SOME RESULTS ON PRIME AND PRIMARY SUBMODULES

This article is devoted to study some properties of prime and primary submodules. First we characterize prime submodules of free modules and give a primality condition for certain submodules in terms of associated prime ideals. Furthermore, by using symmetric algebra of modules we describe Rees algebras associated to prime sub-modules and provide a computational method to check if some primary submodules of a free module have prime radical.


INTRODUCTION
To extend the concepts of prime and primary ideal from the category of rings to the category of modules has stimulated several authors to show that many, but not all, of the results in the theory of rings are also valid for modules.(See, for example, Refs.[1], [5], [7] .) Before we state some results let us introduce some notation and terminology.Throughout this note all rings are commutative with identity and all modules are unital.Let R be a ring and let M be an R-module.A proper submodule N of M is said to be prime (resp.primary) if for every a ∈ R, the induced homothety h a : M/N → M/N, h a (n) = a•n, is either injective or null (resp.nilpotent).In light of this definition, it turns out that if N is a prime (resp.primary) submodule of M then the set of homotheties of R vanishing on M/N, i.e., (N : M ) = {a ∈ R / aM ⊆ N } = Ann(M/N ) is a prime (resp.primary) ideal of R. Furthermore, if N is a primary submodule of M, the radical of the primary ideal (N : M ), denoted by p (N : M ), is a prime ideal of R formed by all nilpotent homotheties of R on M/N, i.e., p N is a prime submodule of M with p = (N : M ) we shall call N p-prime submodule and if N is a primary submodule of M being p = p (N : M ) we will say that N is a p-primary submodule.Note that a p-primary submodule N of M is p -prime if and only if (N : M ) = p ∈ SpecR.
It is easily seen that a submodule N of M is called to be p -prime (resp.p-primary) if N 6 = M and, given r ∈ R, m ∈ M , then rm ∈ N implies m ∈ N or r ∈ p = (N : M ) (resp.r ∈ p = p (N : M )).A special class of prime submodules of a module over a domain is given by the ones whose prime ideal in R is the ideal (0).According to [2, Definition 2.1], these prime submodules are called 0−submodules and it can be proved that a proper submodule N of a module M over a domain R is a 0−submodule if and only if M/N is a torsion-free module, or equivalently if and only if for every non-zero a ∈ R, n ∈ M, the relation a • n ∈ N implies that n ∈ N .
For any submodule N of an R-module M the radical, rad M (N ), of N is defined to be the intersection of all prime submodules of M containing N and rad M (N ) = M if N is not contained in any prime submodule of M .The radical of the module M is defined to be rad M (0).The set of all prime submodules of M is called the Spectrum of M and denoted by Spec(M ).
The purpose of this paper is to explore some basic facts of these class of submodules.First we turn our attention to prime submodules of free modules of finite rank.Assume that R is a domain and let K be its quo-tient field.Let N be a submodule of the free R-module F = R n .In [5] and [1], based upon the relationship between KN and KF , several results concerning prime submodules of free modules were obtained.By applying an useful relationship between Spec(M ) and Spec(M S ) we provide a characterization of prime submodules of free modules in order to show how is their structure.In section 2 we prove a primality condition for some submodules of a finitely generated module in terms of associated prime ideals of the corresponding quotient module.
In the next sections we apply our study of prime submodules, begun in [3], to investigate two interesting questions: on the one hand, given a prime submodule of a finitely generated module we describe the Rees algebra of the corresponding quotient module and, on the other, we give a simple condition for detecting whether a primary submodule of a finitely generated module has prime radical and it is then used to determine, by using the computer algebra system Macaulay2, when some primary submodules of free modules have prime radical.

A CHARACTERIZATION OF PRIME SUBMODULES OF FREE MODULES
This section is devoted to describing prime submodules of free modules and to do so we first need to introduce some notation.Given an R-module M and a multiplicative set S, let us consider the canonical map f : M −→ M S .If N is a submodule of M then the submodule f −1 (N S ) is denoted by S (N ) and called the saturation of N with respect to S. Clearly N ⊂ S (N ) and S (N ) = {m ∈ M : sm ∈ N for some s ∈ S}.Note that if N is a submodule of M such that (N : M ) = p ∈ Spec(R) then the saturation S p (N ) of N with respect to p, i.e., S = R \ p , is a p-prime submodule of M.
Theorem 1.Let R be a ring, let F be a free R-module of finite rank and let M be a submodule of F .Then M is p-prime if and only if M = S(N + pF ), where N is the inverse image of a direct summand of F p by the canonical map F → F p .
Proof.Clearly S(N + pF ) is a p-prime submodule of F (See [6, Lemma 1.7]).Conversely, assume that M is a p-prime submodule of F. Thus we have pF ⊆ M. Let us consider the following composition of morphisms: where as above f stands for the canonical map.Choose a basis {n 1 , ..., n s } of g(M p ) and a set of elements m 1 , ..., m s ∈ M p ⊆ F p such that g(m i ) = n i for 1 ≤ i ≤ s.Then it is esay to check that M p = (m 1 , ..., m s ) + pF p .
Because pF p ⊆ M p we have an exact sequence Now by applying Nakayama's lemma it follows that (m 1 , ..., m s ) must be a direct summnad of F p .If we set N = f −1 ((m 1 , ..., m s )), it is clear that M ⊆ N + F ⊆ S(N + pF ) ∈ Spec(F ).Finally, by applying [1, Corollary 3] there exists a bijective correspondence between the p-prime submodules of F and those F p .Since M p = N p + pF p = S(N + pF ) p , it is obtained, undoubtedly, that M = S(N + pF ) and the theorem is proved.2

A PRIMALITY CONDITION FOR SOME SUBMOD-ULES IN TERMS OF ASSOCIATED PRIME IDEALS
Let R be a Noetherian ring and M a finitely generated R-module.Let Ass(M ) be the set of associated prime ideals of M , that is As it is well-known a submodule Q ⊂ M is called p-primary if Ass(M/Q) consists of the p ideal only (See[4, Theorem 6.6]).The aim of this section is to obtain an analogue result for some prime submodules.
Theorem 2. Let R be a Noetherian ring, P a prime ideal of R and let M be a finitely generated R-module.Assume that N is a submodule of M such that P M ⊂ N .Then N is a p-prime submodule of M if and only if Ass(M/N ) = p Proof.Since every prime submodule is also primary the sufficient condition is clear.Assume that Ass(M/N ) consists of a unique ideal p.Let b ∈ R an element which induces a non injective homothety on M/N .This implies that there exists an element m ∈ M such that bm = 0. Let now q be a minimal prime ideal among all prime ideals containing Ann(m).Since Supp(m) = {p ∈ Spec(R) / m p 6 = 0} coincides with the closed subset V (Ann(m)) = {p ∈ Spec(R) / p ⊃ Ann(m)} it follows that q is also a minimal prime ideal of the set Supp(m).From [4, Theorem 6.5] q is an associated prime ideal to (m).By definition there exists an element m 0 ∈ (m) such that (m 0 ) ' R/q.This implies that q is an associated prime ideal of M/N, so that q = p.Hence b ∈ p and since pM ⊂ N it follows that b induces a vanishing homothety on M/N.Therefore, N is a prime submodule of M.

REES ALGEBRAS ASSOCIATED TO PRIME SUBMOD-ULES
Throughout this section we deal with the definition of the Rees algebra of a module suggested by W. Vasconcelos in [8, Pag.3].As it is well-known, for an ideal I of a commutative ring R there is a canonical epimorphism between the symmetric algebra of I and its Rees algebra.If, further, the ring R is a domain, then the Kernel of φ is just the R-torsion submodule of S(I).This suggests the definition of the Rees algebra <(M ) of an R-module M as S(M )/T , where T is the prime ideal of the R-torsion elements of S(M ).Given a prime submodule of a finitely generated module, the aim of this section is to describe the Rees algebra of the corresponding quotient module but to do so we need some results obtained in [3, Section 2].Let M be a finitely generated R−module.Then to each p−prime submodule N of M , we associate a prime ideal of the symmetric algebra of M , called the expansion E N of N and defined to be the set of all elements b ∈ S(M ) for which there exists an a ∈ R , a / ∈ p such that a • b ∈ (p, N ) • S(M ).Let N be a p-prime submodule of a finitely generated R−module M. The following theorem describes the Rees algebra of the quotient module M/N over R/p.
Proof.As it is well-known, the exact sequence Hence, taking into account the natural map Now it turns out that the R− torsion elements of S R (M/N ) are just the image of the expansion E N of N by the map π, which will be denoted by E N .Indeed, if m is a R− torsion element of S R (M/N ), there exists an a ∈ R such that a• m = 0.But this implies that a • m ∈ (p, N ) • S R (M ), so that m ∈ E N .
Conversely, if m ∈ E N we choose an inverse image m ∈ E N by π.Then by definition there exists an Finally, since (p, N ) • S R (M ) ⊆ E N , it follows that < R (M/N) ' S R (M )/E N and the theorem is proved.

A CONDITION FOR DETECTING WHEN A PRIMARY SUBMODULE HAS PRIME RADICAL
Contrary to what happens in rings, the radical of a primary submodule is not, in general, a prime submodule (See [7, Example 1.12]).So the aim of this section is to give a condition which allows us to determine whether primary submodules of a module have prime radical.
Proposition 1.Let R be a Noetherian ring, let M be an R-module and let Q be a p-primary submodule of M. Then rad M (Q) is a p -prime submodule of M if and only if S p (rad M (Q)) = rad M (Q) Proof.
One direction is obvious.Assume now that S p (rad [7,Lemma1.4]).Hence by applying [1, Corollary 3] and taking into account that (S p (rad The importance of this condition is given by the fact that since we know to compute some radicals of submodules of free modules (See [3, Section 3, 3.2]) we also are able to determine if S p (rad M (Q)) = rad M (Q) by using the computer algebraic system Macaulay2.Therefore we can decide if some radicals of submodules of free modules have prime radical.To illustrate this we present the following example:

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. On the other hand, by using Macaulay2 we obtain that (rad F (Q) : M y) = he 1 , xe 2 i 6 = rad F (Q), so that Q has not prime radical.