Co-coatomically supplemented modules
It is shown that if a submodule $N$ of $M$ is co-coatomically supplemented and $M/N$ has no maximal submodule, then $M$ is a co-coatomically supplemented module. If a module $M$ is co-coatomically supplemented, then every finitely $M$-generated module is a co-coatomically supplemented module. Every...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507593166815232 |
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| author | Alizade, R. Güngör, S. Алізаде, Р. Гюнгер, С. |
| author_facet | Alizade, R. Güngör, S. Алізаде, Р. Гюнгер, С. |
| author_sort | Alizade, R. |
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| datestamp_date | 2019-12-05T09:25:34Z |
| description | It is shown that if a submodule $N$ of $M$ is co-coatomically supplemented and $M/N$ has no maximal submodule, then $M$ is a co-coatomically supplemented module. If a module $M$ is co-coatomically supplemented, then every finitely $M$-generated
module is a co-coatomically supplemented module. Every left $R$-module is co-coatomically supplemented if and only if the ring $R$ is left perfect. Over a discrete valuation ring, a module $M$ is co-coatomically supplemented if and only if the basic submodule of $M$ is coatomic. Over a nonlocal Dedekind domain, if the torsion part $T(M)$ of a reduced module $M$ has a weak supplement in $M$, then $M$ is co-coatomically supplemented if and only if $M/T (M)$ is divisible and $TP (M)$ is bounded for each maximal ideal $P$. Over a nonlocal Dedekind domain, if a reduced module $M$ is co-coatomically amply supplemented, then $M/T (M)$ is divisible and $TP (M)$ is bounded for each maximal ideal $P$. Conversely, if $M/T (M)$ is divisible and $TP (M)$ is bounded for each maximal ideal $P$, then $M$ is a co-coatomically supplemented module. |
| first_indexed | 2026-03-24T02:11:46Z |
| format | Article |
| fulltext |
UDC 512.5
R. Alizade (Yaşar Univ., Turkey),
S. Güngör (Izmir Inst. Technology, Turkey)
CO-COATOMICALLY SUPPLEMENTED MODULES
КО-КОАТОМНО ПОПОВНЕНI МОДУЛI
It is shown that if a submodule N of M is co-coatomically supplemented and M/N has no maximal submodule, then M is
a co-coatomically supplemented module. If a module M is co-coatomically supplemented, then every finitely M -generated
module is a co-coatomically supplemented module. Every left R-module is co-coatomically supplemented if and only if
the ring R is left perfect. Over a discrete valuation ring, a module M is co-coatomically supplemented if and only if the
basic submodule of M is coatomic. Over a nonlocal Dedekind domain, if the torsion part T (M) of a reduced module M
has a weak supplement in M, then M is co-coatomically supplemented if and only if M/T (M) is divisible and TP (M)
is bounded for each maximal ideal P. Over a nonlocal Dedekind domain, if a reduced module M is co-coatomically amply
supplemented, then M/T (M) is divisible and TP (M) is bounded for each maximal ideal P. Conversely, if M/T (M) is
divisible and TP (M) is bounded for each maximal ideal P, then M is a co-coatomically supplemented module.
Показано, що у випадку, коли субмодуль N модуля M є ко-коатомно поповненим, а M/N не має максимального
субмодуля, модуль M є ко-коатомно поповненим. Якщо модуль M є ко-коатомно поповненим, то кожен скiнченно
M -породжений модуль є ко-коатомно поповненим. Кожний лiвий R-модуль є ко-коатомно поповненим тодi i
тiльки тодi, коли кiльце R є лiвим досконалим. Поза дискретним метризацiйним кiльцем модуль M є ко-коатомно
поповненим тодi i тiльки тодi, коли базовий субмодуль M є коатомним. Поза нелокальною дедекiндовою областю у
випадку, коли торсiонна частина T (M) зведеного модуля M має слабке поповнення в M, модуль M є ко-коатомно
поповненим тодi i тiльки тодi, коли M/T (M) є подiльним, а TP (M) — обмеженим для кожного максимального
iдеалу P. Поза нелокальною дедекiндовою областю у випадку, коли зведений модуль M є ко-коатомно широко
поповненим, M/T (M) є подiльним, а TP (M) — обмеженим для кожного максимального iдеалу P. Навпаки, якщо
M/T (M) є подiльним, а TP (M) — обмеженим для кожного максимального iдеалу P, то модуль M є ко-коатомно
поповненим.
1. Introduction. Throughout this paper R denotes an associative ring with identity and all modules
are left unitary R-modules (RM ) unless otherwise stated. Let U be a submodule of M. A submodule
V of M is called a supplement of U in M if V is minimal element in the set of submodules L \leq M
with U+L = M. V is a supplement of U in M if and only if U+V = M and U\cap V \ll V. A module
M is called supplemented if every submodule of M has a supplement in M (see [9], Section 41, or
[5], Chapter 4). Semisimple, artinian and hollow (in particular local) modules are supplemented. A
module M is called coatomic if every proper submodule of M is contained in a maximal submodule
(see [12]). Let N be a submodule of a module M. We say that N is a co-coatomic submodule in
M if M/N is coatomic. Semisimple, finitely generated and local modules are coatomic modules.
Since every factor module of a coatomic module is coatomic, every submodule of semisimple, finitely
generated and local modules is co-coatomic. A module M is said to be co-coatomically supplemented
module if every co-coatomic submodule of M has a supplement in M. A submodule N of M is
called cofinite if M/N is finitely generated. M is called a cofinitely supplemented module if every
cofinite submodule of M has a supplement in M (see [1]). Clearly a co-coatomically supplemented
module is cofinitely supplemented and a coatomic module is co-coatomically supplemented if and
only if it is a supplemented module. A module M is called co-coatomically weak supplemented
c\bigcirc R. ALIZADE, S. GÜNGÖR, 2017
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 7 867
868 R. ALIZADE, S. GÜNGÖR
if every co-coatomic submodule N of M has a weak supplement in M, i.e., N + K = M and
N \cap K \ll M for some submodule K of M. It is clear that a co-coatomically supplemented module
is co-coatomically weak supplemented. A submodule U of an R-module M has ample supplements
in M if, for every submodule V of M with U + V = M, there exists a supplement V \prime of U
with V \prime \leq V (see [5, p. 237]). A module M is called co-coatomically amply supplemented if
every co-coatomic submodule of M has ample supplements in M. Clearly a co-coatomically amply
supplemented module is co-coatomically supplemented.
In Section 2, we show that if a submodule N of M is co-coatomically supplemented and M/N
has no maximal submodule, then M is co-coatomically supplemented. Every left R-module is
co-coatomically supplemented if and only if the ring R is left perfect.
In Section 3, we study on co-coatomically supplemented modules over a discrete valuation ring.
We show that a module M is co-coatomically supplemented if and only if the basic submodule of
M is coatomic if and only if M = T (M) \oplus X, where the reduced part of T (M) is bounded and
X/\mathrm{R}\mathrm{a}\mathrm{d}(X) is finitely generated.
In Section 4, we study on co-coatomically supplemented modules over nonlocal Dedekind do-
mains. A torsion module M is co-coatomically weak supplemented if and only if it is co-coatomically
supplemented. We show that for a reduced module M, if the torsion part T (M) of M has a weak
supplement in M, then M is co-coatomically supplemented if and only if M/T (M) is divisible and
TP (M) is bounded for each maximal ideal P. For a reduced module M, if M is co-coatomically
amply supplemented, then M/T (M) is divisible and TP (M) is bounded for each maximal ideal P
of R. Conversely, if M/T (M) is divisible and TP (M) is bounded for each maximal ideal P of R,
then M is a co-coatomically supplemented module.
2. Co-coatomically supplemented modules. For any module M, \mathrm{S}\mathrm{o}\mathrm{c}(M) denotes the socle of
M and \mathrm{R}\mathrm{a}\mathrm{d}(M) denotes the radical of M. The Jacobson radical of RR is denoted by \mathrm{J}\mathrm{a}\mathrm{c}(R).
Let \{ M\lambda \} \lambda \in \Lambda be the family of simple submodules of M that are direct summands of M.
\mathrm{S}\mathrm{o}\mathrm{c}\oplus (M) will denote the sum of M\lambda s for all \lambda \in \Lambda . That is \mathrm{S}\mathrm{o}\mathrm{c}\oplus (M) =
\sum
\lambda \in \Lambda
M\lambda . Clearly
\mathrm{S}\mathrm{o}\mathrm{c}\oplus (M) \leq \mathrm{S}\mathrm{o}\mathrm{c}(M).
Theorem 2.1. Let R be a ring. The following are equivalent for an R-module M :
1. Every co-coatomic submodule of M is a direct summand of M.
2. Every cofinite submodule of M is a direct summand of M.
3. Every maximal submodule of M is a direct summand of M.
4. M/ \mathrm{S}\mathrm{o}\mathrm{c}\oplus (M) does not contain a maximal submodule.
5. M/ \mathrm{S}\mathrm{o}\mathrm{c}(M) does not contain a maximal submodule.
Proof. (1) \Rightarrow (2) is clear since every cofinite submodule is co-coatomic.
(2) \Rightarrow (3). Clear.
(3) \Rightarrow (4). Suppose M/\mathrm{S}\mathrm{o}\mathrm{c}\oplus (M) contains a maximal submodule K/\mathrm{S}\mathrm{o}\mathrm{c}\oplus (M). It follows that
K is a maximal submodule of M. By hypothesis, M = K\oplus K \prime and K \prime is simple. K \prime \leq \mathrm{S}\mathrm{o}\mathrm{c}\oplus (M) \leq
\leq K. Contradiction.
(4) \Rightarrow (5). Clear because \mathrm{S}\mathrm{o}\mathrm{c}\oplus (M) \leq \mathrm{S}\mathrm{o}\mathrm{c}(M).
(5) \Rightarrow (1). Let N be a co-coatomic submodule of M. Since
M/(N + \mathrm{S}\mathrm{o}\mathrm{c}(M)) \sim = (M/N)/
\bigl(
(N + \mathrm{S}\mathrm{o}\mathrm{c}(M))/N
\bigr)
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 7
CO-COATOMICALLY SUPPLEMENTED MODULES 869
and M/N is coatomic, M/(N + \mathrm{S}\mathrm{o}\mathrm{c}(M)) is also coatomic. Since M/\mathrm{S}\mathrm{o}\mathrm{c}(M) has no maximal
submodule, M/(N + \mathrm{S}\mathrm{o}\mathrm{c}(M)) also has no maximal submodule, therefore M = N + \mathrm{S}\mathrm{o}\mathrm{c}(M). It
follows that M = N \oplus N \prime for any submodule N \prime such that \mathrm{S}\mathrm{o}\mathrm{c}(M) =
\bigl(
N \cap \mathrm{S}\mathrm{o}\mathrm{c}(M)
\bigr)
\oplus N \prime .
A supplemented module is co-coatomically supplemented but a co-coatomically supplemented
modules need not be supplemented as it is shown in the following example.
Example 2.1. The \BbbZ -module \BbbQ is co-coatomically supplemented since the only co-coatomic
submodule is \BbbQ itself. But the \BbbZ -module \BbbQ is not supplemented since \BbbQ is not torsion (see [10],
Theorem 3.1).
Proposition 2.1. Let M be a semilocal module with small radical \mathrm{R}\mathrm{a}\mathrm{d}(M). Then M is co-
coatomically supplemented if and only if M is supplemented.
Proof. Let N be a submodule of M. Since M is semilocal, M/\mathrm{R}\mathrm{a}\mathrm{d}(M) is semisimple, i.e.,
coatomic. Consider the following statement:
M/
\bigl(
N +\mathrm{R}\mathrm{a}\mathrm{d}(M)
\bigr) \sim = \bigl(
M/\mathrm{R}\mathrm{a}\mathrm{d}(M)
\bigr)
/
\bigl(
(N +\mathrm{R}\mathrm{a}\mathrm{d}(M))/\mathrm{R}\mathrm{a}\mathrm{d}(M)
\bigr)
.
Since M/\mathrm{R}\mathrm{a}\mathrm{d}(M) is coatomic, M/(N + \mathrm{R}\mathrm{a}\mathrm{d}(M)) is coatomic. Therefore N + \mathrm{R}\mathrm{a}\mathrm{d}(M) has a
supplement in M, say K. Then M = N + \mathrm{R}\mathrm{a}\mathrm{d}(M) + K and (N + \mathrm{R}\mathrm{a}\mathrm{d}(M)) \cap K \ll K. Since
\mathrm{R}\mathrm{a}\mathrm{d}(M) \ll M, it follows that M = N +K and N \cap K \leq (N +\mathrm{R}\mathrm{a}\mathrm{d}(M)) \cap K \ll K. Thus M is
supplemented.
A co-coatomically supplemented module is cofinitely supplemented but the example below show
that a cofinitely supplemented module need not be co-coatomically supplemented.
A ring R is said to be a semiperfect if R/ \mathrm{J}\mathrm{a}\mathrm{c}(R) is semisimple and idempotents in R/ \mathrm{J}\mathrm{a}\mathrm{c}(R)
can be lifted to R (see [9], 42.6).
A ring is called left perfect if R/ \mathrm{J}\mathrm{a}\mathrm{c}(R) is left semisimple and \mathrm{J}\mathrm{a}\mathrm{c}(R) is right t-nilpotent
(see [9], 43.9).
RR
(\BbbN ) denotes the direct sum of R-module R by index set \BbbN . Note that \BbbN denotes the set of all
positive integers.
Any direct sum of cofinitely supplemented modules is cofinitely supplemented [1] (Corollary 2.4).
Example 2.2. Let p be a prime integer and consider the following ring:
R = \BbbZ (p) =
\Bigl\{ a
b
| a, b \in \BbbZ , b \not = 0, (b, p) = 1
\Bigr\}
which is the localization of \BbbZ at (p). In this case, the R-module R is supplemented. Then the
R-module R(\BbbN ) is cofinitely supplemented by [1] (Corollary 2.4). Furthermore, R is a semiperfect
ring and therefore R/ \mathrm{J}\mathrm{a}\mathrm{c}(R) is semisimple (see [9], 42.6). Hence R is semilocal. However, R
is not a perfect ring since its Jacobson radical is not t-nilpotent by [9] (43.9). \mathrm{R}\mathrm{a}\mathrm{d} (RR
(\BbbN )) is a
co-coatomic submodule of RR
(\BbbN ) but \mathrm{R}\mathrm{a}\mathrm{d} (RR(\BbbN )) does not have supplement in RR
(\BbbN ) since R is
not a perfect ring (see [3], Theorem 1). Hence RR
(\BbbN ) is not co-coatomically supplemented.
Example 2.2 shows that over semiperfect rings and discrete valuation rings, cofinitely supple-
mented modules and co-coatomically supplemented modules need not coincide.
Proposition 2.2. A factor module of a co-coatomically supplemented module is co-coatomically
supplemented.
Proof. Let M be a co-coatomically supplemented module and N be a submodule of M. Then
any co-coatomic submodule of M/N is a submodule of the form L/N where L is co-coatomic
submodule of M. By hypothesis, L has a supplement in M, say K. It follows that (K +N)/N is a
supplement of L/N in M/N by [9] (41.1(7)).
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 7
870 R. ALIZADE, S. GÜNGÖR
Proposition 2.3. Let M be a co-coatomically supplemented module. Then every co-coatomic
submodule of the module M/\mathrm{R}\mathrm{a}\mathrm{d}(M) is a direct summand.
Proof. Any co-coatomic submodule of M/\mathrm{R}\mathrm{a}\mathrm{d}(M) has the form N/\mathrm{R}\mathrm{a}\mathrm{d}(M) where N is a
co-coatomic submodule of M. Since M is co-coatomically supplemented, there exists a submodule
K of M such that M = N +K and N \cap K \ll K. It follows that N \cap K \leq \mathrm{R}\mathrm{a}\mathrm{d}(M). Thus
M/\mathrm{R}\mathrm{a}\mathrm{d}(M) =
\bigl(
N/\mathrm{R}\mathrm{a}\mathrm{d}(M)
\bigr)
+
\bigl(
(K +\mathrm{R}\mathrm{a}\mathrm{d}(M))/\mathrm{R}\mathrm{a}\mathrm{d}(M)
\bigr)
,\bigl(
N/\mathrm{R}\mathrm{a}\mathrm{d}(M)
\bigr)
\cap
\bigl(
(K +\mathrm{R}\mathrm{a}\mathrm{d}(M))/\mathrm{R}\mathrm{a}\mathrm{d}(M)
\bigr)
=
\bigl(
N \cap K +\mathrm{R}\mathrm{a}\mathrm{d}(M)
\bigr)
/\mathrm{R}\mathrm{a}\mathrm{d}(M) = 0.
Hence
M/\mathrm{R}\mathrm{a}\mathrm{d}(M) =
\bigl(
N/\mathrm{R}\mathrm{a}\mathrm{d}(M)
\bigr)
\oplus
\bigl(
(K +\mathrm{R}\mathrm{a}\mathrm{d}(M))/\mathrm{R}\mathrm{a}\mathrm{d}(M)
\bigr)
.
To prove that a finite sum of co-coatomically supplemented modules is a co-coatomically supple-
mented module, we use the following standard lemma (see [9], 41.2).
Lemma 2.1. Let N and L be submodules of an R-module M such that N is co-coatomic, L is
co-coatomically supplemented and N +L has a supplement in M. Then N has a supplement in M.
Proof. Let K be a supplement of N + L in M. Note that
L/
\bigl(
L \cap (N +K)
\bigr) \sim = (N +K + L)/(N +K) = M/(N +K).
The last module is coatomic, therefore there is a supplement H of L \cap (N +K) in L, i.e.,
L = H + L \cap (N +K) and H \cap L \cap (N +K) \ll H.
Now
M = N + L+K = N +K +H + L \cap (N +K) = N +K +H,
N \cap (H +K) \leq H \cap (N +K) +K \cap (N +H) \leq
\leq H \cap (N +K) +K \cap (N + L) \ll H +K.
Therefore H +K is a supplement of N in M.
A (direct) sum of infinitely many co-coatomically supplemented modules need not be co-
coatomically supplemented by Example 2.2 but a finite sum of co-coatomically supplemented modules
is always co-coatomically supplemented.
Theorem 2.2. A finite sum of co-coatomically supplemented modules is co-coatomically supple-
mented.
Proof. Clearly it is sufficient to prove that the sum M = M1 + M2 of two co-coatomically
supplemented modules M1 and M2 is co-coatomically supplemented. Let U be a co-coatomic
submodule of M. Then M = M1 +M2 + U. Since M2 + U is co-coatomic submodule of M and
M1 is co-coatomically supplemented, M2 + U has a supplement in M by Lemma 2.1. Since M2 is
co-coatomically supplemented and U is co-coatomic, then again by Lemma 2.1, U has a supplement
in M. Thus M is co-coatomically supplemented.
Let M and N be R-modules. If there is an epimorphism f : M (\Lambda ) \rightarrow N for some finite set \Lambda ,
then N is called a finitely M -generated module.
The following corollary follows from Proposition 2.2 and Theorem 2.2.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 7
CO-COATOMICALLY SUPPLEMENTED MODULES 871
Corollary 2.1. If M is co-coatomically supplemented module, then any finitely M -generated
module is a co-coatomically supplemented module.
A ring R is called a left V -ring if every simple R-module is injective (see [9, p. 192]). A
commutative ring R is a V -ring if and only if R is a von Neumann regular ring (see [9], 23.5).
Proposition 2.4. A module M over a V -ring R is co-coatomically supplemented if and only if
M is semisimple.
Proof. (\Leftarrow ) Clear.
(\Rightarrow ) Since M is a co-coatomically supplemented module, M/ \mathrm{S}\mathrm{o}\mathrm{c}(M) has no maximal submo-
dule by Theorem 2.1. It follows from [9] (23.1) that M/\mathrm{S}\mathrm{o}\mathrm{c}(M) = \mathrm{R}\mathrm{a}\mathrm{d}(M/\mathrm{S}\mathrm{o}\mathrm{c}(M)) = 0 since R
is a V -ring. Thus M is semisimple.
Corollary 2.2. Over a left V -ring, any direct sum co-coatomically supplemented modules is
co-coatomically supplemented.
Proof. By Proposition 2.4, co-coatomically supplemented and semisimple modules coincide over
left V -rings.
Theorem 2.3. Let N be a co-coatomically supplemented submodule of an R-module M such
that M/N has no maximal submodule. Then M is a co-coatomically supplemented module.
Proof. Let L be a submodule of M such that M/L is coatomic. Clearly M/(N + L) is also
coatomic. Since M/N has no maximal submodule, M/(N + L) also has no maximal submodule,
therefore M = N + L. By Lemma 2.1, L has a supplement in M. Thus M is a co-coatomically
supplemented module.
The following corollary is a direct result of Theorem 2.3.
Corollary 2.3. Let M be a module and M/ \mathrm{S}\mathrm{o}\mathrm{c}(M) have no maximal. Then M is co-
coatomically supplemented.
Proposition 2.5. Let M be a co-coatomically supplemented R-module. If M contains a maximal
submodule, then M contains a local submodule.
Proof. Let L be a maximal submodule of M. Then L is a co-coatomic submodule of M. Since
M is a co-coatomically supplemented module, there exists a submodule K of M such that K is a
supplement of L in M, i.e., M = K + L and K \cap L \ll K. It follows from [9] (41.1(3)) that K is
local.
A module M is called linearly compact if for every family of cosets \{ xi +Mi\} \bigtriangleup , xi \in M, and
submodules Mi \leq M (with M/Mi finitely cogenerated), the intersection of any finitely many of
these cosets is not empty, then the intersection is also not empty (see [9], 29.7(c)).
The following proposition gives a characterization of a co-coatomically supplemented module by
a linearly compact submodule.
Proposition 2.6. Let K be a linearly compact submodule of an R-module M. Then M is
co-coatomically supplemented if and only if M/K is co-coatomically supplemented.
Proof. (\Rightarrow ) By Proposition 2.2.
(\Leftarrow ) Let N be a co-coatomic submodule of M. Then (N +K)/K is co-coatomic submodule of
M/K since N +K is co-coatomic submodule of M. Since M/K is co-coatomically supplemented,
(N + K)/K has a supplement in M/K. K has a supplement in every submodule L of M with
K \leq L since K is linearly compact (see [8], Lemma 2.3). K is supplemented by [9] (29.8(2))
and [8] (Lemma 2.3). Therefore N has a supplement in M by [8] (Corollary 2.7). Thus M is
co-coatomically supplemented.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 7
872 R. ALIZADE, S. GÜNGÖR
Remark 2.1. A module M is called \Sigma -selfprojective if for each index set I, the module M (I)
is selfprojective. For an R-module M, if M is \Sigma -selfprojective and U \leq \mathrm{R}\mathrm{a}\mathrm{d}(M), then the
following holds: U has a supplement in M, so U is small in M [11] (Satz 4.1). Clearly RR
(\BbbN ) is
\Sigma -selfprojective and \mathrm{R}\mathrm{a}\mathrm{d} (RR
(\BbbN )) \leq \mathrm{R}\mathrm{a}\mathrm{d} (RR
(\BbbN )), therefore if \mathrm{R}\mathrm{a}\mathrm{d} (RR
(\BbbN )) has a supplement in
RR
(\BbbN ), then \mathrm{R}\mathrm{a}\mathrm{d} (RR
(\BbbN )) \ll RR
(\BbbN ).
Theorem 2.4. Every left R-module is co-coatomically supplemented if and only if the ring R is
left perfect.
Proof. (\Leftarrow ) Clear.
(\Rightarrow ) By hypothesis, every left R-module is co-coatomically supplemented, so every left R-
module is cofinitely supplemented. Then R is semiperfect by [1] (Theorem 2.13). Therefore
R/ \mathrm{J}\mathrm{a}\mathrm{c}(R) is semisimple by [9] (42.6). It follows that RR
(\BbbN )/\mathrm{R}\mathrm{a}\mathrm{d} (RR
(\BbbN )) is semisimple. Thus
\mathrm{R}\mathrm{a}\mathrm{d} (RR
(\BbbN )) is a co-coatomic in RR
(\BbbN ). By hypothesis, \mathrm{R}\mathrm{a}\mathrm{d} (RR(\BbbN )) has a supplement in RR
(\BbbN ). By
Remark 2.1, \mathrm{R}\mathrm{a}\mathrm{d} (RR(\BbbN )) \ll RR
(\BbbN ). Since R/ \mathrm{J}\mathrm{a}\mathrm{c}(R) is semisimple and \mathrm{R}\mathrm{a}\mathrm{d} (RR
(\BbbN )) \ll RR
(\BbbN ),
RR is perfect by [9] (43.9). Thus the ring R is left perfect.
3. Co-coatomically supplemented modules over discrete valuation rings. Throughout this
section R will be a discrete valuation ring. An R-module M is called radical-supplemented if
\mathrm{R}\mathrm{a}\mathrm{d}(M) has a supplement in M (see [11]). A module M is radical supplemented if and only if the
basic submodule of M is coatomic (see [11], Satz 3.1). A module M is coatomic if and only if M
is reduced and supplemented (see [10], Lemma 2.1).
Proposition 3.1. Let M be an R-module. Then M is co-coatomically supplemented module if
and only if the basic submodule of M is coatomic.
Proof. (\Rightarrow ) M/\mathrm{R}\mathrm{a}\mathrm{d}(M) = M/pM is semisimple and therefore coatomic. Since M is co-
coatomically supplemented module, pM has a supplement. Thus M is a radical-supplemented
module. Then the basic submodule of M is coatomic by [11] (Satz 3.1).
(\Leftarrow ) Let X be a submodule of M such that M/X is coatomic and B be the basic submodule of
M. Then M/(X +B) is also coatomic. Furthermore, M/(X +B) is reduced by [10] (Lemma 2.1).
On the other hand, M/(X + B) is divisible since M/B is divisible. Therefore M/(X + B) = 0,
that is M = X+B. By hypothesis, B is coatomic, so supplemented by [10] (Lemma 2.1). Therefore
X has a supplement in M by Lemma 2.1. Hence M is a co-coatomically supplemented module.
Corollary 3.1. Co-coatomically supplemented modules and radical supplemented modules coin-
cide.
The following corollary follows from [11] (Satz 3.1) and Corollary 3.1.
Corollary 3.2. A module M is co-coatomically supplemented if and only if M = T (M) \oplus X
where the reduced part of T (M) is bounded and X/\mathrm{R}\mathrm{a}\mathrm{d}(X) is finitely generated.
The following properties are given in [11] (Lemma 3.2) for radical-supplemented modules over
a discrete valuation ring. Since co-coatomically supplemented modules and radical-supplemented
modules coincide, clearly they hold for co-coatomically supplemented modules.
Corollary 3.3. For an R-module M the following hold:
1. The class of co-coatomically supplemented modules is closed under pure submodules and
extensions.
2. If M is co-coatomically supplemented and M/U is reduced, then U is also co-coatomically
supplemented.
3. Every submodule of M is co-coatomically supplemented if and only if T (M) is supplemented
and M/T (M) has a finite rank.
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CO-COATOMICALLY SUPPLEMENTED MODULES 873
4. Co-coatomically supplemented modules over nonlocal Dedekind domains. Throughout
this section R will be a nonlocal Dedekind domain unless otherwise stated.
Theorem 4.1. Let R be a Dedekind domain and M be an R-module. M is a module whose
co-coatomic submodules are direct summand if and only if
1) T (M) = M1 \oplus M2 where M1 is semisimple and M2 is divisible,
2) M/T (M) is divisible.
Proof. By Theorem 2.1 and [4] (Theorem 6.11).
A submodule N of a module M has(is) a weak supplement in M if M = N+K and N\cap K \ll M
for some submodule K of M. Clearly every supplement is a weak supplement.
Recall that over an arbitrary ring R, a module M is called co-coatomically weak supplemented
if every co-coatomic submodule has a weak supplement in M.
Proposition 4.1. Over an arbitrary ring, a small cover of a co-coatomically weak supplemented
module is co-coatomically weak supplemented.
Proof. Let M be a small cover of a co-coatomically weak supplemented module N. Then
N \sim = M/K for some K \ll M. Take a co-coatomic submodule L of M. (L + K)/K is co-
coatomic submodule of M/K since L+K is co-coatomic submodule of M. By hypothesis, M/K
is co-coatomically weak supplemented so (L+K)/K has a weak supplement in M/K, say X/K.
Since K \ll M, (X \cap L) + K = X \cap (L + K) \ll M (see [5], 2.2(3)). Therefore M = L + X
and L \cap X \ll M, i.e., X is a weak supplement of L in M. Thus M is co-coatomically weak
supplemented.
Proposition 4.2. Over an arbitrary ring, a factor module of a co-coatomically weak supple-
mented module is co-coatomically weak supplemented.
Proof. Let M be a co-coatomically weak supplemented module and N be a submodule of M.
Then any co-coatomic submodule of M/N is a submodule of the form L/N where L is co-coatomic
submodule of M. By hypothesis, L has a weak supplement in M, say K. It follows that (K+N)/N
is a weak supplement of L/N in M/N by [5] (2.2(5)).
Let M be a module and K be a submodule of M. A submodule L of M is called complement of
K in M if it is maximal in the set of all submodules N of M with K \cap N = 0. A submodule L of
M is called a complement submodule if it is a complement of some submodule of M (see [5], 1.9).
A submodule of M is a complement if and only if it is closed (see [5], 1.10). A submodule L of M
is called coclosed in M if L has no proper submodule K for which L/K \ll M/K (see [5], 3.6).
Over a Dedekind domain, a submodule N of M is closed if and only if N is coclosed (see [10],
Lemma 3.3). Over a domain R, torsion submodule T (M) of a module M is a closed submodule
of M (see [7], Example 6.34). Therefore over a Dedekind domain, torsion submodule T (M) of a
module M is a coclosed submodule of M.
Proposition 4.3. Let M be a torsion R-module. Then M is co-coatomically weak supplemented
if and only if it is co-coatomically supplemented.
Proof. (\Leftarrow ) Clear.
(\Rightarrow ) Let K be a submodule of M such that M/K is coatomic. Since M is co-coatomically
weak supplemented K has a weak supplement in M, say N. Then M = K +N and K \cap N \ll M.
Since M is torsion, N is also torsion so it is coclosed. Therefore K \cap N \ll N by [5] (3.7(3)).
Hence M is co-coatomically supplemented.
Let R be a Dedekind domain and \scrP be the set of all maximal ideals of R. For some P \in \scrP , the
submodule
\bigl\{
m \in M | Pnm = 0 for some integer n \geq 1
\bigr\}
is said to be the P -primary component of
M. This submodule is denoted by TP (M).
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 7
874 R. ALIZADE, S. GÜNGÖR
Over a discrete valuation ring, if a module M is torsion, reduced and radical of M has a
supplement in M, then M is bounded (see [10, p. 48], 2nd Folgerung).
Theorem 4.2. Let M be a reduced R-module. If T (M) has a weak supplement in M, then
M is co-coatomically supplemented if and only if M/T (M) is divisible and TP (M) is bounded for
each maximal ideal P.
Proof. (\Rightarrow ) Let M be a co-coatomically supplemented reduced R-module. Then the module
M/T (M) is radical: Suppose K is a maximal submodule of M with T (M) \subseteq K. Since M is co-
coatomically supplemented, K has a supplement, say V. Since K is maximal, V is local, therefore
V is cyclic, i.e., V \sim = R/I (see [9], 41.1(3)). On the other hand, R is nonlocal so I \not = 0, i.e., V
is torsion so V \subseteq T (M), contradiction. Hence M/T (M) has no maximal so M/T (M) is divisible
(see [1], Lemma 4.4). T (M) is closed by [7] (Example 6.34), i.e., it is coclosed by [10] (Lemma
3.3). Since T (M) has a weak supplement, it is a supplement by [5] (20.2). Therefore there is a
submodule N in M such that T (M) +N = M and T (M) \cap N \ll T (M). Then
T (M)/T (M) \cap N \sim = (T (M) +N)/N = M/N.
Since M is co-coatomically supplemented, it is co-coatomically weak supplemented so
T (M)/T (M) \cap N is co-coatomically weak supplemented. By Proposition 4.1, T (M) is co-
coatomically weak supplemented. By Proposition 4.2, TP (M) is also co-coatomically weak sup-
plemented for each P as it is direct summand of T (M). TP (M) is co-coatomically supplemented
module by Proposition 4.3. Thus TP (M) is bounded for each maximal ideal P (see [10, p. 48], 2nd
Folgerung).
(\Leftarrow ) Each TP (M) is bounded so it is supplemented by [10] (Lemma 2.1). Therefore T (M) is
supplemented by [10] (Theorem 3.1). Now let K be a submodule of M such that M/K is coatomic.
Then M/(K+T (M)) is also coatomic. By hypothesis, M/T (M) is divisible, i.e., it has no maximal
submodule (see [1], Lemma 4.4). Therefore M = K + T (M). By Lemma 2.1, K has a supplement
in M. Hence M is co-coatomically supplemented.
Remark 4.1. We see that “if” part of the theorem is true without the condition that “T(M) has a
weak supplement in M ”. We do not know if this condition is necessary for the “only if" part.
Corollary 4.1. Let R be a nonlocal Dedekind domain and M be a reduced R-module. If
\mathrm{R}\mathrm{a}\mathrm{d}(T (M)) \ll T (M), then M is co-coatomically supplemented if and only if M/T (M) is divisible.
Proof. (\Rightarrow ) Clear by the proof of Theorem 4.2.
(\Leftarrow ) By [2] (Corollary 4.1.2.), T (M)/\mathrm{R}\mathrm{a}\mathrm{d}(T (M)) is semisimple, so it is co-coatomically weak
supplemented. Then T (M) is co-coatomically weak supplemented since \mathrm{R}\mathrm{a}\mathrm{d}(T (M)) \ll T (M)
by Proposition 4.1. Therefore T (M) is co-coatomically supplemented by Proposition 4.3. Since
M/T (M) is divisible, M/T (M) has no maximal submodule. Therefore M is co-coatomically
supplemented by Theorem 2.3.
Theorem 4.3. Let R be a nonlocal Dedekind domain and M be a reduced R-module. If
M is co-coatomically amply supplemented then M/T (M) is divisible and TP (M) is bounded for
each P \in \scrP .
Conversely, if M/T (M) is divisible and TP (M) is bounded for each maximal ideal P of R then
M is co-coatomically supplemented.
Proof. Let R be a nonlocal Dedekind domain and M be a co-coatomically amply supplemented
reduced R-module. Then by the proof of Theorem 4.2, M/T (M) is divisible. Now suppose that
TP (M) is not bounded for some P \in \scrP . If basic submodule Bp(M) is bounded then by [6] (Theo-
rem 5), TP (M) = BP (M)\oplus D where D is divisible. Therefore M is not reduced, a contradiction.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 7
CO-COATOMICALLY SUPPLEMENTED MODULES 875
Therefore Bp(M) is not bounded. We will prove that BP (M) is co-coatomically supplemented. Let
K be a co-coatomic submodule of BP (M), i.e., BP (M)/K is coatomic. Therefore BP (M)/K is
bounded by [10, p. 48] (2nd Folgerung). We have the following commutative diagram with exact
rows and columns:
K
i
��
E : 0 // BP (M)
pure
//
\sigma
��
M //
��
X // 0
E\prime : 0 // BP (M)/K
pure
// M \prime // X // 0
Since E is pure E\prime is also pure. Hence E\prime is splitting since BP (M)/K is bounded (see [6],
Theorem 5). By applying \mathrm{E}\mathrm{x}\mathrm{t}, we obtain exact sequence
\rightarrow \mathrm{E}\mathrm{x}\mathrm{t}R(X,K)
i\ast \rightarrow \mathrm{E}\mathrm{x}\mathrm{t}R(X,BP (M))
\sigma \ast \rightarrow \mathrm{E}\mathrm{x}\mathrm{t}R(X,BP (M)/K) \rightarrow .
Since \mathrm{E}\mathrm{x}\mathrm{t}(X,BP (M)/K) = 0, \sigma \ast (E) = 0 and therefore E \in \mathrm{K}\mathrm{e}\mathrm{r}\sigma \ast = \mathrm{I}\mathrm{m} i\ast . Thus there is a
short exact sequence
E\prime \prime : 0 \rightarrow K \rightarrow N \rightarrow X \rightarrow 0
such that i\ast (E\prime \prime ) = E. Therefore we obtain the following diagram:
0
��
0
��
0 // K //
��
N //
��
X // 0
0 // BP (M) //
��
M //
��
X // 0
0 // BP (M)/K
��
BP (M)/K
��
0 0
Without loss of generality, we can assume that K, BP (M) and N are submodules of M. In this
diagram BP (M) \cap N = K, BP (M) +N = M (see [9], Noether isomorphism theorem). Moreover
M/N is coatomic. Since M is co-coatomically amply supplemented there exists a submodule
L of BP (M) such that N + L = M and N \cap L \ll L. Therefore BP (M) = BP (M) \cap (N +
+ L) = L + (BP (M) \cap N) = L + K and L \cap K \leq L \cap N \ll L. Thus K has a supplement
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 7
876 R. ALIZADE, S. GÜNGÖR
in BP (M) and so BP (M) is co-coatomically supplemented. Therefore BP (M) is bounded by
[10, p. 48] (2nd Folgerung). This is a contradiction. Thus TP (M) is bounded for each P \in \scrP .
The converse is clear by Theorem 4.2.
Acknowledgements. The authors would like to express their gratitude to Engin Büyükaşık for
his support during the preparation of this paper.
References
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2001. – 29. – P. 2389 – 2405.
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Univ., 2005.
3. Büyükaşık E., Lomp C. Rings whose modules are weakly supplemented are perfect. Applications to certain ring
extensions // Math. Scand. – 2009. – 105. – P. 25 – 30.
4. Büyükaşık E., Pusat-Yılmaz D. Modules whose maximal submodules are supplements // Hacettepe J. Math. Statist. –
2010. – 39, № 4. – P. 477 – 487.
5. Clark J., Lomp C., Vanaja N., Wisbauer R. Lifting modules. – Birkhäuser-Verlag, 2006.
6. Kaplansky I. Modules over Dedekind rings and valuation rings // Trans. Amer. Math. Soc. – 1952. – 72. – P. 327 – 340.
7. Lam T. Y. Lectures on modules and rings. – Springer, 1999.
8. Smith P. F. Finitely generated supplemented modules are amply supplemented // Arab. J. Sci. Eng. – 2000. – 25. –
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Received 30.04.12,
after revision — 03.04.17
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 7
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| id | umjimathkievua-article-1742 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:11:46Z |
| publishDate | 2017 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/a7/f8655ebe4e0e0203716fcbfeee067ea7.pdf |
| spelling | umjimathkievua-article-17422019-12-05T09:25:34Z Co-coatomically supplemented modules Ко-коатомно поповненi модулi Alizade, R. Güngör, S. Алізаде, Р. Гюнгер, С. It is shown that if a submodule $N$ of $M$ is co-coatomically supplemented and $M/N$ has no maximal submodule, then $M$ is a co-coatomically supplemented module. If a module $M$ is co-coatomically supplemented, then every finitely $M$-generated module is a co-coatomically supplemented module. Every left $R$-module is co-coatomically supplemented if and only if the ring $R$ is left perfect. Over a discrete valuation ring, a module $M$ is co-coatomically supplemented if and only if the basic submodule of $M$ is coatomic. Over a nonlocal Dedekind domain, if the torsion part $T(M)$ of a reduced module $M$ has a weak supplement in $M$, then $M$ is co-coatomically supplemented if and only if $M/T (M)$ is divisible and $TP (M)$ is bounded for each maximal ideal $P$. Over a nonlocal Dedekind domain, if a reduced module $M$ is co-coatomically amply supplemented, then $M/T (M)$ is divisible and $TP (M)$ is bounded for each maximal ideal $P$. Conversely, if $M/T (M)$ is divisible and $TP (M)$ is bounded for each maximal ideal $P$, then $M$ is a co-coatomically supplemented module. Показано, що у випадку, коли субмодуль $N$ модуля $M$ є ко-коатомно поповненим, а $M/N$ не має максимального субмодуля, модуль $M$ є ко-коатомно поповненим. Якщо модуль $M$ є ко-коатомно поповненим, то кожен скiнченно $M$-породжений модуль є ко-коатомно поповненим. Кожний лiвий $R$-модуль є ко-коатомно поповненим тодi i тiльки тодi, коли кiльце $R$ є лiвим досконалим. Поза дискретним метризацiйним кiльцем модуль $M$ є ко-коатомно поповненим тодi i тiльки тодi, коли базовий субмодуль $M$ є коатомним. Поза нелокальною дедекiндовою областю у випадку, коли торсiонна частина $T(M)$ зведеного модуля $M$ має слабке поповнення в $M$, модуль $M$ є ко-коатомно поповненим тодi i тiльки тодi, коли $M/T (M)$ є подiльним, а $TP (M)$ — обмеженим для кожного максимального iдеалу $P$. Поза нелокальною дедекiндовою областю у випадку, коли зведений модуль $M$ є ко-коатомно широко поповненим, $M/T (M)$ є подiльним, а $TP (M)$ — обмеженим для кожного максимального iдеалу $P$. Навпаки, якщо $M/T (M)$ є подiльним, а $TP (M)$ — обмеженим для кожного максимального iдеалу $P$, то модуль $M$ є ко-коатомно поповненим. Institute of Mathematics, NAS of Ukraine 2017-07-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1742 Ukrains’kyi Matematychnyi Zhurnal; Vol. 69 No. 7 (2017); 867-876 Український математичний журнал; Том 69 № 7 (2017); 867-876 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1742/724 Copyright (c) 2017 Alizade R.; Güngör S. |
| spellingShingle | Alizade, R. Güngör, S. Алізаде, Р. Гюнгер, С. Co-coatomically supplemented modules |
| title | Co-coatomically supplemented modules |
| title_alt | Ко-коатомно поповненi модулi |
| title_full | Co-coatomically supplemented modules |
| title_fullStr | Co-coatomically supplemented modules |
| title_full_unstemmed | Co-coatomically supplemented modules |
| title_short | Co-coatomically supplemented modules |
| title_sort | co-coatomically supplemented modules |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1742 |
| work_keys_str_mv | AT alizader cocoatomicallysupplementedmodules AT gungors cocoatomicallysupplementedmodules AT alízader cocoatomicallysupplementedmodules AT gûngers cocoatomicallysupplementedmodules AT alizader kokoatomnopopovnenimoduli AT gungors kokoatomnopopovnenimoduli AT alízader kokoatomnopopovnenimoduli AT gûngers kokoatomnopopovnenimoduli |