V \ (V ∩K). Then v 6∈ K and K + Rv = M.
We get intersection by V in two side, by using Modular law we have K ∩ V + Rv = V ∩M = V
and then V ∩K is obtained to be maximal in V.
Theorem 2.1. Let V be a supplement of a submodule U in M. Then it is possible to define a
bijective map between maximal submodules of V and maximal submodules of M which contain U.
Proof. Let Γ = {K | U ≤ K,K is maximal in M}, Λ = {T | T is maximal in V }. We
can define a map f : Γ → Λ, K → f(K) = K ∩ V. Since U ≤ K and K is maximal in M for
every K ∈ Γ, V 6⊂ K and then by Lemma 2.8 K ∩ V is a maximal submodule of V. That is, f is a
function.
Let T ∈ Λ. Since T is maximal in V, then by Lemma 2.7 U + T ∈ Γ and f(U + T ) =
= (U + T ) ∩ V = T. Thus f is surjective.
Let f(K) = f(L) for K,L ∈ Γ. Then K ∩ V = L ∩ V. Since U ≤ K and U ≤ L, then by
Modular law K = M ∩K = (U +V )∩K = U +V ∩K = U +V ∩L = (U +V )∩L = M ∩L = L.
Hence f is bijective.
The Theorems 2.2 and 2.3 are in [14]. We prove these theorems by different ways.
Theorem 2.2. Let U be a submodule which has a supplement in M which is distinct from zero,
and Rad(M) << M. Then U is contained in a maximal submodule of M.
Proof. Let V be a supplement of U which distinct from zero in M. If V is contained in all
maximal submodules of M, because U+V = M, U+Rad(M) = M and then because Rad(M) <<
<< M, we get U = M. This contradicts V 6= 0. Hence there exists a maximal submodule K of M
which doesn’t contain V. By Lemma 2.8 V ∩K is a maximal submodule of V. Then by Lemma 2.7
U + V ∩K is a maximal submodule of M which contains U.
Theorem 2.3. Let V be a supplement submodule in M. Then Rad(V ) = V ∩ Rad(M).
Proof. Let V be a supplement of U in M. If V ≤ Rad(M), then V has no maximal submodules,
because if K were a maximal submodule of V then U + K would be a maximal submodule of M
and by V ≤ U + K, M = U + V ≤ U + K ≤M and then K = V. Hence if V ≤ Rad(M), then V
has no maximal submodules. In this case Rad(V ) = V = V ∩ Rad(M).
Let V 6⊂ Rad(M). This case clearly we can prove that V has at least one maximal submod-
ule. Clearly we can see that Rad(V ) = ∩{K | K is maximal in V } = ∩{V ∩ (U + K) |
K is maximal in V } = V ∩ [∩{(U + K) | K is maximal in V }]. At the end of this equal-
ity because U + K is maximal in M (by Lemma 2.7), by definition of Rad(M), Rad(M) = ∩{N |
N is maximal in M} ≤ ∩{(U + K) | K is maximal in V }. Thus V ∩ Rad(M) ≤ Rad(V ).
At the end of the equality V ∩ Rad(M) = V ∩ [∩{N | N is maximal in M}] = ∩{V ∩ N |
N is maximal in M}, because N is maximal in M, by Lemma 2.8 V ∩N = V or V ∩N is maximal
in V. Thus Rad(V ) ≤ V ∩Rad(M). Since V ∩Rad(M) ≤ Rad(V ) and Rad(V ) ≤ V ∩Rad(M),
Rad(V ) = V ∩ Rad(M).
A submodule U of M has a weak supplement V in M if U + V = M and U ∩ V << M. M is
called weakly supplemented if every submodule of M has a weak supplement in M. A submodule
V of M is called weak supplement in M if V is a weak supplement of a submodule of M.
A submodule L of M is said to lie above a submodule N of M if N ≤ L and L/N << M/N.
Some properties of weakly supplemented modules are investigated in [10]. Some properties of
lying above are in [11]. We investigate some relations between lying above and weak supplement
submodules.
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964 C. NEBIYEV, A. PANCAR
Lemma 2.9. Let L and N are submodules of M and N ≤ L. Then L lies above N if and only
if N + T = M for every submodule T of M such that L + T = M.
Proof. See [4].
Lemma 2.10. Let M = U + V and M = T + U ∩ V. Then M = U + T ∩ V = V + T ∩ U.
Proof. See [4].
Theorem 2.4. Let U ≤M, L ≤ U and U lies above L. If U and L have weak supplements in
M, then they have the same weak supplements in M.
Proof. Let V be a weak supplement of U in M. Then U+V = M and by Lemma 2.9 L+V = M.
Since V is a weak supplement of U and L ≤ U, L∩V ≤ U∩V << M. Thus V is a weak supplement
of L.
Let T be a weak supplement of L in M. Then L + T = M and by L ≤ U, U + T = M. Let
U ∩ T + S = M. Then by Lemma 2.10 U + T ∩ S = M and by Lemma 2.9 L + T ∩ S = M. By
also Lemma 2.10 L ∩ T + S = M and because L ∩ T << M, S = M. Thus U ∩ T << M and T
is a weak supplement of U in M.
Theorem 2.5. Let U ≤ M, L ≤ U and U lies above L. If U and L have supplements in M
then they have the same supplements in M.
Proof. Let V be a supplement of U in M. Then U + V = M and by Lemma 2.9 L + V = M.
Since V is a supplement of U and L ≤ U, L ∩ V ≤ U ∩ V << V. Thus V is a supplement of L.
Let T be a supplement of L in M. Then L+T = M and by L ≤ U, U +T = M. Let U +S = M
for some S ≤ T. Then by Lemma 2.9 L + S = M and since T is a supplement of L in M, S = T.
Thus T is a supplement of U in M.
Lemma 2.11. Let M be an R-module, U ≤ M, V be a weak supplement of U and K ≤ V.
Then K is a weak supplement of U if and only if V lies above K in M.
Proof. (⇒) Let K be a weak supplement of U. Then by definition U+K = M and U∩K << M.
Since K ≤ V, by Modular law V = V ∩M = V ∩ (U + K) = K + U ∩ V. Let V + T = M for
some submodule T of M. Then K + U ∩ V + T = M and since U ∩ V << M, K + T = M. Thus
by Lemma 2.9 V lies above K.
(⇐) Because V lies above K and M = U + V, then by Lemma 2.9 M = U + K. Since
M = U + K and U ∩K ≤ U ∩ V << M, K is a weak supplement of U in M.
Lemma 2.12. Let M be an R-module, T ≤ U ≤M and V be a weak supplement of T in M.
Then V is a weak supplement of U if and only if U lies above T in M.
Proof. (⇒) Let V be a weak supplement of U in M. Then U is a weak supplement of V in M.
Since T is a weak supplement of V in M and T ≤ U, then by Lemma 2.11 U lies above T.
(⇐) Since V is a weak supplement of T in M, then M = T + V and T ∩ V << M. Since
T ≤ U, then M = U + V. Let S be any submodule of M such that U ∩ V + S = M. Then by
Lemma 2.10 U + S ∩ V = M and since U lies above T, T + S ∩ V = M. Since V + S = M and
T + S ∩ V = M, T ∩ V + S = M. Then by T ∩ V << M we obtain S = M. Thus U ∩ V << M
and V is a weak supplement of U in M.
Corollary 2.5. Let M be a weakly supplemented module and L ≤ U ≤M. Then U and L have
the same weak supplements in M if and only if U lies above L.
Corollary 2.6. Let V be a supplement of U in M and L ≤ U. Then V is a supplement of L in
M if and only if U lies above L.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 7
ON SUPPLEMENT SUBMODULES 965
Corollary 2.7. Let V be a weak supplement of U in M. Then V is a supplement of U if and
only if V lies above no proper submodule.
Corollary 2.8. Let M be an R-module. If every submodule of M has a weak supplement that
is a direct summand of M, then M is ⊕-supplemented.
Proof. Let U has a weak supplement V in M and let M = V ⊕ X. Then V is a supplement
of X and by Corollary 2.7 V lies above no proper submodule. Then also by Corollary 2.7 V is a
supplement of U. Thus M is ⊕-supplemented.
Theorem 2.6. An R-module M is weakly supplemented if and only if every submodule of M
lies above a weak supplement in M.
Proof. (⇒) Since M is weakly supplemented, every submodule of M is a weak supplement in M.
Since every submodule of M lies above itself, every submodule of M lies above a weak supplement
in M.
(⇐) Let U ≤ M. Then by hypothesis U lies above a weak supplement T in M. Since T is a
weak supplement in M, there exists a submodule V of M such that T is a weak supplement of V in
M. Since U lies above T, then by Lemma 2.12 V is also a weak supplement of U in M.
Theorem 2.7. An R-module M is amply supplemented if and only if every submodule of M
lies above a supplement in M.
Proof. (⇒) Let U ≤ M. Since M is amply supplemented, then M is supplemented and U has
a supplement V in M. Since V is a supplement of U in M, then M = U + V. Since M is amply
supplemented, then V has a supplement T in M such that T ≤ U. Since T is a supplement of V in
M, then V is a weak supplement of T in M. Since V is a supplement of U in M, then V is a weak
supplement of U in M. Thus by Lemma 2.12 U lies above T. Hence U lies above a supplement in M.
(⇐) Let every submodule of M be lie above a supplement in M. Let U ≤M and M = U + V.
Then by hypothesis U ∩ V lies above a supplement submodule T in M. Let T be a supplement of
K in M. Then K is a weak supplement of T in M. Since U ∩ V lies above T then by Lemma 2.12
K is a weak supplement of U ∩ V in M and then U ∩ V ∩K << M. Since M = U ∩ V + K then
by Modular law V = V ∩M = V ∩ (U ∩ V + K) = U ∩ V + V ∩ K. Hence M = U + V =
= U + U ∩ V + V ∩ K = U + V ∩ K. Since U ∩ V ∩ K << M, V ∩ K is a weak supplement
of U in M. By hypothesis V ∩ K lies above a supplement submodule S in M. Since V ∩ K is
a weak supplement of U in M then by Lemma 2.11 S is a weak supplement of U in M. Hence
M = U + S and U ∩ S << M. Since S is a supplement in M and U ∩ S << M then by Lemma
2.5 U ∩ S << S and then S is a supplement of U in M with S ≤ V. Thus every submodule of M
has ample supplements in M and M is amply supplemented.
Theorem 2.8. Let M be an R-module. Then the following statements are equivalent:
(a) Every submodule of M lies above a direct summand of M.
(b) M is amply supplemented and every supplement submodule of M is a direct summand.
(c) For every submodules U and V of M such that U + V = M, there is a supplement X of U
in M such that X ≤ V and X is a direct summand of M.
Proof. (a)⇔ (b) is proved in [12].
(b) ⇒ (c) Clear.
(c)⇒ (a) Let U ≤M. By hypothesis U has a supplement V in M. Then U is a weak supplement
of V in M. Also by hypothesis V has a supplement X in M such that X ≤ U and X is a direct
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 7
966 C. NEBIYEV, A. PANCAR
summand of M. Because U and X are weak supplements of V and X ≤ U, then by Lemma 2.11 U
lies above X. Thus every submodule of M lies above a direct summand of M.
Lemma 2.13. Let M be an R-module. Then the following statements are equivalent:
(a) M is semisimple.
(b) Every submodule of M is a direct summand of M.
(c) Every submodule of M is a supplement in M.
Proof. (a)⇔ (b) is proved in [6].
(b) ⇒ (c) Clear, because every direct summand of M is a supplement in M.
(c) ⇒ (b) Let U ≤ M. Then by hypothesis U is a supplement in M. Let U be a supplement of
X in M. Then X +U = M and X ∩U << U. Also by hypothesis X ∩U is a supplement in M. Let
X ∩ U be a supplement of T in M. Then X ∩ U + T = M. And then by X ∩ U << M, T = M.
Thus U ∩X is a supplement of M in M. Hence U ∩X = 0 and M = U ⊕X.
Theorem 2.9. Let M be a weakly supplemented module. Then every weak supplement is a
supplement in M if and only if M is semisimple.
Proof. (⇒) Let U ≤ M. By hypothesis U has a weak supplement V in M. Then U is a weak
supplement of V in M. By hypothesis U is a supplement in M. Thus every submodule of M is a
supplement in M. Then by Lemma 2.13 M is semisimple.
(⇐) Since M is semisimple, every submodule of M is a supplement in M. Thus every weak
supplement is a supplement in M.
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Received 24.05.12,
after revision — 05.12.12
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| resource_txt_mv | umjimathkievua/90/54a700f6bbc5bb78f0da9ef836d8f690.pdf |
| spelling | umjimathkievua-article-24812020-03-18T19:16:28Z On Supplement Submodules Про доповнюючі пвдмодулі Nebiyev, C. Pancar, A. Небієв, С. Пансар, А. We investigate some properties of supplement submodules. Some relations between lying-above and weak supplement submodules are also studied. Let V be a supplement of a submodule U in M. Then it is possible to define a bijective map between the maximal submodules of V and the maximal submodules of M that contain U. Let M be an R-module, U ≤ M, let V be a weak supplement of U, and let K ≤ V. In this case, K is a weak supplement of U if and only if V lies above K in M. We prove that an R-module M is amply supplemented if and only if every submodule of M lies above a supplement in M. We also prove that M is semisimple if and only if every submodule of M is a supplement in M. досліджєно дєякі властивості доповнюючих підмодулів. Також вивчено дєякі співвідношення між вищерозміще-ними та слабкими доповнюючими підмодулями. Нехай $V$ — доповнення підмодуля $U$ в $M$. Тоді можна означити бієкцію між максимальними підмодулями $V$ та максимальними підмодулями $M$, що містять $U$. Нехай $M$ — $R$-модуль, $U ≤ M$, $V$ — слабке доповнення $U$ i $K ≤ V$. У цьому випадку $K$ є слабким доповненням $U$ тоді i тільки тоді, коли $V$ лежить вище $K$ у $M$. Доведено, що $R$-модуль $M$ є достатньо доповненим тоді i тільки тоді, коли кожен підмодуль модуля $M$ лежить вище доповнення в $M$. Також доведено, що $M$ є напівпростим тоді i тільки тоді, коли кожен підмодуль модуля $M$ є доповненням у $M$. Institute of Mathematics, NAS of Ukraine 2013-07-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2481 Ukrains’kyi Matematychnyi Zhurnal; Vol. 65 No. 7 (2013); 961–966 Український математичний журнал; Том 65 № 7 (2013); 961–966 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2481/1728 https://umj.imath.kiev.ua/index.php/umj/article/view/2481/1729 Copyright (c) 2013 Nebiyev C.; Pancar A. |
| spellingShingle | Nebiyev, C. Pancar, A. Небієв, С. Пансар, А. On Supplement Submodules |
| title | On Supplement Submodules |
| title_alt | Про доповнюючі пвдмодулі |
| title_full | On Supplement Submodules |
| title_fullStr | On Supplement Submodules |
| title_full_unstemmed | On Supplement Submodules |
| title_short | On Supplement Submodules |
| title_sort | on supplement submodules |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2481 |
| work_keys_str_mv | AT nebiyevc onsupplementsubmodules AT pancara onsupplementsubmodules AT nebíêvs onsupplementsubmodules AT pansara onsupplementsubmodules AT nebiyevc prodopovnûûčípvdmodulí AT pancara prodopovnûûčípvdmodulí AT nebíêvs prodopovnûûčípvdmodulí AT pansara prodopovnûûčípvdmodulí |