Southeast Asian Bulletin of Mathematics (2013) 37: 259–268
Southeast Asian
Bulletin of
Mathematics
c SEAMS. 2013
⃝
Generalized Power Series over Zip and Weak Zip Rings
R.M. Salem
Department of Mathematics, Faculty of Science, Al-Azhar University, Nasr City 11884,
Cairo, Egypt
Email: rsalem 02@hotmail.com
Received 15 July 2011
Accepted 21 March 2012
Communicated by N.V. Sanh
AMS Mathematics Subject Classification(2000): 06F05, 16W60, 16D25, 16P60
Abstract. In this paper we show that, if 𝑅 is a generalized Armendariz (semicommutative right Noetherian ) ring and 𝑆 is a strictly (strictly totally) ordered monoid, then
the generalized power series ring Λ = [[𝑅𝑆,≤ ]] is a right zip (weak zip) ring if and only
if 𝑅 is.
Keywords: Strictly ordered monoid; Artinian and narrow subset; Generalized power
series rings; Zip rings, Weak zip rings; Armendari rings; Semicommutative rings.
1. Introduction
The concept of zip rings, was initiated by Zelmanowitz [16] although it wasn’t
so called zip at that time. However, Faith in [4] called a ring 𝑅 right zip if the
right annihilator 𝑟𝑅 (𝑋) of a subset 𝑋 ⊆ 𝑅 is zero, then 𝑟𝑅 (𝑋0 ) = 0 for a finite
subset 𝑋0 of 𝑋, equivalently for a left ideal 𝐿 of 𝑅, if the 𝑟𝑅 (𝐿) = 0, then there
exists a finitely generated left ideal 𝐿1 ⊆ 𝐿 such that 𝑟𝑅 (𝐿1 ) = 0. 𝑅 is called a
zip ring if it is both right and left zip. In [16] Zelmanowitz showed that any ring
satisfying the descending chain condition on right annihilators is a right zip ring
but the converse doesn’t hold. Extensions of zip rings were studied by several
authors, Beach and Blair [1] showed that if 𝑅 is a commutative zip ring, then
𝑅[𝑥] is a zip ring.
∑
∑
∑𝑞
∑𝑙
𝑗
′
𝑖
′
𝑗
Now, let 𝑓 = 𝑛𝑖=0 𝑎𝑖 𝑥𝑖 , 𝑔 = 𝑚
𝑗=0 𝑏𝑗 𝑥 (𝑓 =
𝑖=𝑠 𝑎𝑖 𝑥 , 𝑔 =
𝑗=𝑡 𝑏𝑗 𝑥 ) be
polynomials over a ring 𝑅. Rege and Chhawchharia in [10] introduced the notion
of an Armendariz ring, a ring 𝑅 is called Armendariz if whenever 𝑓, 𝑔 ∈ 𝑅[𝑥]
such that 𝑓 𝑔 = 0 then 𝑎𝑖 𝑏𝑗 = 0 for each 0 ≤ 𝑖 ≤ 𝑛 and 0 ≤ 𝑗 ≤ 𝑚. In [8,
260
R.M. Salem
Theorem 1] Hong et al showed that if 𝑅 is an Armendariz ring, then 𝑅 is a right
zip ring if and only if 𝑅[𝑥] is a right zip ring.
In [3, Theorem 2.8], W. Cortes studied skew polynomial extension over zip
rings, in general he showed that:
(i) If 𝜎 is an automorphism of 𝑅 and 𝑓, 𝑔 ∈ 𝑅[𝑥, 𝜎] (𝑓 ′ , 𝑔 ′ ∈ 𝑅[𝑥, 𝑥−1 , 𝜎]) such
that 𝑓 𝑔 = 0 (𝑓 ′ 𝑔 ′ = 0), implies 𝑎𝑖 𝜎 𝑖 (𝑏𝑗 ) = 0 for each 0 ≤ 𝑖 ≤ 𝑛 and
0 ≤ 𝑗 ≤ 𝑛 (𝑠 ≤ 𝑖 ≤ 𝑞 and 𝑡 ≤ 𝑗 ≤ 𝑙). Then the following are equivalent:
(a) 𝑅 is a right zip ring
(b) 𝑅[𝑥, 𝜎] is a right zip ring
(c) 𝑅[𝑥, 𝑥−1 , 𝜎] is a right zip ring
(ii) If 𝜎 is an automorphism of 𝑅 and 𝑓, 𝑔 ∈ 𝑅[[𝑥, 𝜎]] (𝑓 ′ , 𝑔 ′ ∈ 𝑅[[𝑥, 𝑥−1 , 𝜎]])
such that 𝑓 𝑔 = 0, implies 𝑎𝑖 𝜎 𝑖 (𝑏𝑗 ) = 0 for each 𝑖 ≥ 0 and 𝑗 ≥ 0 (𝑖 ≥ 𝑠 and
𝑗 ≥ 𝑡). Then the following are equivalent:
(a) 𝑅 is right zip
(b) 𝑅[[𝑥, 𝜎]] is right zip
(c) 𝑅[[𝑥, 𝑥−1 , 𝜎]] is right zip
Motivated by the above L. Ouyang in [9] introduced the notion of right weak
zip rings, i.e., rings provided that if the right weak annihilator of a subset 𝑋
of 𝑅, 𝑁 𝑟𝑅 (𝑋) ⊆ 𝑛𝑖𝑙(𝑅), then there exists a finite subset 𝑋0 ⊆ 𝑋 such that
𝑁 𝑟𝑅 (𝑋0 ) ⊆ 𝑛𝑖𝑙(𝑅), where 𝑛𝑖𝑙(𝑅) is the set of all nilpotent elements of 𝑅 and
𝑁 𝑟𝑅 (𝑋) = {𝑎 ∈ 𝑅∣𝑥𝑎 ∈ 𝑛𝑖𝑙(𝑅) for each 𝑥 ∈ 𝑋}. The author in [9] studied the
relationship between right (left) weak zip property of 𝑅 and the Ore extension
𝑅[𝑥, 𝜎, 𝛿]. He showed the following results:
(i) A ring 𝑅 is right (left) weak zip if and only if the upper triangular matrix
𝑇𝑛 (𝑅) is right (left) weak zip, where 𝑛 ≥ 2.
(ii) If 𝑅 is a (𝜎, 𝛿)-compatible and reversible ring, then 𝑅 is a right (left) weak
zip ring if and only if 𝑅[𝑥, 𝜎, 𝛿] is right (left) weak zip. A ring 𝑅 is called 𝜎compatible if for 𝑥, 𝑦 ∈ 𝑅, 𝑥𝑦 = 0 ⇔ 𝑥𝜎𝑦 = 0 and 𝑅 is called 𝛿-compatible
if 𝑥𝑦 = 0 ⇒ 𝑥𝛿𝑦 = 0. If 𝑅 is both 𝜎-compatible and 𝛿-compatible, then 𝑅
is called (𝜎, 𝛿)-compatible.
We recall the following notions for a ring 𝑅.
(i) 𝑅 is reduced if for 𝑥 ∈ 𝑅 such that 𝑥2 = 0, then 𝑥 = 0
(ii) 𝑅 is reversible if for 𝑥, 𝑦 ∈ 𝑅 such that 𝑥𝑦 = 0, then 𝑦𝑥 = 0
(iii) 𝑅 is semicommutative if for 𝑥, 𝑦 ∈ 𝑅 such that 𝑥𝑦 = 0, then 𝑥𝑅𝑦 = 0
The above definitions implies that, the class of reduced rings is a subclass of
the class of reversible rings which is a subclass of the class of semicommutative
rings.
In [11, 12], P. Ribenboim studied extensively the rings of generalized power
series.
The motivation of this paper is to continue the studying of the transfer of
some algebraic properties between the base ring 𝑅 and the generalized power
series ring [[𝑅𝑆,≤ ]] see [6, 13] and to extend W. Cortes & L. Ouyang results to
the generalized power series over zip and weak zip rings.
Generalized Power Series
261
Throughout this paper 𝑅 is an associative ring with identity and (𝑆, ≤) is
a strictly ordered monoid (i.e. a monoid 𝑆 with a compatible order ≤ such
that if 𝑎 < 𝑏 in 𝑆, then 𝑎 + 𝑠 < 𝑏 + 𝑠 for all 𝑠 ∈ 𝑆). By a generalized power
series ring Λ = [[𝑅𝑆,≤ ]] we mean the set of all functions 𝑓 : 𝑆 → 𝑅 such that
𝑠𝑢𝑝𝑝(𝑓 ) = {𝑠 ∈ 𝑆∣𝑓 (𝑠) ∕= 0} is artinian and narrow, i.e. every strictly decreasing
sequence of elements of 𝑠𝑢𝑝𝑝(𝑓 ) is finite and every subset of pairwise orderincomparable elements of 𝑠𝑢𝑝𝑝(𝑓 ) is finite, with pointwise addition and product
operation called convolution defined by
∑
(𝑓 𝑔)(𝑠) =
𝑓 (𝑢)𝑔(𝑣)
for each 𝑓, 𝑔 ∈ Λ
(𝑢,𝑣)∈𝑋𝑠 (𝑓,𝑔)
where
𝑋𝑠 (𝑓, 𝑔) = {(𝑢, 𝑣) ∈ 𝑆 × 𝑆∣𝑢 + 𝑣 = 𝑠 and 𝑓 (𝑢) and 𝑔(𝑣) are nonzero}
is a finite set see [11]. By [11, 2.1] Λ = [[𝑅𝑆,≤ ]] becomes a ring called a generalized
power series ring with coefficients in 𝑅 and exponents in 𝑆, with the identity
map 𝑒 : 𝑆 → 𝑅 defined by 𝑒(0) = 1 and 𝑒(𝑠) = 0 for each nonzero 𝑠 ∈ 𝑆.
The mapping 𝑒𝑠 ∈ Λ defined by 𝑒𝑠 (𝑠) = 1 and 𝑒𝑠 (𝑡) = 0 for each 𝑡 ∕= 𝑠 is an
embedding of 𝑆 into Λ. Also, 𝑅 is canonically embedded as a subring of Λ via
𝑟 7→ 𝑐𝑟 such that 𝑐𝑟 (0) = 𝑟 and 𝑐𝑟 (𝑠) = 0 for each nonzero 𝑠 ∈ 𝑆. So, 𝑅 ≃ 𝑐𝑅
and we can identify 𝑟 ∈ 𝑅 with 𝑐𝑟 ∈ Λ.
Let 𝜋(𝑓 ) denote the set of all minimal elements of 𝑠𝑢𝑝𝑝(𝑓 ). If (𝑆, ≤) is strictly
totally ordered, then 𝜋(𝑓 ) consists of only one element which is still denoted by
𝜋(𝑓 ).
So, the structure of generalized power series rings generalizes the structure
of some classical rings such as polynomial rings, formal power series rings, group
rings and semigroup rings.
If 𝑅 is a ring and 𝑆 is a strictly ordered monoid, then the ring 𝑅 is called
a generalized Armendariz ring if for each 𝑓, 𝑔 ∈ Λ = [[𝑅𝑠,≤ ]] such that 𝑓 𝑔 = 0
implies that 𝑓 (𝑢𝑖 )𝑔(𝑣𝑗 ) = 0 for each 𝑢𝑖 ∈ 𝑠𝑢𝑝𝑝(𝑓 ) and 𝑣𝑗 ∈ 𝑠𝑢𝑝𝑝(𝑔). In [14]
Zhongkui called such ring 𝑠-Armendariz ring.
Let 𝑇 = 𝐶(𝑓 ) be the content of 𝑓 , i.e. 𝐶(𝑓 ) = {𝑓 (𝑠)∣𝑠 ∈ 𝑠𝑢𝑝𝑝(𝑓 )} ⊆ 𝑅.
Since, 𝑅 ≃ 𝑐𝑅 we can identify, the content of 𝑓 with
𝑐𝐶(𝑓 ) = {𝑐𝑓 (𝑢𝑖 ) ∣𝑢𝑖 ∈ 𝑠𝑢𝑝𝑝(𝑓 )} ⊆ Λ
2. Generalized Power Series Over Right Zip Ring
Hirano [7] and W. Cortes [3] observed relations between the right (left) annihilators in the ring 𝑅 and the right (left) annihilators in 𝑅[𝑥] and 𝑅[𝑥, 𝜎]
respectively.
In this note we investigate the relations between the right (left) annihilators in
the ring 𝑅 and the right (left) annihilators in the ring Λ = [[𝑅𝑆,≤ ]] of generalized
power series.
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R.M. Salem
Lemma 2.1. Let 𝑅 be a ring, 𝑆 a strictly ordered monoid, Λ = [[𝑅𝑆,≤ ]] the
generalized power series ring and 𝑈 ⊆ 𝑅. Then 𝑟Λ (𝑈 ) = 𝑟𝑅 (𝑈 )Λ.
Proof. Let 𝑔 ∈ 𝑟Λ (𝑈 ). Then for each 𝑢 ∈ 𝑈 , 𝑐𝑢 𝑔 = 0. Thus, 0 = (𝑐𝑢 𝑔)(𝑣) =
𝑢𝑔(𝑣) = 0, for each 𝑣 ∈ 𝑠𝑢𝑝𝑝(𝑔) and we obtain that 𝑔(𝑣) ∈ 𝑟𝑅 (𝑢), for each
𝑣 ∈ 𝑠𝑢𝑝𝑝(𝑔). Consequently, 𝑔 ∈ 𝑟𝑅 (𝑢)Λ and we have that 𝑟Λ (𝑈 ) ⊆ 𝑟𝑅 (𝑈 )Λ.
On the other hand, let 𝑔 ∈ 𝑟𝑅 (𝑈 )Λ. Then for each 𝑣 ∈ 𝑠𝑢𝑝𝑝(𝑔) we have that
𝑔(𝑣) ∈ 𝑟𝑅 (𝑈 ). Thus for each 𝑢 ∈ 𝑈 we have that 0 = 𝑢𝑔(𝑣) = (𝑐𝑢 𝑔)(𝑣) which
implies that 𝑔 ∈ 𝑟Λ (𝑢). Hence, 𝑔 ∈ 𝑟Λ (𝑈 ). So, 𝑟𝑅 (𝑈 )Λ ⊆ 𝑟Λ (𝑈 ).
With the above lemma we have a map 𝜙 = 𝑟𝑅 (2𝑅 ) → 𝑟Λ (2Λ ) defined by
𝜙(𝐼) = 𝐼Λ for every 𝐼 ∈ 𝑟𝑅 (2𝑅 ) = {𝑟𝑅 (𝑈 )∣𝑈 ⊆ 𝑅}. Obviously, 𝜙 is injective.
In the following lemma we show that 𝜙 is a bijective map if and only if 𝑅 is a
generalized Armendariz ring.
Lemma 2.2. Let 𝑅 be a ring, 𝑆 a strictly ordered monoid and Λ = [[𝑅𝑠,≤ ]] the
ring of generalized power series. Then the following are equivalent:
(1) 𝑅 is a generalized Armendariz ring.
(2) The function 𝜙 : 𝑟𝑅 (2𝑅 ) → 𝑟Λ (2Λ ) is bijective, where 𝜙(𝐼) = 𝐼Λ
Proof. Let 𝑌 ⊆ Λ and 𝑇 = ∪𝑓 ∈𝑌 𝐶(𝑓 ). From Lemma 2.1 it is sufficient to show
that 𝑟Λ (𝑓 ) = 𝑟𝑅 (𝐶(𝑓 ))Λ for all 𝑓 ∈ 𝑌 . In fact, let 𝑔 ∈ 𝑟Λ (𝑓 ). Then 𝑓 𝑔 = 0
and by assumption 𝑓 (𝑢𝑖 )𝑔(𝑣𝑗 ) = 0 for each 𝑢𝑖 ∈ 𝑠𝑢𝑝𝑝(𝑓 ) and each 𝑣𝑗 ∈ 𝑠𝑢𝑝𝑝(𝑔).
Then for a fixed 𝑢𝑖 and each 𝑣𝑗 ∈ 𝑠𝑢𝑝𝑝(𝑔), 0 = 𝑓 (𝑢𝑖 )𝑔(𝑣𝑗 ) = (𝑐𝑓 (𝑢𝑖 ) 𝑔)(𝑣𝑖 ) and it
follows that 𝑔 ∈ 𝑟𝑅 ∪𝑢𝑖 ∈𝑠𝑢𝑝𝑝(𝑓 ) 𝑐𝑓 (𝑢𝑖 ) Λ = 𝑟𝑅 𝐶(𝑓 )Λ. So 𝑟Λ (𝑓 ) ⊆ 𝑟𝑅 𝐶(𝑓 )Λ.
Conversely,
( let )𝑔 ∈ 𝑟𝑅 𝐶(𝑓 )Λ, then 𝑐𝑓 (𝑢𝑖 ) 𝑔 = 0 for each 𝑢𝑖 ∈ 𝑠𝑢𝑝𝑝(𝑓 ).
Hence, 0 = 𝑐𝑓 (𝑢𝑖∑
) 𝑔 (𝑣𝑗 ) = 𝑓 (𝑢𝑖 )𝑔(𝑣𝑗 ) for each 𝑢𝑖 ∈ 𝑠𝑢𝑝𝑝(𝑓 ) and 𝑣𝑗 ∈ 𝑠𝑢𝑝𝑝(𝑔).
Thus, (𝑓 𝑔)(𝑠) = (𝑢𝑖 ,𝑣𝑗 )∈𝑋𝑠 (𝑓,𝑔) 𝑓 (𝑢𝑖 )𝑔(𝑣𝑗 ) = 0 and it follows that 𝑔 ∈ 𝑟Λ (𝑓 ).
Hence 𝑟𝑅 𝐶(𝑓 )Λ ⊆ 𝑟Λ (𝑓 ) and it follows that 𝑟𝑅 𝐶(𝑓 )Λ = 𝑟Λ (𝑓 ). So 𝑟Λ (𝑌 ) =
∩𝑓 ∈𝑌 𝑟Λ (𝑓 ) = ∩𝑓 ∈𝑌 𝑟𝑅 𝐶(𝑓 )Λ = 𝑟𝑅 (𝑇 )Λ.
Now, let 𝑓, 𝑔 ∈ Λ be such that 𝑓 𝑔 = 0. Then 𝑔 ∈ 𝑟Λ (𝑓 ) and by assumption
𝑟Λ (𝑓 ) = 𝑇 Λ for some right ideal 𝑇 of 𝑅. Consequently, 0 = 𝑓 𝑐𝑔(𝑣𝑗 ) and for
any 𝑢𝑖 ∈ 𝑠𝑢𝑝𝑝(𝑓 ), 0 = (𝑓 𝑐𝑔(𝑣𝑗 ) )(𝑢𝑖 ) = 𝑓 (𝑢𝑖 )𝑔(𝑣𝑗 ) for each 𝑢𝑖 ∈ 𝑠𝑢𝑝𝑝(𝑓 ) and
𝑣𝑗 ∈ 𝑠𝑢𝑝𝑝(𝑔). Hence, 𝑅 is a generalized Armendariz ring.
Now, we are able to prove the main theorem of this section.
Theorem 2.3. Let 𝑅 be a generalized Armendarinz ring, 𝑆 a strictly ordered
monoid and Λ = [[𝑅𝑠,≤ ]] the ring of generalized power series. Then Λ is a right
zip ring of and only if 𝑅 is.
Proof. Suppose that Λ is a right zip ring and 𝑋 ⊆ 𝑅 such that 𝑟𝑅 (𝑋) = 0. Let
𝑌 = {𝑐𝑥 ∣𝑥 ∈ 𝑋}. Then it can be easily shown that 𝑟Λ (𝑌 ) = 0. Since, Λ is a
right zip ring, then there exists a finite subset 𝑌0 ⊆ 𝑌 such that 𝑟Λ (𝑌0 ) = 0. Let
Generalized Power Series
263
𝑌0 = 𝑐𝑋0 = {𝑐𝑥𝑖 ∣𝑖 = 1, ⋅ ⋅ ⋅ , 𝑛} and 0 ∕= 𝑎 ∈ 𝑟𝑅 (𝑋0 ). Then 0 = 𝑥𝑖 𝑎 = (𝑐𝑥𝑖 𝑐𝑎 )(0)
for each 𝑖 ∈ {1, ⋅ ⋅ ⋅ , 𝑛}. Hence 0 = 𝑐𝑋0 𝑐𝑎 and it follows that 𝑐𝑎 ∈ 𝑟Λ 𝑐𝑋0 =
𝑟Λ 𝑌0 = 0. So, 𝑎 = 0 and 𝑅 is a right zip ring.
Conversely, suppose that 𝑅 is a right zip ring and 𝑌 ⊆ Λ be such that
𝑟Λ (𝑌 ) = 0 and 𝑇 = 𝐶(𝑌 ) be the content of each 𝑓 ∈ 𝑌 , i.e., 𝑇 = ∪𝑓 ∈𝑌 {𝑓 (𝑢)∣𝑢 ∈
𝑠𝑢𝑝𝑝(𝑓 )}. Then by Lemma 2.2 𝑟Λ (𝑌 ) = 𝑟𝑅 (𝑇 )Λ = 0 and it follows that 𝑟𝑅 𝑇 =
0. Since, 𝑅 is a right zip ring there exists a finite subset 𝑇0 ⊆ 𝑇 such that
𝑟𝑅 (𝑇0 ) = 0. Let 𝑎𝑖 ∈ 𝑇0 , where 𝑇0 can be written as 𝑇0 = {𝑎𝑖 ∣𝑖 ∈ {1, ⋅ ⋅ ⋅ , 𝑛}},
𝑓𝑎𝑖 ∈ 𝑌 be such that 𝑎𝑖 ∈ {𝑓𝑎𝑖 (𝑢)∣𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓𝑎𝑖 )}, 𝑌0 be a minimal subset of 𝑌
which contains {𝑓𝑎𝑖 ∈ 𝑌 ∣𝑎𝑖 ∈ 𝑇0 } and let 𝑇1 = ∪𝑓𝑎𝑖 {𝑓𝑎𝑖 (𝑢)∣𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓𝑎𝑖 )}. So,
𝑇0 ⊆ 𝑇1 and it follows that 𝑟𝑅 𝑇1 ⊆ 𝑟𝑅 𝑇0 = 0. So, by lemma 2.2 we have that
𝑟Λ (𝑌0 ) = 𝑟𝑅 (𝑇1 )Λ = 0 and it follows that Λ is a right zip ring.
3. Generalized Power Series Over Right Weak Zip Ring
Proposition 3.1. Let 𝑅 be a semicommutative ring, (𝑆, +, ≤) a strictly totally
ordered monoid and Λ = [[𝑅𝑠,≤ ]] the generalized power series ring. If 𝑓 ∈ Λ is
nilpotent, then 𝑓 (𝑢) ∈ 𝑅 is nilpotent for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ).
Proof. Let 𝑓 ∈ Λ be a nilpotent element. Then there exists 𝑘 ∈ ℕ such that
𝑓 𝑘 = 0. Since, 𝑆 is a strictly totally
∑ ordered monoid, then 𝜋(𝑓 ) = 𝑢0 . By the fact
that, 0 = 𝑓 𝑘 (𝑘𝑢0 ) = (𝑓 (𝑢0 ))𝑘 + (𝑡1 ,⋅⋅⋅ ,𝑡𝑘 )∈𝑋𝑘𝑢 (𝑓,⋅⋅⋅ ,𝑓 )−{(𝑢0 ,⋅⋅⋅ ,𝑢0 )} 𝑓 (𝑡1 ) ⋅ ⋅ ⋅ 𝑓 (𝑡𝑘 )
0
and 𝜋(𝑓 ) = 𝑢0 , there exists 𝑖 ∈ {1, ⋅ ⋅ ⋅ , 𝑘} such that 𝑡𝑖 ≥ 𝑢0 . Thus, 𝑢0 +⋅ ⋅ ⋅+𝑢0 ≤
𝑡1 + ⋅ ⋅ ⋅ + 𝑡𝑘 = 𝑢0 + ⋅ ⋅ ⋅ + 𝑢0 = 𝑘𝑢0 a contradiction. So, 𝑡𝑖 = 𝑢0 for each
𝑖 ∈ {1, ⋅ ⋅ ⋅ , 𝑘} and it follows that 0 = 𝑓 𝑘 (𝑘𝑢0 ) = (𝑓 (𝑢0 ))𝑘 and 𝑓 (𝑢0 ) is nilpotent.
Consider now, 𝑓 = (𝑓 − 𝑐𝑓 (𝑢0 ) 𝑒𝑢0 ) + 𝑐𝑓 (𝑢0 ) 𝑒𝑢0 = (𝑓 − 𝑓0′ ) + 𝑓0′ = 𝑓0 + 𝑓0′ where
𝑠𝑢𝑝𝑝(𝑓0 ) = 𝑠𝑢𝑝𝑝(𝑓 ) − {𝑢0 } and 𝑠𝑢𝑝𝑝(𝑓0′ ) = {𝑢0 }.
Hence,
0 =𝑓 𝑘 = (𝑓0 + 𝑓0′ )𝑘 = (𝑓0 + 𝑓0′ )(𝑓0 + 𝑓0′ ) ⋅ ⋅ ⋅ (𝑓0 + 𝑓0′ )
=𝑓0𝑘 + 𝑓0 𝑓0′𝑘−1 + ⋅ ⋅ ⋅ + 𝑓0′𝑘−2 𝑓0 𝑓0′ + 𝑓0′𝑘−1 𝑓0 + 𝑓02 𝑓0′𝑘−2 + 𝑓0 𝑓0′𝑘−2 𝑓0 + ⋅ ⋅ ⋅
+ 𝑓0′𝑘−2 𝑓02 + ⋅ ⋅ ⋅ + 𝑓0′𝑘 = 𝑓0𝑘 + Δ + 𝑓0′𝑘 ,
where 𝑠𝑢𝑝𝑝𝑓0′𝑘 ⊆ 𝑠𝑢𝑝𝑝𝑓0′ + ⋅ ⋅ ⋅ + 𝑠𝑢𝑝𝑝𝑓0′ 𝑘-times
.
⊆ 𝑘 𝑠𝑢𝑝𝑝𝑓0′ = {𝑘𝑢0 }.
Thus, 0 = 𝑓0′𝑘 (𝑘𝑢0 ) = (𝑓 (𝑢0 ))𝑘 and 𝑓0′ is nilpotent.
Now, it is clear that Δ is the sum of monomials each monomial is the product
of ℓ copies of 𝑓0 and (𝑘 − ℓ) copies of 𝑓0′ , where 𝑠𝑢𝑝𝑝 each monomial ⊆ the sum
of ℓ copies of 𝑠𝑢𝑝𝑝𝑓0 and (𝑘 − ℓ) copies of 𝑠𝑢𝑝𝑝𝑓0′ .
Since, 𝑓 ′ (𝑢0 ) = 𝑓 (𝑢0 ) is nilpotent and 𝑅 is a semicommutative ring, then by
[13, lemma 3.1] we obtain that each monomial is a nilpotent element of Λ. Thus
𝑓0𝑘 is also nilpotent.
If 𝑓 = 𝑓0′ , then 𝑓 ∈ Λ is a nilpotent element of Λ and 𝑓0′ (𝑢0 ) =
(𝑐𝑓 (𝑢0 ) 𝑒𝑢0 )(𝑢0 ) = 𝑓 (𝑢0 ) is a nilpotent element of 𝑅 and there is nothing to
264
R.M. Salem
prove.
Now, suppose that 0 ∕= 𝑓0 = 𝑓 − 𝑓0′ and let 𝜋(𝑓0 ) = 𝜋(𝑓 − 𝑓0′ ) = 𝑢𝑖 .
Since, 0 ∕= (𝑓 − 𝑓0′ )(𝑢𝑖 ) = (𝑓 − 𝑐𝑓 (𝑢0 ) 𝑒𝑢0 )(𝑢𝑖 ) = 𝑓 (𝑢𝑖 ) and 𝑓0 (𝑢0 ) = (𝑓 −
′
𝑓0 )(𝑢0 ) = (𝑓 − 𝑐𝑓 (𝑢0 ) 𝑒𝑢0 )(𝑢0 ) = 𝑓 (𝑢0 ) − 𝑓 (𝑢0 )𝑒𝑢0 (𝑢0 ) = 𝑓 (𝑢0 ) − 𝑓 (𝑢0 ) = 0 then
𝑢0 < 𝑢𝑖 .
By the fact that, 𝑓0𝑘 is nilpotent, then there exists a positive integer 𝑛 such
that (𝑓0𝑘 )𝑛 = 0.
′
Using the same procedure above it can be easily shown that 0 = 𝑓0𝑘 (𝑘 ′ 𝑢𝑖 ) =
′
(𝑓 (𝑢𝑖 ))𝑘 where 𝑘 ′ = 𝑘𝑛. Continuing on this process 𝑓 = 𝑓𝜇 + 𝑓𝜇′ , where 𝑓𝜇′ :
𝑆 → 𝑅 is defined by 𝑓𝜇′ (𝑢𝑚 ) = (𝑐𝑓 (𝑢𝑚 ) 𝑒𝑢𝑚 )(𝑢𝑚 ) = 𝑓 (𝑢𝑚 ) which is nilpotent for
each 𝑢𝑚 ∈ 𝑠𝑢𝑝𝑝(𝑓 ), 𝑚 ≤ 𝜇 and 𝑓𝜇 is a nilpotent element of Λ. Let 𝜋(𝑓𝜇 ) =
𝜋(𝑓 − 𝑓𝜇′ ) = 𝜋(𝑓 − 𝑐𝑓 (𝑢𝜇 ) 𝑒𝑢𝜇 ) = 𝑢𝜃 , 𝑢𝜇 < 𝑢𝜃 .
Using [12, 5.3] we can define a relation on Λ called section relation for 𝑓𝜇′ and
′
𝑓𝜈 ∈ Λ as follows:
i)
ii)
iii)
iv)
𝑓𝜇′ ⪯ 𝑓𝜈′ if 𝜇 < 𝜈
𝜋(𝑓 − 𝑓𝜇′ ) < 𝜋(𝑓 − 𝑓𝜈′ ) where 𝜇 < 𝜈
𝑢 < 𝜋(𝑓 − 𝑓𝜇′ ) for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓𝜇′ )
𝑓𝜇 = 𝑓 − 𝑓𝜇′ ∈ Λ is nilpotent and 𝑓𝜇′ (𝑢𝑚 ) ∈ 𝑅 is nilpotent for each 𝑢𝑚 ∈
𝑠𝑢𝑝𝑝(𝑓 ), 𝑚 ≤ 𝜇.
Let ∗ denotes the section relation ⪯ with the above properties. Let 𝛼 be an
ordinal such that card 𝛼 > 𝑐𝑎𝑟𝑑 ∣ 𝑠𝑢𝑝𝑝𝑓 ∣ and Γ be the set of all ordinals 𝜆 < 𝛼.
We show that for each 𝜆 ∈ Γ there exists 𝑓𝜆′ ∈ Λ such that ∗ is satisfied. In fact
let 𝜆 ∈ Γ and assume that we have already found the element 𝑓𝜇′ ∈ Λ for every
𝜇 < 𝜆 satisfying ∗ for ordinals 𝜇 < 𝜈 < 𝜆.
Now, we will determine an element 𝑓𝜆′ where ∗ is satisfied for 𝜇 < 𝜈 ≤ 𝜆.
Suppose that there exists an ordinal 𝜂 such that 𝜆 = 𝜂 + 1. If 𝑓 − 𝑓𝜂′ = 0, then
𝑓 = 𝑓𝜂′ is nilpotent. Thus 𝑓𝜂′ (𝑢𝑚 ) = (𝑐𝑓 (𝑢𝑚 ) 𝑒𝑢𝑚 )(𝑢𝑚 ) = 𝑓 (𝑢𝑚 ) ∈ 𝑅 is nilpotent
for each 𝑢𝑚 ∈ 𝑠𝑢𝑝𝑝(𝑓 ), 𝑚 ≤ 𝜂 and there is nothing to prove.
Hence, suppose that 𝑓𝜂 = 𝑓 − 𝑓𝜂′ ∕= 0. Let 𝜋(𝑓𝜂 ) = 𝜋(𝑓 − 𝑓𝜂′ ) = 𝑢𝜆 and
𝑓𝜆′ : 𝑆 → 𝑅 be defined by 𝑓𝜆′ = 𝑓𝜂′ + 𝑐𝑓 (𝑢𝜆 ) 𝑒𝑢𝜆 . So 𝑓𝜆′ ∈ Λ and we show that
𝑓𝜂′ ⪯ 𝑓𝜆′ which implies that 𝑓𝜇′ ⪯ 𝑓𝜆′ for every 𝜇 < 𝜆.
In fact 0 ∕= (𝑓 − 𝑓𝜂′ )(𝑢𝜆 ) = 𝑓 (𝑢𝜆 ) and it follows that 𝑠𝑢𝑝𝑝(𝑓𝜆′ − 𝑓𝜂′ ) =
𝑠𝑢𝑝𝑝(𝑐𝑓 (𝑢𝜆 ) 𝑒𝑢𝜆 ) = {𝑢𝜆 }. If 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓𝜂′ ), then by ∗ 𝑢 < 𝜋(𝑓 − 𝑓𝜂′ ) = 𝑢𝜆 ∈
𝑠𝑢𝑝𝑝(𝑓𝜆′ − 𝑓𝜂′ ). Thus 𝑓𝜂′ ⪯ 𝑓𝜆′ , if 𝑓𝜂′ = 𝑓𝜆′ , and we have that 𝑐𝑓 (𝑢𝜆 ) 𝑒𝑢𝜆 = 0 which
is a contradiction. If 𝑓𝜇′ = 𝑓𝜆′ , then 𝑓𝜇′ ⪯ 𝑓𝜂′ ⪯ 𝑓𝜆′ = 𝑓𝜇′ and 𝑓𝜆′ = 𝑓𝜂′ which is
again a contradiction. Hence, 𝑓𝜇′ ∕= 𝑓𝜆′ for each 𝜇 < 𝜂 ≤ 𝜆, and it can be easily
shown that 𝑓𝜆 = 𝑓 −𝑓𝜆′ is nilpotent and 𝑓𝜆′ (𝑢𝑚 ) = (𝑐𝑓 (𝑢𝑚 ) 𝑒𝑢𝑚 )(𝑢𝑚 ) = 𝑓 (𝑢𝑚 ) ∈ 𝑅
is nilpotent for each 𝑚 ≤ 𝜆 .
If 𝑓𝜆 = 𝑓 − 𝑓𝜆′ = 0, there is nothing to prove, otherwise there exists 𝑢𝜉 ∈
𝑠𝑢𝑝𝑝(𝑓 ) such that 𝜋(𝑓 − 𝑓𝜆′ ) = 𝑢𝜉 where 𝑓𝜆′ = 𝑓𝜂′ + 𝑐𝑓 (𝑢𝜆 ) 𝑒𝑢𝜆 . Since, (𝑓 − 𝑓𝜆′ ) =
𝑓 − 𝑓𝜂′ − 𝑐𝑓 (𝑢𝜆 ) 𝑒𝑢𝜆 and (𝑓 − 𝑓𝜆′ )(𝑢𝜆 ) = (𝑓 − 𝑓𝜂′ − 𝑐𝑓 (𝑢𝜆 ) 𝑒𝑢𝜆 )(𝑢𝜆 ) = 𝑓 (𝑢𝜆 ) −
𝑓𝜂′ (𝑢𝜆 ) − 𝑓 (𝑢𝜆 ) = 0 then 𝑢𝜆 < 𝑢𝜉 . By the fact that 0 ∕= (𝑓 − 𝑓𝜆′ )(𝑢𝜉 ) =
(𝑓 − 𝑓𝜂′ − 𝑐𝑓 (𝑢𝜆 ) 𝑒𝑢𝜆 )(𝑢𝜉 ) = (𝑓 − 𝑓𝜂′ )(𝑢𝜉 ) we have that 𝑢𝜉 ∈ 𝑠𝑢𝑝𝑝(𝑓 − 𝑓𝜂′ ). Hence,
Generalized Power Series
265
𝑢𝜆 < 𝑢𝜉 and 𝜋(𝑓 − 𝑓𝜂′ ) < 𝜋(𝑓 − 𝑓𝜆 ) and this implies that 𝜋(𝑓 − 𝑓𝜇′ ) < 𝜋(𝑓 − 𝑓𝜆′ )
for each 𝜇 < 𝜆.
Now, we show that 𝑢 < 𝜋(𝑓 − 𝑓𝜆′ ) for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓𝜆′ ). In fact 𝑠𝑢𝑝𝑝(𝑓𝜆′ ) =
𝑠𝑢𝑝𝑝(𝑓𝜂′ + 𝑐𝑓 (𝑢𝜆 ) 𝑒𝑢𝜆 ) ⊆ 𝑠𝑢𝑝𝑝(𝑓𝜂′ ) ∪ 𝑠𝑢𝑝𝑝(𝑐𝑓 (𝑢𝜆 ) 𝑒𝑢𝜆 ). If 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓𝜂′ ), then 𝑢 <
𝜋(𝑓 − 𝑓𝜂′ ) and if 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑐𝑓 (𝑢𝜆 ) 𝑒𝑢𝜆 ), then 𝑢 = 𝑢𝜆 = 𝜋(𝑓 − 𝑓𝜂′ ) < 𝜋(𝑓 − 𝑓𝜆′ ).
Now, let 𝜆 be a limit ordinal for the family {𝑓𝜆′ ∣𝜇 < 𝜆} of elements 𝑓𝜇′ ∈ Λ and
it was proved in [12, 5.3] that there exists an element 𝑏 =⪯ − sup(𝑓𝜇′ )𝜇<𝜆 ∈ Λ
such that
i) 𝑓𝜇′ ⪯ 𝑏 for every 𝜇 < 𝜆
ii) If 𝑏′ ∈ Λ and 𝑓𝜇′ ⪯ 𝑏′ for every 𝜇 < 𝜆, then 𝑏 ⪯ 𝑏′ .
Let 𝑓𝜆′ = 𝑏 =⪯ − sup(𝑓𝜇′ )𝜇<𝜆 . Then by i) we know that 𝑓𝜇′ ⪯ 𝑓𝜆′ for every
𝜇 < 𝜆, 𝑓𝜆 = 𝑓 − 𝑓𝜆′ is a nilpotent element of Λ and that 𝑓𝜆′ (𝑢𝑚 ) ∈ 𝑅 is nilpotent
′
′
for each 𝑢𝑚 ≤ 𝑢𝜆 . If 𝑓𝜇′ = 𝑓𝜆′ , then 𝑓𝜇′ ⪯ 𝑓𝜇+1
⪯ 𝑓𝜆′ = 𝑓𝜇′ and 𝑓𝜇′ = 𝑓𝜇+1
which
′
′
is a contradiction. Hence, 𝑓𝜇 ∕= 𝑓𝜆 for every 𝜇 < 𝜆.
Since, 𝑓 − 𝑓𝜆′ = (𝑓 − 𝑓𝜇′ ) − (𝑓𝜆′ − 𝑓𝜇′ ) for every 𝜇 < 𝜆, then by [12] 𝑢𝜉 =
𝜋(𝑓 − 𝑓𝜆′ ) ≥ min{𝜋(𝑓 − 𝑓𝜇′ ), 𝜋(𝑓𝜆′ − 𝑓𝜇′ )}. Note that,
′
− 𝑓𝜇′ )
𝜋(𝑓𝜆′ − 𝑓𝜇′ ) = 𝜋(𝑓𝜇+1
′
= 𝜋(𝑓 − 𝑓𝜇′ ) − (𝑓 − 𝑓𝜇+1
)
≥ min{𝑢𝜃 , 𝑢𝜃+1 }
Hence, 𝑢𝜉 ≥ 𝑢𝜃 for all 𝜇 < 𝜆 and 𝑢𝜃 ≤ 𝑢𝜃+1 ≤ 𝑢𝜉 , then 𝑢𝜃 < 𝑢𝜉 .
We now show that 𝑢 < 𝜋(𝑓 −𝑓𝜆′ ) for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓𝜆′ ). In fact, if 𝑠𝑢𝑝𝑝(𝑓𝜆′ ) =
∪𝜇<𝜆 𝑠𝑢𝑝𝑝(𝑓𝜇′ ), then there exists an ordinal 𝜇 < 𝜆 such that 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓𝜇′ ), thus
𝑢 < 𝑢𝜇 < 𝑢𝜆 .
Hence, for 𝜇, 𝜈 ∈ Γ, 𝜇 < 𝜈, then 𝑢𝜇 < 𝑢𝜈 and we have that ∣{𝑢𝜆 } such that
𝜆 ∈ Γ∣ = ∣Γ∣ > ∣𝑆∣ which is a contradiction. So, 𝑓 = 𝑓𝜆′ and the Proposition is
proved.
Proposition 3.2. Let 𝑅 be a semicommutative right Noetherian ring, (𝑆, +, ≤)
a strictly ordered monoid and Λ = [[𝑅𝑆,≤ ]] the ring of generalized power series.
If 𝑓 ∈ Λ is such that 𝑓 (𝑢) is nilpotent for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ), then 𝑓 ∈ Λ is
nilpotent.
Proof. Let 𝑓 ∈ Λ be such that 𝑓 (𝑢) is a nilpotent element for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 )
and 𝐼 the ideal generated by {𝑓 (𝑢)∣𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 )}. Since, 𝑅 is a semicommutative right Noetherian ring, then by [13, lemma 3.1] 𝐼 is a finitely generated
nilpotent ideal.
Thus, there exists an 𝑚 ∈ ℕ such that 𝐼 𝑚 = (0). Since
∑
𝑚
𝑓 (𝑢) =
(𝑢1 ,⋅⋅⋅ ,𝑢𝑚 )∈𝑋𝑢 (𝑓,⋅⋅⋅ ,𝑓 ) 𝑓 (𝑢1 ) ⋅ ⋅ ⋅ 𝑓 (𝑢𝑚 ) for each 𝑢 ∈ 𝑆. Note that
for each summand belongs to ∈ 𝐼 𝑚 , then each summand is equal zero. So,
𝑠𝑢𝑝𝑝(𝑓 𝑚 ) = 𝜙. Thus 𝑓 𝑚 = 0 and 𝑓 is nilpotent.
The following example is due to D. Fields [5, Example 1] which shows us that
the right Noetherian condition can’t be deleted.
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R.M. Salem
Example 3.3. Let 𝑆 = ℤ/(𝑝) where 𝑝 is prime, {𝑋𝑖 }𝑖∈𝑁 a countable collection of
indeterminates over∑
𝑆, 𝑅 = 𝑆[𝑋0 , 𝑋1 , ⋅ ⋅ ⋅ , 𝑋𝑡 , ⋅ ⋅ ⋅ ]/(𝑋0𝑝 , 𝑋1𝑝 , ⋅ ⋅ ⋅ , 𝑋𝑡𝑝 , ⋅ ⋅ ⋅ ), 𝑓𝑖 =
∞
𝑋 𝑖 and let 𝑓 (𝑋) = 𝑖=0 𝑓𝑖 𝑋 𝑖 ∈ 𝑅[[𝑋]]. Then for each 𝑘 ∈ ℕ (𝑓 (𝑘) )𝑝 = 0. But
for each natural number 𝑛, 𝑓0 𝑓1 ⋅ ⋅ ⋅ 𝑓𝑛−1 ∈ (𝐴𝑓 )𝑛 and 𝑓0 𝑓1 ⋅ ⋅ ⋅ 𝑓𝑛−1 ∕= 0. Hence,
𝐴𝑓 isn’t nilpotent.
We combine Proposition 3.1 and Proposition 3.2 to get the following.
Theorem 3.4. Let 𝑅 be a semicommutative right Noetherian ring, (𝑆, +, ≤) a
strictly totally ordered monoid and Λ = [[𝑅𝑠,≤ ]] the ring of generalized power
series. Then 𝑓 ∈ Λ is a nilpotent element if and only if 𝑓 (𝑢) ∈ 𝑅 is a nilpotent
element for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ).
Proof. Suppose that 𝑓 is a nilpotent element of Λ. Then by Proposition 3.1 𝑓 (𝑢)
is nilpotent for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ).
Conversely if 𝑓 (𝑢) is a nilpotent element of 𝑅 for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ), then by
Proposition 3.2 𝑓 is nilpotent.
Theorem 3.5. Let 𝑅 be a semicommutative right Noetherian ring, (𝑆, +, ≤) a
strictly totally ordered monoid and Λ = [[𝑅𝑆,≤ ]] the generalized power series
ring. Then 𝑅 is a right weak zip ring if and only if Λ is a right weak zip ring.
Proof. Suppose that Λ is a right weak zip ring and 𝑋 ⊆ 𝑅 with 𝑁 𝑟𝑅 (𝑋) ⊆
𝑛𝑖𝑙(𝑅). Let 𝑌 = {𝑐𝑥 ∈ Λ∣𝑥 ∈ 𝑋} and 0 ∕= 𝑓 ∈ 𝑁 𝑟Λ (𝑌 ). Then 𝑐𝑥 𝑓 ∈ 𝑛𝑖𝑙(Λ)
for each 𝑐𝑥 ∈ 𝑌 and by Proposition 3.1 𝑥𝑓 (𝑢) ∈ 𝑛𝑖𝑙(𝑅) for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ).
Thus, 𝑓 (𝑢) ∈ 𝑛𝑖𝑙(𝑅) for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ) and by Proposition 3.2 we obtain that
𝑓 ∈ 𝑛𝑖𝑙(Λ). So, 𝑁 𝑟Λ (𝑌 ) ⊆ 𝑛𝑖𝑙(Λ).
Since Λ is a right weak zip ring, then there exists a finite subset 𝑌0 ⊆ 𝑌
such that 𝑁 𝑟Λ (𝑌0 ) ⊆ 𝑛𝑖𝑙(Λ), where 𝑌0 = 𝑐𝑋0 , and 𝑋0 = {𝑥𝑖 ∣𝑖 = 1, ⋅ ⋅ ⋅ , 𝑛}.
Let 𝑇 = ∪𝑓 ∈𝑁 𝑟Λ (𝑌0 ) {𝑓 (𝑢)∣𝑢 ∈ 𝑠𝑢𝑝𝑝𝑓 }. Then by Proposition 3.1 𝑇 ⊆ 𝑛𝑖𝑙(𝑅).
Therefore, 𝑇 = 𝑁 𝑟𝑅 (𝑋0 ) and 𝑅 is a right weak zip ring.
Conversely, assume that 𝑅 is a right weak zip ring and 𝑌 ⊆ Λ with 𝑁 𝑟Λ (𝑌 ) ⊆
𝑛𝑖𝑙(Λ). Let 𝑇 = 𝐶(𝑌 ) be the content of 𝑌 , i.e. 𝐶(𝑌 ) = ∪𝑓 ∈𝑌 {𝑓 (𝑢)∣𝑢 ∈
𝑠𝑢𝑝𝑝(𝑓 )}. Then 𝑓 (𝑢)𝑎 ∈ 𝑛𝑖𝑙(𝑅) for each 𝑓 (𝑢) ∈ 𝑇 and 𝑎 ∈ 𝑁 𝑟𝑅 (𝑇 ). So, for
each 𝑓 ∈ 𝑌 and 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ), 𝑓 (𝑢)𝑎 = (𝑓 𝑐𝑎 )(𝑢) ∈ 𝑛𝑖𝑙(𝑅) and by Proposition 3.2
𝑓 𝑐𝑎 ∈ 𝑛𝑖𝑙(Λ). Thus, 𝑐𝑎 ∈ 𝑁 𝑟Λ (𝑌 ) ⊆ 𝑛𝑖𝑙(Λ). Then by Proposition 3.1 𝑎 ∈ 𝑛𝑖𝑙(𝑅)
and it follows that 𝑁 𝑟𝑅 (𝑇 ) ⊆ 𝑛𝑖𝑙(𝑅).
Since, 𝑅 is a right weak zip ring, then there exists a finite subset 𝑇0 ⊆ 𝑇
such that 𝑁 𝑟𝑅 (𝑇0 ) ⊆ 𝑛𝑖𝑙(𝑅). So, for each 𝑡 ∈ 𝑇0 , there exists 𝑓𝑡 ∈ 𝑌 such that
𝑡 ∈ {𝑓𝑡 (𝑢)∣𝑓𝑡 ∈ 𝑌 for some 𝑢 ∈ 𝑠𝑢𝑝𝑝𝑓𝑡 }.
Let 𝑌0 be a minimal subset of 𝑌 such that 𝑓𝑡 ∈ 𝑌0 for each 𝑡 ∈ 𝑇0 . Then 𝑌0
is a finite subset of 𝑌 . Let 𝑇1 = ∪𝑓𝑡 ∈𝑌0 {𝑓𝑡 (𝑢)∣𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓𝑡 )}. Thus, 𝑇0 ⊆ 𝑇1 and
we obtain that 𝑁 𝑟𝑅 (𝑇1 ) ⊆ 𝑁 𝑟𝑅 (𝑇0 ) ⊆ 𝑛𝑖𝑙(𝑅).
Let 𝑔 ∈ 𝑁 𝑟Λ 𝑌0 . Then 𝑓 𝑔 ∈ 𝑛𝑖𝑙(Λ) for each 𝑓 ∈ 𝑌0 and by Proposition 3.1
Generalized Power Series
267
(𝑓 𝑔)(𝑤) ∈ 𝑛𝑖𝑙(𝑅) for each 𝑤 ∈ 𝑠𝑢𝑝𝑝(𝑓 𝑔).
∑
Since, (𝑓 𝑔)(𝑤) =
(𝑢𝑖 ,𝑣𝑗 )∈𝑋𝑤 (𝑓,𝑔) 𝑓 (𝑢𝑖 )𝑔(𝑣𝑗 ) ∈ 𝑛𝑖𝑙(𝑅) and 𝑆 is a totally
ordered monoid, then there exists 𝑢0 , 𝑣0 ∈ 𝑆 such that 𝜋(𝑓 ) = 𝑢0 , and 𝜋(𝑔) = 𝑣0 .
Thus
∑ by the same procedure used in Proposition 3.1 we have that (𝑓 𝑔)(𝑢0 +𝑣0 ) =
(𝑢𝑖 ,𝑣𝑖 )∈𝑋𝑢 +𝑣 𝑓 (𝑢𝑖 )𝑔(𝑣𝑖 ) = 𝑓 (𝑢0 )𝑔(𝑣0 ) is nilpotent.
0
0
Now, suppose that 𝑓 (𝑢)𝑔(𝑣) is nilpotent, so 𝑔(𝑣)𝑓 (𝑢) is also nilpotent for
each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ) and 𝑣 ∈ 𝑠𝑢𝑝𝑝(𝑔) such that 𝑢 + 𝑣 < 𝑤. Now, we show that
𝑓 (𝑢)𝑔(𝑣) and 𝑔(𝑣)𝑓 (𝑢) are nilpotent for each 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ) and 𝑣 ∈ 𝑠𝑢𝑝𝑝(𝑔) with
𝑢 + 𝑣 = 𝑤.
Since, 𝑋𝑤 (𝑓, 𝑔) = {(𝑢, 𝑣)∣𝑢 + 𝑣 = 𝑤, 𝑢 ∈ 𝑠𝑢𝑝𝑝(𝑓 ) 𝑎𝑛𝑑 𝑣 ∈ 𝑠𝑢𝑝𝑝(𝑔)} is a finite
subset, then 𝑋𝑤 (𝑓, 𝑔) = {(𝑢𝑖 , 𝑣𝑖 )∣𝑖 = 1, ⋅ ⋅ ⋅ , 𝑛}.
Since 𝑆 is a totally ordered monoid, in fact a cancellative monoid, then let
𝑢1 < 𝑢2 < ⋅ ⋅ ⋅ < 𝑢𝑛 . If 𝑢1 = 𝑢2 and 𝑢1 + 𝑣1 = 𝑢2 + 𝑣2 , then 𝑣1 = 𝑣2 and as <
is strictly order, if 𝑢1 < 𝑢2 and 𝑢1 + 𝑣1 = 𝑢2 + 𝑣2 , it must be 𝑣1 > 𝑣2 . Thus,
𝑣𝑛 < 𝑛𝑛−1 < ⋅ ⋅ ⋅ < 𝑣1 , now
∑
(𝑓 𝑔)(𝑤) =
𝑓 (𝑢𝑖 )𝑔(𝑣𝑖 ) ∈ 𝑛𝑖𝑙(𝑅)
(𝑢1 ,𝑣𝑖 )∈𝑋𝑤 (𝑓,𝑔)
= 𝑓 (𝑢1 )𝑔(𝑣1 ) + 𝑓 (𝑢2 )𝑔(𝑣2 ) + ⋅ ⋅ ⋅ + 𝑓 (𝑢𝑛 )𝑔(𝑣𝑛 )
(**)
For 𝑖 ≥ 2 𝑢1 + 𝑣𝑖 < 𝑢𝑖 + 𝑣𝑖 = 𝑤, then by induction hypothesis, we have
𝑓 (𝑢1 )𝑔(𝑣𝑖 ) and 𝑔(𝑣𝑖 )𝑓 (𝑢1 ) are nilpotent elements. Now multiplying ** from the
right by 𝑓 (𝑢1 ), then
𝑓 (𝑢1 )𝑔(𝑣1 )𝑓 (𝑢1 ) = (𝑓 𝑔)(𝑤)𝑓 (𝑢1 ) − 𝑓 (𝑢2 )𝑔(𝑣2 )𝑓 (𝑢1 ) − ⋅ ⋅ ⋅ − 𝑓 (𝑢𝑛 )𝑔(𝑣𝑛 )𝑔(𝑢1 ).
Since, 𝑅 is semicommutative, then nil(𝑅) is a nilpotent ideal and by induction
𝑓 (𝑢1 )𝑔(𝑣1 )𝑓 (𝑢1 ), hence 𝑓 (𝑢1 )𝑔(𝑣1 ) and 𝑔(𝑣1 )𝑓 (𝑢1 ) are nilpotent.
Therefore, by multiplying (**) from the right by 𝑓 (𝑢2 ), ⋅ ⋅ ⋅ , 𝑓 (𝑢𝑛 ) respectively yields 𝑓 (𝑢𝑖 )𝑔(𝑣𝑖 ) and 𝑔(𝑣𝑖 )𝑓 (𝑢𝑖 ) are nilpotent for each 𝑢𝑖 ∈ 𝑠𝑢𝑝𝑝(𝑓 ) and
𝑣𝑖 ∈ 𝑠𝑢𝑝𝑝(𝑔). Consequently, 𝑔(𝑣) ∈ 𝑁 𝑟𝑅 (𝑇1 ) ⊆ 𝑛𝑖𝑙(𝑅) for each 𝑣𝑖 ∈ 𝑠𝑢𝑝𝑝(𝑔).
Thus by Proposition 3.2 𝑔 ∈ 𝑛𝑖𝑙(Λ). So, 𝑁𝑟Λ (𝑌0 ) ⊆ 𝑛𝑖𝑙(Λ) and Λ is a right weak
zip ring.
Acknowledgement. I’d like to express my deepest gratitude to Prof. M.H. Fahmy
for reading this manuscript and his valuable discussion, also I’m so grateful to
the referee for his/her valuable suggestions which helped me to improve the
presentation considerably.
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