Armendariz and SFT Properties in Subring Retracts

Armendariz and SFT Properties in Subring Retracts

Chahrazade Bakkari

Abstract. In this work, we investigate the transfer of Armendariz and SFT properties between a commutative ring and its subring retract. The article includes a brief discussion of the scope and precision of our results.

Mathematics Subject Classification (2000).Primary 13Agg; Secondary 13D05, 13B02.

Keywords.Armendariz ring, SFT ring, subring retract, trivial ring extension.

1. Introduction

Throughout this paper all rings are assumed to be commutative with identity element and the dimension of a ring means its Krull dimension.

Let R be a commutative ring. The contentC(f) of a polynomialf ∈R[X] is the ideal of R generated by all coefficients of f. One of its properties is that C(.) is semi-multiplicative, that isC(f g)⊆C(f)C(g); and a polynomialf ∈R[X] is said to be Gaussian over R if C(f g) = C(f)C(g) holds for any polynomial g overR. A polynomialf ∈R[X] is Gaussian providedC(f) is locally principal by [11, Remark 1.1]. A ringR is said to be Gaussian if every polynomial inR[X] is Gaussian. And a domain is Gaussian if and only if it is a Pr¨ufer domain. See for instance [1],[5],[9],[11].

A ringRis called an Armendariz ring if whenever polynomialsf= m i=0

a_iXⁱ

and g= n i=0

b_iXⁱ ∈R[X] satisfyf g = 0, we haveC(f)C(g) = 0 (that isa_ib_j = 0 for every i and j). It is easy to see that subrings of Armendariz rings are also Armendariz. E. Armendariz [2, Lemma 1] noted that any reduced ring (i.e., ring

© 2009 Birkhäuser Verlag Basel/Switzerland

1660-5446/09/030339-7, published online September 10, 2009 DOI 10.1007/s00009-009-0012-9

241

of Mathematics

without non-zero nilpotent elements) is an Armendariz ring. Also, D.D. Ander- son and V. Camillo [1] showed that a ring R is Gaussian if and only if every homomorphic image of Ris Armendariz. See for instance [1],[2],[19],[20].

An ideal I is called anSF T-ideal if there exists a naturel number k and a finitely generated idealJ ⊆I such thata^k ∈J for eacha∈I. An SF T ring is a ring in wich every ideal is anSF T-ideal.

In [3], Arnold studies the Krull dimension of a power series ringR[[x]] over a ringR and showed that the dimension is infinite unlessRis anSF T ring, which forces us to consider onlySF T rings when we study finite-dimensional power series extensions.

For any ringA with finite Krull dimension, we have:

ANoetherian =⇒dimA[[X]]<∞=⇒A SF T ring.

One important family of SF T rings is that ofSF T Pr¨ufer domains, which are also called generalized Dedekind domains. The beautiful discovery of Arnold is that, for D a finite-dimensional SF T-Pr¨ufer domain, dimD[[x₁, . . . , x_n]] = n(dimD) + 1, and soD[[x₁, . . . , x_n]] is anSF T ring [4]. In 2007, Kang and Park [14, Theorem 10] extend Arnold’s result to the infinite-dimensional case, thus prov- ing that over an infinite-dimensionalSF T Pr¨ufer domainD, the power series ring D[[x₁, . . . , x_n]] is anSF T ring.

SF T rings are similar to Noetherian rings and they have many nice proper- ties. It had been a long-standing open question if the power series extension of an SF T ring is also anSF T ring. Coykendall’s counterexample to this appears in [6].

Remark that these rings are coherent.

For two ringsA⊆B, we say thatAis a module retract (or a subring retract) ofBif there exists anA-module homomorphismφ:B−→Asuch thatφ|A=id|A; φ is called a module retraction map. If such a map φexists, B contains Aas an A-module direct summand.

Considerable works have been concerned with the descent and ascent of a variety of finiteness and related homological properties between a ring and its subring retract. See for instance [8],[17].

A special application of subring retract is the notion of trivial ring extension.

LetAbe a ring,EanA-module andR=A∝E, the set of pairs (a, e) witha∈A and e∈E, under coordinatewise addition and under an adjusted multiplication defined by (a, e)(a, e) = (aa, ae+ae),for alla, a∈A, e, e∈E. ThenRis called the trivial ring extension of A byE. It is clear that A is a module retract ofR, where the module retraction mapφis defined byφ(x, e) =xand so [Ker(φ)]²= 0.

Trivial ring extensions have been studied extensively; the work is summarized in Glaz [8] and Huckaba [12]. See for instance [8],[12],[13],[18].

In this work, we study the transfer of Armendariz and SFT properties to subring retract. A special applications is devoted to trivial ring extensions. Our results generate new and original examples which enrichy the current literature with new families of Armendariz rings and non-coherent SFT rings.

2. Transfer of Armendariz Property in Subring Retracts

This section develops a result of the transfer of Armendariz property to subring retract. And so, we will construct a new class of Armendariz rings (with zero- divisors). Recall thatN il(R) is the set of nilpotent elements in the ringR.

Theorem 2.1. Let R be a ring and Aa subring retract of R.

1) If R is an Armendariz ring, then so isA.

2) Assume that (A, M)is a local ring and the module retraction mapφverifies (Ker(φ))² ⊆Ker(φ),Ker(φ)⊆ N il(R) andKer(φ) is a simple A-module such that M Ker(φ) = 0. Then R is an Armendariz ring if and only if so is A.

Before proving Theorem 2.1, we establish the following Lemma.

Lemma 2.2. Let Rbe a ring and Aa subring retract of R. IfR is an Armendariz ring, then so is A.

Proof. Assume that R is an Armendariz ring and let f(X) = n i=0

a_iXⁱ, g(X) = m

i=0

b_iXⁱ be two polynomials of A[X], where n and m are two positive integers such thatf g= 0. Our aim is to prove thatC_A(f)C_A(g) = 0. Letαbe an element ofC_A(f)C_A(g), we haveα∈C_R(f)C_R(g) so, sinceRis Armendariz,α∈C_R(f g), i.e., α=

n+m

k=0

⎛

⎝

i+j=k

a_ib_j

⎞

⎠r_k where r_k is an element ofR for any 0 ≤k ≤nm.

Then,α=φ(α) =

n+m

k=0

⎛

⎝

i+j=k

a_ib_j

⎞

⎠φ(r_k) whereφis the module retraction map,

which prove thatα∈C_A(f g) = 0. Thus,Ais an Armendariz ring.

Proof of Theorem 2.1.

1) If Ris an Armendariz ring, then so isAby Lemma 2.2.

2) Assume that (A, M) is a local ring, the module retraction mapφverifies (Ker(φ))² ⊆Ker(φ), Ker(φ) ⊆N il(R) and Ker(φ) is a simple A-module such that M Ker(φ) = 0 and A is an Armendariz ring. Set V = Ker(φ), and let f(X) =

n i=0

(a_i+v_i)Xⁱ(= 0) andg(X) = m j=0

(b_j+w_j)X^j(= 0) be two polynomials of R[X], where a_i, b_j ∈ A and v_i, w_j ∈ V for each i, j such that f g = 0. Our aim is to show thatC_R(f)C_R(g) = 0. Without loss of generality, we may assume that C_R(f)=R and C_R(g)= R. Hence, a_i+v_i and b_j+w_j are not invertibles in R for each i, j and so a_i and b_j are not invertibles in A for each i, j since v_i, w_j ∈V ⊆N il(R) (as the sum of a nilpotent element and unit is unit). Hence,

a_i, b_j∈M since (A, M) is local and soa_iV =b_jV = 0 for eachi, j. Then:

C_R(f) = n i=0

(a_i+v_i)A+ n i=0

v_iV

C_R(g) = m j=0

(b_j+w_j)A+ m j=0

w_jV.

Set the polynomials of A[X]f_A(X) = n i=0

a_iXⁱ andg_A(X) = m j=0

b_jX^j. We claim that v_iV = 0 for eachiorw_jV = 0 for eachj. Indeed, assume that there existsi₀ and j₀ such thatv_i0V = 0 andw_j0V = 0. We may assume thati₀ andj₀ are the smallest such integer. Then,v_i0V =w_j0V =V sincev_i0V = 0 andw_j0V = 0 and V is a simpleA-module; thusv_i0w_j0V =V. Therefore,

C_R(f g) =

n+m

k=0

⎡

⎣

⎛

⎝

i+j=k

(a_i+v_i)(b_j+w_j)

⎞

⎠(A+V)

⎤

⎦

=

n+m

k=0

⎡

⎣

⎛

⎝

i+j=k

(a_ib_j+v_iw_j)

⎞

⎠(A+V)

⎤

⎦=C_A(f_Ag_A) +V =V

(since f g = 0 and for k = i₀+j₀, we have

i+j=k

v_iw_jV = v_i0w_j0V = V), a contradiction since f g = 0. Hence, v_iV = 0 for each i or w_jV = 0 for each j.

Without loss of generality, we may assume thatv_iV = 0 for eachi. Hence,

C_R(f)C_R(g) = _n

i=0

(a_i+v_i)A ⎡

⎣^m

j=0

(b_j+w_j)A+ m j=0

w_jV

⎤

⎦

= _n

i=0

a_iA ⎡

⎣^m

j=0

b_jA

⎤

⎦=C_A(f_A)C_A(g_A) = 0

since a_i, b_j ∈M,v_iV = 0 for each i, j, A is an Armendariz ring and f_Ag_A = 0;

and this completes the proof of Theorem 2.1.

In [16], Lee and Zhou characterized trivial extensions that are Armendariz rings. The following Corollary is a particular case of this work.

Corollary 2.3. Let (A, M) be a local Armendariz ring. Then the trivial ring ex- tension R :=A ∝ (A/M) of A by A/M is an Armendariz ring. In particular, R:=A∝ (A/M) is an Armendariz ring for every local domainA.

Proof. By Theorem 2.1 and since each domain is Armendariz.

In [15], Kim and Lee proved that the trivial extension of a ringAby itself is Armendariz if and only if A is reduced. The following Examples are a particular case of this work.

The next two examples prove that the condition M V = 0 imposed in Theo- rem 2.1 is necessary even if Ais Armendariz andB=A.

Example. LetK be a field,A=K∝K be the trivial ring extension ofK byK, and letR=A∝Abe the trivial ring extension ofA byA. Then:

1) A is an Armendariz ring.

2) R is not an Armendariz ring.

Proof.

1) The ring Ais Gaussian by [5, Example 2.3 (1.b)]. In particular, A is an Armendariz ring.

2) Our aim is to show that R is not Armendariz. Let f = ((0,1),(0,0)) + ((0,0),(1,0))X and g = ((0,1),(0,0)) + ((0,0),(−1,0))X be two polynomials in R[X]. We easily check thatf g= 0 and

C(f)C(g) = [R((0,1),(0,0)) +R((0,0),(1,0))][R((0,1),(0,0)) +R((0,0),(−1,0))]

=R((0,0),(0,1))= 0,

as desired.

Example. Let A:=Z/(2ⁱZ), where i≥2 be an integer, and let R :=A ∝A be the trivial ring extension of AbyA. Then:

1) A is an Armendariz ring.

2) R is not an Armendariz ring.

Proof.

1) Clear.

2) Letf := ( ¯2ⁱ⁻¹,0) + ( ¯2ⁱ⁻¹,1)X ∈R[X]. We havef²= 0 (and soC_R(f²) = 0) and (C_R(f))²=R(0,2ⁱ⁻¹¯ )(= 0_R). Therefore,Ris not an Armendariz ring.

3. Transfer of SFT Property in Subring Retracts

This section develops a result of the transfer of SFT property to subring retract, specially, to trivial ring extensions. And so, we will construct a new class of SFT rings (with zero-divisors).

Theorem 3.1. Let R be a ring and Aa subring retract of R. Then:

1) If R is an SFT ring, then so isA.

2) Assume that R:=A∝E be the trivial ring extension of Aby E. ThenR is an SF T ring if and only if so isA.

Proof.

1) Assume that Ris anSF T ring and letI:=

iAb_i be a nonzero ideal of A. SetJ :=

iR(b_i,0) be an ideal of R. Then, there exists a finitely generated

ideal ofR:J:=_n

i=1R(a_i, e_i)(⊆J) and there existsk∈N^∗such that (a, e)^k∈J for each (a, e)∈J. Hence,I:=_n

i=1Aa_i is a finitely generated ideal ofA,I ⊆I anda^k∈Ifor eacha∈I. Therefore,Iis anSF T- ideal and soAis anSF T-ring.

2) Assume that R:=A∝E is the trivial ring extension ofA byE. IfR is an SF T ring, then so isAby 1). Conversely, assume thatA is anSF T-ring. Let P :=q∝Ebe a prime ideal ofR, where q is a prime ideal ofA. Our aim is to show that P is anSF T-ideal. Sinceqis anSF T-ideal, there exists a finitely generated ideal of A : I := _n

i=1a_iA(⊆ q) and there exists k ∈ N^∗ such that a^k ∈ I for each a ∈q. Set J :=_n

i=1R(a_i,0)(=I ∝E ⊆P). Let (a, e) ∈P, where a∈ q and e∈E. We have (a, e)^k+1 = (a^k+1,(k+ 1)a^ke)∈I ∝IE =J and soP is an SF T-ideal. Therefore,R is anSF T-ideal and this complets the proof of theorem

3.1.

Now, we are able to give new examples of non coherent SF T rings.

Example. LetR:=Z∝Qbe the trivial ring extension ofZbyQ. Then:

1. R is anSF T ring by Theorem 3.1.

2. R is not a coherent ring by [13, Theorem 2.8(1)]. In particular, R is not a Noetherian ring.

Example. Let (A, M) be a localSF T ring with maximal idealM,E:= (A/M)^∞, and Rbe the trivial ring extension ofA byE. Then:

1. R is anSF T ring by Theorem 3.1.

2. R is not a coherent ring by [18, Theorem 2.1]. In particular, R is not a Noetherian ring.

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Acknowledgment

I would like to thank the referee for his/her useful suggestions and comments, which have greatly improved this paper.

Chahrazade Bakkari Department of Mathematics

Faculty of Science and Technology of Fez Box 2202, University S. M. Ben Abdellah Fez

Morocco

e-mail:cbakkari@hotmail.com

Received: August 28, 2008.

Revised: October 18, 2008.

Accepted: Oct ber 30, 2008.o