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arXiv:1202.0377v1 [math.AC] 2 Feb 2012
Modules Satisfying the Prime Radical Condition
and a Sheaf Construction for Modules I∗†‡
M. Behboodia,b§and M. Sabzevaria
aDepartment of Mathematical Sciences, Isfahan University of Technology
P.O.Box: 84156-83111, Isfahan, Iran
bSchool of Mathematics, Institute for Research in Fundamental Sciences (IPM)
P.O.Box: 19395-5746, Tehran, Iran
mbehbo od@cc.iut.ac.ir
sabzevari@math.iut.ac.ir
Abstract
The purpose of this paper and its sequel, is to introduce a new class of modules over a
commutative ring R, called P-radical modules (modules Msatisfying the prime radical
condition “( p
√PM:M) = P” for every prime ideal P ⊇ Ann(M), where p
√PMis the
intersection of all prime submodules of Mcontaining PM). This class contains the
family of primeful modules properly. This yields that over any ring all free modules
and all finitely generated modules lie in the class of P-radical modules. Also, we
show that if Ris a domain (or a Noetherian ring), then all pro jective modules are
P-radical. In particular, if Ris an Artinian ring, then all R-modules are P-radical
and the converse is also true when Ris a Noetherian ring. Also an R-module Mis
called M-radical if ( p
√MM:M) = M; for every maximal ideal M ⊇ Ann(M). We
show that the two concepts P-radical and M-radical are equivalent for all R-modules
if and only if Ris a Hilbert ring. Semisimple P-radical (M-radical) modules are also
characterized. In Part II we shall continue the study of this construction, and as an
application, we show that the sheaf theory of spectrum of P-radical modules (with the
Zariski topology) resembles to that of rings.
∗The research of first author was in part supported by a grant from IPM (No. 90160034).
†Key Words: Prime submodule, prime spectrum, P-radical module, Zariski topology, sheaf of rings,
sheaf of modules.
‡2010 Mathematics Subject Classification: 13C13, 13C99, 13A99, 14A25.
§Corresponding author.
1
1 Introduction
All rings in this article are commutative with identity and modules are unital. For a ring
Rwe denote by dim(R) the classical Krull dimension of Rand for a submodule Nof
an R-module Mwe denote the annihilator of the factor module M/N by (N:M), i.e.,
(N:M) = {r∈R|rM ⊆N}. We call Mfaithful if (0 : M) = 0.
The theory sheaf of rings (with the Zariski toplogy) on the spectrum of prime ideals
of a commutative ring is one of the main tools in Algebraic Geometry. Recall that the
spectrum Spec(R) of a ring Rconsists of all prime ideals of Rand is non-empty. For
each ideal Iof R, we set V(I) = {P ∈ Spec(R) : I⊆ P}. Then the sets V(I), where Iis
an ideal of R, satisfy the axioms for the closed sets of a topology on Spec(R), called the
Zariski topology of R. The distinguished open sets of Spec(R) are the open sets of the
form D(f) = {P ∈ Spec(R) : f6∈ P} = Spec(R)\V(f), where V(f) = V(Rf ). These sets
form a basis for the Zariski topology on Spec(R) (see for examples, Atiyah and Macdonald
[1] and Hartshorne [14]).
We recall that a sheaf of rings OXon a topological space Xis an assignment of a ring
OX(U) to each open set Uin X, together with, for each inclusion U⊆Varestriction
homomorphism resV,U :OX(V)→ OX(U), subject to the following conditions:
(i) OX(∅) = 0.
(ii) resU,U =idU.
(iii) If U⊆V⊆W, then resV ,U ◦resW,V =resW,U .
(iv) For each open cover {Uα}of U⊆Xand for each collection of elements fα∈ OX(Uα)
such that for all α, β if resUα,Uα∩Uβ(fα) = resUβ,Uα∩Uβ(fβ), then there is a unique
f∈ OX(U) such that for all α,fα=resU,Uα(f).
We think of the elements of OX(U) as “functions” defined on U. The restriction
homomorphisms correspond to restricting a function on a big open set to a smaller one.
It is well-known that for any commutative ring R, there is a sheaf of rings on Spec(R),
called the structure sheaf, denoted by OSpec(R), defined as follows: for each prime ideal
Pof R, let RPbe the localization of Rat P. For an open set U⊆Spec(R), we define
OSpec(R)(U) to be the set of functions s:U→`P∈URP, such that s(P)∈RP, for each
P ∈ U, and such that sis locally a quotient of elements of R: to be precise, we require that
for each P ∈ U, there is a neighborhood Vof P, contained in U, and elements a, f ∈R,
such that for each Q ∈ V,f6∈ Q, and s(Q) = a
fin RQ(see for example Hartshorne [14],
for definition and basic properties of the sheaf OSpec(R)).
Let Rbe a ring. Then an R-module M6= 0 is called a prime module if rm = 0 for
m∈M,r∈Rimplies that m= 0 or rM = (0) (i.e., r∈Ann(M)). We call a proper
2
submodule Pof an R-module Mto be a prime submodule of Mif M/P is a prime module,
i.e., whenever rm ∈P, then either m∈Por rM ⊆Pfor every r∈R,m∈M. Thus Pis
a prime submodule (or P-prime submodule) of Mif and only if P= Ann(M/P ) is a prime
ideal of R and M/P is a torsion free R/P-module. This notion of prime submodule was
first introduced and systematically studied in [8, 11] and recently has received a good deal
of attention from several authors; see for examples [3, 5, 6, 15, 19, 20, 21, 23]. The set of
all prime submodules of Mis called the Spectrum of Mand denoted by Spec(M). For any
submodule Nof M, we have a set V(N) = {P∈Spec(M)|(N:M)⊆(P:M)}. Then
the sets V(N), where Nis a submodule of Msatisfy the axioms for the closed sets of a
topology on Spec(M), called the Zariski topology of M(see for examples [16, 17, 22, 25]).
For an R-module Mlet ψ: Spec(M)→Spec(R/Ann(M)) defined by ψ(P) = (P:
M)/Ann(M) for every P∈Spec(M). ψis called the natural map of Spec(M). An
R-module Mis called primeful if either M= (0) or M6= (0) and the natural map
ψ: Spec(M)→Spec(R/Ann(M)) is surjective. This notion of primeful module has been
extensively studied by Lu [18]).
Let Mbe a primeful faithful R-module. In [25], the author obtained an R-module
OX(U) for each open set Uin X= Spec(M) such that OXis a sheaf of modules over X.
In fact, OXis a generalization of the structure sheaf of rings to primeful faithful modules.
The purpose of this paper and its sequel, is to develop the structure sheaf of rings
to a wider class of modules called P-radical modules ((modules Msatisfying the prime
radical condition “( p
√PM:M) = P” for every prime ideal P ⊇ Ann(M), where p
√PM
is the intersection of all prime submodules of Mcontaining PMbut, if Mhas no prime
submodule, then p
√PM=M). In Section 1, we introduce and study P-radical modules.
Several characterizations of P-radical are given in Proposition 2.1. It is shown that the
class of P-radical modules contains the class of primeful R-modules properly (Proposition
2.3 and Example 2.4). This yields over any ring all free modules, all finitely generated
modules and all homogeneous semisimple modules lie in the class of P-radical modules.
Moreover, if Ris a Noetherian ring (or an integral domain), then every projective R-
modules is P-radical (see Theorem 2.5 and Corollary 2.4(iii)). In particular, if Ris an
Artinian ring, then all R-modules are P-radical (Theorem 2.13). Although the converse is
not true in general (Example 2.14), but we show that the converse is true for certain classes
of rings like Noetherian rings and integral domains (see Theorem 2.16 and Corollary 2.17).
Also, an R-module Mis called M-radical module if ( p
√MM:M) = M; for every maximal
ideal Mcontaining Ann(M). Thus, we have the following chart of implications for M:
Mis finitely generated ⇒Mis primeful ⇒Mis P-radical ⇒Mis M-radical
We show that the two concepts P-radical and M-radical are equivalent for all R-modules if
3
and only if Ris a Hilbert ring (Theorem 2.11). Also the two concepts primful and M-radical
are equivalent for all R-modules if and only if dim(R) = 0 (Theorem 2.12). Recall that an
R-module Mis called a multiplication module if for every submodule Nof Mthere exists
an ideal Iof Rsuch that N=IM (see for example [2, 10] for more details). Proposition
2.18, suggests that for a multiplication module Mthe four concepts “finitely generated”,
“primeful, “P-radical” and “M-radical” are equivalent. Also we proof the Proposition 2.19
that is analogue of Nakayamas Lemma for P-radical modules. In Section 3 semisimple
primful modules, semisimple P-Radical modules and semisimple M-Radical modules, are
fully investigated. For instance, it is shown that a semisimple R-module Mis a M-radical
if and only if there exists a submodule Nof Msuch that N∼
=LAnn(M)⊆M∈Max(R)R/M)
(Proposition 3.5.) Also, a semisimple R-module Mis a P-radical if and only if Mis
aM-radical module and R/Ann(M) is a Hilbert ring (Proposition 3.6). In Part II we
shall continue the study of this construction and, as an application, we show that the
sheaf theory of spectrum of P-radical modules (with Zariski topology) resembles to that
of rings.
2P-Radical and M-Radical Modules
Unlike the rings with identity, not every R-module contains a prime submodule; for ex-
ample Zp∞as Z-module does not contain a prime submodule (see [5] or [11]). Let Nbe a
proper submodule of an R-module M. Then the prime radical p
√Nis the intersection of
all prime submodules of Mcontaining Nor, in case there are no such prime submodules,
p
√Nis M. Clearly V(N) = V(p
√N). We note that, for each ideal Iof R,p
√I=√I(the
intersection of all prime ideals of Rcontaining I). The prime radical of submodules are
studied by several authors; see for example [4, 5, 19].
The problem asking which R-module Msatisfies the equality p
√IM =√IM for every
ideal Icontaining Ann(M), had been investigated for finitely generated modules in [19].
Also, in [18], the author, extend the investigation to primeful flat content modules (e.g.
free modules), and primeful flat modules over rings with Noetherian spectrum.
In this article we introduce a slightly different of the above equality; that is the equality
(p
√IM :M) = √Ifor every ideal Icontaining Ann(M). This prime radical condition of
modules play a key role in our investigation for give a sheaf construction for modules.
The following proposition offer several characterizations of R-modules Msatisfies the
equality ( p
√IM :M) = √Ifor every ideal Icontaining Ann(M).
Proposition 2.1 For an R-module Mthe following statements are equivalent:
4
(1) ( p
√IM :M) = √Ifor each ideal I⊇Ann(M).
(2) ( p
√PM:M) = Pfor each prime ideal P ⊇ Ann(M).
(3) √I=TP∈V(IM )(P:M), for every ideal I⊇Ann(M).
(4) P=TP∈V(PM)(P:M), for every prime ideal P ⊇ Ann(M).
Proof. (1) ⇒(2) is trivial.
(2) ⇒(1). Let I⊇Ann(M). Clearly, I⊆(IM :M)⊆(p
√IM :M) and p
√IM :M) is a
radical ideal (i.e., is an intersection of all prime ideals). Thus √I⊆(p
√IM :M). On the
other hand,
(p
√IM :M)⊆\
P∈V(I)
(p
√PM:M) = \
P∈V(I)P=√I .
Thus, ( p
√IM :M) = √I.
(1) ⇒(3). Let I⊇Ann(M). It is easy to check that
√I= ( p
√IM :M) = (( \
P∈V(IM )
P) : M) = \
P∈V(IM )
(P:M).
(3) ⇒(4) is trivial.
(4) ⇒(2). Suppose that Pis a prime ideal of Rsuch that P ⊇ Ann(M). Since
TP∈V(PM)P=p
√PMand TP∈V(PM)(P:M) = (TP∈V(PM)P:M), we conclude that
P= ( p
√PM:M).
Definition 2.2 .Let Mbe an R-module M. Then Mis called a P-radical module if M
satisfies the equivalent conditions listed in the above Proposition 2.1.
The following proposition shows that for any ring R, all primeful modules are P-radical.
Proposition 2.3 Let Rbe a ring. Then any primeful R-module Mis a P-radical module.
Proof. Let P ⊇ Ann(M). Since Mis a primeful module, there exists P∈Spec(M) such
that (P:M) = P. It follows that p
√PM⊆Pand P=√P ⊆ (p
√PM:M). Thus
P=√P ⊆ (p
√PM:M)⊆(P:M) = P,
and hence ( p
√PM:M) = P. Thus Mis a P-radical module.
Let Rbe a ring. It is easy to see that every free R-module Mis primeful. Also in
[16, Theorem 2.2], it is shown that every finitely generated R-module Mis primeful. In
particular, if Ris a domain, then every projective R-module is primeful (see [16, Corollary
4.3]). Thus by using these facts and above proposition, we have the following corollary.
5
Corollary 2.4 Let Rbe a ring.
(i) Every free R-module is P-radical.
(ii) Every finitely generated R-module is P-radical.
(iii) If Ris a domain, then every projective R-module is P-radical.
Next, we show that Corollary 2.4 (iii) is also true when we replace “Ris a domain”
with “Ris a Noetherian ring”.
Theorem 2.5 Let Rbe a Noetherian ring. Then every projective R-module is P-radical.
Proof. Suppose that Mis a projective R-module and P ⊇ Ann(M). We claim that that
PM6=M. If PM=M, then we put A={IDR|IM 6=Mand Ann(M)⊆I⊆ P}.
Then Ann(M)∈ A and so A 6=∅. Since Ris Noetherian, Ahas a maximal element, say
P0. Then P0is a prime ideal of R, otherwise, there exist a,b∈R\P0such that ab ∈ P0. It
follows that (P0+Ra)M= (P0+Rb)M=Mand so M= (P0+Ra)(P0+Rb)M⊆ P0M,
a contradiction. It is well-known that for each projective R-module Mand each ideal I
of R, the factor module M/IM is also projective as an R/I-module. Thus ¯
M:= M/P0M
is projective as an ¯
R:= R/P0-module. Since ¯
Ris a domain, by Corollary 2.4 (iii), ¯
M
is a P-radical ¯
R-module. If r∈R\ P0, then (Rr +P0)M=Mand it follows that
r+P06∈ Ann ¯
R(¯
M), i.e., Ann ¯
R(¯
M) = (0). Since ¯
P:= P/P0is a prime ideal of ¯
Rwe
must have ( p
√¯
P¯
M:¯
M) = ¯
P. But, the equality PM=Mimplies that ¯
P¯
M=¯
Mand
so ( p
√¯
P¯
M:¯
M) = R, a contradiction. Thus PM6=Mand so PMis a proper sub-
module of M. Suppose F=M⊕Lwere Fis a free R-module and Lis a submodule
of F. Clearly PFis a prime submodule of F, i.e., F/PFis a prime R-module and
Ann(F/PF) = P. Since F/PF∼
=M/PM⊕L/PL, we conclude that M/PMis also a
prime R-module with Ann(M/PM) = Ann(F/PF) = P, i.e., PMis a prime submodule
of Mwith (PM:M) = P. Thus ( p
√PM:M) = (PM:M) = P. Therefore Mis a
P-radical R-module.
The following example shows that the converse of Proposition 2.3 is not true in general
(see also next Proposition 2.7). Thus, the class of P-radical modules contains the class of
primeful R-modules properly.
Example 2.6 (See also [18], P. 136, Example 1.) Let M=Lp∈ΩZ/pZas a Z-module
where Ω is the set of prime integers. Clearly Ann(M) = 0 and according to [18], for any
non-zero prime ideal (p) of Z,pM is a (p)-prime submodule of M, but it doesn’t have any
(0)-prime submodule. Thus Mis not a primeful Z-module. Clearly the zero submodule
(0) of Mis an intersection of maximal submodules of Mand hence p
p(0) = (0). Thus
6
(p
p(0)M:M) = ((0) : M) = Ann(M) = (0). Also, if (q) is a nonzero prime (maximal)
ideal of Z, then (q)M=Lq6=p∈ΩZ/pZ6=M. It follows that (q) = ( p
p(q)M:M). Thus
Mis a P-radical module and it is not a primeful module.
Proposition 2.7 Let Rbe a ring. Then there exists an R-module Mwhich is a P-radical
module but it is not a primeful module if and only if there exist prime ideals Pand {Pi}i∈I
of Rsuch that P$Pi(for each i∈I) and P=Ti∈IPi.
Proof. (⇒) Let Mbe an R-module which is a P-radical module but it is not a primeful
module. Assume that Pis a prime ideal of Rsuch that Mhas not any P-prime submodule.
Thus for any prime submodule Nof Mwith PM⊆N, we must have P$PN:= (N:M).
Since Mis a P-radical module,
P= ( p
√PM:M) = ( \
PM⊆N∈Spec(M)
N:M) = \
PM⊆N∈Spec(M)
(N:M) = \
PM⊆N∈Spec(M)PN.
(⇐) Without loss of generality we can assume that P=Ti∈IPiand for each prime ideal
Q%Pof Rthere exists i∈Isuch that Q=Pi. Let M=Li∈IR/Pias R-module.
Clearly Ann(M) = Ti∈IPi=Pand for each j∈I,Nj=Lj6=i∈IR/Piis a Pj-prime
submodule of Mwith PM⊆ PjM⊆Nj. Thus for each prime ideal Pj%Ann(M) = P,
(p
pPjM:M) = Pj(since Pj⊆(p
pPjM:M)⊆(Nj:M) = Pj). On the other hand
(p
√PM:M)⊆(\
i∈I
Ni:M) = \
i∈I
(Ni:M) = \
i∈IPi=P.
Thus Mis a P-radical module. We claim that Mdoesn’t have any P-prime submodule,
for if not, let Nbe a P-prime submodule of Mwith (N:M) = P. Since N6=M, there
exists j∈Isuch that (···,0,1 + Pj,0,···)6∈ N. Since Pj(···,0,1 + Pj,0,···)∈N, we
must have PjM⊆N, i.e., Pj⊆ P, a contradiction. Thus Mis not a primeful R-module.
By Proposition 2.1 and Definition 2.2, an R-module Mis a P-radical module if and
only if ( p
√PM:M) = Pfor each prime ideal P ⊇ Ann(M). Now we have to adapt the
notion of M-radical modules which generalized P-radical modules.
Definition 2.8 .Let Mbe an R-module M. Then Mis called an M-radical module (or
aMaxful module) if ( p
√MM:M) = Mfor each maximal ideal M ⊇ Ann(M).
The following evident proposition offers several other characterizations of M-radical
modules.
7
Proposition 2.9 Let Rbe a ring and Ma nonzero R-module. Then the following state-
ments are equivalent:
(1) Mis an M-radical module.
(2) MM6=Mfor every maximal ideal M ⊇ Ann(M).
(3) PM6=Mfor every prime ideal P ⊇ Ann(M).
(4) For every maximal ideal M ⊇ Ann(M)there exists a maximal submodule Pof M
such that (P:M) = M.
(5) For every maximal ideal M ⊇ Ann(M)there exists a prime submodule Pof Msuch
that (P:M) = M.
Proof. (1) ⇒(2). Clearly for each maximal ideal M ⊇ Ann(M), the equality ( p
√MM:
M) = Mimplies that MM6=M.
(2) ⇒(1). Suppose MM6=Mfor every maximal ideal M ⊇ Ann(M). Then for each
maximal ideal M ⊇ Ann(M), MMis a prime submodule of Mand so p
√MM=MM.
It follows that M ⊆ (p
√MM:M) = (MM:M) = M, i.e., ( p
√MM:M) = M. Thus M
is an M-radical module.
(2) ⇔(3) is clear.
(2) ⇒(5). Suppose that M ⊇ Ann(M) is a maximal ideal. Then MM6=Mimplies that
MMis a prime submodule with (MM:M) = M.
(5) ⇒(2) is clear.
(4) ⇒(5) is clear (since every maximal submodule is a prime submodule).
(5) ⇒(4). Suppose that (P:M) = Mwhere Pis a prime submodule of Mand
M ⊇ Ann(M) is a maximal ideal. Then M/P is an R/M-vector space and hence M/P
has a maximal R/M-subspace say K/P . Clearly K⊆Mis a maximal R-submodule and
(K:M) = M.
A commutative ring Ris called a Hilbert ring, also Jacobson or Jacobson-Hilbert ring,
if every prime ideal of Ris the intersection of maximal ideals. The class of commutative
Hilbert rings is closed under forming finite polynomial rings. For an integral extension
R⊆S, the ring Sis Hilbert if and only if Ris a Hilbert ring. The main interest
in Hilbert rings in commutative algebra and algebraic geometry is their relation with
Hilbert’s Nullstellensatz (see Goldman [12], Cortzen and Small [7], Theorem 1, and [9],
Theorem 4.19, for more ditals).
Next, we will to show that the two concepts P-radical and M-radical are equivalent
for all R-modules if and only if Ris a Hilbert ring. As we have observed earlier some
modules Mhave no prime submodules (for example Zp∞as Z-module) and we call them
primeless. We recall that a module Mover a domain Ris called divisible if rM =Mfor
8
each 0 6=r∈Rand Mis called torsion if Ann(m)6= 0 for each m∈M.
We need the following evident lemma.
Lemma 2.10 Let Rbe a domain. Then any torsion divisible R-module is primeless.
Theorem 2.11 Let Rbe a ring. Then the two concepts P-radical and M-radical are
equivalent for all R-modules if and only if Ris a Hilbert ring.
Proof. (⇒). Let every M-radical module is a P-radical module. To obtain a contradiction,
suppose that Pis a prime ideal of Rsuch that it is not an intersection of maximal ideals,
i.e., P 6=TP⊆M∈M ax(R)M. It is easy to check that every M-radical R/P-module is also
aP-radical R/P-module. Let S=R/Pand Qbe the field of fraction of S. Since P
is not a maximal ideal, Sis not a field and so S6=Q. Suppose Kis a nonzero proper
S-submodule of Q. Then L=Q/K is a torsion divisible S-module and so by Lemma 2.10,
Lis a primeless S-module. We put
M= ( M
M∈Max(S)
S/M)⊕L
as an S-module. Clearly Mis a M-radical S-module, since for each M1∈Max(S),
M1M= ( M
M16=M∈Max(S)
S/M)⊕L
is a prime S-module with (M1M:M) = M1. On the other hand we claim that every
prime S-submodule of Mis of the above form. To see this, let Pbe a prime S-submodule
of M. We claim that LM∈Max(S)S/M 6⊆ P, otherwise we must have M/P ∼
=L/T for
some proper submodule Tof L, a contradiction (since Lis a primeless S-module). Thus
there exists M1∈Max(S) such that (0,··· ,0,1 +M1,0···)6∈ P. Since M1(0,··· ,0,1+
M1,0···)⊆P,M1M⊆N, and hence M1M=N. It follows that
p
p(0) = \
M∈Max(R)MM=L.
Since Ann(SL) = (0), we conclude that Ann(SM) = (0). Thus
(p
p(0)M:M) = ( p
p(0) : M) = (L:M) = ((0) : M
M∈Max(S)
S/M) = \
M∈Max(R)M.
But TM∈Max(R)M 6= (0) since P 6=TP⊆M∈Max(R)M. Thus ( p
p(0)M:M)6= (0), which
is impossible (since Mis a P-radical module). Thus Ris a Hilbert ring.
9
(⇐). Since every P-radical module is a M-radical module, it suffices to show that every
M-radical module is a P-radical module. Assume that Mis an M-radical module and Pis
a prime ideal of Rwith P ⊇ Ann(M). Since Ris a Hilbert ring, P=TP ⊆M∈M ax(R)M.
On the other hand, since Mis an M-radical module, (MM:M) = Mfor each P ⊆ M ∈
Max(R). It follows that
(p
√PM:M)⊆(\
P⊆M∈M ax(R)MM:M) = \
P⊆M∈M ax(R)
(MM:M) = \
P⊆M∈M ax(R)M=P.
Since P ⊆ (PM:M)⊆(p
√PM:M), we conclude that ( p
√PM:M) = P. Thus Ma
P-radical module.
Let Mbe an R-module. By Proposition 2.3 and Proposition 2.9, we have the following
chart of implications for M:
Mis a primeful module ⇒Mis a P-radical module ⇒Mis a M-radical module
In general, none of the implications is reversible. However, for zero dimensional rings,
the implications can be replaced by equivalences, as we shall now show.
Theorem 2.12 Let Rbe a ring. Then the two concepts primful and M-radical are equiv-
alent for all R-modules if and only if dim(R) = 0.
Proof. Assume that the two concepts primful and M-radical are equivalent for all R-
modules. Then by Theorem 2.11, Ris a Hilbert ring. Suppose, contrary to our claim,
that dim(R)≥2 and Pis a non-maximal prime ideal of R. Thus P=TP⊆M∈M ax(R)M.
Therefore, by Proposition 2.7, there exists an R-module Mwhich is a P-radical (M-radical)
module but it is not a primeful module, a contradiction. Thus dim(R) = 0. The converse
is clear.
The following result shows over an Artinian ring, every module is P-radical (conse-
quently, every module is primeful and also every module is M-radical).
Theorem 2.13 Let Rbe an Artinian ring. Then every R-module is a P-radical module.
Proof. Since dim(R) = 0, by Theorem 2.12, the three concepts primful, P-radical and
M-radical are equivalent for all R-modules. Since Ris Artinian, R=R1× ··· × Rn,
where n∈Nand each Riis an Artinian local ring. First give the proof for the case
n= 1, i.e., Ris a local ring with maximal ideal M. Suppose Mis a nonzero R-module.
Then Ann(M)6=Rand so Ann(M)⊆ M. By Proposition 2.9, it is suffices to show that
10
MM6=M. If Ris a domain (field), then M= (0) and so the proof is complete. Hence,
to obtain a contradiction, suppose that Ris not a domain and also MM=M. The
there are nonzero elements a,b∈Rsuch that ab = 0. Thus (Ra)(Rb)M= (0) and so
either RaM 6=Mor RbM 6=M. It follows that A:= {IDR|IM 6=M and I 6= (0)}
is a nonempty set of ideals of R. Since Ris Noetherian, Ahas a maximal element, say
P. Since PM6=M,Pis not a prime (maximal) ideal of R(i.e., P 6=M). Thus there
exist ideals A,Bof Rsuch that P$A,P$Band AB ⊆ P. Thus the equalities
AM =BM =Mimplies that M=ABM ⊆ PM, a contradiction. Therefore, in the
case Ris a local ring every R-module is a P-radical module. Now assume n≥2 and for
each i(1 ≤i≤n) let Mibe the maximal ideal of the local ring Ri. Suppose Mis a
nonzero R-module and Ann(M)⊆ M where Mis a maximal ideal of R. Clearly Mis
the form R1× · ·· × Ri−1× Mi×Ri+1 · ·· × Rnfor some i. Without loss of generality we
can assume that i= 1, i.e., M=M1×R2× · ·· × Rn. Again, by Proposition 2.9, it is
suffices to show that MM= (M1×R2×···×Rn)M6=M. On the contrary, suppose that
(M1×R2×·· · × Rn)M=M. Let I=R1×(0) × · · · ×(0) and J= (0) ×R2× ·· · × Rn.
Then I,Jare ideals of Rwith J= Ann(I). Thus R1∼
=R/J and so ¯
M=IM is an
unitary R1-module (in fact, ¯
M= (R1×0× ··· × 0)Mis an unitary R1-module with
r1¯mdefined to be r1(1,0,···,0) ¯mfor r1∈R1and ¯m∈¯
M). We claim that ¯
M6= (0),
otherwise, R1×(0) × ··· × (0) ⊆Ann(M)⊆ M1×R2× ··· × Rn, a contradiction.
Thus ¯
Mis a nonzero R1-module and so by using case n= 1, we have M1¯
M6=¯
M, i.e.,
(M1×0× ··· × 0)M6= (R1×0× ··· × 0)M. On the other hand, for each m∈M,
(1,0,···,0)m∈M= (M1×R2× · ·· × Rn)M. Thus for each m∈M,
(1,0,···,0)m=
k
X
j=1
(p1j, r2j,···, rnj )mj
for some k∈N,mj∈M,p1j∈ M1,rij ∈Riwere 2 ≤i≤nand 1 ≤j≤k. Thus
(1,0,···,0)m= (1,0,··· ,0)2m=
k
X
j=1
(p1j,0,···,0)mj∈(M1×(0) × · ·· × (0))M,
for each m∈M. It follows that (R1×0× ··· × 0)M⊆(M1×(0) × ··· × (0))M, i.e.,
M1¯
M=¯
M, a contradiction.
The following example shows that in general the converse of Theorem 2.13 is not true.
Example 2.14 Let Kbe a field, D:= K[{xi:i∈N}] (a unique factorization domain)
and R=K[{xi:i∈N}]/({xixj:i, j ∈N}) where ({xixj:i, j ∈N}) is the ideal of D
11
generated by the subset {xixj:i, j ∈N} ⊆ D. Let ¯xk=xk+ ({xixj:i, j ∈N}) for
each k∈Nand M= ({¯xk:k∈N}). The ideal Mis simply the image of the maximal
ideal N= ({xk:k∈N}) of D. Clearly M2= (0) and so Ris a local zero-dimensional
ring, but is not Artinian (Noetherian). Since M2= (0), for each nonzero R-module M,
MM6=M. Thus by Proposition 2.9 and Theorem 2.12, Mis a P-radical module. Thus
every R-module is P-radical, but Ris not Artinian.
A ring Ris called a Max-ring (or a Bass ring) if every nonzero R-module has a maximal
submodule. Also, a ring Ris called a P-ring if every nonzero R-module has a prime
submodule. In [6], Theorem 3.9, it is shown that commutative P-rings coincide with Max-
rings. Moreover, we have the following lemma which is essentially Theorem 2 of [13].
Lemma 2.15 For a commutative ring R, the following conditions are equivalent:
(1) Ris a max ring;
(2) R/J(R)is a regular ring and J(R)is a t-nilpotent ideal.
The following theorem offer several characterizations of Noetherian rings Rover which
every module is P-radical.
Theorem 2.16 Consider the following statements for a ring R:
(1) Ris an Artinian ring.
(2) Every R-module is primeful.
(3) Every R-module is P-radical.
(4) Every R-module is M-radical.
(5) Ris a Max-ring.
(6) Ris a P-ring.
(7) dim(R) = 0.
Then (1) ⇒(2) ⇔(3) ⇔(4) ⇒(5) ⇔(6) ⇒(7). Consequently, when Ris a Noetherian
ring (or a domain), all the six statements are equivalent.
Proof. (1) ⇒(2) is by Theorem 2.13.
(2) ⇒(3) is by Proposition 2.3.
(3) ⇒(4) is clear.
(4) ⇒(5). Assume that every R-module is M-radical. Let Mbe a nonzero R-module.
12
Then Ann(M)6=Rand so there exists a maximal ideal Mof Rsuch that Ann(M)⊆ M.
Since Mis maxful, there exists a prime submodule Pof Mwith (P:M) = M. Thus
M/P is an R/M-module (R/M-vector space), and hence, M/P has a maximal R/M-
submodule, say K/P . It is easy to see that KMis a maximal R-submodule. Thus
every nonzero R-module has a maximal submodule, i.e., Ris a Max-ring.
(5) ⇔(6) is by [6], Theorem 3.9.
(6) ⇒(7). Let Rbe a P-ring. Suppose that Pis a non-maximal prime ideal of R. Set
R′=R/Pand let Kbe the field of fractions of R′. Since R′is not a field, R′6=Kand
Kis a divisible R′-module. It follows that K/R′is a nonzero torsion divisible R′-module.
Thus by Lemma 2.10, K/R′is a primeless R′-module. Now it is easy to see that K/R′is
a primeless R-module. Thus Ris not a P-ring, a contradiction.
(4) ⇒(2). Assuming (4) to hold. We conclude from (4) ⇒(7) that dim(R) = 0. Theorem
2.12 now shows that every R-module is primeful.
Finally, if Ris a Noetherian ring, then dim(R) = 0 if and only if Ris an Artinian ring.
Thus (6) ⇒(1), when Ris a Noetherian ring.
The following is now immediate.
Corollary 2.17 Let Rbe a domain. Then the following statements are equivalent:
(1) Every R-module is primeful.
(2) Every R-module is P-radical.
(3) Every R-module is M-radical.
(4) Ris a field (i.e., Ris an Artinan domain).
The following proposition suggests that for a multiplication module Mthe four con-
cepts “finitely generated”, “primeful, “P-radical” and “M-radical” are equivalent.
Proposition 2.18 Consider the following statements for a nonzero R-module M:
(1) Mis finitely generated.
(2) Mis primeful.
(3) Mis a P-radical module.
(4) (PM:M) = Pfor every prime ideal P ⊇ Ann(M).
(5) Mis a M-radical module.
Then (1) ⇒(2) ⇒(3) ⇒(4) ⇒(5). When Mis a multiplication module, then (5) ⇒(1).
Proof. (1) ⇒(2) is by [18], Proposition 3.8.
13
(2) ⇒(3) is by Proposition 2.3.
(3) ⇒(4). Since P ⊆ (PM:M)⊆(p
√PM:M) for every prime ideal P ⊇ Ann(M), the
proof is clear.
(4) ⇒(5). For every prime ideal P ⊇ Ann(M) the equality (PM:M) = Pimplies that
PM6=M. Thus Mis a M-radical module.
When Mis a multiplication module, then (5) ⇒(1) is by [18], Proposition 3.8.
We conclude this section with the next analogue of Nakayama’s Lemma.
Proposition 2.19 Let Mbe a M-radical R-module. Then Msatisfies the following con-
dition (NAK): If Iis an ideal of Rcontained in the Jacobson radical J(R)such that
IM =M, then M= (0).
Proof. Suppose that M6= (0). Then Ann(M)6=R. If Mis any maximal ideal containing
Ann(M), then I⊆ M and I M =M=MM, a contradiction.
3 Characterization of Semisimple P-Radical Modules
Recall that for a module M, the socle of M(denoted by soc(M)) is the sum of all simple
(minimal) submodules of M. If there are no minimal submodules in Mwe put soc(M) =
(0). Thus Mis a semisimple module if soc(M) = M. Also, a semisimple module Mis
called homogeneous if any two simple submodules of Mare isomorphic. It is easy to check
that an R-module Mis homogeneous semisimple if and only if Ann(M) is a maximal ideal.
Next, we aim to characterize P-radical semisimple modules. First we need the following
definition.
Definition 3.1 .Let Rbe a ring. A semisimple R-module Mis called full semisimple
if for each maximal ideal M ⊇ Ann(M) the simple R-module R/Mmay be embedded in
M(i.e., there exists a submodule Nof Msuch that N∼
=LAnn(M)⊆M∈Max(R)R/M).
Example 3.2 Let M=Lp∈ΩZ/pZas a Z-module where Ω is the set of prime inte-
gers. Clearly Mis a full semisimple Z-module. But the semisimple Z-module M1=
L26=p∈ΩZ/pZis not full semisimple since Ann(M1) = (0) ⊆2Zand Z/2Zis not a sub-
module of M1.
The proof of the following result is straightforward and left to the reader.
14
Proposition 3.3 Let Mbe a semisimple R-module such that Ann(M)is a finite inter-
section of maximal ideals. Then Mis full semisimple. In particular, all finitely generated
semisimple modules as well as all homogenous semisimple modules are full semisimple.
Lemma 3.4 Let Mbe an R-module with nonzero socle. Then Mis a prime module if
and only if Mis a homogeneous semisimple module (i.e., Ann(M)is a maximal ideal).
Proof. Let Mbe a prime module with nonzero socle and let Rm be a simple submod-
ule of M. Then Ann(m) = Ann(M) = Pand hence Pis a maximal ideal of R. Since
Ann(m) = Ann(m′) for each 0 6=m′∈M,Mis a homogeneous semisimple R-module.
The converse is evident.
Now we are in position to show that the two concepts M-radical and full semisimple
are equivalent for semisimple modules.
Proposition 3.5 Let Mbe a semisimple R-module. Then Mis M-radical if and only if
Mis full semisimple.
Proof. Since Mis a semisimple R-module, we can assume that M=Li∈IR/Miwhere
Iis an index set and each Miis a maximal ideal of R.
(⇒). Clearly Ann(M) = Ti∈IMi. Suppose M ⊇ Ann(M) is a maximal ideal of R.
If M 6=Mifor each i∈I, then M(R/Mi) = R/Mifor each i∈I. It follows that
MM=M, contrary to (2) (see Proposition 2.9). Thus M=Mifor some i∈Iand
hence R/Mmay be embedded in M.
(⇐). Suppose M ⊇ Ann(M) is a maximal ideal of R. Since Mis full semisimple, M=Mi
for each i∈Iand so MM6=M. Thus by Proposition 2.9, Mis a M-radical module.
Proposition 3.6 Let Mbe a semisimple R-module. Then the following statement are
equivalent:
(1) Mis a P-radical module.
(2) Mis a M-radical module and R/Ann(M)is a Hilbert ring.
(3) Mis full semisimple and R/Ann(M)is a Hilbert ring.
Proof. Since Mis a semisimple R-module, we can assume that M=Li∈IR/Miwhere
Iis an index set and each Miis a maximal ideal of R.
(1) ⇒(2). Since every P-radical module is M-radical, it is suffices to show that R/Ann(M)
15
is a Hilbert ring. Suppose P ⊇ Ann(M) is a prime ideal of R. Thus ( p
√PM:M) = Pand
hence p
√PM 6=M. We can assume that p
√PM =Tλ∈ΛPλ, where Λ is an index set and
each Pλis a prime submodule of Mcontaining PM. Thus for each Pλ, the factor module
M/Pλis a prime semisimple module and hence by Lemma 3.4, M/Pλis a homogenous
semisimple R-module, i.e., Mλ:= (Pλ:M) is a maximal ideal of R. Thus
P= ( p
√PM:M) = ( \
λ∈Λ
Pλ:M) = \
λ∈Λ
(Pλ:M) = \
λ∈ΛMλ.
Thus every prime ideal P ⊇ Ann(M) is an intersection of maximal ideals of R, i.e.,
R/Ann(M) is a Hilbert ring.
(2) ⇔(3) is by Proposition 3.5.
(2) ⇒(1) is by Theorem 2.11.
Corollary 3.7 Let Rbe a Hilbert ring or a domain with dim(R) = 1. For a semisimple
R-module Mthe following statement are equivalent:
(1) Mis a P-radical module.
(2) Mis a M-radical module.
(3) Mis full semisimple.
Proof. If Ris a Hilbert ring then by Proposition 3.5 and Theorem 2.11, the proof is com-
plete. Thus we can assume that Ris a domain with dim(R) = 1. Since Mis a semisimple
R-module, we can assume that M=Li∈IR/Miwhere Iis an index set and each Miis
a maximal ideal of R.
(1) ⇒(2) is clear.
(2) ⇒(3) is by Proposition 3.5.
(3) ⇒(1). Suppose P ⊇ Ann(M) is a prime ideal of R. Let M ⊇ P be a maximal
ideal. Since Ris a domain with dim(R) = 1, either P= (0) or M=P. If M=P,
then P=Mifor some iand so PM=MiM6=M. Clearly PMis a prime submod-
ule of Mwith (PM:M) = P. It follows that ( p
√PM:M) = (PM:M) = P. Now
assume that P= (0). Then Ann(M) = (0) = Ti∈IMi. Since every proper submodule
of a simisimple module is an intersection of maximal submodules and each maximal sub-
module is a prime submodule, we conclude that p
√PM =p
p(0) = (0). It follows that
(p
p(0)M:M) = ((0) : M) = Ann(M) = (0) = p(0). Thus Mis a P-radical module.
We conclude this paper with the following result that offer several characterizations
for semisimple primeful modules.
16
Corollary 3.8 Let Mbe a semisimple R-module. Then the following statement are equiv-
alent:
(1) Mis a primeful module.
(2) Mis a P-radical module and dim(R/Ann(M)) = 0.
(3) Mis a M-radical module and dim(R/Ann(M)) = 0.
(4) Mis a full semisimple module and dim(R/Ann(M)) = 0.
Proof. (1) ⇒(2). Since every primful module is P-radical, it is suffices to show that
dim(R/Ann(M)) = 0. Suppose P ⊇ Ann(M) is a prime ideal of R. Thus there exists
a prime submodule Pof Msuch that (P:M) = P. Since M/P is a prime semisimple
R-module, by Lemma 3.4, Pis a maximal ideal. Thus dim(R/Ann(M)) = 0.
(2) ⇒(3) is by by Theorem 2.12.
(3) ⇒(4) is by by Proposition 3.5.
Note: In Part II we shall continue the study of this construction.
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