Article
Kyungpook Mathematical Journal 2020; 60(2): 307-318
Published online June 30, 2020
Copyright © Kyungpook Mathematical Journal.
Rings which satisfy the Property of Inserting Regular Elements at Zero Products
Hong Kee Kim, Tai Keun Kwak*, Yang Lee, Yeonsook Seo
Department of Mathematics and RINS, Gyeongsang National University, Jinju 52828, Korea
e-mail : hkkim@gsnu.ac.kr
Department of Mathematics, Daejin University, Pocheon 11159, Korea
e-mail : tkkwak@daejin.ac.kr
Department of Mathematics, Yanbian University, Yanji 133002, China and Institute of Basic Science, Daejin University, Pocheon 11159, Korea
e-mail : ylee@pusan.ac.kr
Department of Mathematics, Pusan National University, Busan 46241, Korea
e-mail : ysseo0305@pusan.ac.kr
Received: October 22, 2019; Revised: January 21, 2020; Accepted: February 10, 2020
Abstract
This article concerns the class of rings which satisfy the property of inserting regular elements at zero products, and rings with such property are called
Keywords: regular-IFP ring, regular element, IFP ring, polynomial ring, generalized reduced ring
1. Introduction
Throughout this article every ring is an associative ring with identity. Let
2. Regular-IFP Rings
In this section we study the properties of regular-IFP rings as well as the relations between regular-IFP rings and ring properties that play important roles in noncommutative ring theory. Due to Bell [4], a ring is called
Definition 2.1
A ring
Regular-IFP rings are clearly unit-IFP (hence Abelian), but not conversely by the following example.
Example 2.2
There exists a unit-IFP ring that is not regular-IFP. Let
Remark 2.3
-
(1) The following conditions are equivalent, which can be proved by applying the regular-IFPness iteratively:
-
(i) A ring
R is regular-IFP; -
(ii)
a 1C (R )a 2C (R )a 3 ⋯a n −1C (R )a n = 0 whenevera 1a 2 ⋯a n = 0 fora 1,a 2, …,a n ∈R .
-
-
(2)
D 3(R ) is (regular-)IFP over a reduced ringR by [15, Proposition 2.1]. HoweverM n (R ) andT n (R ), over any ringR forn ≥ 2, cannot be regular-IFP since they are not Abelian, noting that unit-IFP (or regular-IFP) rings are Abelian. -
(3) There exists an Abelian ring that is not regular-IFP. Set
R =D n (S ) forn ≥ 4 over an Abelian ringS . ThenR is Abelian by [10, Lemma 2]. LetA =E 12,B =E 34 ∈R . ThenAB = 0. ConsiderC =I n +E 23 ∈R . ThenC ∈C (R ) clearly. ButACB =E 14 ≠ 0, so thatR is not regular-IFP. -
(4) Let
R be a regular-IFP ring such thatR =C (R ) ∪N (R ). Letab = 0 fora ,b ∈R . ThenaC (R )b = 0 sinceR is regular-IFP. MoreoveraN (R )b = 0 by [15, Lemma 1.2(2)]. ThusR is IFP.
Based on Armendariz [3, Lemma 1], a ring
By Goodearl [7], a ring
Following [11], a ring is called
Proposition 2.4
-
(1)
Let R be a locally finite ring .-
(i)
If R is an Armendariz ring, then it is regular-IFP . -
(ii)
If R is a regular-IFP ring, then R /J (R )is strongly regular with J (R ) =N (R ).
-
-
(2)
Let R be a right or left Artinian ring. If R is regular-IFP, then R /J (R )is a strongly regular ring with J (R ) =N (R ).
(1)–(i) Let
(1)–(ii) Since regular-IFP rings are Abelian, we obtain the result by [11, Proposition 2.5].
(2) It is well-known that
The class of regular-IFP rings is not closed under homomorphic images as can be seen by the ring
Proposition 2.5
-
(1)
Let R λ (λ ∈ Λ) be Abelian rings. Then R λ is regular-IFP for each λ ∈ Λif and only if Πλ ∈ΛR λ is regular-IFP if and only if the subring of Πλ ∈ΛR λ generated by ⊕λ ∈ΛR λ and 1Πλ ∈ΛR λ is regular-IFP . -
(2)
Let R be an Abelian ring and e 2 =e ∈R. Then R is regular-IFP if and only if both eR and (1 −e )R are regular-IFP .
(1) Suppose that the subring of Π
(2) The proof is obtained from (1) since
For a given ring
Corollary 2.6
Suppose that
Conversely assume that
Recall that homomorphic images of regular-IFP rings need not be regular-IFP. Considering this fact, one may ask whether a ring
Example 2.7
There exists a non-regular-IFP ring
3. Extensions of Regular-IFP Rings
In this section we examine the regular-IFP property of ring extensions that play roles in noncommutative ring theory.
Regarding Remark 2.3(3), we have the following.
Proposition 3.1
-
(1)
R is a reduced ring; -
(2)
D 3(R )is an IFP ring; -
(3)
D 3(R )is a regular-IFP ting; -
(4)
D 3(R )is a unit-IFP ring; -
(5)
AN (D 3(R ))B = 0whenever AB = 0for A ,B ∈D 3(R ).
The equivalences of the conditions (1), (2), and (4) are proved by [15, Proposition 2.1], and so they are equivalent to (3).
(4) ⇒ (5): Suppose that (4) holds and let
(5) ⇒ (1): Suppose that (5) holds. Assume on the contrary that there exists 0 ≠
in
Following Cohn [6], a ring
Example 3.2
We refer to the construction and argument in [16, Example 2.1]. Let
be the free algebra generated by noncommuting indeterminates
where the constant terms of
with their images in
Now, consider
Then
because
Remark 3.3
-
(1) Note that
D 2(R ) over a reduced ringR is IFP by [16, Proposition 1.6] and so it is regular-IFP. Moreover, there exists a non-reduced non-commutative reversible ringR over whichD 2(R ) is regular-IFP by [15, Example 2.2]. However, the ringS is always regular-IFP whenD 2(S ) is regular-IFP. For, suppose thatD 2(S ) is regular-IFP and letab = 0 fora ,b ∈S . For, we have AB = 0 and soAC (D 2(S ))B = 0 by assumption. Setfor any c ∈C (S ). ThenC ∈C (D 2(S )) andACB = 0, entailingacb = 0. ThusS is regular-IFP. -
(2) Related to (1) above, there exists a reversible ring
R such thatD 2(R ) is not regular-IFP. Let ℍ be the Hamilton quaternions over ℝ andR =D 2(ℍ). ThenR is reversible [16, Proposition 1.6]. We refer to the argument in [16, Example 1.7]. Considerin
D 2(R ). ThenAB = 0.Note that
by [13, Lemma 2.1] because
. But ACB ≠ 0, henceD 2(R ) is not regular-IFP. -
(3) For a ring
R andn ≥ 2, letV n (R ) be the ring of all matrices (a ij ) inD n (R ) such thata st =a (s +1)(t +1) fors = 1, …,n − 2 andt = 2, …,n − 1. Note that. If R is a reduced ring, thenV n (R ) is (regular)-IFP by [17, Lemma 2.3 and Proposition 3.3], but the converse does not hold in general as can be seen by the commutative ringV n (R ) over a non-reduced commutative ring (e.g., ℤn l forn ,l ≥ 2)R forn ≥ 2.
Proposition 3.4
-
Let M be a multiplicatively closed subset of a ring R consisting of central regular elements. Then R is regular-IFP if and only if M −1R is regular-IFP . -
Let R be a ring. Then R [x ]is regular-IFP if and only if R [x ,x −1]is regular-IFP .
(1) It comes from the fact that
(2) Recall the ring of
In [15, Example 2.7], we see an IFP ring
Proposition 3.5
-
{
c +c 1x + … +c t x t ∈R [x ] |c ∈C (R )for t ≥ 1} ⊆C (R [x ])and {d 0 +d 1x + … +d s −1x s −1 +dx s ∈R [x ] |d ∈C (R )for s ≥ 1} ⊆C (R [x ]). -
If R [x ]is regular-IFP, then so is R . -
Let R be a regular-IFP ring such that C (R [x ]) = {c +xN (R )[x ] |c ∈C (R )}. If R is Armendariz then R [x ]is regular-IFP .
(1) Consider
(2) It is routine.
(3) Suppose that
The next example shows that the condition “
Example 3.6
We use the ring and argument in [12, Example 2]. Let
where
Notice that (
because
Considering Proposition 3.5, it is natural to ask whether
Example 3.7
Let
This implies
We consider next some equivalent conditions to the regular-IFP property in relation to the sum of coefficients of polynomials which satisfy some property of inserting regular polynomials. For
Proposition 3.8
-
(1)
R is regular-IFP; -
(2)
If f 1(x )f 2(x ) ⋯f n (x ) = 0for f 1(x ),f 2(x ), …,f n (x ) ∈R [x ],then the sum of all coefficients of every polynomial in is zero; -
(3)
If f (x )g (x ) = 0for f (x ),g (x ) ∈R [x ],then the sum of all coefficients of every polynomial in f (x )C (R )[x ]g (x )is zero; -
(4)
If f (x )g (x ) = 0for linear polynomials f (x ),g (x )in R [x ],then the sum of all coefficients of every polynomial in f (x )C (R )[x ]g (x )is zero; -
(5)
f (x )g (x ) = 0implies f (x )C (R )[x ]g (x ) = 0for linear polynomials f (x ),g (x )in R [x ].
The procedure of the proof is almost similar to one of [15, Proposition 2.8], but we write it here for completeness. (1) ⇒ (2): Assume that the condition (1) holds. Let
By Remark 2.3(1), we have
is zero.
(2) ⇒ (3), (3) ⇒ (4), and (5) ⇒ (1) are obvious.
(4) ⇒ (5): Assume that the condition (4) holds. Let
From
for all
4. Related Topic
Based on Proposition 3.1(5), a ring
Theorem 4.1
-
(1)
N 0(R ) =N *(R ) =N *(R ). -
(2)
J (R [x ]) =N 0(R [x ]) =N *(R [x ]) =N *(R [x ]) =N 0(R )[x ] =N *(R )[x ] =N *(R )[x ] =N 0(R )[x ] ⊆J (R )[x ]. Moreover, J (R [x ]) =J (R )[x ]when J (R )is nil .
(1) Let
and hence (
(2) By help of [1, Theorem 1] and [5, Corollary 4], we have
and so
Moreover,
Notice that
On the other hand, Hong et al. [9] consider the duo property on the monoid of regular elements as follows. They call a ring
Proposition 4.2
Suppose that
Notice that the converse of Proposition 4.2 does not hold in general by the next example.
Example 4.3
Let
Consider the group ring
Corollary 4.4
-
(1)
R is DR; -
(2)
R is regular-IFP; -
(3)
R is unit-IFP; -
(4)
R is nilpotent-IFP; -
(5)
R is Abelian; -
(6)
The equation 1 +x 2 +y 2 = 0has no solutions in K; -
(7)
R is isomorphic to a finite direct product of division rings; -
(8)
R is reduced .
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