Semi-commutative Modules and Armendariz Modules As Published in Arab Journal of Mathematical Sciences, Vol.8 (June

2002), pp.53-65. (Two very minor corrections have been added.)

By Ardeline Mary Buhphang and M.B.Rege Department of Mathematics, North-Eastern Hill University, Shillong-793 022, Meghalaya, India

e-mail addresses: (1) Buhphang: ardelineb@hotmail.com (2) Rege: mb29rege@yahoo.co.in (corresponding author)

Abstract. A ring R is semi-commutative if whenever elements a, b in R satisfy ab = 0, then acb = 0 for each element c of R. Armendariz rings are defined through polynomial rings over them. The two concepts are closely related and have straight-forward extensions to modules. We prove a number of results concerning semi-commutative modules and Armendariz modules. Motivated by a characteristic property of commutative Gaussian rings, a ring R is called left Gaussian if all cyclic left R-modules are Armendariz.Basic properties of Gaussian rings are given.

Key words and phrases. Semi-commutative ring, Armendariz ring, flat module, Gaussian ring.

2000 Mathematics Subject Classification: Primary 16S36, 16U99.

§1. Introduction. A ring R (associative with identity) is abelian if each idempotent in R is central; it is reduced if it has no nonzero nilpotent elements; it is semi-commutative if whenever elements a, b in R satisfy ab = 0, then acb = 0 for each element c of R. It is an elementary exercise to show that reduced rings are abelian. It was recorded in [5,§4] that the class of semicommutative rings lies between the classes of reduced rings and abelian rings.

Closely related to semi-commutative rings are Armendariz rings. The notion of an Armendariz ring was introduced by Rege and Chhawchharia in [9]. A ring R is Armendariz if given polynomials f(X) = Σa_{i}X^{i} and g(X) = Σb_{j}X^{j} with coefficients in R, the condition f(X)g(X) = 0 implies a_{i}b_{j} = 0 for every i and j. A number of examples of Armendariz rings were recorded by Anderson and Camillo [1] and Kim and Lee[7]; they also extended several results of [9]. Huh, Lee and Smoktunowicz [6] made a comparative study of Armendariz rings and semi-commutative rings. The class of Armendariz rings also lies between the classes of reduced rings and abelian rings by [3, Lemma 1] and [6, Corollary 8]. Semi-commutative rings need not be Armendariz by [9, 3.2] and Armendariz rings need not be semi-commutative by [6, Example 14]. Both concepts are defined through ‘vanishing conditions’ on elements and have straight-forward extensions to modules. In §2 of this paper we initiate the study of basic properties of semi-commutative modules and Armendariz modules. All left ideals of the ring R are two-sided if and only if all ( cyclic ) left R-modules are semi-commutative (Proposition 2.11). A consequence of Theorems 2.15 and 2.16 is the result that all flat modules over a reduced ring are both Armendariz and semi-commutative. In §3 we study rings over which all cyclic left modules are Armendariz; these rings have been called as ‘left Gaussian rings’. We also collect ( in 3.12 ) a number of questions which arise in the course of our study.

§2. Basic properties and examples. This section is devoted to developing basic properties and furnishing a number of examples of the concepts under consideration. Easier proofs are omitted. It is hoped that the omission of ‘respectively’ in parentheses does not cause confusion.

By a ring we mean an associative ring with identity. Ring homomorphisms are unitary; modules are unitary left modules. Throughout R denotes a ring and D a commutative domain. All our left-sided concepts and results have right-sided counterparts.

2.1. Definition. A module M over a ring R is semi-commutative if it satisfies the following condition: whenever elements a ∈ R and m ∈ M satisfy am = 0, then acm = 0 for each element c of R.

The concept of an Armendariz module was mentioned in [9, 4.7].

2.2. Definition. Let M be a module over a ring R. Let M[X] be the corresponding polynomial module over R[X]. The module M is an Armendariz module if whenever polynomials f(X)=Σa_{i}X^{i} ∈ R[X], g(X)=Σm_{j}X^{j} ∈ M[X] satisfy f(X)g(X)=0, we have a_{i}m_{j} = 0 for every i and j.

For ease of reference we recall the following result ( see [1, Theorem 12] ).

2.3. Proposition. If M is a torsion free module over a commutative domain D then M is Armendariz as a D-module.

2.4. Remark. A ring R is semi-commutative (Armendariz) if and only if the module _{R}R is semi-commutative (Armendariz).

We next record a ‘change of rings’ result.

2.5. Proposition. Let θ:R⟶A be a ring homomorphism and let M be an A-module. Regard M as a left R-module via θ. Then we have:

(i) if _{A}M is semi-commutative (Armendariz), then _{R}M is semi-commutative (Armendariz); (ii) if θ is onto, then the converses of the statements in (i) hold; (iii) if A is a semi-commutative ( Armendariz ) ring , then A is semicommutative ( Armendariz ) as a left R-module.

Proof. For the sake of notational simplicity we prove the result in the semi-commutative case. The proof in the Armendariz case is similar.

(i) Assume that r, s ∈ R, m ∈ M, and rm = 0. Then θ(r)m = 0 implies

(rs)m = θ(rs)m = θ(r)θ(s)m = 0

since the left A-module M is semi-commutative. (ii) The proof is similar to that of part (i). (iii) This is a consequence of 2.4 and part (i).□

The following assertion is immediate.

2.6. Proposition. The class of semi-commutative( Armendariz ) Rmodules is closed under direct sums, direct products and submodules.

An R-module is torsionless if it is a submodule of a direct product of copies of R. If M is a faithful R-module then R is a submodule of a direct product of copies of M. An application of 2.4 and 2.6 yields the following theorem.

2.7. Theorem. The following conditions are equivalent.

(1) R is a semi-commutative ( Armendariz ) ring.

(2) Every torsionless R-module is semi-commutative ( Armendariz ).

(3) Every submodule of a free R-module is semi-commutative (Armendariz).

(4) There exists a faithful R-module which is semi-commutative ( Armendariz ).

2.8. Remark. If M is a left R-module \overline{R} denotes the ring R/ann(M) and E(M) = End _{R}(M) the ring of endomorphisms of M. With this notation consider the following conditions.

(1) The left R-module M is semi-commutative ( Armendariz ).

(2) The left \overline{R}-module M is semi-commutative (Armendariz).

(3) \overline{R} is a semi-commutative (Armendariz) ring.

(4) The right E(M)-module M is semi-commutative (Armendariz).

(5) The ring E(M) is semi-commutative ( Armendariz).

An application of 2.5 yields the equivalence of conditions (1) and (2); since the left R-, right E(M)-bimodule M is faithful as a left \overline{R}-module and also faithful as a right E(M)-module, applying (4)⇒(1) of Theorem 2.7 we get (2)⇒(3) and (4)⇒(5).

Since the definition of a semi-commutative ( Armendariz ) module involves a single element ( finitely many elements) of the module the following result is obvious.

2.9. Proposition. A module M is semi-commutative ( Armendariz ) if and only if every cyclic ( finitely generated ) submodule of M is semicommutative ( Armendariz ).

A ring R is left invariant if every left ideal of R is two-sided. In 2.11 we characterize left invariant rings as rings over which all left modules are semi-commutative.

2.10. Proposition. If the cyclic left R-module R/J is semi-commutative, then J is an ideal of R.

Proof. Let x ∈ J, r ∈ R. Then we have (for \overline{1} = 1 + J ∈ R/J), x\overline{1} = 0 which implies xr\overline{1} = 0, yielding xr ∈ J.□

2.11. Proposition. The following conditions are equivalent.

(1) R is left invariant.

(2) Every left R-module is semi-commutative.

(3) Every cyclic left R-module is semi-commutative.

Proof. (1) ⇒ (2). Let M be a left R-module. Suppose that a ∈ R and m ∈ M satisfy am = 0. Let c ∈ R. Since the left ideal Ra is two-sided we have aR ⊆ Ra. It follows that acm ∈ Ram = 0. The implication (2) ⇒ (3) is trivial, while (3) ⇒ (1) follows from 2.10.□

2.12. Examples. For a field K and an integer n ≥ 2, let A = M_{n}(K), the full matrix ring over K. The ring A is non-abelian, non-semi-commutative and non-Armendariz. (Indeed the zero module is the only A-module which is semi-commutative ( Armendariz ). Notice also that the example of the zero ideal in A shows that the converse of 2.10 does not hold.) The ring A is the homomorphic image of a domain, namely, the ring R of polynomials in sufficiently many non-commuting indeterminates over K. Let θ : R ⟶ A be an onto ring homomorphism. By 2.5(ii) A is neither semi-commutative nor Armendariz as an R-module. However, it is a factor module of _{R}R which is semi-commutative and Armendariz. Further, the left R-module M = R⊕A is R-faithful, but M ( which has A as a submodule ) is neither semi-commutative nor Armendariz as a left R-module; this shows that (3)⇒ (1) does not hold in ( either case of ) Theorem 2.8.

In the rest of this section we present a few results related to Proposition 2.3. Proposition 2.13 is a straight-forward extension. Theorem 2.15 asserts that flat modules over Armendariz rings are Armendariz, and 2.16 is a semi-commutative analogue of this result. (Flat modules are torsion free, even over arbitrary rings [10, Example 1, §10, Chapter 1].) In Proposition 2.20 we extend 2.3 in another direction. Later we show in Proposition 3.8 that over Prüfer domains all modules are Armendariz.

2.13. Proposition. Let D be a commutative domain and M a D-module. The module M is Armendariz if and only if its torsion submodule T(M) is Armendariz.

Proof. Let f(X)=Σa_{i}X^{i} ∈ D[X], g(X)=Σm_{j}X^{j} ∈ M[X] satisfy f(X)g(X) = 0. We have the system of equations a_{0}m_{0} = 0, a_{0}m_{1} + a_{1}m_{0} = 0, a_{0}m_{2} + a_{1}m_{1} + a_{2}m_{0} = 0, …, a_{s}m_{t} = 0 ( for some s, t ). We can assume a_{0} ≠ 0.

Now the second of these equations yields (on multiplying by a_{0} and using a_{0}m_{0} = 0) a_{0}^{2}m_{1} = 0. Thus a_{0}^{2} annihilates both m_{0} and m_{1}. The third equation now implies a_{0}^{3}m_{2} = 0. Continuing we get g(X) ∈ T(M)[X]. Since T(M) is Armendariz as a D-module, we conclude that a_{i}m_{j} = 0 for all i, j, proving the ‘if’ part. The other implication is trivial.□.

Next we consider flat modules. We first recall a well-known result [10,Corollary 11.4, Chapter 1]; module homomorphisms are written on the right.

2.14. Proposition. Let M be a flat R-module and 0→K→F→M→0 be an exact sequence with F free over R. For each finite family u_{1}, u_{2}, …, u_{m} of elements of K there exists an R-homomorphism v: F→K such that (u_{i})v = u_{i} for each i.

2.15. Theorem. Flat modules over Armendariz rings are Armendariz. Proof. Let R be an Armendariz ring and M a flat R-module. Let 0 → K

→ F → M → 0 be an exact sequence with F free over R. ( In what follows for an element y of F, we denote \overline{y} = y + K in M.) Let f(X) = Σ_{i=0}^{s}a_{i}X^{i} ∈ R[X] and g(X) = Σ_{j=0}^{t}\overline{y_{j}}X^{j} ∈ M[X] satisfy f(X)g(X) = 0. Then we have

a_{0}\overline{y_{0}} = 0

a_{0}\overline{y_{1}} + a_{1}\overline{y_{0}} = 0
⋮
a_{s}\overline{y_{t}} = 0

Therefore the elements a_{0}y_{0}, a_{0}y_{1} + a_{1}y_{0}, …, a_{s}y_{t} all belong to K. Proposition 2.14 ensures the existence of an R-module homomorphism v:F→K such that (a_{0}y_{0})v = a_{0}y_{0}, …, (a_{s}y_{t})v = a_{s}y_{t}. Write w_{j} := (y_{j})v − y_{j} for j = 0, …, t. Each w_{j} is an element of F and therefore the polynomial h(X) = Σ_{j=0}^{t}w_{j}X^{j} ∈ F[X]. Now F is free, and therefore Armendariz as an R-module. Since f(X)h(X) = 0 we have a_{i}w_{j} = 0 for each i and j. It follows that a_{i}y_{j} ∈ K for each i and j. Hence a_{i}\overline{y_{j}} = 0 in M proving that the module M is Armendariz. □

2.16. Theorem. Flat modules over semi-commutative rings are semicommutative.

Proof. The proof is similar to ( and notationally simpler than ) the proof of Theorem 2.15. We omit it.□

By a regular ring we mean a von Neumann regular ring. For later reference we record the following immediate consequence of 2.15, 2.16 and the well-known result that over a regular ring all modules are flat.

2.17. Remark. If R is a reduced, regular ring (in particular, a commutative regular ring) all R-modules are Armendariz as well as semi-commutative.

Next we study localizations. Let M be an R-module and let C be the centre of R. If S is a multiplicatively closed subset of C then S^{−1}M has a S^{−1}R-module structure. Let Spec(C) ( respectively, Max(C) ) denote the set of all prime (respectively, maximal) ideals of C. The module M is S-torsion free if whenever s is an element of S and m is a nonzero element of M, we have sm ≠ 0. Applying standard localization techniques we can prove Propositions 2.18 and 2.19.(See [1] and [6] for related results.)

2.18. Proposition. Let M be S-torsion free. The R-module M is semicommutative ( Armendariz ) if and only if the S^{−1}R-module S^{−1}M is semicommutative (Armendariz).

2.19. Proposition. Let M be an R-module and let C = Centre(R). Then the following conditions are equivalent.

(1) M is semi-commutative ( Armendariz ).

(2) S^{−1}M is a semi-commutative ( Armendariz ) S^{−1}R-module for each multiplicatively closed subset S of C.

(3) M_{P} is a semi-commutative ( Armendariz ) R_{P}-module for each P ∈ Spec(C).

(4) M_{Q} is a semi-commutative ( Armendariz ) R_{Q}-module for each Q ∈ Max (C).

Let S_{0} denote the set of all non-zero-divisors of a commutative ring R. The ring R is quasi − regular if its classical quotient ring S_{0}^{−1}R is regular. It is well-known that commutative pp-rings (i.e., rings in which every principal ideal is projective) are quasi-regular. Since commutative domains are trivially quasi-regular, the following result extends Proposition 2.3.

2.20. Proposition. If R is a commutative quasi-regular ring, and M is an S_{0}-torsion free R-module, then M is Armendariz.

Proof. Since A = S_{0}^{−1}R is a commutative regular ring, S_{0}^{−1}M is Armendariz as an A-module by 2.17. It follows, by 2.18, that the R-module M is Armendariz.□

§3. Gaussian rings. Let R be a ring and let M be a left R-module. For g ∈ M[X], its content A_{g} is the R-submodule of M generated by the coefficients of g; a commutative ring R is Gaussian [11] if A_{fg} = A_{f}A_{g} for all f, g ∈ R[X] .

We have the following classical result due to Tsang ([11], [4, Exercise 7 on page 351] and Theorem 1.3 in [2].)

3.1. Theorem. A commutative domain D is Gaussian if and only if it is Prüfer (i.e. every finitely generated nonzero ideal of D is invertible ).

If ‘P’ is a term applicable to rings, a ring R is completely P if every homomorphic image of R is P. We have the following reformulation of Theorem 8 of [1].

3.2. Proposition. A commutative ring is Gaussian if and only if it is completely Armendariz.

Gaussian/Prüfer domains play a significant role in multiplicative ideal theory [4]. We have not come across a non-commutative generalization of the concept of a Gaussian ring. We attempt to bridge this gap ( see Definitions 3.4 ). Our motivation comes from the following result.

3.3. Proposition. Let R be a commutative ring. The ring R is Gaussian if and only if every cyclic R-module is Armendariz.

Proof. By 2.4 and 2.5 the cyclic R-module R/J is Armendariz if and only if the ring R/J is Armendariz. Now apply 3.2.□

3.4. Definitions. A ring R is left Gaussian if every cyclic left R-module is Armendariz; it is left strongly Gaussian if every left R-module is Armendariz.

We can use 2.5 to prove the following remarks.

3.5. Remarks.(a) The class of left ( strongly ) Gaussian rings is closed under homomorphic images and the formation of finite direct products; consequently left Gaussian rings are completely Armendariz.

(b) Left invariant, completely Armendariz rings are left Gaussian.

Example 3.6 shows: (1) the Gaussian property is not left-right symmetric, and (2) completely Armendariz rings need not be left Gaussian.

3.6. Example. Let L be a field and let K = L( Y, T ) be the field of rational functions in two commuting indeterminates Y and T. Let h be the endomorphism of the field K defined via h(Y) = Y^{2} and h(T) = T^{2}. With notation as in [9], the ring R = K ( + )_{h}K has the following properties:

(i) Apart from R and 0 the only right ideal of R is 0 ⊕ K, which is also a left ideal. Since the ring R_{0} = R/(0 ⊕ K) is a field, R_{0} is Armendariz as a right R-module. By Proposition 2.8 of [9] R itself is an Armendariz ring. The ring R is thus a right Gaussian and completely Armendariz ring.

(ii) Let W = h(K) = L(Y^{2}, T^{2}), a subfield of K. Then B = 0 ⊕ W is a left ideal of R. We verify that the cyclic left R-module R/B is not Armendariz, and thus the ring R is not left Gaussian.

Write f(X) = (0, Y) + (0, T) X ∈ R[X], and g(X) = \overline{(Y, 0)} + \overline{(−T, 0)} X ∈ (R/B)[X]. Then f(X)g(X) = 0; however, we have (0, Y) \overline{(−T, 0)} ≠ 0 in R/B.

A commutative ring is arithmetical if its lattice of ideals is distributive, equivalently, if every finitely generated ideal is locally principal. Prüfer domains, principal ideal rings and regular rings are arithmetical. Finite products of arithmetical rings are arithmetical. It was pointed out in [1] (and follows from Theorem 1.3 of [2]) that arithmetical rings are Gaussian. In Proposition 3.8 we show that a commutative ring is arithmetical if and only if it is strongly Gaussian. We first recall a result implicit in Remark 1.5 of [2].

3.7. Lemma. Let R be a commutative ring and let f(X) ∈ R[X]. The ideal A_{f} is locally principal if and only if for each R-module M and each g(X) ∈ M[X] we have A_{f}A_{g} = A_{fg}.

3.8. Proposition. The commutative ring R is strongly Gaussian if and only if R is arithmetical.

Proof. Let R be an arithmetical ring, let M be an R-module and let f(X) ∈ R[X] and g(X)∈ M[X] satisfy f(X)g(X) = 0. Since A_{f} is locally principal, by Lemma 3.7 we have A_{f}A_{g} = A_{fg} = 0 implying that M is Armendariz. This proves the ‘if’ part. Next, let R be a strongly Gaussian ring and let B be a finitely generated ideal of R. Clearly B = A_{f} for some polynomial f(X) ∈ R[X]. Let M be an R-module, let g(X) ∈ M[X], let N = A_{fg}, let \overline{M} be the factor module M/N and let \overline{g}(X) denote the residue class of g(X) in \overline{M}[X] = M[X]/N[X]. We have A_{f\overline{g}} = 0 implying ( since \overline{M} is Armendariz ) 0 = A_{f}A\overline{g} = A_{f}A_{g}/A_{fg}. Hence A_{f}A_{g} = A_{fg} implying ( by 3.7 again ) that B = A_{f} is locally principal. □

3.9. Remarks. (a) As a special case of Proposition 3.8 all abelian groups are Armendariz as Z-modules.

(b) Using 3.1, 3.8 and Theorem 12 of [1] we can deduce:if D is a domain then D is Prüfer ⇔ D is Gaussian ⇔ D is strongly Gaussian ⇔ for each Dmodule M the idealisation D( + )M is an Armendariz ring. ( cf. Theorems 2.2, 2.3 and Corollary 2.9 of [9].)

(c) It is easy to give examples of commutative Gaussian rings which are not arithmetical and therefore not strongly Gaussian. Let K be a field and V a K-vector space of dimension ≥ 2. Then the idealisation K( + )V is a local ring which is Gaussian [1] with a non-distributive lattice of ideals. Choosing K to be the field of two elements and V a vector space of dimension 2 we thus get a Gaussian ring of 8 elements which is not strongly Gaussian.

3.10. Remarks. Some back-ground to the example studied in 3.11 is provided by the following remarks.

(a) Commutative P.I.R.s are completely Armendariz ( ≡ Gaussian); this follows from 3.8 and was recorded in the remarks after Corollary 9 of [1]. Any semi-simple (artinian) non-reduced ring is a left and right P.I.R. which is not Armendariz. Let H = R + Ri + Rj + Rk be the division ring of real quaternions. In Example 3.11 we verify that the ring H[X] (which is a left and right P.I.D.) is not completely abelian, and therefore is neither completely Armendariz nor completely semi-commutative.

(b) Theorem 16 of [1] asserts that a commutative ring R is regular if and only if R[X] is completely Armendariz. Example 3.11 shows that this result does not extend to non-commutative rings.

(c) It was shown by Anderson and Camillo [1, Theorem 5] that if R is a reduced ring then for each positive integer n the ring R[X]/(X^{n}) is an Armendariz ring. The example below also shows that a conjectured generalization of this result, namely, ‘if R is a reduced ring then for each polynomial f(X) in the centre of R[X] the ring R[X]/(f(X)) is an Armendariz ring’ is false.

3.11. Example. We show that the ring T = H[X]/(X^{2}+1) has infinitely many non-central idempotents, and is, in particular, non-abelian. Thus H[X] is not completely abelian.

(i) Claim. Let α, β be real quaternions. The residue class of β X + α in T is an idempotent element if and only if conditions (1) below hold.

Proof. The equality

β X + α = (βX + α)^{2} (mod X^{2} + 1)

holds if and only if

βX + α = β^{2}X^{2} + βαX + αβX + α^{2} (mod X^{2} + 1)

⇔ 0 = −β^{2}X^{2} + (β − βα − αβ)X + α − α^{2} (mod X^{2} + 1) ⇔ 0 = −β^{2}(X^{2} + 1) + (β − βα − αβ)X + α − α^{2} + β^{2} (mod X^{2} + 1) ⇔ 0 = (β − βα − αβ)X + (α − α^{2} + β^{2}) (mod X^{2}+1).

Necessary and sufficient conditions for the last equality to hold are

β − βα − αβ = 0 and α − α^{2} + β^{2} = 0  (1)

(ii) Claim. There are infinitely many idempotents in T of the form the residue class of kX + α.

Proof. Assume β = k. Then the equations in (1) reduce to

k − kα − αk = 0 and α − α^{2} − 1 = 0  (2)

Let α = α_{0} + α_{1}i + α_{2}j + α_{3}k, where the α_{m} (0 ≤ m ≤ 3) are real numbers. Then easy computations show that conditions (2) hold if and only if α_{0} = 1/2, α_{3} = 0 and α_{1} and α_{2} satisfy α_{1}^{2} + α_{2}^{2} = 3/4. Since there are infinitely many pairs (α_{1}, α_{2}) of real numbers satisfying this last equation our claim is proved.

(iii) Since j (kX + α)− (kX + α) j is never a multiple of X^{2} +1 it follows that for each α the residue class of kX + α does not belong to the centre of T.

In the next paragraph we collect some questions which arise naturally in the context of our study. A ring R is left quasi-invariant if every maximal left ideal of R is two-sided; it is semi-primitive if its Jacobson radical vanishes.

3.12. Remark. It would be of interest to get examples (if they exist) of (i) a left (strongly)Gaussian ring which is not left ( quasi-)invariant (ii) a semi-primitive left Gaussian ring which is not reduced and (iii) a reduced, left Gaussian ring which is not left strongly Gaussian.

3.13. Remark. (a) Clearly an example satisfying the conditions in (ii) will also satisfy those of (i) since quasi-invariant, semi-primitive rings are reduced by [8, Corollary 4.5].

(b) While in 3.12(i) we ask for an example of a left Gaussian ring which is not left quasi-invariant, the domain R considered in 2.12 is an Armendariz ring which is neither left nor right quasi-invariant (since it has the ring M_{n}(K) as a homomorphic image).

(c) Left Gaussian rings are Armendariz, and therefore abelian; abelian rings are Dedekind finite (i.e., for elements a, b of the ring ab = 1 implies ba = 1 ). Left quasi-invariant rings are also Dedekind finite. It follows that left Gaussian rings and left quasi-invariant rings are sub-classes of the class of completely Dedekind finite rings. Thus we have some (albeit meagre) evidence for suspecting that( under fairly general conditions) the implication left Gaussian ⇒ left quasi-invariant may hold.

Acknowledgements. Some results in this paper were obtained during a collaboration with Indrakshi Choudhuri. Thanks are also due to the referees for valuable suggestions.

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