Preradicals and submodules
Some collections of submodules of a module defined by certain conditions are studied.
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| Цитувати: | Preradicals and submodules / Y. Maturin // Algebra and Discrete Mathematics. — 2010. — Vol. 10, № 1. — С. 88–96. — Бібліогр.: 4 назв. — англ. |
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irk-123456789-1548742019-06-17T01:31:51Z Preradicals and submodules Maturin, Y. Some collections of submodules of a module defined by certain conditions are studied. 2010 Article Preradicals and submodules / Y. Maturin // Algebra and Discrete Mathematics. — 2010. — Vol. 10, № 1. — С. 88–96. — Бібліогр.: 4 назв. — англ. 1726-3255 000 Mathematics Subject Classification:16D90, 16D10. http://dspace.nbuv.gov.ua/handle/123456789/154874 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Some collections of submodules of a module defined by certain conditions are studied. |
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Article |
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Maturin, Y. |
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Maturin, Y. Preradicals and submodules Algebra and Discrete Mathematics |
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Maturin, Y. |
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Maturin, Y. |
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Preradicals and submodules |
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Preradicals and submodules |
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Preradicals and submodules |
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Preradicals and submodules |
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Preradicals and submodules |
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preradicals and submodules |
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Інститут прикладної математики і механіки НАН України |
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2010 |
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http://dspace.nbuv.gov.ua/handle/123456789/154874 |
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Preradicals and submodules / Y. Maturin // Algebra and Discrete Mathematics. — 2010. — Vol. 10, № 1. — С. 88–96. — Бібліогр.: 4 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT maturiny preradicalsandsubmodules |
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2025-07-14T06:56:28Z |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 10 (2010). Number 1. pp. 88 – 96
c© Journal “Algebra and Discrete Mathematics”
Preradicals and submodules
Yuriy Maturin
Communicated by M. Ya. Komarnytskyj
Abstract. Some collections of submodules of a module
defined by certain conditions are studied.
Throughout the whole text, all rings are considered to be associative
with unit 1 6=0 and all modules are left unitary.
Let R be a ring. The category of left R-modules will be denoted by
R−Mod. We shall write N ≤ M if N is a submodule of M .
Let a ∈ R, I ⊆ R. Put
(I : a) = {x ∈ R|xa ∈ I} .
Let M be an R-module. Let End(M) be the set of all endomorphisms
of the R-module M . A submodule N of M is said to be fully invariant in
case
∀f ∈ End(M) : f(N) ≤ N.
Let N ≤ M and f ∈ End(M). Put
(N : f)M = {x ∈ M |f(x) ∈ N} .
It is clear that (N : f)M ≤ M . Put
End(M)N = {f ∈ End(M)|f(M) ⊆ N} .
Let F (M) be some non-empty collection of submodules of a left R-
module M .
We shall consider the following conditions:
2000 Mathematics Subject Classification: 16D90, 16D10.
Key words and phrases: ring, module, preradical.
Yu. Maturin 89
C1. L ∈ F (M), L ≤ N ≤ M ⇒ N ∈ F (M);
C2. L ∈ F (M), f ∈ End(M) ⇒ (L : f)M ∈ F (M);
C3. N,L ∈ F (M) ⇒ N ∩ L ∈ F (M);
C4. N ∈ F (M), N ∈ Gen(M), L ≤ N ≤ M ∧ ∀g ∈ End(M)N : (L :
g)M ∈ F (M) ⇒ L ∈ F (M);
C5. N,K ∈ F (M), N ∈ Gen(M) ⇒ t(K⊆M)(N) ∈ F (M).
Remark 1. Let F be a non-empty set of left ideals of R.
(1) Then F is a preradical filter if and only if F satisfies C1, C2, C3.
(2) Then F is a radical filter if and only if F satisfies C1, C2, C4.
Proof. (1) (⇒)Let F be a preradical filter.
(C1) This is clear.
(C2) Let f ∈ End(R). Then there exists a ∈ R such that
∀r ∈ R : f(r) = ra.
Therefore for L ∈ F we obtain (L : f)R = {x ∈ R|f(x) ∈ L} = {x ∈
R|xa ∈ L} = (L : a). Since F is a preradical filter, (L : f)R = (L : a) ∈ F .
(C3) This is clear.
(⇐) Let F satisfy (C1), (C2), (C3). Then it satisfies (a2) and (a3) [1,
p.36].
(a1) (See [1, p.34]. Let a ∈ R and L ∈ F . Define f : R → R, where
∀x ∈ R : f(x) = xa.
It is easy to see that f ∈ End(R). Obtain (L : f)R = (L : a). But
(L : f)R ∈ F . Hence (L : a) ∈ F .
(2) (⇒)Let F be a radical filter.
(C1),(C2) This is clear (see (1) (⇒)).
(C4) Let N be a left ideal of R and g ∈ End(R)N . Then there exists
a ∈ R such that
∀r ∈ R : g(r) = ra.
It follows from this that a = 1a = g(1) ∈ N . Taking into account (a5)
(see [1, p.36]), it is obvious that F satisfies C4.
(⇐) Let F satisfy C1, C2, C4. Then it is easy to see that it satisfies
(a1) and (a2).
(a5) Let N ∈ F,L ≤ N ≤ R∧∀a ∈ N : (L : a) ∈ F . Then N ∈ Gen(R)
because R is a generator. Let g ∈ End(R)N . It means that there exists
a ∈ N such that ∀r ∈ R : g(r) = ra. But (L : g)R = (L : a). Therefore
(L : g)R ∈ F . By (C4), L ∈ F . Hence F satisfies (a5).
Remark 2. Let M and H be left R-modules and q : M → H be an
isomorphism, U be a non-empty set of submodules of M and q(U) =
= {q(L)|L ∈ U } . Then q(U) satisfies (Ci) if and only if U satisfies (Ci)
for every i ∈ {1, 2, 3, 4, 5}.
90 Preradicals and submodules
Proof. It is suffice to verify that q(U) satisfies (Ci) if U satisfies (Ci) for
every i ∈ {1, 2, 3, 4, 5}.
(1) Let U satisfy (C1). Consider L ∈ q(U), L ≤ N ≤ H. Hence
q−1(L) ∈ U, q−1(L) ≤ q−1(N) ≤ M . By (C1), q−1(N) ∈ U . Whence
N = q(q−1(N)) ∈ q(U).
(2) Let U satisfy (C2). Consider L ∈ q(U), f ∈ End(H). Hence
q−1fq ∈ End(M), q−1(L) ∈ U . By (C2), q−1(L : f)H = (q−1(L) :
q−1fq)M ∈ U . Hence (L : f)H ∈ q(U).
(3) Let U satisfy (C3). Consider N,L ∈ q(U). Hence q−1(N), q−1(L) ∈
U . By (C3), q−1(N ∩L) = q−1(N)∩ q−1(L) ∈ U . Therefore N ∩L ∈ q(U).
(4) Let U satisfy (C4) and let
N ∈ q(U), N ∈ Gen(H), L ≤ N ≤ H ∧ ∀g ∈ End(H)N : (L : g)H ∈ q(U).
Then q−1(N) ∈ U, q−1(N) ∈ Gen(M), q−1(L) ≤ q−1(N) ≤ M . Let
f ∈ End(M)q−1(N). Hence qfq−1 ∈ End(H)N . Since ∀g ∈ End(H)N :
(L : g)H ∈ q(U),
q(q−1(L) : f)M = (L : qfq−1)H ∈ q(U).
Hence (q−1(L) : f)M ∈ U . By (C4), q−1(L) ∈ U . Hence L ∈ q(U).
(5) Let U satisfy (C5) and N,K ∈ q(U), N ∈ Gen(H). Hence
q−1(N), q−1(K) ∈ U, q−1(N) ∈ Gen(M).
By (C5), t(q−1(K)⊆M)(q
−1(N)) ∈ U . Hence q(t(q−1(K)⊆M)(q
−1(N))) ∈
q(U). But
q(t(q−1(K)⊆M)(q
−1(N))) = q
∑
g∈End(M)
q−1(N)
g(q−1(K))
=
=
∑
g∈End(M)
q−1(N)
q(g(q−1(K))) =
∑
g∈End(M)
q−1(N)
(qgq−1)(K) =
=
∑
f∈End(H)N
f(K) = t(K⊆H)(N). Therefore t(K⊆H)(N) ∈ q(U).
Remark 3. Let M be a left R-module. Then the sets
{M} and {L|L ≤M}
satisfy (C1), (C2), (C3), (C4), (C5).
Proposition 1. If F(M) satisfies (C1), (C2), (C4), then it satisfies (C5).
Proof. Let N,K ∈ F (M), N ∈ Gen(M). Then
t(K⊆M)(N) =
∑
g∈End(M)N
g(K)
Yu. Maturin 91
(see [3, p.40]).
It is easy to see that
∀h ∈ End(M)N : K ≤ (t(K⊆M)(N) : h)M .
By (C1), it follows from this that
∀h ∈ End(M)N : (t(K⊆M)(N) : h) ∈ F (M).
It is obvious that t(K⊆M)(N) ≤ N . Therefore
N ∈ F (M), N ∈ Gen(M), t(K⊆M)(N) ≤ N ≤ M∧
∧∀h ∈ End(M)N : (t(K⊆M)(N) : h)M ∈ F (M).
Taking into consideration (C4), it follows from this that t(K⊆M)(N) ∈
F (M).
Put
kerF (M) :=
⋂
L∈F (M)
L.
Proposition 2. If F (M) satisfies (C2), then kerF (M) is a fully invariant
submodule of M .
Proof. Let f ∈ End(M),m ∈ kerF (M). By (C2),
⋂
L∈F (M)
L ⊆
⋂
L∈F (M)
(L : f)M .
Then m ∈
⋂
L∈F (M)
(L : f)M . Hence f(m) ∈
⋂
L∈F (M)
L. Thus
f(kerF (M)) ⊆ kerF (M).
Let M be a left R-module. Let r be a preradical in R-Mod. Put
Fr(M) = {L ≤ M |M/L ∈ T (r)} ,
where T (r) = {M ∈ R−Mod|r(M) = M} (see [1, 3]).
Lemma 1. Let N ≤ M and f ∈ End(M). Then f(M) ∩ N = f((N :
f)M )and f(M)/(f(M) ∩N) ∼= M/(N : f)M .
92 Preradicals and submodules
Proof. It is obvious that f(M) ∩N = f((N : f)M ). Let
g :
{
M/(N : f)M → f(M)/(f(M) ∩N);
m+ (N : f)M 7→ f(m) + f(M) ∩N.
Since
m1 + (N : f)M = m2 + (N : f)M ⇒ m1 −m2 ∈ (N : f)M ⇒
⇒ f(m1−m2) ∈ f(M)∩N ⇒ f(m1)+ f(M)∩N = f(m2)+ f(M)∩N,
the correspondence g is a mapping.
It is easy to see that g is an epimorphism.
Let f(m1)+f(M)∩N = f(m2)+f(M)∩N . Then m1−m2 ∈ (N : f)M .
Hence m1 + (N : f)M = m2 + (N : f)M . Thus g is a monomorphism.
Therefore g is an isomorphism.
Theorem 1. Let r be a hereditary preradical in R − Mod and M be a
left R-module. Then the system Fr(M)satisfies (C1), (C2), (C3).
Proof. (C1). Let L ∈ Fr(M), L ≤ N ≤ M . Since M/N ∼= (M/L)/(N/L)
[2],N ∈ Fr(M) because the class T (r) is closed under epimorphic images
[3] ( see also [1, p.36]).
(C2). Let L ∈ Fr(M), f ∈ End(M). Then M/L ∈ T (r). Since the
class T (r) is closed under submodules, (f(M)+L)/L ∈ T (r). But (f(M)+
L)/L ∼= f(M)/(f(M) ∩ L) [2]. Hence f(M)/(f(M) ∩ L) ∈ T (r). Since
f(M)/(f(M) ∩ L) ∼= M/(L : f)M ( See Lemma 1), (L : f)M ∈ Fr(M).
(C3). Let N,L ∈ Fr(M). Then M/N,M/L ∈ T (r). Hence M/N ⊕
M/L ∈ T (r) because the class T (r) is closed under direct sums [1, 3].
Consider the homomorphism w :
{
M → M/N ⊕M/L,
m 7→ (m+N,m+ L).
Then im(w) ∈ T (r) because T (r) is closed under submodules [1, 3].
It is clear that ker(w) = N ∩L. But im(w) ∼= M/ ker(w) [2]. Therefore
N ∩ L ∈ Fr(M) (see also [1, p.36]).
Theorem 2. Let r be a hereditary radical in R −Mod and M be a left
R-module. Then the system Fr(M) satisfies (C1), (C2), (C3), (C4).
Proof. Taking into account Theorem 1, we shall prove only (C4).
Let
N ∈ Fr(M), N ∈ Gen(M), L ≤ N ≤ M∧
∧∀g ∈ End(M)N : (L : g)M ∈ Fr(M).
By Lemma 1,
∀g ∈ End(M)N : (g(M) + L)/L ∼= M/(L : g)M .
Yu. Maturin 93
It follows from ∀g ∈ End(M)N : (L : g)M ∈ Fr(M) that
∀g ∈ End(M)N : M/(L : g)M ∈ T (r).
Therefore
∀g ∈ End(M)N : (g(M) + L)/L ∈ T (r).
Since the class T (r) is closed under direct sums and factor modules
[1, 3],
∑
g∈End(M)N
(g(M) + L)/L ∈ T (r).
Since N ∈ Gen(M),
∑
g∈End(M)N
g(M) = N
(see [2]).
Thus N/L = (N + L)/L =
∑
g∈End(M)N
(g(M) + L)/L ∈ T (r).
Since the class T (r) is closed under extensions [1, 3], taking into
account that N/L ∈ T (r), (R/L)/(N/L) ∼= R/N ∈ T (r), we have that
R/L ∈ T (r).
Thus L ∈ Fr(M).
Corollary 1. Let r be a hereditary radical in R−Mod and M be a left
R-module. Then the system Fr(M)satisfies (C5).
Proof. By Theorem 2, Proposition 1.
Corollary 2. Let r be a hereditary preradical in R −Mod and M be a
left R-module. Then kerFr(M) is a fully invariant submodule of M .
Proof. By Theorem 1, Proposition 2.
Proposition 3. Let M be a left R-module and K be a fully invariant
submodule of M . Then U = {L|K ≤ L ≤ M} satisfies (C1), (C2), (C3).
Proof. (C1) This is clear.
(C2) Let L ∈ U, f ∈ End(M). Since K is a fully invariant submodule
of M , f(K) ≤ K ≤ L. Therefore K ≤ (L : f)M . By (C1), (L : f)M ∈ U .
(C3) This is clear.
Theorem 3. Let M be a left R-module and K be a fully invariant sub-
module of M such that t(K⊆M)(K) = K. Then U = {L|K ≤ L ≤ M}
satisfies (C1), (C2), (C3), (C4).
94 Preradicals and submodules
Proof. By Proposition 3, U satisfies (C1), (C2), (C3).
(C4) Let
N ∈ U ∧ L ≤ N ≤ M ∧ (∀g ∈ End(M)N : (L : g)M ∈ U)
and k be an arbitrary element of K. Since t(K⊆M)(K) = K,
k = p1(k1) + p2(k2) + ...+ ps(ks),
where s ∈ {1, 2, ...}, p1, p2, ..., ps ∈ End(M)K , k1, k2, ..., ks ∈ K. Since
K ≤ N , End(M)K ≤ End(M)N . Thus p1, p2, ..., ps ∈ End(M)N . Now
we obtain that ∀j ∈ {1, 2, ..., s} : (L : pj)M ∈ U . Hence ∀j ∈ {1, 2, ..., s} :
pj(K) ≤ L. It means that ∀j ∈ {1, 2, ..., s} : pj(kj) ∈ L. Taking this into
consideration, we have
k = p1(k1) + p2(k2) + ...+ ps(ks) ∈ L.
Therefore K ≤ L. It follows from this that L ∈ U .
Theorem 4. Let F be a field and V be a vector space over F with
dimF V < ∞. If U 6= {V } is a non-empty set of subspaces of V satisfying
(C1), (C2), (C3), then U = {L|L ≤ V } .
Proof. By Proposition 2, kerU is a fully invariant subspace of V . It is easy
to see that every fully invariant subspace of V is either {0} or V . Since
U 6= {V }, kerU = {0}. Let P = {dimF
(
⋂
L∈D
L
)
|D ⊆ U&|D| < ∞}.
Hence ∅ 6= P ⊆ {0, 1, 2, ...}. Therefore there exists t = minP ∈ {0, 1, 2, ...}.
It follows from this that there exists D0 ⊆ U such that |D0| < ∞ and
dimF
(
⋂
L∈D0
L
)
= t. Hence
∀B ∈ U : t ≤ dimF
⋂
L∈D0
L
∩B
≤ dimF
⋂
L∈D0
L
= t.
Hence ∀B ∈ U : dimF
((
⋂
L∈D0
L
)
∩B
)
= dimF
(
⋂
L∈D0
L
)
&
&
(
⋂
L∈D0
L
)
∩B ≤
⋂
L∈D0
L. It follows from this that ∀B ∈ U :
(
⋂
L∈D0
L
)
∩
∩ B =
(
⋂
L∈D0
L
)
. Then we obtain ∀B ∈ U :
⋂
L∈D0
L ⊆ B. It means
Yu. Maturin 95
that
⋂
L∈D0
L ⊆ kerU . But kerU ⊆
⋂
L∈D0
L. Therefore kerU =
⋂
L∈D0
L. Now
we have that {0} =
⋂
L∈D0
L. Taking into account |D0| < ∞, by (C3),
{0} =
⋂
L∈D0
L ∈ U . Now apply (C1).
Example 1. Let F be a field and V be a vector space over F with
dimF V = k0 and k be a non-finite cardinal number such that k ≤ k0.
Then
Uk = {L|L ≤ V, dimF (V/L) < k}
satisfies (C1), (C2), (C3), (C4), (C5).
Proof. (C1) Let L ∈ Uk, L ≤ N ≤ V . It is obvious that there exists an
epimorphism π : V/L → V/N . Hence V/L = H ⊕ T for some subspaces
H ∼= V/N, T of V/L. It follows from this that dimF (V/N) = dimF (H) ≤
dimF (H ⊕ T ) = dimF (V/L) < k. Whence dimF (V/N) < k. Now we
obtain N ∈ Uk.
(C2) Let L ∈ Uk, f ∈ End(V ). By Lemma 1, V/(L : f)V ∼=
∼= f(V )/(f(V ∩ L). By Corollary 3.7 (3) [2, p.46], f(V )/(f(V ) ∩ L) ∼=
∼= (f(V ) + L)/L. It follows from this that V/(L : f)V ∼= (f(V ) + L)/L.
Since (f(V ) + L)/L ≤ V/L&dimF (V/L) < k, dimF (V/(L : f)V ) < k.
Thus (L : f)V ∈ Uk.
(C3) Let L,M ∈ Uk. Hence dimF (V/L) < k&dimF (V/M) < k. It is
easy to see that
dimF (V/(L ∩M)) ≤ dimF (V/L) + dimF (V/M).
Therefore dimF (V/(L∩M)) < k+k = k [4, p.417]. Thus L∩M ∈ Uk.
(C4) Let N ∈ Uk ∧ L ≤ N ≤ V ∧ (∀g ∈ End(V )N : (L : g)V ∈
Uk). Thus V = N ⊕ W for some subspace W . Consider the following
homomorphism:
g : V → V, g(n+ w) = n, (n ∈ N,w ∈ W ).
Then, by Lemma 1, V/(L : g)V ∼= g(V )/(g(V ) ∩ L) = N/L. Hence
dimF (N/L) = dimF (V/(L : g)V ) < k.
It is obvious that V/L = N/L⊕K for some subspace K. But K ∼=
∼= (V/L)/(N/L) ∼= V/N . Since N ∈ Uk, dimF K = dimF (V/N) < k.
Hence dimF (V/L) = dimF (N/L) + dimF K < k + k = k. Thus L ∈ Uk.
(C5) Now apply Proposition 1.
96 Preradicals and submodules
Definition 1. A non-empty collection F (M)of submodules of a left R-
module M satisfying (C1), (C2), (C3) is said to be a (preradical) filter of
M .
Example 2. Let M be a left R-module and f ∈ End(M) such that
fn(M) is a fully invariant submodule of M for any n ∈ {1, 2, ...}. Then
{L ≤ M |∃n ∈ {1, 2, ...} : fn (M) ⊆ L} is a collection satisfying (C1), (C2),
(C3), (C4), (C5).
Proof. (C1) This is clear.
(C2) Let fn(M) ⊆ L and g ∈ End(M). Then g(fn(M)) ⊆ fn(M).
Hence g(fn(M)) ⊆ L. Therefore fn(M) ⊆ (L : g)M .
(C3) Let fn(M) ⊆ L,fm(M) ⊆ N , and n ≤ m. fm(M) ⊆ fn(M).
Hence fm(M) ⊆ L ∩N .
(C4) Let fm(M) ⊆ N ,N ∈ Gen(M), L ≤ N ≤ M , and
∀g ∈ End(M)N∃n(g) : fn(g)(M) ⊆ (L : g)M .
Then it is easily seen that fm ∈ End(M)N . Put n0 := n(fm). Hence
fn0(M) ⊆ (L : fm)M . Therefore fn0+m(M) ⊆ L. (C5) Apply Proposi-
tion 1.
References
[1] A.I. Kashu, Radicals and torsions in modules, Chisinau: Stiintsa, 1983. 154 p.
[2] F.W. Anderson, K.R. Fuller, Rings and categories of modules, Berlin-Heidelberg-New
York: Springer, 1973. 340 p.
[3] L. Bican, T. Kepka, P. Nemec, Rings, modules, and preradicals, Lect. Notes Appl.
Math. 75, New York, Marcell Dekker, 1982. 241 p.
[4] W. Sierpinski, Cardinal and ordinal numbers, Warsaw: PWN, 2nd edition, 1965.
492 p.
Contact information
Yuriy Maturin Institute of Physics, Mathematics and Com-
puter Science,Drohobych Ivan Franko State
Pedagogical University, Stryjska, 3, Dro-
hobych 82100, Lviv Region, Ukraine
E-Mail: yuriy_maturin@hotmail.com
URL: www.drohobych.net/ddpu/
Received by the editors: 03.11.2010
and in final form 03.11.2010.
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