SUBMODULES OF SECONDARY MODULES

Let R be a commutative ring with nonzero identity. Our objective is to investigate representable modules and to examine in particular when submodules of such modules are representable. Moreover, we establish a connection between the secondary modules and the pure-injective, the Σ-pure-injective, and the prime modules.


Introduction.
In this paper, all rings are commutative rings with identity and all modules are unital.The notion of associated prime ideals and the related one of primary decomposition are classical.In a dual way, we define the attached prime ideals and the secondary representation.This theory is developed in the appendix to Section 6 in Matsumura [6] and in Macdonald [5].Now we define the concepts that we will need.
Let R be a ring and let 0 ≠ M be an R-module.Then M is called a secondary module (second module) provided that for every element r of R the homothety M r • → M is either surjective or nilpotent (either surjective or zero).This implies that nilrad(M) = P (Ann(M) = P ) is a prime ideal of R, and M is said to be P -secondary (P -second), so every second module is secondary (the concept of second module is introduced by Yassemi [14]).A secondary representation for an R-module M is an expression for M as a finite sum of secondary modules (see [5]).If such a representation exists, we will say that M is representable.
If R is a ring and N is a submodule of an R-module M, the ideal {r ∈ R : r M ⊆ N} will be denoted by (N : M).Then (0 : M) is the annihilator of M, Ann(M).A proper submodule N of a module M over a ring R is said to be prime submodule (primary submodule) if for each r ∈ R the homothety M/N r • → M/N is either injective or zero (either injective or nilpotent), so (0 : M/N) = P (nilrad(M/N) = P ) is a prime ideal of R, and N is said to be P -prime submodule (P -primary submodule).So N is prime in M if and only if whenever r m ∈ N, for some r ∈ R, m ∈ M, then m ∈ N or r M ⊆ N. We say that M is a prime module (primary module) if zero submodule of M is prime (primary) submodule of M, so N is a prime submodule of M if and only if M/N is a prime module.Moreover, every prime module is primary.
Let R be a ring, and let N be an R-submodule of M. Then N is pure in M if for any finite system of equations over N which is solvable in M, the system is also solvable in N. A module is said to be absolutely pure if every embedding of it into any other modules is pure embedding.A submodule N of an R-module M is called relatively divisible (or an RD-submodule) if r N = N ∩r M for all r ∈ R. Every RD-submodule of a P -secondary module over a commutative ring R is P -secondary (see [2,Lemma 2

.1]).
A module M is pure-injective if and only if any system of equations in M which is finitely solvable in M, has a global solution in M [7,Theorem 2.8].The module N is a pure-essential extension of M if M is pure in N and for all nonzero submodules L of N, if M ∩ L = 0, then (M ⊕ L)/L is not pure in N/L.A pure-injective hull H(M) of a module M is a pure essential extension of M which is pure-injective.Every module M has a pure-injective hull which is unique to isomorphism over M [12].
Given an R-module M and index set I, the direct sum of the family {M i : i ∈ I} where M i = M for each i ∈ I will be denoted by M (I) .Given a module property ᏼ, we will say that a module M is -ᏼ if M (I) satisfies ᏼ for every index set I.
Let R be a commutative ring.An element a ∈ R is said to be regular if there exists b ∈ R such that a = a 2 b, and R is said to be regular if each of its elements is regular.An important property of regular rings is that every module is absolutely pure (see [13,Theorem 37.6]).
Let R be a ring and M an R-module.A prime ideal P of R is called an associated prime ideal of M if P is the annihilator Ann(x) of some x ∈ M. The set of associated primes of M is written Ass(M).For undefined terms, we refer to [6,7].

Secondary submodules.
In general, a nonzero submodule of a representable (even secondary) R-module is not representable (secondary), but we have the following results.Proof.This follows from Lemma 2.1.

Theorem 2.3. Let R be a commutative regular ring. Then every nonzero submodule of a representable R-module is representable.
Proof.Let M be a representable R-module and let M = n i=1 M i be a minimal secondary representation with nilrad(M i ) = P i .There is an element r 1 ∈ P 1 such that r 1 ∉ ∪ n i=2 P i .Otherwise P 1 ⊆ ∪ n i=2 P i , so by [10, Theorem 3.61], P 1 ⊆ P j for some j, and hence P 1 = P j , a contradiction.Thus there exists a positive integer m 1 such that r m 1 1 ∈ Ann(M 1 ) and the module r 1 M i is representable.By using this process for the ideals P 2 ,...,P n−1 , there are integers m 2 ,...,m n−1 and elements for some b i ∈ M, i = 1,...,n.By assumption, there exists t 1 ,...,t n ∈ R such that for each i, s i = s 2 i t i .As 0 ≠ a, s i b i ≠ 0 for some i and s i t i a = s 2 i t i b i = s i b i , so s i N ≠ 0. We can assume that s i 1 N ≠ 0,...,s i k N ≠ 0, where {i 1 ,...,i k } ⊆ {1,...,n}.By a similar argument as above, if a ∈ N, then a = k j=1 s i j t i j a ∈ k j=1 s i j N, and hence N = k j=1 s i j N. Since for each j, where j = 1,...,k, s i j N is pure in the P i j -secondary module M i j , it is P i j -secondary by [2, Lemma 2.1], as required.
Theorem 2.4.Let R be a commutative ring and let N be a prime submodule of secondary R-module of M. Then N is (N : M)-secondary.
Proof.Suppose that M is a P -secondary module over R. Let r ∈ R. If r ∈ P , then r n N ⊆ r n M = 0 for some n.If r ∉ P , then r M = M. Suppose that n ∈ N, so there is an element m ∈ M such that n = r m.As N is a prime submodule of M and N ≠ r M = M, m ∈ N, so r N = N, hence N is P -secondary.
By [4, Lemma 1], the ideal P = (N : M) = {r ∈ R : r M ⊆ N} is prime.Clearly, P ⊆ P .Let s ∈ P .Then s n N = s n M = 0 for some n.There is an element m ∈ M such that m ∉ N and s n m = 0 ∈ N, so s n ∈ P , hence s ∈ P .Thus P = P , as required.Theorem 2.6.Let M be a P -second module over a commutative ring R, and let N be a prime submodule of M. Then every submodule of M properly containing N is an RD-submodule.In particular, it is P -second.
Proof.Let K be a submodule of M properly containing N. Then K/N is a prime submodule of prime and P -second module M/N, so by Proposition 2.5, K/N is an RDsubmodule of M/N.Now the assertion follows from [3, Consequences 18-2.2(c)] and Proposition 2.5.

Lemma 2.7. Let M be a nonzero module over a commutative domain R. Then M is (0)-second if and only if M is (0)-secondary.
Proof.The proof is completely straightforward.
Corollary 2.8.Let R be a commutative ring.
(i) Every Artinian primary module over R is secondary.(ii) Every Noetherian secondary module over R is primary.(iii) Every finitely generated secondary module is primary.every integer n ≥ 1. However no A n is a prime submodule of E, for if m is any positive integer, then p m ∉ (A n : R E) = 0 and a n+m ∉ A n , but p m a m+n = a n ∈ A n .Theorem 2.12.Let R be a Dedekind domain, and let M be an R-module.Then M is 0 ≠ P -second if and only if M is P -prime.
Proof.By Proposition 2.11, it is enough to show that if M is P -prime, then M is P -second.Since (0 : M) = P is a maximal ideal in R, so M is a vector space over R/P , hence M is P -second.Proposition 2.13.Let R be a Dedekind domain.Then any 0 ≠ P -prime R-module is a direct sum of copies of R P /P R P R/P .Proof.By the proof of Proposition 2.11, every element of R − P acts invertibly on M, so the R-module structure of M extends naturally to a structure of M as a module over the localisation R P of R at P .Therefore, we can assume that R is a commutative local Dedekind domain with maximal ideal P = Rp.Let M j denote the indecomposable summand of M, so M j is P -prime.Let m j be a nonzero element of M j , hence (0 : m j ) = (0 : M) = P .Then Rm j R/P is pure in M j since m j is not divisible by p in M j , but by [1, Proposition 1.3], the module R/P is itself pure-injective, so Rm j is a direct summand of M j , and hence M j Rm j , as required.

Pure-injective modules
Proposition 3.1.Let M be a P -secondary module over a commutative ring R. Then H = H(M), the pure-injective hull, is P -secondary.
Proof.Let r ∈ R. If r ∉ P , then r M = M, so M satisfies the sentence for all x there exists y (x = r y), and hence so does H (because any module and its pure-injective hull satisfy the same sentences [7,Chapter 4]).If r ∈ R, then r n M = 0, so M satisfies the sentence for all x (r n x = 0), hence so does in H, as required.
Theorem 3.2.The following conditions are equivalent for a Prufer domain R: (i) the ring R is a Dedekind domain; (ii) every secondary R-module is pure-injective.
Proof.Let R be a Dedekind domain and M a secondary R-module.If Ann(M) = 0, then M is divisible, hence injective.If Ann(M) ≠ 0, then M is a torsion R-module of bounded order, so that M is Σ-pure-injective (see [15]).In both cases, M is Σ-pureinjective (so pure-injective).
Conversely, let R be a Prufer domain with the property that every secondary module is pure-injective.In order to prove that R is Dedekind domain, it suffices to show that every divisible R-module is injective.Let M be a divisible R-module.Then M is secondary, Hence pure-injective.Since R is Prufer, pure-injective modules are RDinjective (see [7]).The embedding of M in its injective envelope E(M) is an RD-pure monomorphism, because for every nonzero r ∈ R we have that M = r M, so that r E(M) ∩ M ⊆ M ⊆ r M. Since M is the RD-injective, M is a direct summand of E(M).Thus M is injective.This shows that R is a Dedekind domain.

Remarks. (i)
There is a module over a commutative regular ring which is injective but not secondary (see [9,Theorem 2.3]).The commutative regular ring R = F × F , F a field, is an Artinian Gorenstein, that is, R is injective (so pure-injective) as an Rmodule.But R is not secondary, because multiplication by (1, 0) is neither nilpotent nor surjective.
(ii) The above consideration thus leads us to the following question: are secondary modules pure-injective?The answer is yes because of the following reason.Every non-Noetherian Prufer domain has secondary modules that are not pure-injective.For instance, every non-Noetherian valuation domain has secondary modules that are not pure-injective.Proposition 3.3.Let M be an R-module.
(i) M is -secondary if and only if M is secondary.(ii) Let M be a direct sum of modules M i (i ∈ I) where for each i, M i is secondary and Ann(M i ) = Ann(M j ) for all i, j ∈ I. Then M is secondary.
Proof.(i) The necessity is immediate by the definition.Conversely, suppose that M is P -secondary.Given an index set J, and let r ∈ R. If r ∈ P , then r n M = 0 for some n, so r n M (J) = 0.If r ∉ P then r M = M, so r M (J) = M (J) , as required.
(ii) Since the annihilators of all direct summands coincide, we can assume that M i is P -secondary (say) for all i ∈ I. Now the proof of (ii) is similar to that (i) and we omit it.

Lemma 2 . 1 .Corollary 2 . 2 .
Let R be a commutative ring and let 0 ≠ N be an RD-submodule of R-module M. Then M is P -secondary if and only if N and M/N are P -secondary.Proof.If M is P -secondary, then N and M/N are P -secondary by [2, Lemma 2.1] and [5, Theorem 2.4], respectively.Conversely, suppose that r ∈ R. If r ∈ P , then r n (M/N) = 0 and r n N = 0 for some n, hence r n M ⊆ N and 0 = r n N = r n M ∩ N = r n M. If r ∉ P , then r M + N = M, r N = N, and N = r N = r M ∩ N, so we have r M = M, as required.Let R be a commutative regular ring, and let 0 ≠ N be a submodule of R-module M. Then M is P -secondary if and only if N and M/N are P -secondary.

Proposition 2 . 5 .
Let R be a commutative ring and let N be a prime submodule of P -second R-module of M. Then N is an RD-submodule of M. Proof.Let r ∈ R. If r ∈ P , then r N ⊆ r M = 0, so r N = N ∩ r M = 0.If r ∉ P , then r M = M, so the homothety M/N r • → M/N is not zero since N is prime.It follows that the above homothety is injective.If a ∈ N ∩r M, then there is b ∈ M such that a = r b.Since r (b + N) = 0, so b ∈ N, hence r N = N ∩ r M, as required.

Corollary 3 . 4 .Corollary 3 . 5 .
Let M be an indecomposable Σ-pure-injective module over a commutative Prufer ring R. Then M is secondary.Proof.Set P = {r ∈ R : Ann M r ≠ 0} and P = ∩ n P n .Then P and P are prime ideals in R by [8, Fact 3.1 and Lemma 2.1].By [8, Fact 3.2], M is either P -secondary or P -secondary, as required.Every Σ-pure-injective module over a Prufer ring is representable.Proof.Suppose M is a Σ-pure-injective module over a commutative Prufer ring R. By [8, page 967], we can write M = M 1 ⊕ ••• ⊕ M m where M i is secondary for all i by Proposition 3.3 and Corollary 3.4, as required.