The radical formula for noncommutative rings
UDC 512.5 We determine some classes of left modules satisfying the radical formula in a noncommutative ring. We also show that, under a certain condition, a finitely generated module over an $HNP$-ring (the generalization of Dedekind domain) both satisfies the radical formula and can be decompos...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507308564414464 |
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| author | Alkan, M. Ӧneş, O. Алкан, М. Онес, О. |
| author_facet | Alkan, M. Ӧneş, O. Алкан, М. Онес, О. |
| author_sort | Alkan, M. |
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| datestamp_date | 2019-12-05T08:57:49Z |
| description | UDC 512.5
We determine some classes of left modules satisfying the radical formula in a noncommutative ring.
We also show that, under a certain condition, a finitely generated module over an $HNP$-ring (the generalization of Dedekind domain) both satisfies the radical formula and can be decomposed into a direct sum of torsion modules and extending modules. |
| first_indexed | 2026-03-24T02:07:15Z |
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| fulltext |
UDC 512.5
O. Öneş, M. Alkan (Akdeniz Univ., Antalya, Turkey)
THE RADICAL FORMULA FOR NONCOMMUTATIVE RINGS*
РАДИКАЛЬНА ФОРМУЛА ДЛЯ КIЛЕЦЬ, ЩО НЕ КОМУТУЮТЬ
We determine some classes of left modules satisfying the radical formula in a noncommutative ring. We also show that,
under a certain condition, a finitely generated module over an HNP -ring (the generalization of Dedekind domain) both
satisfies the radical formula and can be decomposed into a direct sum of torsion modules and extending modules.
Визначено деякi класи лiвих модулiв, якi задовольняють радикальну формулу в кiльцi, що не комутує. Також
показано, що (за деякої умови) скiнченнопороджений модуль над HNP -кiльцем (що є узагaльненням областi
Дедекiнда) не лише задовольняє радикальну формулу, але й може бути розкладений у пряму суму модулiв кручення
та модулiв розтягу.
1. Introduction. It is well known that the set of nilpotent elements of a commutative ring R with
unity forms an ideal which is equal to the intersection of all the prime ideals. This notion has been
generalized in [9] to modules. Let N be a proper submodule of an R-module M. The radical of
N in M, denoted by \mathrm{r}\mathrm{a}\mathrm{d}M (N), is defined to be the intersection of all prime submodules of M
containing N. If there is no prime submodule containing N, then we put \mathrm{r}\mathrm{a}\mathrm{d}M (N) = M. The
envelope submodule REM (N) of N in M is a submodule of M generated by the set
EM (N) = \{ rm : r \in R and m \in M such that rnm \in N for some n \in \BbbN \} .
Then N is said to satisfy the radical formula in M if \mathrm{r}\mathrm{a}\mathrm{d}M (N) = REM (N). Unfortunately, not
every module satisfies the radical formula. The radical formula and relations between Dedekind
domain and the radical formula were studied in many papers (see, for example, [1, 2, 4, 10, 11,
14 – 16]). Hence, by the use of these concepts, some characterizations for Dedekind domains and
modules were obtained in many results. Unfortunately, in noncommutative case, there are not enough
useful results between the radical formula and an HNP -ring, which is one of the generalizations of
Dedekind domian.
Let R be a noncommutative ring with unity. Then an ideal P of R is called prime if, for
a, b \in R, either a or b is in P whenever aRb \subseteq P. For an ideal I of R, the radical of I, denoted
by \mathrm{R}\mathrm{a}\mathrm{d} (I) , is defined as the intersection of all prime ideals containing I. The radical of an ideal is
characterized by m-system in [6] (Theorem 10.7). A nonempty set S \subseteq R is called an m-system
if, for any a, b \in S, there exists r \in R such that arb \in S. Then, for any ideal I of any ring R, it
follows that
\mathrm{R}\mathrm{a}\mathrm{d} (I) = \{ s \in R : every m-system containing s meets I\} .
The concepts of prime ideals and radicals of a noncommutative ring have been generalized to
modules in [13]. Let M be a module over a noncommutative ring. A proper submodule P of M is
prime if, for any r \in R and m \in M such that rRm \subseteq P, either rM \subseteq P or m \in P. Then similarly,
\mathrm{r}\mathrm{a}\mathrm{d}M (N) is defined as the intersection of all prime submodules containing N for a submodule N
* This research was supported by the Scientific Research Project Administration of Akdeniz University.
c\bigcirc O. ÖNEŞ, M. ALKAN, 2019
ISSN 1027-3190. Укр. мат. журн., 2019, т. 71, № 9 1241
1242 O. ÖNEŞ, M. ALKAN
of M. As in commutative ring theory, it is natural to ask whether the radical of a submodule of a
module over a noncommutative ring has a simple description and there is any relation among the
radical formula, the HNP -ring.
In this paper, we are interested in the radical of a submodule of a module M over a noncom-
mutative ring and describe a submodule WM (N) generated by a strongly nilpotent element on a
submodule N. In fact, this submodule is the generalization of the envelope of a submodule over a
commutative ring. Then we prove that the radical of a projective module has a simple description
and so show that a projective module satisfies the radical formula. Moreover, we give a simple de-
scription for radical of a submodule N of a module M that \mathrm{r}\mathrm{a}\mathrm{d}M (N) = WM (N) = WR(0)M +N
for a ring R such that \=R = R/\mathrm{r}\mathrm{a}\mathrm{d}R(0) is semisimple.
In the last section, we deal with an HNP -ring to show that under a condition, a finitely generated
module over an HNP - ring, the generalization of Dedekind domain, both satisfies the radical formula
and can be decomposed into a direct sum of a torsion module and an extending module. (The notion
of extending modules is one of the other important concepts in module theory and many authors
focus on this topic [5, 12].) Moreover, by using the prime submodule, we also prove that the left
socle of a left extending ring is in the Jacobson radical under the condition.
2. The radical of a submodule. Throughout the paper R will denote a ring with identity and
M be an unital left module over R. We start to prove some properties of a prime submodule of RR.
Lemma 2.1. Let P be a left ideal of R. Then:
(i) If a nonzero homomorphism f of \mathrm{E}\mathrm{n}\mathrm{d}R(R/P ) is injective, then P is a prime submodule
of RR.
(ii) If P is a prime submodule such that xRy \subseteq P whenever xy \in P, then every homomorphism
f of \mathrm{E}\mathrm{n}\mathrm{d}R(R/P ) is injective.
Proof. Let xRy \subseteq P for y \in R and x \in R. If y is not in P, define a homomorphism
f \in \mathrm{E}\mathrm{n}\mathrm{d}R(R/P ) such that f(l + P ) = ly + P so f(xr + P ) = 0 for all r \in R. This means that
xR \subseteq P.
Let f \in \mathrm{E}\mathrm{n}\mathrm{d}R(R/P ). Then there is y \in R such that f(1 +P ) = y+P. If x+P \in \mathrm{K}\mathrm{e}\mathrm{r} f, then
xy \in P and so xRy \subseteq P. Hence, y \in P or xR \subseteq P and so \mathrm{K}\mathrm{e}\mathrm{r} f = 0.
Corollary 2.1. Let P be a left ideal of R such that xRy \subseteq P whenever xy \in P. If P is a prime
submodule of RR, then R/P is an indecomposable R-module.
Proof. Let R/P = A/P \oplus B/P for some A,B \subseteq R. Consider the projection homomorphism
from f(a, b) = a where a = a+ P. Then \mathrm{K}\mathrm{e}\mathrm{r} f = 0 by Lemma 2.1.
Now we define the concept of strongly nilpotent element on a submodule. Let N be a submodule
of a module M. An element x = am of an R-module M on N is called strongly nilpotent if, for
each sequence
\{ ai \in R : ai+1 \in aiRai and a0 = a, i \in \BbbZ \} ,
there is a positive integer k such that akRm \subseteq N. We use the notation SM (N) to denote the set of
strongly nilpotent elements of M on N. In generally, SM (N) does not need to be a submodule. Then
the submodule generated by SM (N) is denoted by WM (N). In the commutative ring, it is easy
prove that WM (N) = REM (N). Therefore, it is a generalization of the envelope of a submodule
and so the radical of a submodule N of a module may not equal to WM (N) (see, for example, [11]).
Similarly, N is said to satisfy the radical formula when WM (N) = \mathrm{r}\mathrm{a}\mathrm{d}M (N). If every submodule
ISSN 1027-3190. Укр. мат. журн., 2019, т. 71, № 9
THE RADICAL FORMULA FOR NONCOMMUTATIVE RINGS 1243
of a module M satisfies the radical formula, then M is said to satisfy the radical formula. If every
R-module M satisfies the radical formula, then R is said to satisfy the radical formula.
Let I be a submodule of RR. It is clear that a is a strongly nilpotent element of RR on I if, for
each sequence
\{ ai \in R : ai+1 \in aiRai and a0 = a, i \in \BbbZ \} ,
there is a positive integer k such that akR \subseteq I. It is also easy to see that every element of a left
T -nilpotent ideal L of R is strongly nilpotent and so L is in WR(0).
Lemma 2.2. Let M be any finitely generated R-module and N, L be submodules of M. Then
WM (N) +WM (L) = M if and only if N + L = M.
Proof. We know that SM (N) \cup SM (L) generates WM (N) + WM (L). Let N + L \not = M.
Since M is a finitely generated R-module, there exists a maximal submodule T of M such that
N + L \subseteq T. Since T is also a prime submodule of M, we have WM (N) \subseteq T and WM (L) \subseteq T
and so WM (N) +WM (L) \subseteq T. This is a contradiction. Then N + L = M.
Since N \subseteq WM (N), L \subseteq WM (L) and N + L = M, it follows that WM (N) +WM (L) = M.
Lemma 2.3. Let M, N be R-modules and B be a submodule of M. Then for an R-module
homomorphism f : M \rightarrow N, f(WM (B)) \subseteq WN (f(B)).
In particular, if f is an epimorphism and \mathrm{k}\mathrm{e}\mathrm{r} f \subseteq B, the inverse inclusion holds.
Proof. Let x = am \in SM (B) where m \in M and a \in R. Then, for each sequence a0, a1, a2, . . .
such that a0 = a and ai+1 \in aiRai, there is a positive integer k such that akRm \subseteq B. Hence,
f(akRm) = akRf(m) \subseteq f(B) and so f(x) \in SN (f(B)). Therefore, f(WM (B)) \subseteq WN (f(B)).
Let x = tn \in SN (f(B)) where n \in N and t \in R. Let n = f(b) for some b \in B. Then, for
each sequence a0, a1, a2, . . . such that a0 = t and ai+1 \in aiRai, there is a positive integer k such
that akRn \subseteq f(B). Thus we get akrn = f(br) where br \in B and r \in R. Then f(akrb - br) = 0
and so, for all r \in R, it follows that akRb \subseteq B. We have x \in f(SM (B)). This means that
SN (f(B)) \subseteq f(SM (B)). Therefore, WN (f(B)) \subseteq f(WM (B)).
Lemma 2.4. Let M be an R-module and let N be a submodule of M. Then
WR(N : M)M \subseteq WM (N).
Proof. Let x = am where a \in SR(N : M) and m \in M. Since a \in SR(N : M), for each
sequence a0, a1, a2, . . . , where a0 = a and ai+1 \in aiRai, there is a positive integer k such that
akR \subseteq (N : M). Therefore, we get that akRM \subseteq N and so akRm \subseteq N. Then am \in SM (N) and
so it follows that WR(N : M)M \subseteq WM (N).
Lemma 2.5. Let N be a submodule of a module M. Then we have WM (N) \subseteq \mathrm{r}\mathrm{a}\mathrm{d}M (N).
Proof. Let x = a0m \in SM (N) and assume that x /\in \mathrm{r}\mathrm{a}\mathrm{d}M (N). Then there exists a prime
submodule P of M containing N such that a0m /\in P . Since P is a prime submodule of M, we
get that a0Ra0m \subseteq P and so there exists an element a1 \in a0Ra0 such that a1m /\in P. Similarly, by
the hypothesis on P, we also get that a1Ra1m \subseteq P. So there exists an element a2 \in a1Ra1 such
that a2m /\in P. Therefore, we obtain the sequence a0, a1, a2, . . . such that a = a0 and ai+1 \in aiRai,
i = 0, 1, 2, 3, . . . , but there does not exist any positive integer k such that akm is in P and so akRm
is not in N. This means that a0m is not a strongly nilpotent element of M on N, a contradiction.
Then SM (N) \subseteq \mathrm{r}\mathrm{a}\mathrm{d}M (N) and the proof is completed.
Theorem 2.1. Let I be a left ideal of a ring R. Then
WR(I) \subseteq \mathrm{r}\mathrm{a}\mathrm{d}R(I) \subseteq \mathrm{r}\mathrm{a}\mathrm{d}R(IR) = \mathrm{R}\mathrm{a}\mathrm{d}(IR) = WR(IR). (2.1)
ISSN 1027-3190. Укр. мат. журн., 2019, т. 71, № 9
1244 O. ÖNEŞ, M. ALKAN
Proof. Since every prime ideal of R is a prime submodule of RR, we get that \mathrm{r}\mathrm{a}\mathrm{d}R(I) \subseteq
\subseteq \mathrm{R}\mathrm{a}\mathrm{d}(IR). Therefore, we have WR(I) \subseteq \mathrm{r}\mathrm{a}\mathrm{d}R(I) \subseteq \mathrm{R}\mathrm{a}\mathrm{d}(IR) and WR(IR) \subseteq \mathrm{r}\mathrm{a}\mathrm{d}R(IR) \subseteq
\subseteq \mathrm{R}\mathrm{a}\mathrm{d}(IR).
Let x /\in SR(IR). Then there is a sequence
\{ xi \in R : xi+1 \in xiRxi and x0 = x, i \in \BbbZ \}
such that, for all positive integer k, xkR \subseteq IR and so xk /\in IR. It is clear that S = \{ x0, x1, x2, . . . \}
is m-system but S does not meet IR. Therefore, we get that x /\in \mathrm{R}\mathrm{a}\mathrm{d}(IR). Thus \mathrm{R}\mathrm{a}\mathrm{d}(IR) \subseteq
\subseteq WR(IR) and this completes the proof.
Corollary 2.2. Let I be a submodule of RR. If I satisfies one of the following conditions:
(i) every element of IR is strongly nilpotent on I,
(ii) I is an ideal of a ring R,
then \mathrm{r}\mathrm{a}\mathrm{d}R(I) = WR(I).
Proof. (i) Let x be in WR(IR). Then, for each sequence \{ xi \in R : xi+1 \in xiRxi and x0 = x,
i \in \BbbZ \} , there is a positive integer k such that xkR \subseteq IR and so xk \in IR. Hence, by the hypothesis,
for each sequence \{ ai \in R : ai+1 \in aiRai and a0 = xk, i \in \BbbZ \} , there is a positive integer t such
that atR \subseteq I. This means that x is strongly nilpotent element on I and so x \in WR(I).
(ii) It is clear from (i).
Theorem 2.2. Let M = Rm be an R-module. Then we have:
(i) WM (0) = W (\mathrm{a}\mathrm{n}\mathrm{n}(m))m,
(ii) if every element of \mathrm{a}\mathrm{n}\mathrm{n}(m)R is strongly nilpotent on \mathrm{a}\mathrm{n}\mathrm{n}(m), then \mathrm{r}\mathrm{a}\mathrm{d}M (0) = WM (0).
Proof. (i) Let M = Rm and I = \mathrm{a}\mathrm{n}\mathrm{n}(m). Then I is an left ideal of R. Let x = rdm \in SM (0).
Then, for each sequence a0, a1, a2, . . . , where a0 = rd and ai+1 \in aiRai, there is a positive
integer k such that akRm = 0 and so akR \subseteq \mathrm{a}\mathrm{n}\mathrm{n}(m). This means that rd is in SR(\mathrm{a}\mathrm{n}\mathrm{n}(m)) and
so WM (0) \subseteq W (\mathrm{a}\mathrm{n}\mathrm{n}(m))m. The converse is clear and we get the first equality.
(ii) Since M = Rm is isomorphic to R/I, it follows that WM (0) is isomorphic to WR/I(0) =
= WR(I)/I. Then similarly, we get that \mathrm{r}\mathrm{a}\mathrm{d}M (0) is isomorphic to \mathrm{r}\mathrm{a}\mathrm{d}R/I(0) = \mathrm{r}\mathrm{a}\mathrm{d}R(I)/I. On
the other hand, by Corollary 2.2, we have that \mathrm{r}\mathrm{a}\mathrm{d}R(I) = WR(I) and so \mathrm{r}\mathrm{a}\mathrm{d}M (0) is isomorphic
to WM (0). Therefore, by Lemma 2.3, we get \mathrm{r}\mathrm{a}\mathrm{d}M (0) = WM (0).
Lemma 2.6. Let M be an R-module and N be a submodule of M. Then we have:
(i) WM/N (0) = (WM (N))/N,
(ii) \mathrm{r}\mathrm{a}\mathrm{d}M/N (0) = \mathrm{r}\mathrm{a}\mathrm{d}M (N)/N.
Proof. (i) It is sufficient to show that SM/N (0) = \{ r(x+N) : rx \in SM (N)\} .
Let \=m = r(m + N) \in SM/N (0). Then, for each sequence a0, a1, a2, . . . , where a0 = r and
ai+1 \in aiRai, there is a positive integer k such that akR \=m = \=0. Thus, akR \=m = (akRm+N)/N = 0
and so akRm \subseteq N. This means that rm \in SM (N). Therefore, \=m = rm + N \in \{ rx + N :
rx \in SM (N)\} .
Let \=m = m +N \in \{ rx +N : rx \in SM (N)\} . We may assume that m = rx \in SM (N). Then,
for each sequence a0, a1, a2, . . . , where a0 = r and ai+1 \in aiRai, there is a positive integer k such
that akRx \subseteq N and so we have akR(x+N) = 0+N. Hence it follows that r(x+N) \in SM/N (0).
(ii) It is clear.
Lemma 2.7. Let M and M\ast be R-modules. Then we have:
(i) WM (0)\oplus WM\ast (0) = WM\oplus M\ast (0),
(ii) \mathrm{r}\mathrm{a}\mathrm{d}M (0)\oplus \mathrm{r}\mathrm{a}\mathrm{d}M\ast (0) = \mathrm{r}\mathrm{a}\mathrm{d}M\oplus M\ast (0).
ISSN 1027-3190. Укр. мат. журн., 2019, т. 71, № 9
THE RADICAL FORMULA FOR NONCOMMUTATIVE RINGS 1245
Proof. It is enough to prove (i).
It is clear that the set A = \{ (ra, 0) : ra \in SM (0)\} (the set B = \{ (0, kb) : kb \in SM\ast (0)\} )
generates the submodule WM (0)\oplus 0 ( 0\oplus WM\ast (0), respectively).
Let x \in A \cup B. We may assume that x = (rm, 0) \in A and so rm \in SM (0). Then, for each
sequence a0, a1, a2, . . . , such that a0 = r and ai+1 \in aiRai, there is a positive integer k such
that akRm = 0 and so akR(m, 0) = 0. Then r(m, 0) \in SM\oplus M\ast (0). This means that WM (0) \oplus
\oplus WM\ast (0) \subseteq WM\oplus M\ast (0).
Let x = r(m,n) \in SM\oplus M\ast (0). Then, for every sequence a0, a1, a2, . . . such that a0 = r and
ai+1 \in aiRai, there is a positive integer k such that akR(m,n) = 0, and it follows that akRm = 0
and akRn = 0. Therefore, r(m,n) = r(m, 0) + r(0, n) \in WM (0) \oplus WM\ast (0). This means that
WM\oplus M\ast (0) \subseteq WM (0)\oplus WM\ast (0).
By using the fact that every module is a homorphic image of a free module and Lemma 2.7, we
get the following result.
Theorem 2.3. Let R be a ring. If every free R-module satisfies the radical formula, then so
does every R-module.
Proposition 2.1. Let M be a projective R-module. Then we have
WR(0)M = WM (0) = \mathrm{r}\mathrm{a}\mathrm{d}M (0) = \mathrm{r}\mathrm{a}\mathrm{d}R(0)M.
Proof. Let M be a projective R-module. Then there exists a free R-module F and an R-
module A such that F = M \oplus A.
Firstly, we prove that our claim is true for F. Let \{ xi : i \in I\} be a basis for F. Then F = \oplus Rxi
and so each x \in F has a unique expansion x =
\sum
i\in I rixi, where ri \in R and almost all ri = 0.
Define a homomorphism \varphi i from F to R by \varphi i(x) = ri. Then \varphi i is an epimorphism for all i \in I
and we obtain x =
\sum
i\in I \varphi i(x)xi.
Let u =
\sum
i\in I rixi \in WF (0) where ri \in R and almost all ri = 0. Thus, u =
\sum
i\in I \varphi i(u)xi
and, by Lemma 2.3, we have u =
\sum
i\in I \varphi i(u)xi \in WR(0)F. Now, we get WF (0) \subseteq WR(0)F and
so WF (0) = WR(0)F.
Take m \in WM (0). By Lemma 2.7, it follows that WF (0) = WM (0) \oplus WA(0) and we have
m \in WF (0) = WR(0)F = WR(0)M \oplus WR(0)A. This implies that m =
\sum
rimi +
\sum
kjaj
where ri, kj \in WR(0), mi \in M and aj \in A. Therefore, we get m =
\sum
rimi \in WR(0)M and
WR(0)M = WM (0).
By the using the similar argument, we get \mathrm{r}\mathrm{a}\mathrm{d}M (0) = \mathrm{r}\mathrm{a}\mathrm{d}R(0)M. Therefore, we complete the
proof since \mathrm{r}\mathrm{a}\mathrm{d}R(0) = WR(0).
Theorem 2.4. Let M/N be a projective R-module and N be a submodule of M. Then we get
\mathrm{r}\mathrm{a}\mathrm{d}M (N) = WM (N) = WR(0)M +N.
Proof. We know that \mathrm{r}\mathrm{a}\mathrm{d}M/N (0) = (\mathrm{r}\mathrm{a}\mathrm{d}M (N))/N and WM (N)/N = WM/N (0). Since M/N
is a projective R-module, we have \mathrm{r}\mathrm{a}\mathrm{d}M/N (0) = WM/N (0) = WR(0)(M/N) and so WM (N)/N =
= (WR(0)M +N)/N. Therefore, we obtain WM (N) = WR(0)M +N = \mathrm{r}\mathrm{a}\mathrm{d}M (N).
Corollary 2.3. Let N be a submodule of an R-module M such that M/N is a projective
R-module and WR(0)M \subseteq N. Then we have \mathrm{r}\mathrm{a}\mathrm{d}M (N) = WM (N) = N.
Proof. It is clear by Theorem 2.4.
Theorem 2.5. Let R be a ring such that \=R = R/\mathrm{r}\mathrm{a}\mathrm{d}R(0) is semisimple and let N be a sub-
module of an R-module M. Then we have \mathrm{r}\mathrm{a}\mathrm{d}M (N) = WM (N) = WR(0)M +N.
ISSN 1027-3190. Укр. мат. журн., 2019, т. 71, № 9
1246 O. ÖNEŞ, M. ALKAN
Proof. Assume that N = 0. Then it will be enough to show that \mathrm{r}\mathrm{a}\mathrm{d}M (0) = WR(0)M. Since \=R
is semisimple, we know that \=M = M/WR(0)M is a semisimple and so WR(0) \=M = \mathrm{r}\mathrm{a}\mathrm{d} \=M (0) = \=0.
On the other hand, since WR(0)M \subseteq \mathrm{r}\mathrm{a}\mathrm{d}R(0)M = \mathrm{r}\mathrm{a}\mathrm{d}M (0), we get
\mathrm{r}\mathrm{a}\mathrm{d} \=M (0) = \mathrm{r}\mathrm{a}\mathrm{d}M (0)/WR(0)M = \=0.
This means that \mathrm{r}\mathrm{a}\mathrm{d}M (0) = WR(0)M.
Now let N \not = 0. Then we obtain
\mathrm{r}\mathrm{a}\mathrm{d}M/N (0) = WR(0)(M/N).
Therefore, \mathrm{r}\mathrm{a}\mathrm{d}M (N) = WR(0)M +N = WM (N).
Corollary 2.4. Let R be a ring such that \=R = R/WR(0) is semisimple. Then R satisfies the
radical formula.
Proof. Let M be any R-module. Then, by Theorem 2.5, we get \mathrm{r}\mathrm{a}\mathrm{d}M (0) = WR(0)M. If
WR(0)M \subseteq WM (0) \subseteq \mathrm{r}\mathrm{a}\mathrm{d}M (0), we have WR(0)M = WM (0) = \mathrm{r}\mathrm{a}\mathrm{d}M (0). This means that M
satisfies radical formula.
3. Modules over \bfitH \bfitN \bfitP -rings. Let M be a finitely generated module over an HNP -ring R
and M = \oplus n
i=1Rmi \oplus K where K is a projective module for some element mi \in M. Assume that
\mathrm{a}\mathrm{n}\mathrm{n}(mi) = \mathrm{a}\mathrm{n}\mathrm{n}(rmi) for all non zero elements r \in R and i \in \{ 1, . . . , n\} .
Theorem 3.1. With the above notations, we have
WR
\Biggl(
n\bigcap
i=1
\mathrm{a}\mathrm{n}\mathrm{n}(mi)
\Biggr) \Biggl(
n\bigoplus
i=1
Rmi
\Biggr)
\oplus WR(0)K = WM (0).
Proof. Let M be a module. Then M = \oplus n
i=1Rmi \oplus K for some elements m1, . . . ,mt of M
and a projective module K.
We show that WR
\bigl( \bigcap n
i=1 \mathrm{a}\mathrm{n}\mathrm{n}(mi)
\bigr) \bigl( \bigoplus n
i=1Rmi
\bigr)
= W(\oplus n
i=1Rmi)(0).
We may assume that n = 2.
Let rm \in S(Rm1\oplus Rm2)(0). Then, for each sequence a0, a1, a2, . . . such that a0 = r and
ai+1 \in aiRai, there is a positive integer k such that akR(d1, e1) = 0 where m = (d1, e1). It follows
akRd1 = 0 and akRe1 = 0 and, by the hypothesis, we get that akR \subseteq \mathrm{a}\mathrm{n}\mathrm{n}(m1) \cap \mathrm{a}\mathrm{n}\mathrm{n}(m2).
Therefore, it follows that r \in SR(\mathrm{a}\mathrm{n}\mathrm{n}(m1) \cap \mathrm{a}\mathrm{n}\mathrm{n}(m2)) and so W(Rm1\oplus Rm2)(0) \subseteq WR(\mathrm{a}\mathrm{n}\mathrm{n}(m1) \cap
\cap \mathrm{a}\mathrm{n}\mathrm{n}(m2))(Rm1 \oplus Rm2).
For the converse, take an element n = a(d1, e1) \in SR(\mathrm{a}\mathrm{n}\mathrm{n}(m1) \cap \mathrm{a}\mathrm{n}\mathrm{n}(m2))(Rm1 \oplus Rm2)
where a \in SR(\mathrm{a}\mathrm{n}\mathrm{n}(m1) \cap \mathrm{a}\mathrm{n}\mathrm{n}(m2)). Then, for each sequence a0, a1, . . . such that a0 = a
and ai+1 \in aiRai, there is a positive integer k such that akR \subseteq \mathrm{a}\mathrm{n}\mathrm{n}(m1) \cap \mathrm{a}\mathrm{n}\mathrm{n}(m2) and so
akR(m1,m2) = 0. Therefore, we get the equality.
By Proposition 2.1, we get
WM (0) = W(\oplus n
i=1Rmi)(0)\oplus WK(0) = W(\oplus n
i=1Rmi)(0)\oplus WR(0)K.
An element m of a module M over a ring R is called a torsion element of the module if there
exists a regular element r of the ring (an element that is neither a left nor a right zero divisor) that
annihilates m (i.e., rm = 0) and if T (M) = M, then M is called torsion module and if T (M) = 0,
then M is called torsion free module.
ISSN 1027-3190. Укр. мат. журн., 2019, т. 71, № 9
THE RADICAL FORMULA FOR NONCOMMUTATIVE RINGS 1247
As well known, the concept of Dedekind domain is important for the commutative ring theory
and it is well known that a Dedekind domain satisfies radical formula. In noncommutative ring
theory, one of its generalization is called an HNP -ring (a right and left Noetherian prime ring in
which every right and every left ideal are a projective module). From [7] (Lemma 7.4), we note
that if M is a finitely generated R-module over an HNP -ring, then M = T (M)\oplus K where K is
isomorphic to M/T (M).
To recall the definition of an extending module, we need the definition of an essential submodule.
A submodule N of a module M is called essential in M if, for all m \in M, there is an element
r \in R such that 0 \not = rm \in N. A submodule N is said to be closed in M if N has no proper
essential extension in M. Let N and K be submodules of M such that N is essential in K. If K
is closed in M, then K is called the closure of N. A module M is called extending if every closed
submodule of M is a direct summand of M.
Lemma 3.1. Let R be a ring such that every element of R is regular and let N be a submodule
of a module M such that T (M) is a submodule of N. Then the closure L of N is of the form
T (M/N) = L/N.
In particular, if M is finitely generated over an HNP -ring, then L is a direct summand of M.
Proof. Let N and K be submodules of M such that N is essential in K. Then it is clear that
K/N is a submodule of T (M/N).
Let N and L be submodules such that T (M) is a submodule of N and L/N = T (M/N). Take
an element 0 \not = x + N \in T (M/N) and so there is an element r \in R such that 0 \not = rx \in N.
Otherwise x+N = 0. Hence, L is an essential extension of N and also L is a closed submodule of
M. Moreover, we derive that M/L is torsion free since M/L is isomorphic to (M/N) /T (M/N).
If M is a finitely generated module over an HNP -ring, then M/L is projective and so L is a direct
summand of a module M.
For any prime ideal P of R, we say that P has the properties S if xRy \subseteq P whenever xy \in P
where x, y \in R. We call that a ring R is HNPS -ring if R is HNP -ring and the zero ideal satisfies
S -property. It is clear that HNPS -ring is a generalization of Dedekind domain.
Corollary 3.1. Let R be a left extending ring. Then:
(i) Let P be a prime submodule of RR with the S -property. Then P is essential in R.
(ii) If every prime ideal has the S -property, then \mathrm{S}\mathrm{o}\mathrm{c} (RR) \subseteq \mathrm{r}\mathrm{a}\mathrm{d}R(0) \subseteq J(R) where J(R) is
the Jacobson radical of R.
Proof. (i) Let P be a prime submodule of RR and L be a closure of P. Then there is a
decomposition R = L \oplus K for a submodule K of R since R is extending. On the other hand, by
Corollary 2.1, it follows that R/P is indecomposable and so K = 0. Hence, P is essential in R.
(ii) It is clear since \mathrm{S}\mathrm{o}\mathrm{c}(RR) is the intersection of essential left ideals of R.
Theorem 3.2. Let R be an HNPS -ring. Then a finitely generated module is the direct sum of
a torsion module and an extending module.
Proof. By [7] (Lemma 7.4), it is known that M = T (M) \oplus K where K is isomorphic to
M/T (M) if M is a finitely generated R-module over an HNP -ring. Let A be a submodule of K.
Then L is a closure of A where T (K/A) = L/A and so L is a summand of K. This means that K
is an extending module.
Theorem 3.3. Let M be a finitely generated module over an HNP -ring R. If for any element
m of M, every element of \mathrm{a}\mathrm{n}\mathrm{n}(m)R is strongly nilpotent on \mathrm{a}\mathrm{n}\mathrm{n}(m), then M satisfies the radical
formula.
ISSN 1027-3190. Укр. мат. журн., 2019, т. 71, № 9
1248 O. ÖNEŞ, M. ALKAN
Proof. Let M be a finitely generated module over an HNP -ring R. By [3] (Lemma 7), we get
that M is direct sum of cyclic modules and projective module. Then, by the use of Theorem 2.2 and
Proposition 2.1, the proof is completed.
In the commutative theory, every module over a Dedekind domain satisfies the radical formula.
Therefore, we wonder whether the condition in Theorem 3.3 can be removed or not.
References
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P. 601 – 611.
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Received 09.08.16
ISSN 1027-3190. Укр. мат. журн., 2019, т. 71, № 9
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| id | umjimathkievua-article-1512 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:07:15Z |
| publishDate | 2019 |
| publisher | Institute of Mathematics, NAS of Ukraine |
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| resource_txt_mv | umjimathkievua/0d/bb9b7da35a5fba4c7a305b8aedf3900d.pdf |
| spelling | umjimathkievua-article-15122019-12-05T08:57:49Z The radical formula for noncommutative rings Радикальна формула для кiлець, що не комутують Alkan, M. Ӧneş, O. Алкан, М. Онес, О. UDC 512.5 We determine some classes of left modules satisfying the radical formula in a noncommutative ring. We also show that, under a certain condition, a finitely generated module over an $HNP$-ring (the generalization of Dedekind domain) both satisfies the radical formula and can be decomposed into a direct sum of torsion modules and extending modules. УДК 512.5 Визначено деякі класи лівих модулів, які задовольняють радикальну формулу в кільці, що не комутує. Також показано, що (за деякої умови) скінченнопороджений модуль над $HNP$-кiльцем (що є узагaльненням області Дедекінда) не лише задовольняє радикальну формулу, але й може бути розкладений у пряму суму модулів кручення та модулів розтягу. Institute of Mathematics, NAS of Ukraine 2019-09-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1512 Ukrains’kyi Matematychnyi Zhurnal; Vol. 71 No. 9 (2019); 1241-1248 Український математичний журнал; Том 71 № 9 (2019); 1241-1248 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1512/496 Copyright (c) 2019 Alkan M.; Ӧneş O. |
| spellingShingle | Alkan, M. Ӧneş, O. Алкан, М. Онес, О. The radical formula for noncommutative rings |
| title | The radical formula for noncommutative rings |
| title_alt | Радикальна формула для кiлець, що не комутують |
| title_full | The radical formula for noncommutative rings |
| title_fullStr | The radical formula for noncommutative rings |
| title_full_unstemmed | The radical formula for noncommutative rings |
| title_short | The radical formula for noncommutative rings |
| title_sort | radical formula for noncommutative rings |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1512 |
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