On the primeness of modules and submodules
Prime submodules and prime modules have emerged in various contexts, as researchers aim to extend the concept of prime ideals to right or left modules over arbitrary associative rings By adapting the fundamental properties of prime ideals, scholars have introduced and examined the structures of prime submodules and prime modules While these notions are applicable in certain cases of modules over commutative rings, such as multiplication modules, analogous structures for non-commutative rings remain largely unexplored.
In 1961, Andrunakievich and Dauns introduced the concept of prime modules, defining a left R-module M as prime if, for every ideal I of R and every element m in M where Im = 0, it follows that either m = 0 or IM = 0.
In 1975, Beachy and Blair introduced a new definition of primeness, stating that a left R-module M is considered a prime module if (0 :R M) is contained in (0 : R N) for every nonzero submodule N of M This definition is referenced in the relevant literature.
[48] of Goodearl and Warfield in 1983, McConnel and Robson [77] in 1987.
In 1978, Dauns defined a prime module M as one where the annihilator (0 :R M) equals A(M), with A(M) representing the set of elements a in R such that aRm = 0 for any m in M He also introduced the concepts of prime and semiprime submodules A submodule P of a left R-module M is termed a prime submodule if, for any r in R and m in M where rRm is contained in P, it follows that either m is in P or r is in (P :RM) Additionally, a submodule N of M is classified as a semiprime submodule if N is not equal to M, and for any r in R and m in M where r n m is in N, it results in rm being in N.
A left R-module M is defined as B-prime if it is cogenerated by all its nonzero submodules, as established by Bican It is important to note that being B-prime inherently implies that M is also prime Additionally, it has been highlighted that M qualifies as B-prime if and only if the Hom R (M, N) is nonzero for every pair of nonzero submodules L and N of M.
In 1983, Wisbauer introduced the category σ[M], which is the full subcategory of Mod-R consisting of M-generated modules A left R-module M is classified as a strongly prime module if it is subgenerated by any of its nonzero submodules This means that for every nonzero submodule N of M, the module M is part of σ[N] Equivalently, for any elements x and y in M, there exists a set of elements {a1, a2, , an} in R such that ann R {a1x, a2x, , anx} is contained in ann R {y}.
In 1984, Lu [72] defined that for a left R-module M and a submodule
In the context of module theory, a submodule X of a module M over a ring R is defined as a prime submodule if it satisfies the condition that for any element r in R, if the product rm belongs to X, then either m is in X or r is contained in the set of elements that annihilate X Specifically, X can be represented as the set of elements m in M such that the action of r on M results in elements that remain within X, denoted as X = {m ∈ M | rM ⊂ X} Furthermore, the homothety defined by hr: M/X → M/X, which maps m in M/X to mr, is either injective or zero, indicating that the annihilator (0 : M/X) forms a prime ideal in R.
In 1993, McCasland and Smith defined a prime submodule P of a left R-module M as one where, for any ideal I of R and any submodule X of M that satisfies IX ⊂ P, it follows that either IM ⊂ P or a specific condition holds.
In 2002, Ameri [2] and Gaur, Maloo, Parkash ([42], [43]) examined the structure of prime submodules in multiplication modules over commutative rings. Following them, a left R-module M is a multiplication module if every submodule
X is defined as IM for some ideal I of R, with M categorized as a weak multiplication module if all prime submodules of M also take the form IM for some ideal I of R While the multiplicative ideal theory of rings was initially introduced by Dedekind and Noether in the 19th century, Barnard introduced multiplication modules over commutative rings in 1980 to create a module structure that mirrors ring behavior The exploration of multiplication modules over noncommutative rings began with Tuganbaev in 2003.
In 2004, Behboodi and Koohy introduced the concept of weakly prime submodules A submodule P of a module M is classified as a weakly prime submodule if, for any ideals I and J of the ring R and any submodule X of M, the condition IJ X ⊂ P implies that either I ⊆ P or J ⊆ P.
In 2008, Sanh introduced a new definition of prime submodule within the context of ring theory Specifically, for a ring R and a right R-module M with its endomorphism ring S, a fully invariant submodule X of M is classified as a prime submodule if, for any ideal I of S and any fully invariant submodule U of M, the condition I(U)⊂X necessitates that either I(M)⊂X or U⊂X Furthermore, a fully invariant submodule is deemed semiprime if it can be expressed as an intersection of prime submodules A right R-module M is referred to as a semiprime module if the zero submodule is a semiprime submodule of M Consequently, the ring R qualifies as a semiprime ring if R itself is a semiprime module Additionally, by symmetry, the ring R is also considered a semiprime ring if R is a semiprime left R-module.
In 2008, Sanh explored M-annihilators and Goldie modules to extend the notion of Goldie rings A right R-module M is defined as a Goldie module if it possesses a finite Goldie dimension and adheres to the ascending chain condition for M-annihilators A ring R qualifies as a right Goldie ring when R R is Goldie as a right R-module This definition is equivalent to stating that a ring R is a right Goldie ring if it exhibits finite right Goldie dimension and meets the ascending chain condition for right annihilators Additionally, the study investigates the characteristics of prime modules and Goldie modules to examine the class of prime Goldie modules.
BASIC KNOWLEDGE 5
Generators and cogenerators
Generators and cogenerators are notions in categories They play an important role in Module Theory and in some categories Below we will review these notions.
(a) A module B R is called a generator for Mod-R, if
(a) A module C R is called a cogenerator for Mod-R, if
The property thatB is a generator for Mod-R means that for any right R-moduleM, Im(B, M) is as large as possible for everyM and so equals M.
The property thatC R is a cogenerator for Mod-Rmeans that Ker(M, C) is as small as possible for every M and so equals 0.
An R-module M is called aself-generator (self-cogenerator)if it gener- ates all its submodules (cogenerates all its factor modules).
(a) IfB is a generator andAis a module such that Im(A, B) =B, then
(b) Every module M such that there is an epimorphism fromM to RR is also a generator;
(c) If C is a cogenerator and D is a module such that Ker(C, D) = 0, then D is also a cogenerator.
Generators and cogenerators can be characterized in the following the- orem by properties of homomorphisms.
Injectivity and projectivity
Injective modules can be understood as "complete" modules because any "partial" homomorphism from a submodule of a module B into an injective module A can be extended to a "full" homomorphism from the entirety of B into A.
Injective modules originated in the study of abelian groups, with the broader concept for modules being explored by Baer in 1940 The theory surrounding injective modules was developed prior to the consideration of projective modules The terms "injective" and "projective" were introduced by Cartan and Eilenberg in 1956.
Definition 2.2.1 Let M be a right R-module.
(1) A submodule N of M is called essential or large in M if for any submodule X of M, X ∩N = 0 ⇒ X = 0 If N is essential in M we denote
(2) A submoduleN of M is called superfluous orsmall in M if for any submoduleX of M, N +X =M, then X =M In this case we write N ⊂ > ◦ M.
A right ideal I of a ring R is classified as a large right ideal if it is considered large in the context of R R as a right R-module Conversely, a right ideal I is termed a small right ideal if it is deemed small within the same framework of R R as a right R-module.
(4) A homomorphism α : M R → N R is called large if Imα ⊂ > ∗ N The homomorphism α is called small if Kerα⊂ > ◦ M.
Remark From the definition, we have the following:
(1) For any module M, we have 0⊂ > ◦ M, M ⊂ > ∗ M.
(2) A module M is called semisimple if every submodule is a direct summand IfM is a semisimple module, then only 0 is small in M and only M is essential inM.
(3) In any free Z-module (free abelian group), only 0 is small.
(4) Every finitely generated submodule of QZ is small in QZ.
(4) If α : A →B and β :B → C are small epimorphisms, then βα is also a small epimorphism.
Lemma 2.2.4 ([63], Lemma 5.1.4) For a ∈ M R , the submodule aR of M is not small in M if and only if there exists a maximal submodule C ⊂ > M such that a /∈C.
4) If α :A→B and β :B →C are large monomorphisms, then βα is also a large monomorphism.
Definition 2.2.7 Let M and U be two right R-modules A right R-module U is said to be M-injective if for every monomorphism α : L → M and every homomorphism ψ : L →U, there exists a homomorphism ψ 0 :M → U such that ψ 0 α=ψ.
A right R-module E is injective if it is M-injective, for all right R- module M A right R-module M is called quasi-injective (or self-injective) if it is
The following Theorem gives us characterizations of injective modules.
Theorem 2.2.8 ([63], Theorem 5.3.1) Let M be a right R-module The following conditions are equivalent:
(2) Every monomorphismϕ:M →B splits (i.e Im (ϕ)is a direct summand in B);
(3) For every monomorphism α : A → B of right R-modules and any homo- morphism ϕ: A → M, we can find a homomorphism ϕ: B → M such that ϕα=ϕ;
Hom(α,1 M ) :Hom R (B, M)→Hom R (A, M) is an epimorphism.
A powerful test of injectivity is given as Baer’s Criterion which guar- antees the equivalence between injectivity and R- injectivity.
Theorem 2.2.9([100], 16.4)For a right R-moduleE,the following conditions are equivalent:
(3) For every right ideal I of R and every homomorphism h : I → E, there exists y∈E with h(a) = ya, for all a∈I.
Definition and basic properties of projective modules are dual to those of injective modules.
Definition 2.2.10 A right R-module P is said to be M-projective if for every epimorphism β : M → N and every homomorphism ϕ : P → N, there exists a homomorphism ϕ 0 :P →M such that βϕ 0 =ϕ.
Now we have the following fundamental characterizations of projective modules.
Theorem 2.2.11 ([63], Theorem 5.3.1) The following properties of a right R- module P are equivalent :
(2) Every epimorphism ϕ : M → P splits (i.e Ker(ϕ) is a direct summand in M);
(3) For every epimorphism β : B → C of right R-modules and any homomor- phism ϕ:P →C, there is a homomorphism ϕ:P →B such that βϕ =ϕ;
Hom(1 P , β) :Hom R (P, B)→Hom R (P, C) is an epimorphism.
Theorem 2.2.12 ([63], Theorem 5.4.1)A module is projective if and only if it is isomorphic to a direct summand of a free module.
Proposition 2.2.13 ([3], Proposition 16.10) Let M be a right R-module and (Uα)α∈A be an indexed set of right R-modules Then
Uα isM-projective if and only if each Uα isM-projective.
U α is M-injective if and only if each U α isM-injective.
Proposition 2.2.14 ([3], Corollary 16.11) Let (U α )α∈A be an indexed set of right R-modules Then
U α is projective if and only if each U α is projective.
Uα is injective if and only if each Uα is injective.
Noetherian and Artinian modules and rings
Definition 2.3.1(1) A rightR-moduleMR is calledNoetherianif every nonempty set of its submodules has a maximal element Dually, a module M R is called Artinian if every set of its submodules has a minimal element.
(2) A ring R is called right Noetherian (resp right Artinian) if the module RR is Noetherian (resp Artinian).
(3) A chain of submodules of M R ã ã ã ⊂>Ai−1 ⊂>Ai ⊂>Ai+1 ⊂>ã ã ã (finite or infinite) is called stationary if it contains a finite number of distinctA i
Remarks (a) Clearly, the definitions above are preserved by isomorphisms.
(b) Noetherian modules are calledmodules with maximal condition and Artinian modules are called modules with minimal condition.
Theorem 2.3.2([63], Theorem 6.1.2) Let M be a rightR-module and let A be its submodule.
I The following statements are equivalent:
(3) Every descending chain A1 ⊃ A2 ⊃ ã ã ã ⊃ An−1 ⊃ An ⊃ ã ã ã of submodules of M is stationary;
(4) Every factor module of M is finitely cogenerated;
(5) For every family {A i |i ∈I} 6=∅ of submodules of M, there exists a finite subfamily {A i |i∈I 0 } (i.e., I 0 ⊂I and finite) such that
II The following conditions are equivalent:
(3) Every ascending chain A 1 ⊂ A 2 ⊂ ã ã ã ⊂ An−1 ⊂ A n ⊂ ã ã ã of submodules of M is stationary;
(4) Every submodule of M is finitely generated;
(5) For every family {Ai |i ∈I} 6=∅ of submodules of M, there exists a finite subfamily {A i |i∈I 0 } (i.e., I 0 ⊂I and finite) such that
III The following conditions are equivalent:
(2) M is a module of finite length.
Theorem 2.3.2 establishes two key conditions for modules: the descending chain condition (DCC) and the ascending chain condition (ACC) A module M is classified as Noetherian if it meets the ACC, while it is considered Artinian if it satisfies the DCC.
(1) IfM is a finite sum of Noetherian submodules, then it is Noetherian; if M is a finite sum of Artinian submodules, then it is Artinian.
(2) If the ring R is right Noetherian (resp right Artinian), then every finitely generated right R-moduleM R is Noetherian (resp Artinian).
(3) Every factor ring of right Noetherian (resp Artinian) ring is again right Noetherian (resp Artinian).
Primeness in module category
In this section, before stating our new results we would like to list some basic properties from [48].
Definition 2.4.1 A proper ideal P in a ring R is called a prime ideal of R if for any ideals I, J of R with IJ ⊂ P, then either I ⊂ P or J ⊂ P An ideal I of a ring R is called strongly primeif for any a, b∈R with ab∈I,then either a∈I or b∈I A ringR is called aprime ringif 0 is a prime ideal (Note that a prime ring must be nonzero).
Proposition 2.4.2 ([48], Proposition 3.1) For a proper ideal P of a ring R, the following conditions are equivalent:
(2) If I and J are any ideals of R properly containing P, then IJ *P;
(4) If I and J are any right ideals of R such that IJ ⊂ P, then either
(5) If I and J are any left ideals of R such that IJ ⊂ P, then either
(6) If x, y ∈R with xRy ⊂P,then either x∈P or y∈P.
According to Proposition 2.4.2, if P is a prime ideal in a ring R and J₁, , Jₙ are right ideals of R such that J₁ ⊇ ⊇ Jₙ ⊂ P, then there exists at least one i such that Jᵢ ⊂ P A maximal ideal in a ring is defined as a maximal proper ideal, which is the highest element in the set of proper ideals.
Proposition 2.4.3 ([48], Proposition 3.2) Every maximal ideal of a ring R is a prime ideal.
Proposition 2.4.3 together with Zorn’s Lemma guarantees that every nonzero ring has at least one prime ideal.
Definition 2.4.4 A prime ideal P in a ring R is called a minimal prime ideal if it does not properly contain any other prime ideals For instance, if R is a prime ring, then 0 is the unique minimal prime ideal of R.
Proposition 2.4.5([48], Proposition 3.3)Any prime ideal P in a ring R contains a minimal prime ideal.
Theorem 2.4.6([48], Theorem 3.4)In a right or left Noetherian ringR, there exist only finitely many minimal prime ideals, and there is a finite product of minimal prime ideals (repetitions allowed) that equals zero.
Definition 2.4.7 An ideal P in a ring R is called a semiprime ideal if it is an intersection of prime ideals (By convention, the intersection of the empty family of prime ideals of R is R, so R is a semiprime ideal of itself) A ring R is called a semiprime ring if 0 is a semiprime ideal.
In the ring of integers Z, the intersection of any infinite number of prime ideals results in the zero ideal Conversely, the intersection of a finite list of distinct prime ideals, represented by p₁Z, , pₖZ, leads to the ideal generated by their product, p₁ pₖZ Consequently, the nonzero semiprime ideals in Z include the ring of integers Z itself, along with the ideals nZ, where n is any square-free positive integer.
According to Proposition 3.6, an ideal I in a commutative ring R is semiprime if x ∈ R and x² ∈ I imply that x ∈ I This criterion does not hold in noncommutative rings, as demonstrated by a matrix ring over a field However, an analogous criterion exists in the noncommutative context, as established by Levitzki-Nagata, which will be discussed in the next theorem.
Theorem 2.4.8 ([48], Theorem 3.7) An ideal I in a ring R is semiprime if and only if
Many authors characterize semiprime ideals through the condition outlined in Theorem 2.4.8, which states that an ideal is semiprime if and only if it can be expressed as an intersection of prime ideals.
Corollary 2.4.9 ([48], Corollary 3.8) For an ideal I in a ring R, the following conditions are equivalent:
(2) If J is any ideal of R such that J 2 ⊂I, then J ⊂I;
(3) If J is any ideal of R such that J %I, then J 2 *I;
(4) If J is any right ideal of R such that J 2 ⊂I, then J ⊂I;
(5) If J is any left ideal of R such that J 2 ⊂I, then J ⊂I.
Corollary 2.4.10 ([48], Corollary 3.9) Let I be a semiprime ideal in a ring R If
J is a right or a left ideal of R such that J n ⊂I for some positive integer n, then
Definition 2.4.11An elementxin a ring Ris called anilpotent elementifx n = 0 for some n ∈ N A right or a left ideal I in a ring R is called a nilpotent ideal if
I n = 0 for somen∈N More generally,I is called a nil idealif each of its elements is nilpotent Theprime radicalP(R) of a ringR is the intersection of all the prime ideals of R.
Remarks([48], page 53) (1) In Noetherian rings, all nil one-sided ideals are nilpo- tent.
(2) If R is the zero ring, it has no prime ideals, and so P(R) = R If
R is nonzero, it has at least one maximal ideal, which is prime by Lemma 2.4.3. Thus, the prime radical of a nonzero ring is a proper ideal.
(3) A ring R is semiprime if and only if P(R) = 0 In any case, P(R) is the smallest semiprime ideal of R, and because P(R) is semiprime, it contains all nilpotent one-sided ideals of R.
In a semiprime ring R, if A and B are right ideals with the property that AB = 0, it follows that (BA)² = 0 and (A ∩ B)² = 0, leading to the conclusion that BA = 0 and A ∩ B = 0 Consequently, for any ideal I of R, we have Ir(I) = 0, which implies that r(I)I = 0, and similarly, Il(I) = 0 This results in the equality l(I) = r(I) Furthermore, if I acts as a right annihilator, then it can be expressed as I = r(l(I)) = l(r(I)), confirming that I is also a left annihilator In this context, we refer to I as an annihilator ideal.
We have the following lemmas.
Lemma 2.4.12([100], Proposition 3.13)For a ring R with identity, the following conditions are equivalent:
(2) 0 is the only nilpotent ideal in R;
(3) For ideals I, J in R with IJ = 0 implies I∩J = 0.
Lemma 2.4.13([53], Lemma 1.16)LetR be a semiprime ring with the ACC (equiv- alently DCC) for annihilators ideals, then R has only finite number of minimal prime ideals IfP 1 ,ã ã ã, P n are the minimal prime ideals ofR thenP 1 ∩ã ã ã∩P n = 0. Also a prime ideal of R is minimal if and only if it is an annihilator ideal.
Proposition 2.4.14 ([48], page 54) In any ring R, the prime radical equals the intersection of the minimal prime ideals of R.
Definition 2.4.15LetXbe a subset of a rightR-moduleM.Theright annihilator of X is the set r R (X) ={r ∈ R : xr = 0 for all x ∈ X} which is a right ideal of
R If X is a submodule of M, then r R (X) is a two-sided ideal of R Annihilators of subsets of left R-modules are defined analogously, and are left ideals of R If
M =R, then the right annihilator of X ⊂R is r R (X) ={r∈R |xr= 0 for allx∈X} as well as aleft annihilator of X is l R (X) ={r ∈R|rx= 0 for allx∈X}.
A right annihilator is a right ideal ofR which is of the form r R (X) (or simply r(X)) for some subset X of R and a left annihilator is a left ideal of the forml R (X) (or simply l(X)).
We now give the following basic properties of right and left annihilators which have important consequences.
Properties 2.4.16 ([53]) Let R be a ring and let X, Y be subsets of R Then we have the following properties:
From these relationships it follows easily that the ACC for right anni- hilators is equivalent to the DCC for left annihilators.
Definition 2.4.17 Let M be a right R-module and S = End R (M), its endomor- phism ring A submoduleX ofM is called afully invariant submoduleof M if for any f ∈S,we have f(X)⊂X.
The class of all fully invariant submodules of a module M is nonempty and exhibits closure under both intersections and sums Specifically, for any two fully invariant submodules X and Y of M, and for every function f in S, it follows that f(X+Y) equals f(X)+f(Y), which is contained within X+Y Additionally, we have f(X∩Y) contained within f(X)∩f(Y), which is also a subset of X∩Y Furthermore, for a family of fully invariant submodules {X_i : i ∈ I}, where I is an index set, the properties of closure under intersections and sums remain valid.
X i are fully invariant submodules of M Especially, a right ideal I of a ring R is a fully invariant submodule of R R if it is a two-sided ideal.
Now, let I, J ⊂ S and X ⊂ M For convenience, we denote I(X) P f∈I f(X), Ker(I) = T f∈I
In the context of right R-modules, for any right ideal I of R and a right R-module M, the set MI is identified as a fully invariant submodule of M This sets the stage for the definition of prime submodules, which are essential in understanding the structure of modules.
Definition 2.4.18 Let M be a right R-module and X, a fully invariant proper submodule ofM.Then X is called aprime submoduleofM (we say thatX is prime inM) if for any idealI of S,and any fully invariant submoduleU ofM, I(U)⊂X impliesI(M)⊂XorU ⊂X.A fully invariant submoduleX ofM is calledstrongly prime if for any f ∈S and any m∈M, f(m)∈X implies f(M)⊂X or m∈X.
The following theorem gives some characterizations of prime submod- ules similar to that of prime ideals and we use it as a tool for checking the primeness.
Theorem 2.4.19 ([86], [87]) Let M be a right R-module and P, a proper fully invariant submodule of M Then the following conditions are equivalent:
(2) For any right idealI of S and any submoduleU of M, if I(U)⊂P, then either I(M)⊂P or U ⊂P;
(3) For anyϕ∈S and any fully invariant submoduleU ofM,ifϕ(U)⊂
(4) For any left idealI of S and any subset Aof M,if IS(A)⊂P,then either I(M)⊂P or A⊂P;
(5) For any ϕ ∈ S and any m ∈ M, if ϕ(S(m)) ⊂ P, then either ϕ(M)⊂P or m∈P.
Moreover, if M is quasi-projective, then the above conditions are equivalent to:
In addition, ifM is quasi-projective and a self-generator, then the above conditions are equivalent to:
(7) If I is an ideal of S and U, a fully invariant submodule of M such that I(M) and U properly contain P, then I(U)6⊂P.
Examples 2.4.20(1) LetZ4 ={0,1,2,3}be the additive group of integers modulo
4 Then X = is a prime submodule ofZ 4
(2) IfM is a semisimple module having only one homogeneous compo- nent, then 0 is a prime submodule Especially, if M is simple, then 0 is a prime submodule.
Definition 2.4.21A prime submodule P of a right R-moduleM is called amin- imal prime submodule if it is minimal in the class of prime submodules ofM.
The following proposition gives us a property similar to that of rings (see Lemma 2.4.5).
Proposition 2.4.22 [86] If P is a prime submodule of a right R-module M, then
P contains a minimal prime submodule of M.
Lemma 2.4.23 [86] Let M be a right R-module and S = End R (M) Suppose that
X is a fully invariant submodule of M Then the set IX ={f ∈S |f(M)⊂X} is a two-sided ideal of S.
Theorem 2.4.24 [86] Let M be a right R-module, S = End R (M) and X, a fully invariant submodule of M If X is a prime submodule of M, then I X is a prime ideal ofS.Conversely, if M is a self-generator and ifI X is a prime ideal ofS,then
Definition 2.4.25A fully invariant submoduleX of a rightR-moduleM is called a semiprime submodule if it is an intersection of prime submodules of M.
A rightR-moduleM is called aprime moduleif 0 is a prime submodule of M A ring R is a prime ring if R R is a prime module.
A right R-module M is defined as a semiprime module if its zero submodule is semiprime As a result, a ring R is categorized as a semiprime ring if the right R-module R is semiprime Similarly, the ring R is also considered semiprime if R serves as a semiprime left R-module.
Examples 2.4.26(1) Every semisimple module with only one homogeneous com- ponent is a prime module Especially, every simple module is prime.
(2) Every semisimple module is semiprime.
(3) As a Z-module, the module Z4 is not semiprime.
Theorem 2.4.27 [86] Let M be a prime module Then its endomorphism ring S is a prime ring Conversely, if M is a self-generator and S is a prime ring, then
Lemma 2.4.28 [86] Let M be a quasi-projective module, P be a prime submodule of M, A⊂P be a fully invariant submodule ofM ThenP/Ais a prime submodule of M/A.
Lemma 2.4.29 [86] Let M be a quasi-projective module and A a fully invariant submodule of M If P¯ ⊂ M/A is a prime submodule of M/A, then ν −1 ( ¯P) is a prime submodule of M.
A GENERALIZATION OF HOPKINS-LEVITZKI
ON NIL RADICAL AND LEVITZKI RADICAL
Nil submodules
Definition 4.1.1 Let M be a right R-module and X, a submodule of M X is called anil submodule of M if I X is a nil right ideal of S.
It is clear by the definition that a nilpotent submodule is a nil submod- ule.
Note that, if X is a nil submodule of M, then for any f ∈I X we have
LetM be a rightR-module which is a self-generator IfM is a semiprime module with the ACC for M-annihilators, thenM has no nonzero nil submodules.
Since M is a semiprime module and S is a semiprime ring, M exhibits the ascending chain condition (ACC) for its M-annihilators, which implies that S also has the ACC for its right annihilators For a nonzero submodule X of M, the ideal I_X is a nonzero right ideal of M By selecting a nonzero element f in I_X with the largest possible right annihilator r_S(f), and given that S is semiprime (ensuring fSf ≠ 0), we can find an element g in S such that fgf ≠ 0 This leads to r_S(f) being equal to r_S(fgf), indicating that both f and g are non-zero and that g does not belong to the right annihilators of f or fgf Consequently, we establish that (fg)^3 ≠ 0 for all n ≥ 1, demonstrating that f g is in I_X and is not nilpotent Therefore, I_X is not nil, which confirms that X is not nil.
In the caseM =R R ,we have the following corollary.
Corollary 4.1.3 Let R be a semiprime ring with the ACC for right annihilators. Then R has no nonzero nil right ideals.
Proposition 4.1.4 Let M be a quasi-projective right R-module Let X, Y be sub- modules of M and X ⊂Y Then we have the following:
(1) If Y is a nil submodule of M, then Y /X is a nil submodule of M/X.
(2) If Y is a fully invariant nil submodule of M, then Y /X is a fully invariant nil submodule of M/X.
To prove that ¯ϕ is nilpotent, we start by defining ¯S as End(M/X) and I Y/X as the set of φ in ¯S such that φ(M/X) is contained in Y/X Given ¯ϕ in I Y/X, and knowing that M is quasi-projective, we can find a φ in S satisfying ¯ϕν = νφ This leads us to conclude that ¯ϕ(M/X) equals (φ(M) + X)/X, which implies that (φ(M) + X) is a subset of Y Consequently, we establish that φ(M) is contained in Y, indicating that φ is an element of I Y Since Y is a nil submodule of M, φ must be nilpotent, meaning φ^n equals zero for some positive integer n Therefore, we find that ¯ϕ^n(M/X) equals zero, confirming that ¯ϕ is nilpotent.
(2) If Y is a fully invariant submodule of M, then it is easy to check that Y /X is a fully invariant submodule of M/X From this fact and (1) we have
Y /X is a fully invariant nil submodule ofM/X
Proposition 4.1.5 Let M be a quasi-projective module and X, a fully invariant nil submodule of M Then we have the following:
(1) If Y¯ is a nil submodule of M/X, then ν −1 ( ¯Y) is a nil submodule of M.
(2) If Y¯ is a fully invariant nil submodule of M/X, then ν −1 ( ¯Y) is a fully in- variant nil submodule of M.
Proof (1) Put ¯S = End(M/X) and IY ¯ = {φ ∈ S¯ | φ(M/X) ⊂ Y¯} Put
In the context of module theory, let Y be defined as Y = ν −1(¯Y) with f belonging to the ideal IY Given that X is a fully invariant submodule of M, there exists a function ϕ in S¯ such that ϕν equals νf Consequently, we have ϕ(M/X) = ϕν(M) = νf(M) included in ν(Y) = ¯Y, leading to ϕ being an element of IY ¯ Since ¯Y is identified as a nil submodule of M/X, it follows that ϕ is nilpotent, implying ϕ^n = 0 for some positive integer n This results in 0 = ϕ^n(M/X) = ϕ^nν(M) = νf^n(M), demonstrating that f^n(M) is contained within X, thus indicating f^n belongs to I X Given that X is nil, f^n is also nilpotent, leading to the conclusion that f^n k = 0 for some positive integer k, which ultimately shows that Y is a nil submodule of M.
(2) Put Y = ν −1 ( ¯Y) Take f ∈ S Then, there exist ϕ ∈ S¯ such that ϕν = νf We have νf(Y) = ϕν(Y) = ϕ( ¯Y) ⊂ Y¯ = ν(Y) = Y /X Thus f(Y) +X ⊂ Y, sof(Y)⊂Y This shows that Y is a fully invariant submodule of
M From this fact and (1), we can see thatY is a fully invariant nil submodule of
Proposition 4.1.6 Let M be a quasi-projective, finitely generated right R-module which is self-generator Then any sum of fully invariant nil submodules of M is also fully invariant nil.
Proof Suppose that X, Y are fully invariant nil submodules of M Then I X and
The ideals I_Y and I_X are nil ideals of S, which implies that their sum, I_X + I_Y, is also a nil ideal of S, as stated in Proposition 2.5.10 Consequently, we have I_X + I_Y = I_{X+Y}, indicating that X + Y is a fully invariant nil submodule of M By induction, it follows that a finite direct sum of fully invariant nil submodules of M remains fully invariant nil Furthermore, for a family {X_i | i ∈ Ω} of fully invariant nil submodules of M, each I_X_i is a nil ideal of S, leading to the conclusion that the sum P_{i ∈ Ω} retains this property.
I X i is a nil ideal of S (by Proposition 2.5.10) Note that P i∈Ω
X i is a fully invariant nil submodule of M
Theorem 4.1.7 Let M be a quasi-projective, finitely generated right R-module which is a self-generator Then we have the following:
(1) The sum of two nilpotent submodules of M is again nilpotent.
(2) If M is a Noetherian module, then every fully invariant nil submodule of M is nilpotent.
Proof (1) LetX, Y be nilpotent submodules ofM.ThenI X , I Y are nilpotent right ideals ofS.It follows thatIX+IY is a nilpotent right ideal of S ButIX +Y =IX+IY, so X+Y is nilpotent.
If \( M \) is a Noetherian module, then the ring \( S \) is a right Noetherian ring, which implies that any two-sided nil ideal in \( S \) is nilpotent Given a fully invariant nil submodule \( X \) of \( M \), the ideal \( I X \) forms a two-sided nil ideal in \( S \), leading to the conclusion that \( I X \) is nilpotent This ultimately demonstrates that the submodule \( X \) is also nilpotent.
Nil radical of modules
In this section, we introduce the notion nil radical of a given right R- moduleM and investigate some properties of nil radicals that are similar with the nil radical of rings.
Definition 4.2.1 Let M be a right R-module The nil radical of M, denoted by
N(M), is the sum of all fully invariant nil submodules of M.
By proposition 4.1.6 and the definition, if M is a quasi-projective, finitely generated right R-module which is a self-generator, then N(M) is the largest fully invariant nil submodule of M.
Proposition 4.2.2Let M be a quasi-projective module LetX be a fully invariant nil submodule of M Then N(M/X) = N(M)/X.
Proof Let F ={Y¯ |Y¯is a fully invariant nil submodule ofM/X} and G ={Y |Y is a fully invariant nil submodule ofM}
Corollary 4.2.3 Let M be a quasi-projective, finitely generated right R-module which is a self-generator Then N(M/N(M)) = 0.
Note that, fromN(M/N(M)) = 0,the moduleM/N(M) has no nonzero nilpotent submodules.
Proposition 4.2.4 Let M be a quasi-projective, finitely generated right R-module which is self-generator with N :=N(M) If M is an Artinian module, then N is the largest fully invariant nilpotent submodule of M.
It is evident that N encompasses all fully invariant nilpotent submodules of M, and we aim to demonstrate that N itself is nilpotent Given that M is an Artinian module, S is a right Artinian ring We examine a descending chain of ideals in S, specifically I_N ⊃ I_N^2 ⊃ Due to S being a right Artinian ring, there exists a positive integer k such that I_N^k = I_N^t for all t > k We define K as the set of all f ∈ S such that f I_N^k = 0, and we seek to prove that K equals S If this is not the case, the collection of all right ideals of S that properly contain K must have a minimal element, denoted as J By selecting g ∈ J \ K, we find that J can be expressed as K + gS, leading to the conclusion that K + J I_N = K + (gS) I_N = K + g I_N.
J I N 6⊂K,thenK+J I N =J by the minimality ofJ.Sinceg ∈J,we haveg =f+gh for some f ∈ K, h ∈ I N Hence 1−h is invertible in S So g = f(1−h) −1 ∈ K, a contradiction Thus J I N ⊂ K Then we have J I N k+1 = (J I N )I N k ⊂ KI N k = 0. From I N k+1 =I N k we haveJ I N k = 0, soJ ⊂K, a contradiction Thus K =S.Since
KI N k = 0 we can see that SI N k = 0 and hence I N k = 0 Thus I N is nilpotent, this means N is nilpotent
Proposition 4.2.5 Let M be a quasi-projective, finitely generated right R-module which is a self-generator If N is a nil submodule of M, then N ⊂J(M).
Proof By hypothesis, J(S) =I J(M ) If N is a nil submodule of M, then I N is a nil right ideal of S It follows that I N ⊂ J(S) = I J(M) (by Theorem 2.5.12), and hence N ⊂J(M)
Since N(M) is a nil submodule ofM, we have the following corollary.
Corollary 4.2.6 Let M be a quasi-projective, finitely generated right R-module which is a self-generator Then N(M)⊂J(M).
Proposition 4.2.7 Let M be a quasi-projective, finitely generated right R-module which is a self-generator Then P(M)is a fully invariant nil submodule of M and
Proof We have P(S) =IP (M).SinceP(S) is a nil ideal ofS,we must haveP(M) is a fully invariant nil submodule of M and hence P(M)⊂N(M)
Corollary 4.2.8 Let M be a quasi-projective, finitely generated right R-module which is a self-generator Then P(M)⊂N(M)⊂J(M).
Theorem 4.2.9 Let M be a quasi-projective, finitely generated right R-module which is a self-generator If M is Artinian, then J(M) = N(M).
According to Theorem 3.2.12, J(M) is identified as a nilpotent submodule of M, and it can be expressed as J(S) = I J(M) It is important to note that J(M) is a fully invariant nilpotent submodule of M Furthermore, if K is a fully invariant nilpotent submodule of M, then by Proposition 4.2.5, it follows that K is a subset of J(M) Consequently, J(M) stands out as the largest fully invariant nilpotent submodule within the structure of M.
In Proposition 3.2.13, it was established that for an Artinian quasi-projective, finitely generated right R-module M that acts as a self-generator, the relationship P(M) is a subset of J(M) This finding, in conjunction with Theorem 4.2.9, leads to an important corollary.
Corollary 4.2.10 Let M be a quasi-projective, finitely generated right R-module which is a self-generator If M is Artinian, then P(M) =N(M) =J(M).
From Proposition 4.2.5 and Theorem 4.2.9, we have the following corol- lary.
Corollary 4.2.11 Let M be a quasi-projective, finitely generated right R-module which is a self-generator If M is Artinian module, then any nil submodule of M is nilpotent.
Proposition 4.2.12 Let M, N be right R-modules with M is quasi-projective and let f :M →N be an epimorphism Then we have the following:
(1) If X is a nil submodule of M such that Kerf ⊂ X, then f(X) is a nil submodule of N.
(2) If X is a fully invariant nil submodule of M such that Kerf ⊂X,then f(X) is a fully invariant nil submodule of N.
(3) If X and Kerf are fully invariant nil submodules of M, then f(X) is a fully invariant nil submodule of N.
To demonstrate that \( f(X) \) is nilpotent, we start by defining \( S_0 = End(N) \) and \( I_f(X) = \{ \phi \in S_0 \mid \phi(N) \subset f(X) \} \) Given \( \phi \in I_f(X) \), and since \( M \) is quasi-projective, there exists \( \psi \in S \) such that \( f \psi = \phi f \) This implies \( \phi(N) = \phi f(M) = f \psi(M) \subset f(X) \), leading to \( \psi(M) \subset X + \text{Ker} f = X \) Consequently, \( \psi \in I_X \) and is thus nilpotent, meaning \( \psi^n = 0 \) for some positive integer \( n \) As a result, \( \phi^n(N) = \phi^n f(M) = f \psi^n(M) = 0 \), confirming that \( \phi \) is nilpotent and, ultimately, that \( f(X) \) is nilpotent.
(2) It is easy to check that f(X) is a fully invariant submodule of N. From (1) we have f(X) is a fully invariant nil submodule ofN.
(3) Let φ ∈ I f(X ) Since M is a quasi-projective, there exists ϕ ∈ S such that f ϕ = φf We have φ(N) = φf(M) = f ϕ(M) ⊂ f(X) Thus ϕ(M) ⊂
X+ Kerf Since X and Kerf are fully invariant nil submodules ofM,we see that
X+ Kerf is also a fully invariant nil submodule ofM.It follows thatϕis nilpotent and hence φ is nilpotent Therefore f(X) is nilpotent
Proposition 4.2.13 Let M, N be right R-modules and f : M → N, an epimor- phism such that Kerf is a fully invariant nil submodule of M Then we have the following:
(1) If Y is a nil submodule of N, then X :=f −1 (Y) is a nil submodule of M.
(2) If Y is a fully invariant nil submodule of N, then X := f −1 (Y) is a fully invariant nil submodule of M.
To prove that X is nilpotent, we start with S₀ as End(N) and define Iₓ as the set of φ in S₀ such that φ(N) is contained in X For any φ in Iₓ, it follows that φ(M) is a subset of X Given that Kerf is a fully invariant submodule of M, there exists a φ in S₀ such that φf = fϕ, leading to the conclusion that fϕ(M) is contained in Y This implies φf(M) is also contained in Y, confirming that φ belongs to Iᵧ As a result, φ is nilpotent, indicating that φⁿ equals 0 for some positive integer n Consequently, fϕⁿ(M) equals φⁿf(M) and equals φⁿ(N), which results in ϕⁿ(M) being a subset of Kerf Since Kerf is a nil submodule of M, we establish that ϕⁿ is nilpotent, ultimately proving that ϕⁿₖ equals 0 for some positive integer k and confirming that X is nilpotent.
(2) By Lemma 3.2.15 (2),X is a fully invariant submodule of M.From
(1), X is a fully invariant nil submodule of M
Theorem 4.2.14Let M, N be right R-modules with M,a quasi-projective module and f : M → N be an epimorphism If Kerf ⊂ X for any fully invariant nil submodule of M, then f(N(M))⊂N(N).
Proof LetF ={X |Xis a fully invariant nil submodule ofM} We haveN(M) P
X ∈F f(X) By Proposition 4.2.12, f(X) is a fully in- variant nil submodule of N, where X is a fully invariant submodule of M Thus f(N(M))⊂N(N)
Theorem 4.2.15 Let M, N be right R-modules with M is a quasi-projective and let f : M → N be an epimorphism If Kerf is a fully invariant nil submodule of
Proof LetF ={X |Xis a fully invariant nil submodule ofM} We haveN(M) P
X∈F f(X) If X ⊂ Kerf, then f(X) = 0, other- wise f(X) is a fully invariant nil submodule of N by Proposition 4.2.12 Thus f(N(M))⊂N(N)
Proposition 4.2.16 Let M, N be right R-modules and f : M → N, an epimor- phism such that Kerf is a fully invariant nil submodule of M Then f(N(M))⊃
Proof Let F ={Y | Y is a fully invariant nil submodule ofN} We have N(N) P
By Proposition 4.2.13, f −1 (Y) is a fully invariant nil submodule of M, where Y is a fully invariant nil submodule of M Thus f −1 (N(N)) ⊂ N(M), and hence f(N(M))⊃N(N)
According to Proposition 4.2.15 and Proposition 4.2.16, Corollary 4.2.17 states that for right R-modules M and N, if M is a quasi-projective module and f: M → N is an epimorphism with Kerf being a fully invariant nil submodule of M, then the image of N(M) under f is equal to N(N).
Proposition 4.2.18LetM be a quasi-projective, finitely generated rightR-module which is a self-generator Assume that M satisfies the ACC for M-annihilators. Then we have the following:
(1) Any nil submodule X of M is contained in P(M);
(2) Any nonzero nil submodule X of M contains a nonzero nilpotent submodule.
In particular, if M is also a semiprime module, then every nil submodule of M is zero.
Since M satisfies the ascending chain condition (ACC) for M-annihilators, it follows that S also satisfies the ACC for right annihilators, as established in Lemma 2.4.41 Let X represent a nil submodule of M; consequently, I X becomes a nil right ideal of S According to Lemma 10.29, I X is contained within P(S) Therefore, if P(S) equals I P(M), it follows that I X is also a subset of I P(M), which ultimately demonstrates that X is a subset of P(M).
Let X be a nonzero nil submodule of M, which implies that I X is a nonzero nil right ideal of S According to Lemma 10.29, I X includes a nonzero nilpotent right ideal, denoted as J Consequently, we have X containing J(M), which is also nonzero Given that J equals I J(M) and is nilpotent, it follows that J(M) is nilpotent as well.
Furthermore, if M is a semiprime module, then P(M) = 0 So if N is a nil submodule ofM, then by (1), N = 0
Proposition 4.2.19LetM be a quasi-projective, finitely generated rightR-module which is a self-generator If M is a Noetherian module, then P(M) =N(M) and this is the largest nilpotent submodule of M.
Since M is a Noetherian module and S is a right Noetherian ring, it follows that P(S) equals N(S), establishing P(S) as the largest nilpotent right ideal of S, as stated in Theorem 10.30 Additionally, it is important to note that P(S) corresponds to P(M) and N(S) corresponds to N(M).
P(M) =N(M) Since N(S) is the largest nilpotent right ideal of S, N(M) is the largest nilpotent submodule of M.
Levitzki radical of modules
In this section, we introduce the notion Levitzki radical of a given right R-moduleM as a generalization of Levitzki radical of rings.
Definition 4.3.1 Let I be a right ideal of a ring R I is called locally nilpotent if for any finite subset {s1, , sn} ⊂ I, there exists an integer k such that any product of k elements from {s 1 , , s n } is zero.
Now, let X be a submodule of a right R-module M We say that X is locally nilpotent if I X is a locally nilpotent right ideal of S.
By definition, we see that for any submodule X of M, we have:
X is nilpotent ⇒X is locally nilpotent ⇒X is nil.
In the context of quasi-projective, finitely generated right R-modules, Proposition 4.3.2 states that for a self-generator module M, if N and P are fully invariant submodules with P contained in N, then N exhibits local nilpotency if and only if both P and the quotient module N/P are locally nilpotent.
Proof Put ¯S = End(M/P) and I N/P = {φ ∈ S¯ | φ(M/P) ⊂ N/P} Suppose that N is locally nilpotent Since P ⊂ N, I P ⊂ I N and since I N is a locally nilpotent ideal of S, we have I P is a locally nilpotent ideal of S It follows that
In the context of locally nilpotent ideals, let ¯ϕ 1, , ϕ¯ n belong to I N/P Given that M is quasi-projective, we can find elements ϕ 1, , ϕ n in S such that ¯ϕ i ν = νϕ i for each i from 1 to n This leads to the conclusion that νϕ i (M) is contained in ¯ϕ i ν(M), which in turn is a subset of N/P Consequently, it follows that ϕ i (M) is included in N, indicating that ϕ i belongs to I N for all i Since I N is locally nilpotent, there exists an integer t such that the product ϕ k 1, , ϕ k t equals zero Therefore, we have ¯ϕ k 1 ã ã ãϕ¯ k t (M/P) equating to zero, which confirms that ¯ϕ k 1 ã ã ãϕ¯ k t is indeed zero, thereby demonstrating that N/P is locally nilpotent.
Conversely, assume that both P and N/P are locally nilpotent Let ϕ 1 , , ϕ n ∈I N SinceP is fully invariant submodule ofM,there exist ¯ϕ 1 , ,ϕ¯ n ∈
Given that ¯ϕ i ν = νϕ i for all i = 1, , n, we find that ¯ϕ i (M/P) equals νϕ i (M), which is a subset of ν(N) = N/P, indicating that ¯ϕ i belongs to I N/P Since I N/P is locally nilpotent, there exists an integer s such that the product ¯ϕ k 1 ¯ϕ k s equals zero Consequently, we have νϕ k 1 ϕ k s (M) = ¯ϕ k 1 ¯ϕ k s (M/P) = 0, leading to the conclusion that ϕ k 1 ϕ k s is contained in P, thus implying ϕ k 1 ϕ k s is an element of I P These n products yield a finite set T of elements within I P.
Because I P is locally nilpotent, there is an integer t such that all product of t elements of T are zero Because any product oft elements from T is a product of st element fromI N , we conclude that any product of stelements from I N is zero. Hence IN is locally nilpotent, proving that N is locally nilpotent
Proposition 4.3.3 Let M be a quasi-projective, finitely generated right R-module which is a self-generator Let X, Y be fully invariant submodules of M If X, Y are locally nilpotent, then X+Y is locally nilpotent.
The proof demonstrates that the quotient Y/X ∩ Y is locally nilpotent, given that X ∩ Y is a subset of Y and both are locally nilpotent By Proposition 4.3.2, since both Y and X ∩ Y are locally nilpotent, it follows that Y/X ∩ Y inherits this property Additionally, since X is also locally nilpotent, the sum X + Y is confirmed to be locally nilpotent as well.
By induction, we establish that the finite sum of fully invariant locally nilpotent submodules of M remains locally nilpotent Given that locally nilpotent is a finitary property, it follows that any arbitrary sum of fully invariant locally nilpotent submodules is also locally nilpotent.
Definition 4.3.4LetM be a right R-module TheLevitzki radical of M, denoted byL(M), is the sum of all fully invariant locally nilpotent submodules of M.
By definition, we see that when M is a quasi-projective, finitely gener- ated, self-generator module, then L(M) is the largest fully invariant locally nilpo- tent submodule of M.
Proposition 4.3.5 Let M be a quasi-projective, finitely generated right R-module which is a self-generator IfN is a fully invariant submodule ofM andN ⊂L(M), then L(M/N) = L(M)/N.
Proof LetX be a fully invariant submodule of M with X ⊂L(M) IfX/N is a locally nilpotent submodule of M/N, then X is a locally nilpotent submodule of
M It follows thatX ⊂L(M) and it is easy to see thatL(M/N) =L(M)/N Corollary 4.3.6 Let M be a quasi-projective, finitely generated right R-module which is a self-generator Then L(M/L(M)) = 0.
Theorem 4.3.7 Let M be a quasi-projective, finitely generated right R-module which is a self-generator Then P(M)⊂LM)⊂N(M)⊂J(M).
Proof Let F = {X | Xis fully invariant locally nilpotent submodule ofM} By hypothesis, we have J(S) = I J(M ) , P(S) = I P (M ) , N(S) = I N (M) Now, L(S) sum of locally nilpotent ideals of S = P
THE ZARISKI TOPOLOGY ON THE PRIME
Let R be a ring We denote Spec(R) (or X R ) for the set of all prime ideals of R For any ideal I of R, we define:
Then we have the following properties
(2) If{Ei}i∈J is any family of ideals of R, then T i∈J
(3) IfI, J are ideals of R, then V R (I)∪V R (J) = V R (IJ) = V R (I∩J).
Let Γ(R) represent the collection of all closed sets in the Zariski topology on Spec(R), defined as Γ(R) = {V R (I) | I is an ideal of R} Based on the properties outlined in (1)−(3), we can establish a topology, denoted as Γ R, on Spec(R) that incorporates Γ(R) This structure is known as the Zariski topology, which plays a crucial role in algebraic geometry.
Let R be a ring Then we have the following:
(2) A subset Y of X R is irreducible if and only if J ? (Y) is a prime ideal of R, where J ? (Y) denote the intersection of all elements in Y.
Let M be a right R-module, and we define Spec(M) as the set of all prime submodules of M, known as the prime spectrum of M Additionally, if N is a fully invariant submodule of M, we will explore its implications within this context.
We have the following lemmas.
(b) If{Ni}i∈J is any family of fully invariant submodules ofM,then T i∈J
(c) IfN, Lare fully invariant submodules ofM,thenV ? (N)∪V ? (L)⊂V ? (N∩L).
Proof It is clear from the definition
(ii) If{N i }i∈J is any family of fully invariant submodules of M,then T i∈J
(iii) If N, Lare fully invariant submodules of M, thenV(N)∪V(L) =V(N∩L).
(i) It is clear from the definition.
(ii) LetP ∈Spec(M).Then we haveP ∈ T i∈J
V(N i ) It follows thatI N i ⊂I P ,for alli ∈J and hence I N i (M)⊂P, for all i∈J Therefore P i∈J
I N i ⊂I P , proving thatI N i ⊂I P ,for all i∈J Thus P ∈ T i∈J
(iii) LetP ∈Spec(M) Then we haveP ∈V(N)∪V(L) It givesIN ⊂IP or IL⊂
I P and then IN∩L =I N ∩I L ⊂I P , proving that P ∈V(N ∩L) Conversely, takeP ∈V(N∩L) ThenIN∩L⊂IP and henceIN.IL⊂IN∩IL ⊂IP.Since
P ∈ X M we have I P ∈ X S So I N ⊂ I P or I L ⊂ I P Thus P ∈ V(N) or
P ∈V(L) and from this we getP ∈V(N)∪V(L).ThereforeV(N)∪V(L) V(N ∩L)
Put Γ ? (M) = {V ? (N)|N is a fully invariant submodule of M} Γ(M) ={V(N)|N is a fully invariant submodule of M}
Note that, from (a)-(c), there exists a topology on X M , having Γ ? (M) as the set of all closed sets if and only if Γ ? (M) is closed under finite union.
From (i)-(iii), there exists a topology, say Γ, on X M , having Γ(M) as the family of all closed sets The topology Γ is called the Zariski topology on the Spec(M).
Let M be a rightR-module Let N and Lbe fully invariant submodules of M Then we have the following:
(1) If I N = I L , then V(N) = V(L) The converse is true if both N and L are prime in M;
(2) V(N) = V(I N (M)) = V ? (I N (M)) In particular, V(P M) = V ? (P M) for any ideal P of a ring S.
(1) It is clear that V(N) =V(L) if I N =I L Conversely, suppose that N and L are prime submodules of M Since V(N) = V(L) we get I N ⊂ I L and I L ⊂
(2) First, we will show that V(N) =V(I N (M)) LetP ∈V(N) Then I N ⊂I P
It follows that IN(M) ⊂ IP(M) ⊂ P and hence I I N (M ) ⊂ IP This shows that P ∈ V(I N (M)) Conversely, let P ∈V(I N (M)) Then I I N (M) ⊂ I P It implies that IN ⊂IP, showing thatP ∈V(N).
Now, we will show thatV(I N (M)) = V ? (I N (M)).LetP ∈V(I N (M)) Then
I I N (M ) ⊂ IP It implies that IN ⊂ IP and hence IN(M) ⊂ IP(M) ⊂ P, showing that P ∈ V ? (I N (M)) Conversely, take P ∈ V ? (I N (M)) Then
IN(M) ⊂ P and thus I I N (M ) ⊂ IP It gives P ∈ V(IN(M)) Therefore,
Let P be any ideal of S We finally show that V(P M) = V ? (P M) Let
Q∈V(P M).Then we getI P M ⊂I Q and henceI P M (M)⊂I Q (M)⊂Q.Since
P ⊂ IP M, it follows that P M ⊂ IP M(M) ⊂ Q and therefore Q ∈ V ? (P M). Conversely, take Q ∈ V ? (P M) Then P M ⊂ Q It implies that I P M ⊂ I Q , and hence Q∈V(P M) Therefore V(P M) =V ? (P M)
Theorem 5.5 Let M be right R-module which is self-generator Then V ? (N)∪
V ? (L) = V ? (N ∩L) for any fully invariant submodules N, L of M In this case, we have Γ ? (M) = Γ(M).
Proof Let P ∈ Spec(M) We have P ∈ V ? (N)∪ V ? (L) if and only it N ⊂
P or L ⊂ P It is equivalent to IN ⊂ IP or IL ⊂ IP We see that IN ⊂
The relationship I P or I L is a subset of I P if and only if the intersection of I N and I L is a subset of I P It is important to note that the intersection of I N and L equals I N intersecting with I L, making the condition equivalent to I N intersecting with L being a subset of I P Given that M is a self-generator, the subset condition holds true if and only if N intersecting with L is a subset of P This equivalently indicates that P belongs to the set V ? (N intersecting with L).
By Theorem 5.5, when M is a self-generator, there exists a topology,say Γ ? , on X M , having Γ ? (M) as the collection of all closed sets Moreover the topology Γ ? coincides with the Zariski topology of X M
Assuming that Spec(M) is non-empty, it follows that Spec(S) is also non-empty We define a map, ψ: Spec(M) → Spec(S), by sending a prime ideal P to its extension IP in Spec(S) This mapping is well-defined and is referred to as the natural map of Spec(M) In the following sections, we will explore various properties of the natural map ψ associated with X M.
Proposition 5.6 The natural map of Spec(M) is continuous More precisely, ψ −1 (V S (P)) =V(P M) for any ideal P of S.
Proof Let U be a closed subset in Spec(S) Then U = V S (P) for some ideal
Let N be an element of Spec(M) It follows that N belongs to ψ −1 (V S (P)) if and only if ψ(N) equals I N, which is contained in V S (P) This is equivalent to stating that P is a subset of I N Furthermore, P is a subset of I N if and only if P M is contained in N This condition is also equivalent to I P M being a subset of I N, leading to the conclusion that N is in V(P M) Therefore, we can establish that ψ −1 (V S (P)) is equal to V(P M), indicating that it is a closed set.
Spec(M) and so ψ is continuous
For any prime ideal P of S, we denote
Then we have the following:
Let M be a right R-module and P, Q ∈ Spec(M) Then the following conditions are equivalent:
(1) The natural map ψ : Spec(M)→Spec(S) is injective;
Proof (1)⇒(2) Assume thatV(P) =V(Q).SinceP, Q∈X M ,we haveI P =I Q
It implies that ψ(P) = ψ(Q) and henceP =Qbecause ψ is injective.
(2) ⇒ (3) If Spec P (M) = ∅, then |Spec P (M)| = 0 We now assume that Spec P (M)6=∅ LetP, Q∈Spec P (M) Then I P =I Q =P would imply that
V(P) =V(Q), and hence P =Q by (2) Thus |Spec P (M)| ≤1.
(3) ⇒ (1) Assume that ψ(P) = ψ(Q), where P, Q ∈ Spec(M) Then
IP =IQ =P, a prime ideal of S ThusP, Q ∈Spec P (M) By (3) we get P =Q.
Theorem 5.8 Let M be a right R-module Let ψ :X M →X S be the natural map of X M If ψ is surjective, then ψ is open and closed.
According to Proposition 5.6, the map ψ is continuous, satisfying ψ −1 (V S (P)) = V(P M) for any ideal P of S For a fully invariant submodule N of M, we have ψ −1 (V S (I N )) = V(I N (M)) = V(N) Given that ψ is surjective, it follows that ψ(V(N)) is contained in V S (I N ), which is closed in Spec(S), proving that ψ is closed Additionally, we find that ψ(X M \ V(N)) = ψ(ψ −1 (X S) \ ψ −1 (V S (I N))) = ψ(ψ −1 (X S \ V S (I N))) = X S \ V S (I N), indicating that this set is open in Spec(S) Thus, ψ is also open.
For each f ∈ S, denote X f M = X M \ V(Sf(M)) and X f S = X S \
V S (Sf S) Then X f M is an open set of X M and it is clear that X 0 M = ∅ and
Proposition 5.9 Let M be a right R-module with the natural map ψ :X M →X S and f ∈S Then:
(2) ψ(X f M )⊂X f S The equality holds if ψ is surjective.
(2) From X f M =ψ −1 (X f S ) we get ψ(X f M )⊂X f S Ifψ is surjective then it is clear that we have the equality
Theorem 5.10 Let M be a right R-module Then the set B = {X f M | f ∈ S} forms a basis for the Zariski topology on X M
Proof If X M = ∅ Then B = ∅ In this case, the theorem is true Now, assume that X M 6= ∅ Let U be an open set of X M Then U = X M \V(N), where N is a fully invariant submodule of M Recall that V(N) = V(I N (M)) = V ? (I N (M)).
B is a basis for the Zariski topology of X M
Theorem 5.11 Let M be a right R-module If the natural map ψ of X M is surjective, then X M is compact.
Proof Since the set B = {X f M | f ∈ S} forms a basis for the Zariski topol- ogy on X M , for any open cover of X M , we can write X M = S i∈Ω
X f S i By Lemma 5.1, X S is compact Then there exists a finite set J ⊂ Ω such that X S = S i∈J
Let Y be a subset of X M We denote the intersection of all elements in Y by J(Y), and the closure of Y in X M by cl(Y) Then we have the following proposition:
Proposition 5.12 Let M be a right R-module and Y be a subset of X M Then
V(J(Y)) = cl(Y) Therefore, Y is closed in X M if and only if V(J(Y)) =Y.
Proof LetP ∈Y.Then we getJ(Y)⊂P It follows thatIJ(Y )⊂IP, showing that
P ∈V(J(Y)).ThusY ⊂V(J(Y)).Now, letV(N) be any closed subset ofX M such thatY ⊂V(N).ThenIN ⊂IP,for anyP ∈Y.ThusIN ⊂ T
P =IJ (Y ). Let Q ∈ V(J(Y)) Then I Q ⊃ I J(Y ) ⊃ I N , and hence Q ∈ V(N) Therefore
V(J(Y)) ⊂ V(N) It means that V(J(Y)) is the smallest closed subset of X M containingY Therefore V(J(Y)) =cl(Y)
Proposition 5.13 Let M be a right R-module and P ∈ X M Then we have the following:
(2) For any Q∈X M , Q∈cl({P}) if and only if IP ⊂IQ if and only if V(Q)⊂
(3) Let M be a self-generator module Then the set {P} is closed in X M if and only if P is a maximal prime submodule of M.
(1) It is clear from Proposition 5.12.
(2) We have Q ∈ cl({P}) if and only if Q ∈ V(P) (by (1)) It is equivalent to
IP ⊂IQ This condition is equivalent toV(Q)⊂V(P).
(3) Suppose that{P}is closed Then V(P) =cl({P}) ={P}.LetQbe a prime submodule ofM andQ⊃P.ThenIQ⊃IP It implies thatQ∈V(P) = {P}, showing that Q=P ThereforeP is a maximal prime submodule of M.
Conversely, take Q ∈ cl({P}) Then we get Q ∈ V(P) and hence I P ⊂ I Q SinceM is a self-generator, we haveP ⊂Q SinceP is a maximal prime, we haveP =Q.Thus cl({P}) ={P},proving that {P}is closed
Recall that a topological space A is called irreducible if for any decom- position A = B ∪C with B, C are closed sets of A, we have A = B or A = C.
A subset B of A is deemed irreducible if it cannot be expressed as the union of two closed sets in A, implying that if B is contained within the union of two closed sets B1 and B2, then it must be entirely contained in one of them According to Proposition 5.13, we derive the corollary that for any right R-module M, the variety V(P) is an irreducible closed subset of the space X_M for any point P in X_M.
Proof Since {P} is irreducible, cl({P}) is irreducible But cl({P}) = V(P), we have V(P) is irreducible
Recall that, a topological space X is called a T0 space if for any two distinct points ofX,there is an open set which contains precisely one of the points.
Let M be a rightR-module with the Zariski topology onSpec(M) Then the following conditions are equivalent:
(2) The natural map ψ :X M →X S is injective;
Proof (1) ⇒ (3) Suppose that X M is a T 0 space Let P, Q ∈ X M such that
V(P) = V(Q) Then cl({P}) = cl({Q}) If P 6= Q, then there exists an open set
U of X M such that P ∈ U but Q 6∈ U Then X M \U is a closed set of X M such that Q ∈ X M \U It follows that cl({Q}) ⊂ X M \U Thus cl({P}) ⊂ X M \U and hence P 6∈U,a contradiction Therefore P =Q.
(3) ⇒ (1) Let P, Q be two distinct points of X M By (3), we get
V(P)6=V(Q).If every open set of X M which contains one point will contain the other point, then P ∈cl({Q}) and Q∈cl({P}) Thus P ∈ V(Q) and Q∈ V(P), and hence V(P) =V(Q), a contradiction.
Corollary 5.16 Let M be a right R-module which is a self-generator Then X M is a T 0 space with the Zariski topology.
Proof Let P, Q ∈ X M If V(P) = V(Q), then I P = I Q Since M is a self- generator, we have P =Q.Thus X M is a T 0 space
Let X be a topological space and x, y be distinct points in X.We say thatxand ycan beseperatedif each lies in an open set which does not contain the other point X is called aT 1 spaceif any two distinct points inX can be seperated.
A topological space X is a T1 space if and only if all points of X are closed in X.
Theorem 5.17 Let M be a right R-module which is a self-generator Then X M is a T 1 space if and only if M axp(M) = Spec(M), where M axp(M) is the set of all maximal prime submodule of M.
Let M be a right R-module with surjective natural map ψ :X M →X S Then the following conditions are equivalent:
Proof (1) ⇒ (2) Since ψ is surjective, continuous and X M is connected, we get