Weak annihilator property of Malcev-Neumann rings

Weak annihilator property of Malcev-Neumann rings

Ouyang Lunqun and Liu Jinwang

Department of Mathematics, Hunan University of Science and Technology Xiangtan, Hunan 411201, P.R. China

E-mail: [email protected]

AbstractLetRbe an associative ring with identity,Gan totally ordered group, σ a map fromGinto the group of automorphisms ofR, andta map fromG×G to the group of invertible elements of R. The weak annihilator property of the Malcev-Neumann ring R∗((G)) is investigated in this paper. Letnil(R) denote the set of all nilpotent elements ofR, and for a nonempty subsetXof a ringR, let N_R(X) ={a∈R|Xa⊆nil(R)} denote the weak annihilator ofX inR. Under the conditions that R is an N I ring with nil(R) nilpotent andσ is compatible, we show that: (1) If the weak annihilator of each nonempty subset of R which is not contained in nil(R) is generated as a right ideal by a nilpotent element, then the weak annihilator of each nonempty subset of R∗((G)) which is not contained in nil(R∗((G))) is generated as a right ideal by a nilpotent element.

(2) If the weak annihilator of each nonnilpotent element of R is generated as a right ideal by a nilpotent element, then the weak annihilator of each nonnilpotent element of R∗((G))) is generated as a right ideal by a nilpotent element. As a generalization of left APP-rings, we next introduce the notion of weak APP-rings and give a necessary and sufficient condition under which the ringR∗((G)) over a weak APP-ring R is weak APP.

Keywords and phrases: weak annihilator, weak APP-ring, Malcev-Neumann ring.

2000 Mathematics Subject Classification: 16W60.

1. Introduction

Throughout this paper R denotes an associative ring with identity, and nil(R) denotes the set of all nilpotent elements of R. For a nonempty subsetX of R,l_R(X) = {a ∈ R | aX = 0} and r_R(X) = {a ∈ R | Xa = 0} stand for the left and right annihilator of X inR, respectively. Recall that a ringR is reduced if it has no nonzero nilpotent elements, and a ring R is semicommutative if for all a,b ∈R, ab= 0 implies aRb = 0. Due to Marks [6], a ring R is called NI if nil(R) forms an ideal. Clearly, reduced rings and semicommutative rings are NI rings. An ideal I of R is said to be nilpotent if I^k= 0 for some natural number k.

Let R be a ring, G a totally ordered group, and suppose that σ is a map from G into the group of automorphisms of R,x−→σ_x, t is a map from G×G toU(R), the

group of invertible elements ofR. Then we can form a Malcev-Neumann ringR∗((G)):

an element of R∗((G) is a infinite series f = P

x∈G

rxx with rx ∈ R such that the set supp(f) ={x∈G|r_x 6= 0}, called the support off, is a well ordered subset ofG, and the ring structure is given by componentwise addition and by a multiplication defined as follows:

(X

x∈G

a_xx)(X

y∈G

b_yy) =X

z∈G

( X

{x,y|xy=z}

a_xσ_x(b_y)t(x, y))z.

In order to insure associativity, it is necessary to impose two additional conditions on σ and t, namely that for all x, y, z ∈G,

(i) t(xy, z)σ_z(t(x, y)) = t(x, yz)t(y, z), (ii) σ_yσ_z =σ_yzδ(y, z),

where δ(y, z) denotes the automorphism of R induced by the unit t(y, z) (see [10, Lemma 1.1]). It is now routine to check that R∗((G)) is a ring which we call the Malcev-Neumann ring.

LetU be a subset ofR. We denote byU ∗((G)) the subset of R∗((G)) consisting of those elements whose coefficients lie in U, that is, U ∗((G)) = {f = P

x∈G

axx ∈ R∗((G))|a_x ∈U, x∈supp(f)}.

The Malcev-Neumann construction appeared for the first time in the latter part of 1940’s (the Laurent series ring, a particular case of Malcev-Neumann rings, was used before by Hilbert). Using them, Malcev and Neumann independently showed (in 1948 and 1949 resp.) that the group ring of an ordered group over a division ring can be embedded in a division ring. Since then, the construction has appeared in many papers, mainly in the study of various properties of division rings and related topics.

For instance, Makar-Limanov in [7] used a particular skew Laurent series division ring to prove that the skew field of fractions of the first Weyl-algebra contains a free noncommutative subalgebra. The study of Malcev-Neumann group rings over arbitrary rings was initiated in [5] by Lorenz while investigating properties of group algebras of nilpotent groups. Other results on Malcev-Neumann rings can be found in Musson and Stafford [8] and Sonin [11] and Zhao et al. [14]. In this paper, we investigate the relationship between the weak annihilator NR(X) of a nonempty subset X of R and the weak annihilator N_R∗((G))(V) of a nonempty subset V of R∗((G)). Also as a generalization of left APP-rings, we introduce the notion of weak APP-rings, and study the conditions under which the ring R∗((G)) is a weak APP-ring.

2. Weak annihilator property

As a generalization of annihilators, L. Ouyang and G. F. Birkenmeier in [9] in- troduced the concept of weak annihilators. For a nonempty subset X of a ring R, we define NR(X) = {a ∈ R | Xa ⊆ nil(R)}, which is called the weak annihilator of X in R. If X is a finite set, say X = {r₁, r₂,· · ·, r_n}, we use N_R(r₁, r₂,· · · , r_n) in place of N_R({r₁, r₂,· · · , r_n}). Obviously, for any nonempty subset X of a ring R, NR(X) = {a ∈ R | Xa ⊆ nil(R)} = {b ∈ R | bX ⊆ nil(R)}, rR(X) ⊆ NR(X) and l_R(X)⊆N_R(X).

For example, Let Z be the ring of integers and T₂(Z) the 2×2 upper triangular matrix ring over Z. We consider the subset X =

½µ 2 0 0 2

¶¾

. Then r_T2(Z)(X) =

l_T2(Z)(X) = 0, but N_T2(Z)(X) =

½µ 0 m 0 0

¶

,|m ∈Z

¾

. Thus r_T2(Z)(X) ( N_T2(Z)(X) and l_T2(Z)(X)(N_T2(Z)(X).

If R is reduced, then r_R(X) = N_R(X) = l_R(X) for any nonempty subset X of R.

It is easy to see that for any nonempty subset X ⊆R, NR(X) is an ideal of R in case nil(R) is an ideal. For more details and results of weak annihilators, see [9]. In this section, we mainly discuss the weak annihilator property of the ring R∗((G)).

The next Lemma appears in [9].

Lemma 2.1 Let X, Y be subsets ofR. Then we have the following results:

(1) X ⊆Y impliesN_R(X)⊇N_R(Y).

(2) X ⊆N_R(N_R(X)).

(3) N_R(X) = N_R(N_R(N_R(X))).

Lemma 2.2 Let R be an NI ring. Then we have the following results:

(1) ab∈nil(R) implies RaRbR⊆nil(R) for anya,b ∈R.

(2) Let p∈ R and let p·R denote the principal right ideal of R generated by p.

Then N_R(p) = N_R(p·R).

(3) Let X be a subset of R and let I be the ideal of R generated by the subset X. Then N_R(X) = N_R(I).

Proof (1) Sincenil(R) of anNIring is an ideal, we obtainab∈nil(R)⇒abR⊆ nil(R)⇒bRa⊆nil(R)⇒bRaR⊆nil(R)⇒aRbR⊆nil(R)⇒RaRbR ⊆nil(R).

(2) Since p ∈ p·R, N_R(p·R) ⊆ N_R(p) is clear. Now we show that N_R(p) ⊆ N_R(p·R). If x ∈ N_R(p), then px ∈ nil(R). By (1), we have pRx ⊆ nil(R), and so x∈N_R(p·R). Hence N_R(p)⊆N_R(p·R). Therefore N_R(p) =N_R(p·R).

(3) It suffices to show that N_R(X)⊆ N_R(I). Let r ∈N_R(X). Then xr ∈ nil(R) for all x ∈ X, and so by (1), we obtain sxtr ∈ nil(R) for any s ∈ R and t ∈ R.

Hence for any P_n

i=1s_ix_it_i ∈I, we haveP_n

i=1s_ix_it_ir ∈nil(R), and sor ∈N_R(I). Thus N_R(X)⊆N_R(I) is proved.

Definition 2.3 Let σ be a map from G into the group of automorphisms of R, x −→ σ_x. We say that σ is compatible if for each a, b ∈ R and x ∈ G, ab = 0 ⇔ aσ_x(b) = 0.

Lemma 2.4 Let σ be a map from G into the group of automorphisms of R, x −→ σ_x. If σ is compatible, then for each a, b ∈ R, and each x ∈ G, we have the following results:

(1) ab∈nil(R)⇔aσ_x(b)∈nil(R).

(2) ab∈nil(R)⇔σ_x(a)b ∈nil(R).

Proof (1) (⇒) Suppose ab ∈ nil(R). There exists some positive integer k such that (ab)^k = 0. Since σ is compatible, we have 0 = (ab)^k = abab· · ·ab ⇒ abab· · ·aσ_x(b) = 0 ⇒ abab· · ·abaσ_x(baσ_x(b)) = abab· · ·abaσ_x(b)σ_x(aσ_x(b)) = 0 ⇒ abab· · ·abaσx(b)aσx(b) = 0⇒ · · · ⇒aσx(b)∈nil(R).

(⇐) Assume thataσ_x(b)∈ nil(R). There exists some positive integer k such that (aσ_x(b))^k = 0. In the following computations, we use freely the condition that σ is compatible. (aσx(b))^k = aσx(b)aσx(b)· · ·aσx(b) = 0 ⇒ aσx(b)aσx(b)· · ·aσx(b)ab =

0 ⇒ aσ_x(b)aσ_x(b)· · ·aσ_x(b)σ_x(ab) = 0 ⇒ aσ_x(b)aσ_x(b)· · ·aσ_x(b)aσ_x(bab) = 0 ⇒ aσx(b)aσx(b)· · ·aσx(b)abab= 0 ⇒ · · · ⇒ab∈nil(R).

(2) ab∈nil(R)⇔ba∈nil(R)⇔bσ_x(a)∈nil(R)⇔σ_x(a)b ∈nil(R).

Proposition 2.5 Let R be an NI ring with nil(R) nilpotent, and letσ be com- patible, andf = P

x∈G

a_xx∈R∗((G)). Thenf ∈nil(R∗((G))) if and only ifa_x ∈nil(R) for every x∈supp(f).

Proof (⇒) Suppose that f = P

x∈G

axx ∈nil(R∗((G))). Then there exists some positive integer k such that

(1) f^k = (X

x∈G

a_xx)^k = 0.

We will use transfinite induction on the ordered group (G,≤) to show that a_x ∈ nil(R) for every x ∈ supp(f). Let x0 be the minimal element of supp(f) on the ≤ order. If v₁, v₂, . . ., v_k ∈ supp(f) are such that v₁v₂· · ·v_k = x^k₀, then x₀ ≤ v_i for all 1 ≤i ≤ k. If x₀ < v_i for some 1 ≤ i ≤k, then x^k₀ < v₁v₂· · ·v_k =x^k₀, a contradiction.

Thus x0 =vi for 1≤i≤k. Hence from Eq. (1), it follows that a_x0σ_x0(a_x0)t(x₀, x₀)σ_x²₀(a_x0)t(x²₀, x₀)· · ·σ_x^k−1

0 (a_x0)t(x^k−1₀ , x₀) = 0.

Since σ is compatible and t(x, y) is invertible for all x, y ∈ G, and nil(R) of an NI ring is an ideal, we have

a_x0σ_x0(a_x0)t(x₀, x₀)σ_x²

0(a_x0)t(x²₀, x₀)· · ·σ_x^k−1

0 (a_x0)t(x^k−1₀ , x₀) = 0

⇒ a_x0σ_x0(a_x0)t(x₀, x₀)σ_x²₀(a_x0)t(x²₀, x₀)· · ·σ_x^k−1

0 (a_x0) = 0

⇒ ax0σx0(ax0)t(x0, x0)σ_x²₀(ax0)t(x²₀, x0)· · ·σ_x^k−2

0 (ax0)t(x^k−2₀ , x0)ax0 = 0

⇒ ax0ax0σx0(ax0)t(x0, x0)σ_x²₀(ax0)t(x²₀, x0)· · ·σ_x^k−2

0 (ax0)t(x^k−2₀ , x0)∈nil(R)

⇒ a_x0a_x0σ_x0(a_x0)t(x₀, x₀)σ_x²₀(a_x0)t(x²₀, x₀)· · ·σ_x^k−2

0 (a_x0)∈nil(R)

⇒ a_x0a_x0σ_x0(a_x0)t(x₀, x₀)σ_x²₀(a_x0)t(x²₀, x₀)· · ·σ_x^k−3

0 (a_x0)t(x^k−3₀ , x₀)a_x0 ∈nil(R)

⇒ a_x0a_x0a_x0σ_x0(a_x0)t(x₀, x₀)σ_x²₀(a_x0)t(x²₀, x₀)· · ·σ_x^k−3

0 (a_x0)t(x^k−3₀ , x₀)∈nil(R)

⇒ · · · ⇒a_x0 ∈nil(R).

Now suppose that w ∈ supp(f) is such that for any x ∈ supp(f) with x < w, a_x ∈ nil(R). We will show that a_w ∈ nil(R) for w ∈ supp(f). For convenience, we write

{(u₁, u₂,· · · , u_k)|u₁u₂· · ·u_k =w^k, u_i ∈supp(f), i= 1,2, . . . , k}

as

{(w, w,· · · , w)} ∪ {(u_i1, u_i2,· · · , u_ik)|i= 2,3, . . . , n}, and for each

(u_i1, u_i2,· · · , u_ik)∈ {(u_i1, u_i2,· · · , u_ik)|i= 2,3, . . . , n}, there exists some 1 ≤l ≤k such thatu_il 6=w. Now we show that for each

(u_i1, u_i2,· · · , u_ik)∈ {(u_i1, u_i2,· · · , u_ik)|i= 2,3, . . . , n},

there exists some 1 ≤ p ≤ k such that u_ip < w. If u_il < w, then we are done. So assume that uil > w. If for all 1≤j ≤k,j 6=l,uij ≥w, then w^k < ui1ui2· · ·uik =w^k, a contradiction. Thus for each

(u_i1, u_i2,· · · , u_ik)∈ {(u_i1, u_i2,· · · , u_ik)|i= 2,3, . . . , n},

there exists some 1 ≤ p ≤ k such that u_ip < w. Then by induction hypothesis, we obtain a_uip ∈ nil(R), and so by Lemma 2.4, 1·a_uip ∈ nil(R) implies 1· σ_x(a_uip) = σ_x(a_uip)∈nil(R) for every x∈G. Hence

a_ui1σ_ui1(a_ui2)t(u_i1, u_i2)· · ·σ_{(ui1ui2···ui(k−1))}(a_uik)t(u_i1u_i2· · ·u_i(k−1), u_ik)∈nil(R) for all 2≤i≤n, becausenil(R) of an NI ring is an ideal. Now from Eq.(1), we have

a_wσ_w(a_w)t(w, w)· · ·σ_w^k−1(a_w)t(w^k−1, w)

= −Pⁿ

i=2

a_ui1σ_ui1(a_ui2)t(u_i1, u_i2)· · ·σ_{(ui1ui2···ui(k−1))}(a_uik)t(u_i1u_i2· · ·u_i(k−1), u_ik)∈nil(R).

Then a_wσ_w(a_w)t(w, w)· · ·σ_w^k−1(a_w)t(w^k−1, w)∈nil(R)

⇒ a_wσ_w(a_w)t(w, w)· · ·σ_w^k−1(a_w)∈nil(R)

⇒ a_wσ_w(a_w)t(w, w)· · ·σ_w^k−2(a_w)t(w^k−2, w)a_w ∈nil(R)

⇒ a_wa_wσ_w(a_w)t(w, w)· · ·σ_w^k−2(a_w)t(w^k−2, w)∈nil(R)

⇒ · · · ⇒a_w ∈nil(R).

Therefore by transfinite induction, ax ∈nil(R) for anyx∈supp(f).

(⇐) Assume that a_x ∈ nil(R) for every x ∈ supp(f). By Lemma 2.4, we have σ_z(a_x) ∈ nil(R) for each z ∈ G. Since nil(R) is nilpotent, there exists some positive integer k such that (nil(R))^k= 0. Now we show that

f^k= (X

x∈G

a_xx)^k=X

y∈G

b_yy= 0.

For every y∈supp(f^k), we write

{(u₁, u₂,· · · , u_k)|u₁u₂· · ·u_k =y, u_i ∈supp(f), i= 1,2, . . . , k} as

{(u_i1, u_i2,· · · , u_ik)|i= 1,2, . . . , n}.

Then from f^k = (P

x∈G

a_xx)^k= P

y∈G

b_yy, it follows that

b_y = Xn

i=1

a_ui1σ_ui1(a_ui2)t(u_i1, u_i2)· · ·σ_{(ui1ui2···ui(k−1))}(a_uik)t(u_i1u_i2· · ·u_i(k−1), u_ik).

Since for each 1≤i≤n,

aui1σui1(aui2)t(ui1, ui2)· · ·σui1ui2···u_i(k−1)(auik)t(ui1ui2· · ·u_i(k−1), uik)∈(nil(R))^k= 0, we have b_y = 0. Hencef^k= 0, and so f ∈nil(R∗((G))). Then we finish our proof of Proposition 2.5.

RemarkIn the proof of the implication (⇒) in Proposition 2.5, the condition that nil(R) is nilpotent is not used. Hence if R is an NI ring, and σ is compatible, then nil(R∗((G))) ⊆nil(R)∗((G)).

By Proposition 2.5 we have the following result.

Corollary 2.6 Let R be anNI ring withnil(R) nilpotent, and letσ be compat- ible. Then

(1) R∗((G)) is anNI ring.

(2) nil(R∗((G))) =nil(R)∗((G)).

Proposition 2.7 Let R be an NI ring with nil(R) nilpotent, and letσ be com- patible. If the weak annihilator of each nonempty subset ofRwhich is not contained in nil(R) is generated as a right ideal by a nilpotent element, then the weak annihilator of each nonempty subset ofR∗((G)) which is not contained in nil(R∗((G))) is generated as a right ideal by a nilpotent element.

Proof Let V be a nonempty subset of R∗((G)) with V 6⊆ nil(R∗((G))). We show that N_R∗((G))(V) is generated as a right ideal by a nilpotent element. For any f = P

x∈G

axx∈R∗((G)), let Cf denote the set {ax |x∈ supp(f)}, and for any subset U ⊆R∗((G)), letC_U denote the set S

f∈U

C_f. SinceV 6⊆nil(R∗((G))), by Corollary 2.6, we haveCV 6⊆nil(R). So there exists an elementc∈nil(R) such thatNR(CV) =c·R.

Now we show that

N_R∗((G))(V) =c·(R∗((G))).

Let f = P

x∈G

a_xx∈V and g = P

y∈G

b_yy ∈R∗((G)). Then f·c·g = (X

x∈G

a_xx)·c·(X

y∈G

b_yy) = X

z∈G

( X

{x,y|xy=z}

a_xσ_x(c)t(x,1)σ_x(b_y)t(x, y))z.

Since c∈nil(R) and σ is compatible, for any x∈supp(f) and y∈supp(g), we have c∈nil(R) ⇒ σ_x(c)∈nil(R)⇒a_xσ_x(c)t(x,1)σ_x(b_y)t(x, y)∈nil(R)

⇒ X

{x,y|xy=z}

a_xσ_x(c)t(x,1)σ_x(b_y)t(x, y)∈nil(R).

Thus by Proposition 2.5, we obtain f ·c·g ∈ nil(R∗((G))). Hence N_R∗((G))(V) ⊇ c·(R∗((G))).

Conversely, let g = P

y∈G

b_yy ∈ N_R∗((G))(V). Then f g ∈ nil(R∗((G))) for any f = P

x∈G

a_xx∈V. Letf g = (P

x∈G

a_xx)(P

y∈G

b_yy) = P

z∈G

∆_zz.Then by Proposition 2.5, we have

∆_z ∈nil(R). Note that

(2) ∆_z = X

{x,y|xy=z}

a_xσ_x(b_y)t(x, y).

We will use transfinite induction on the ordered group (G,≤) to show that a_xb_y ∈ nil(R) for every x∈supp(f) and y∈supp(g).

Letx₀ and y₀ be the minimal elements of supp(f) and supp(g) in the order ≤, re- spectively. If x∈supp(f) and y∈supp(g) are such that xy =x0y0, then x0 ≤x, and y₀ ≤y. Ifx₀ < x, then x₀y₀ < xy₀ ≤ xy =x₀y₀, a contradiction. Thus x₀ =x. Sim- ilarly,y=y₀. Then from Eq.(2), we obtain ∆_x0y0 =a_x0σ_x0(b_y0)t(x₀, y₀)∈nil(R). Thus we haveax0σx0(by0)t(x0, y0)∈nil(R)⇒ax0σx0(by0)t(x0, y0)(t(x0, y0))⁻¹ =ax0σx0(by0)∈ nil(R)⇒a_x0b_y0 ∈nil(R).

Now suppose that w ∈ G is such that for any x ∈ supp(f) and y ∈ supp(g) with xy < w, axby ∈ nil(R). We will show that axby ∈ nil(R) for any x ∈ supp(f) and y ∈ supp(g) with xy = w. For convenience, we write {(x, y) | xy = w, x ∈ supp(f), y ∈ supp(g)} as {(x_i, y_i) | i = 1,2, . . . , n, x_i ∈ supp(f), y_i ∈ supp(g)} with x1 < x2 < · · · < xn (Note that if x1 = x2, then from x1y1 = x2y2, it follows that y₁ =y₂, and thus (x₁, y₁) = (x₂, y₂)). Now from Eq.(2), we have

(3) ∆w = X

{x,y|xy=w}

axσx(by)t(x, y) = Xn

i=1

axiσxi(byi)t(xi, yi),

and ∆_w ∈ nil(R). For any 1 ≤ i ≤ n−1, x_iy_n < x_ny_n = w, and thus, by induction hypothesis, we have a_xib_yn ∈ nil(R). Then by Lemma 2.2, a_xiσ_xi(b_yi)t(x_i, y_i)b_yn ∈ nil(R). Hence multiplying Eq.(3) on the right by b_yn, we obtain

a_xnσ_xn(b_yn)t(x_n, y_n)b_yn = ∆_wb_yn− Xn−1

i=1

a_xiσ_xi(b_yi)t(x_i, y_i)b_yn.

Then a_xnσ_xn(b_yn)t(x_n, y_n)b_yn ∈nil(R) because nil(R) of an NI ring is an ideal. Now a_xnσ_xn(b_yn)t(x_n, y_n)b_yn ∈nil(R)⇒b_yna_xnσ_xn(b_yn)t(x_n, y_n)∈nil(R)

⇒ b_yna_xnσ_xn(b_yn)t(x_n, y_n)(t(x_n, y_n))⁻¹ =b_yna_xnσ_xn(b_yn)∈nil(R)

⇒ bynaxnbyn ∈nil(R)⇒axnbyn ∈nil(R).

From Lemma 2.4, it follows that

axnbyn ∈nil(R)⇒axnσxn(byn)∈nil(R)⇒axnσxn(byn)t(xn, yn)∈nil(R).

Now Eq.(3) becomes (4)

Xn−1

i=1

a_xiσ_xi(b_yi)t(x_i, y_i) = ∆_w−a_xnσ_xn(b_yn)t(x_n, y_n)∈nil(R).

Multiplyingb_yn−1 on Eq.(4) from the right-hand side, we obtaina_xn−1b_yn−1 ∈nil(R) by the same way as above. Continuing this process, we can prove that a_xib_yi ∈nil(R) for i = 1,2, . . . , n. Thus axby ∈ nil(R) for all x ∈ supp(f) and y ∈ supp(g) with xy =w.

Therefore, by transfinite induction, a_xb_y ∈ nil(R) for any x ∈ supp(f) and y ∈ supp(g). Thus for any y ∈ supp(g), by ∈ NR(CV) = c·R. So for any y ∈ supp(g), there exists r_y ∈R such that b_y =cr_y. Henceg =c·h whereh= P

y∈G

r_yy∈R∗((G)),

and so N_R∗((G))(V)⊆c·(R∗((G))). ThereforeN_R∗((G))(V) = c·(R∗((G))) wherecis a nilpotent element.

Corollary 2.8 Let R be an NI ring with nil(R) nilpotent, and let σ be com- patible. If the weak annihilator of each ideal of R which is not contained in nil(R) is generated as a right ideal by a nilpotent element, then the weak annihilator of each ideal of R∗((G)) which is not contained innil(R∗((G))) is generated as a right ideal by a nilpotent element.

Proof This is immediate from Lemma 2.2 and Proposition 2.7.

Proposition 2.9 Let R be an NI ring with nil(R) nilpotent, and letσ be com- patible. If the weak annihilator of each nonnilpotent element of R is generated as a right ideal by a nilpotent element, then the weak annihilator of each nonnilpotent element of R∗((G)) is generated as a right ideal by a nilpotent element.

Proof Let f = P

x∈G

a_xx be a nonnilpotent element of R∗((G)). Then by Propo- sition 2.5, there exists some u ∈ supp(f) such that a_u 6∈ nil(R). Hence we can find c∈nil(R) such that N_R(a_u) = c·R. Now we show that

N_R∗((G))(f) = c·(R∗((G))).

For any g = P

y∈G

b_yy∈R∗((G)), we have

f·c·g = (X

x∈G

a_xx)·c·(X

y∈G

b_yy) = X

z∈G

( X

{x,y|xy=z}

a_xσ_x(c)t(x,1)σ_x(b_y)t(x, y))z.

Since c∈nil(R) and σ is compatible, it is easy to see that X

{x,y|xy=z}

a_xσ_x(c)t(x,1)σ_x(b_y)t(x, y)∈nil(R)

for any z ∈supp(f cg). Then by Proposition 2.5, we obtain f cg ∈nil(R∗((G))), and so c·R∗((G))⊆N_R∗((G))(f).

Conversely, let g = P

y∈G

b_yy ∈ N_R∗((G))(f). Then f g ∈ nil(R∗((G))). By analogy with the proof of Proposition 2.7, we obtain a_xb_y ∈ nil(R) for any x ∈ supp(f) and any y ∈ supp(g). Hence by ∈ NR(au) for any y ∈ supp(g). Thus for any y ∈ supp(g), there exists r_y ∈R such that b_y =c·r_y. Then g =ch where h= P

y∈G

r_yy∈R∗((G)), and so N_R∗((G))(f)⊆c·(R∗((G))). Therefore, N_R∗((G))(f) = c·(R∗((G))).

Corollary 2.10 Let R be an NI ring with nil(R) nilpotent, and let σ be com- patible. If the weak annihilator of each principal right ideal ofRwhich is not contained innil(R) is generated as a right ideal by a nilpotent element, then the weak annihilator of each principal right ideal of R ∗((G)) which is not contained in nil(R∗((G))) is generated as a right ideal by a nilpotent element.

Proof This is immediate from Lemma 2.2 and Proposition 2.9.

Example 2.11 Let F be a field and let S denote the F-space on basis {1, c, c², . . . , cⁿ},

where cⁿ⁺¹ = 0. Then nil(S) = {a₁c+a₂c² +· · ·+a_ncⁿ | a_i ∈ F} is an ideal of S. For any m = b₀ +b₁c+· · ·+b_ncⁿ ∈ S, if b₀ = 0, then m ∈ nil(S). If b₀ 6= 0, then m = b₀ +b₁c+· · ·+b_ncⁿ is invertible. For any nonempty subset V 6⊆ nil(S), now we show that N_S(V) is generated as a right ideal by a nilpotent element. Let Ω = {b₀ | b₀ + b₁c + · · ·+ b_ncⁿ ∈ V}. If Ω = {0}, then V ⊆ nil(S). This is contrary to the fact that V 6⊆ nil(S). Thus we have Ω 6= {0}. In this case, we have N_S(V) = nil(S) = c·S, where c ∈ nil(S). Hence S is a ring such that for each nonempty subset V 6⊆nil(S), N_S(V) is generated as a right ideal by a nilpotent element.

LetR be a field. Then the residue ring R[x]/(xⁿ⁺¹) is an R-space on basis {1, x, x², . . . , xⁿ},

where xⁿ⁺¹ = 0. Hence R[x]/(xⁿ⁺¹) is a ring such that for each nonempty subset V 6⊆

nil(R[x]/(xⁿ⁺¹)),N_R[x]/(xⁿ⁺¹₎(V) is generated as a right ideal by a nilpotent element.

LetR be a field and let

R_n=















a1 a2 · · · an

0 a₁ · · · a_n−1

· · · · 0 0 · · · a1





|a_i ∈R









be the subring of n×n upper triangular matrix ring. Then R_n∼=R[x]/(xⁿ). ThusR_n is also a ring that for each nonempty subset V 6⊆ nil(R_n), N_Rn(V) is generated as a right ideal by a nilpotent element.

Example 2.12 Ifpis a prime, the ringZ_pⁿ of integers modulopⁿis a commutative local ring and the Jacobson radicalJ ofZ_pⁿ isJ =nil(Z_pⁿ) = Z_pⁿ·[p]. Hence it is easy to see that for any nonempty subset V 6⊆ nil(Zpⁿ), NZpn(V) is generated as a right ideal by a nilpotent element.

3. Weak APP-rings

An idealI of R is said to be rights-unital if a∈aI for eacha∈I. If I and J are right s-unital ideals, then so is I∩J. It follows from [12, Theorem 1] that I is right s-unital if and only if for any finitely many elements a₁, a₂, . . ., a_n ∈ I, there exists an element x∈I such that a_i =a_ix, i= 1,2, . . . , n. A ring R is called a left APP-ring if the left annihilator l_R(Ra) is right s-unital as an ideal of R for any element a∈ R, right APP-rings may be defined analogously. A ring is biregular if every principal ideal is generated by some idempotent in the center of the ring, and a ring is quasi-Baer if the left annihilator of every left ideal is generated by an idempotent. Thus the class of left APP-rings includes all biregular rings and all quasi-Baer rings. It was shown in [4, Theorem 2] that if R is a ring satisfying descending chain condition on right annihilators, then the skew power series ring R[[x;α]] is left APP if and only if for any sequence (b₀, b₁,· · ·) of elements of R, the ideall_R(P_∞

j=0

P_∞

k=0Rα^k(b_j)) is right s- unital, whereαis an automorphism of R. It was also proved in [13, Theorem 3] that if

(S,≤) is a strictly totally ordered monoid,ω:S −→Aut(R) a monoid homomorphism and Ra ring satisfying descending chain condition on right annihilators, then the skew generalized power series ring [[R^S,≤, ω]] is left APP if and only if for any S-indexed subset Aof R, the ideal l_R(P

a∈A

P

s∈SRω_s(a)) is rights-unital. For more details and properties of left APP-rings, see [2, 3, 4, 13].

As a generalization of left APP-rings, in this section, we introduce the notion of weak APP-rings and investigate its properties. We first briefly develop the definition of weak APP-rings. Also we provide several basic results. Next, we investigate the weak APP-property of Malcev-Neumann rings.

Definition 3.1 LetR be an NI ring. An ideal I ofR is said to be weak s-unital if, for each a∈I, there exists an elementx∈I such that ax−a∈nil(R).

Obviously, for alla,x∈R,ax−a =a(x−1)∈nil(R)⇔(x−1)a=xa−a∈nil(R).

So all right s-unital ideals and all left s-unital ideals are weak s-unital. But the following example shows that the converse is not true in general.

Example 3.2 Let R be a domain and let R₂ =

½µ a b 0 a

¶

|a, b∈R

¾

be the subring of 2×2 upper triangular matrix ring. Consider the ideal I =R₂

µ 0 1 0 0

¶ R₂ generated by

µ 0 1 0 0

¶

.ThenI is neither right s-unital nor left s-unital. But it is easy to see that I is weak s-unital.

Proposition 3.3 LetR be an NI ring. Then the following conditions are equiv- alent:

(1) I is weak s-unital.

(2) For any finitely many elements a1, a2, . . . , an ∈ I, there exists an element x∈I such that a_ix−a_i ∈nil(R),i= 1,2, . . . , n.

Proof (1) =⇒ (2) We prove it by induction on n with the case n = 1 clear.

Now suppose thatn ≥2. From the condition thatI is weak s-unital and the induction hypothesis, it follows that there exist e₁, e₂ ∈ I such that a_ie₁ −a_i ∈ nil(R) for all 1≤i≤n−1, anda_ne₂−a_n ∈nil(R). In the following computations, we use freely the condition thatRis anNIring. For each 1≤i≤n−1,a_ie₁−a_i =a_i(e₁−1)∈nil(R)⇒ (e₁−1)a_i ∈nil(R)⇒(e₁−1)a_i(e₂−1)(e₁−1)∈nil(R)⇒a_i(e₂−1)(e₁−1)(e₁−1) = a_i(e₂e²₁−2e₂e₁+e₂−e²₁+2e₁−1)∈nil(R)⇒a_i(e₂e²₁−2e₂e₁+e₂−e²₁+2e₁)−a_i ∈nil(R), and a_ne₂ − a_n = a_n(e₂ −1) ∈ nil(R) ⇒ a_n(e₂ −1)(e₁ − 1)(e₁ − 1) ∈ nil(R) ⇒ a_n(e₂e²₁ −2e₂e₁ +e₂−e²₁+ 2e₁)−a_n ∈ nil(R). Set x= e₂e²₁ −2e₂e₁ +e₂−e²₁+ 2e₁. Then we obtain a_ix−a_i ∈nil(R) for all 1≤i≤n.

(2)⇒(1) It is straightforward.

Proposition 3.4 Let R be an NI ring and I, J are weak s-unital ideals. Then I ∩J and I+J are weak s-unital.

Proof Leta∈I∩J. Then there existx∈I and y∈J such thatax−a∈nil(R) and ay − a ∈ nil(R). So we can find α, β ∈ nil(R) such that ax = a +α and ay =a+β. Thus axy = (a+α)y =ay+αy =a+β +αy. Hence axy−a ∈ nil(R) with xy ∈IJ ⊆I∩J. Therefore I∩J is weak s-unital.

Now we see that I +J is weak s-unital. Let a1 +a2 ∈ I +J with a1 ∈ I and a₂ ∈ J. Then there exist e₁ ∈ I and e₂ ∈ J such that a₁e₁ − a₁ ∈ nil(R) and a₂e₂ −a₂ ∈ nil(R). By analogy with the proof of Proposition 3.3, we can find x = e2e²₁−2e2e1+e2−e²₁+ 2e1 ∈I+J such thataix−ai ∈nil(R),i= 1,2.Thus we have (a₁ +a₂)x−(a₁+a₂)∈nil(R). This implies that I+J is weak s-unital.

Definition 3.5 An NI ring R is called a weak APP-ring if the weak annihilator N_R(a) is weak s-unital as an ideal of R for any element a∈R.

Example 3.6 Here are some examples of weak APP-rings.

(1) Obviously, all domains and division rings are weak APP-rings. If a ring R is reduced, then for any a ∈R, N_R(a) =r_R(aR) =l_R(Ra). So reduced left (resp. right) APP-rings are weak APP-rings. Since reduced PP-rings and reduced p.q.-Baer rings are left (resp. right) APP-rings (see [3]), they are also weak APP-rings. Hence the class of weak APP-rings includes reduced left (resp. right) APP-rings. In particular, the class of weak APP-rings includes reduced PP-rings and reduced p.q.-Baerrings.

(2) Let R be an NI ring and let T_n(R) be the n ×n upper triangular matrix ring over R. Now we show that R is a weak APP-ring if and only if T_n(R) is a weak APP-ring. Clearly, T_n(R) is an NI ring. Suppose that R is a weak APP-ring. Let A = (a_ij) ∈ T_n(R) and B = (b_ij) ∈ N_Tn(R)(A). Then BA ∈ nil(T_n(R)) and so b_iia_ii ∈ nil(R) for all 1 ≤ i ≤ n. Thus b_ii ∈ N_R(a_ii) for all 1 ≤ i ≤ n. Because R is a weak APP-ring, there exists c_ii ∈ N_R(a_ii) such that b_iic_ii−b_ii ∈ nil(R) for each 1≤i≤n. Now it is easy to see that

B







c₁₁ 0 · · · 0 0 c₂₂ · · · 0

· · · · · · · 0 0 · · · c_nn





−B ∈nil(T_n(R))

and 





c₁₁ 0 · · · 0 0 c₂₂ · · · 0

· · · · · · · 0 0 · · · c_nn





∈N_Tn(R)(A).

Conversely, assume that T_n(R) is a weak APP-ring. Let a, b∈R such that b∈N_R(a).

Set

A=







a 0 · · · 0 0 0 · · · 0

· · · · 0 0 · · · 0





, B =







b 0 · · · 0 0 0 · · · 0

· · · · 0 0 · · · 0





.

Then B ∈ N_Tn(R)(A). Since T_n(R) is a weak APP-ring, there exists C = (c_ij) ∈ NTn(R)(A) such thatBC−B ∈nil(Tn(R)). Now it is easy to see thatbc11−b∈nil(R)