# Skew Polynomial Extensions over Zip Rings

Authors: Wagner Cortes
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

In this article, we study the relationship between left (right) zip
property ofR and skew polynomial extension over R, using the skew
versions of Armendariz rings.


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## Body

## 1. Introduction

Throughout this paperR denotes an
associative ring with
identity and σ:R→R an automorphism
of R, otherwise unless stated. We denote R[[x;σ]] (R[[x,x−1;σ]]) the skew series rings (skew Laurent series rings)
whose elements are the series ∑i≥0aixi (∑j=p∞bjxj), where the addition is defined as usual and the
multiplication is defined by the rule, xa=σ(a)x (xa=σ(a)x and x−1a=σ−1(a)x), for any a∈R. Note that the skew polynomial
rings of automorphism type R[x;σ] (skew Laurent
of polynomial R[x,x−1;σ]) are subrings
of R[[x;σ]] (R[[x,x−1;σ]]) whose elements are ∑i=0naixi (∑j=qmbjxj) where the sum and multiplication
are defined as before.Rege and Chhawchharia in [1] introduced the notion of an
Armendariz ring. A ring R is called
Armendariz if whenever polynomials ∑i=0naixi, ∑j=0mbjxj∈R[x] satisfy f(x)g(x)=0, then aibj=0 for each 0≤i≤n and 0≤j≤m. The name Armendariz ring was chosen because
Armendariz [2] had
shown that a reduced ring (i.e., ring without nonzero nilpotent elements)
satisfies this condition. Some properties of Armendariz rings have been studied
by Rege and Chhawchharia [1], Armendariz [2], Anderson and Camillo [3], and Kim and Lee [4].Faith in [5] called a ring R right zip if
the right annihilator rR(X) of a subset X of R is zero, then rR(Y)=0 for a finite
subset Y⊆X; equivalently, for a left ideal L of R with rR(L)=0, there exists a finitely generated left ideal L1⊆L such that rR(L1)=0. R is zip if it is
right and left zip. The concept of zip rings was initiated by Zelmanowitz
[6] and appeared in
various papers [5, 7–12], and references therein.
Zelmanowitz stated that any ring satisfying the descending chain condition on
right annihilators is a right zip ring (although not so-called at that time),
but the converse does not hold. Extensions of zip rings were studied by several
authors. Beachy and Blair [7]
showed that if R is a
commutative zip ring, then the polynomial ring R[x] over R is zip. The
authors in [13] proved
that R is a right
(left) zip ring if and only if R[x] is a right
(left) zip ring when R is an
Armendariz ring.In this paper, we study skew polynomial extensions over
zip rings by using skew versions of Armendariz rings and we generalized the
results of [13]. Our
skew versions of Armendariz rings follow the ideas of [14, Definition]. Moreover, we
provide some examples to display some of the phenomenas of Section 2.


## 2. Skew Polynomial Extensions over Zip Rings

Throughout this
paperσ is an
automorphism of R unless
otherwise stated and S will denote one
of the following rings: R[x;σ], R[[x;σ]], R[x,x−1σ], and R[[x,x−1;σ]]. A left (right) annihilator of a subset U of R is defined by lR(U)={a∈R:aU=0} (rR(U)={a∈R:Ua=0}). For a ring R, put rAnnR(2R)={rR(U):U⊆R} and lAnnR(2R)={lR(U):U⊆R}.We begin with the following lemma and use it
without further mention.Lemma 2.1.
LetS be one of the
rings above and  U  a subset of  R. The following statements hold:
(i)lS(U)=SlR(U),
(ii)rS(U)=rR(U)S.Proof.
(i) We only prove for the caseS=R[x;σ] because the
other cases are similar. Let f(x)=∑i=0naixi∈R[x;σ] such that f(x)U=0. Then σ−i(ai)U=0 for all 0≤i≤n and it follows
that σ−i(ai)∈lR(U) for all 0≤i≤n. Hence f(x)=∑i=0nxiσ−i(ai)∈R[x;σ]lR(U). So lR[x;σ](U)⊆R[x;σ]lR(U). We clearly have that R[x;σ]lR(U)⊆lR[x;σ](U). Therefore, we have lR[x;σ](U)=R[x;σ]lR(U).
(ii) We only prove for the caseS=R[x;σ] because the
other cases are similar. Let f(x)=∑i=0naixi∈R[x;σ] such that Uf(x)=0. Then Uai=0 for all 0≤i≤n and it follows
that ai∈rR(U) for all 0≤i≤n. Hence f(x)=∑i=0naixi∈rR(U)R[x;σ]. So rR[x;σ](U)⊆rR(U)R[x;σ]. We clearly have that rR(U)R[x;σ]⊆rR[x;σ](U). Therefore, we have rR[x;σ](U)=rR(U)R[x;σ].With the above lemma, we have mapsϕ:rAnnR(2R)→rAnnS(2S) defined by ϕ(I)=IS for every I∈rAnnR(2R) and(2.1)Ψ:lAnnR(2R)→lAnnS(2S) defined by Ψ(I)=SI for every I∈lAnnR(2R). Moreover, we have maps Φ:rAnnS(2S)→rAnnR(2R) defined by Φ(J)=J∩R for every J∈rAnnS(2S) and Γ:lAnnS(2S)→lAnnR(2R) defined by Γ(J)=J∩R for every J∈lAnnS(2S). Obviously, ϕ is injective
and Φ is surjective.
Clearly, ϕ is surjective
if and only if Φ is injective,
and in this case ϕ and Φ are the
inverses of each other. Note that Ψ and Γ satisfy the same relations
as above. The first item of the definition below appears in [14,
Definition].Definition 2.2.
(i) Suppose thatσ is an
endomorphism of R. A ring R satisfies SA1′
if for f(x)=∑i=0naixi and g(x)=∑j=0mbjxj in R[x;σ],  f(x)g(x)=0 implies
that aiσi(bj)=0 for all 0≤i≤n and 0≤j≤m.
(ii)
Suppose thatσ is an
endomorphism of R. A ring R satisfies SA2′
if for f(x)=∑i=0∞aixi and g(x)=∑j=0∞bjxj in R[[x;σ]],  f(x)g(x)=0 implies
that aiσi(bj)=0 for all i≥0, j≥0.(iii)
Suppose thatσ is an
automorphism of R. A ring R satisfies SA3′
if for f(x)=∑i=sqaixi and g(x)=∑j=tnbjxj∈R[x,x−1;σ],  f(x)g(x)=0 implies that aiσi(bj)=0 for all s≤i≤q and t≤j≤n.(iv)
Suppose thatσ is an
automorphism of R. A ring R satisfies SA4′
if for f(x)=∑i=s∞aixi and g(x)=∑j=t∞bjxj∈R[[x,x−1;σ]],  f(x)g(x)=0 implies that aiσi(bj)=0 for all i≥s and j≥t.Note that ifR satisfies one
of the conditions above, then all subrings S of R such that σ(S)⊆S satisfies the same
property. The following implications are easy to verify: SA4′⇒SA3′ and SA2′⇒SA1′. Following [15, Example 2.1] when σ=idR, the last implication is not reversible.Lemma 2.3.
Letσ  be an
automorphism of  R. Then
(i)R satisfies SA1′
if and only if R satisfies SA3′;
(ii)R satisfies SA2′
if and only if R satisfies SA4′.Proof.
Letf(x),g(x)∈R[x,x−1;σ] such that f(x)g(x)=0, where f(x)=∑i=−pqaixi and g(x)=∑j=−tsbjxj. We clearly have xpf(x)∈R[x;σ] and g(x)xt∈R[x;σ], then xpf(x)g(x)xt=0. By assumption, σp(ai)σi+p(bj)=0 for all −p≤i≤q and −t≤j≤s. Hence aiσi(bj)=0 for all −p≤i≤q and −t≤j≤s. Since R[x;σ]⊆R[x,x−1;σ], the converse
follows.
The proof of the other statement is similar.The following definition appears in [16, Definition
2.1].Definition 2.4.
LetR be a ring and σ an endomorphism
of R. Then R is said σ-compatible
like right R-module, if ar=0 if and only if aσ(r)=0 for any a∈R and r∈R.LetR be a ring and α an endomorphism
of R. Following [17], the endomorphism α is said α-rigid if rα(r)=0, then r=0. A ring R is said a rigid
ring if it exists a rigid endomorphism α of R.Proposition 2.5.
Letσ be an
endomorphism of R. If R is a reduced
ring and σ-compatible
like right R-module, then R is a σ-rigid ring and
hence satisfies SA1′ and SA2′.Proof.
We only
prove the case of SA2′ because the other are similar. We claim that R[[x;σ]] is a reduced
ring. In fact, let f(x)=∑i=0∞aixi such that (f(x))2=0. We have that a02=0. Since R is reduced,
then a0=0. Next, we have a1σ(a1)=0, since R is σ-compatible and
reduced, then a1=0. By induction, we get f(x)=0. Hence R[[x;σ]] is reduced.
Using the same ideas of [14, Proposition 3], we have that R is σ-rigid and
using similar ideas of [14, Corollary 4], we obtain that R satisfies SA2′.Without the
assumption thatR is σ-compatible,
Proposition 2.5 is not true. In fact, let R=ℤ2⊕ℤ2 and σ:R→R, defined by σ((a,b))=(b,a). By [14, Example 2], R does not
satisfy SA2′ because R does not
satisfy SA1′. Observe that (1,0)(0,1)=(0,0) but (1,0)σ(0,1)≠(0,0) and so R is not σ-compatible. We
have the following natural questions.Questions
(i)
Letσ be an
endomorphism of R. Suppose that R satisfies SA2′.
Is Rσ-compatible
like right R-module?(ii)
Letσ be an endomorphism
of R. Suppose that R is σ-compatible
like right R-module. Does R satisfy SA2′?The question
(i) is false. LetR0 be any domain
and R=R0[x]. Let σ:R→R be defined by σ(t)=0 and σ|R0=idR0. By [16, Example 4.1], R is not σ-compatible and
using the similar ideas of the proof of [14, Proposition 10], we have that R satisfies SA2′
and consequently R satisfies SA1′.The question (ii) is false. LetR=K[x,y]/(x2,y2), where K is a field of
characteristic 2, and consider T=M2(R). In this case, take σ=idT. By [18, Example 3.6], S does not
satisfy SA2′ because T does not
satisfy SA1′. Moreover, T is σ-compatible
like right T-module.In [19] the authors introduced the following version of skew
Armendariz rings.
(i)
Suppose thatσ is an
endomorphism of R. Let f(x)=∑i=0naixi,g(x)=∑j=0mbjxj∈R[x;σ] such that f(x)g(x)=0 implies aibj=0 for all 0≤i≤n and 0≤j≤m.(ii)
Suppose thatσ is an
endomorphism of R. Let f(x)=∑i≥0aixi, g(x)=∑j≥0bjxj∈R[[x;σ]] such that f(x)g(x)=0 implies aibj=0 for all i≥0 and j≥0.Note that the item (i) above in [20, Definition 1.1] the
authors called it by σ-Armendariz,
the item (ii) above is similar with [20, Definition 1.1] and we call it here by σ-power
Armendariz.In the next proposition, we
give a relationship between the definition above and the skew versions of
Armendariz rings used in this paper. Using [21, Lemma 2.1] and [20, Theorem 1.8], the proof of next proposition is easy
to verify.Proposition 2.6.
Letσ be an
endomorphism of R and suppose
that R is σ-compatible
like right R-module. Then
(i)R satisfies SA1′
if and only if R is σ-Armendariz;
(ii)R satisfies SA2′
if and only if R is σ-power Armendariz.The proposition above without the compatibility
assumption is not true according to [20, Example 1.9] and the
authors in [22,
Theorem 2.2] obtained an approach of the result above without the compatibility
assumption.The following proposition is a generalization of
[18, Proposition
3.4] and partially generalizes [15, Proposition 2.6].Lemma 2.7.
LetS
be any of the
rings  R[x;σ]   and  R[[x;σ]]. The following conditions are equivalent:
(i)R satisfies SA2′(SA1′);
(ii)ϕ:rAnnR(2R)→rAnnS(2S) defined by ϕ(J)=JS is bijective;
(iii)Ψ:lAnnR(2R)→lAnnS(2S) defined by Ψ(J)=SJ is bijective.Proof.
We only prove the proposition in the
case of SA2′ because the equivalence of (i) and (ii) when R satisfies SA1′
was proved in [23,
Proposition 3.2]. The equivalence between (i) and (iii) when R satisfies SA1′
has similar proof.
(i)→(ii). It is only
necessary to show that ϕ is surjective.
For an element f(x)=∑i=0∞aixi∈R[[x;σ]], define Cf(x)={σ−i(ai),i≥0}, and for a subset T of R[[x;σ]], we denote the
set ∪f(x)∈TCf(x) by CT. We show that rR[[x;σ]](f(x))=rR[[x;σ]](Cf(x)). In fact, given g(x)=∑j=0∞bjxj in rR[[x;σ]](f(x)), we have f(x)g(x)=0. Since R satisfies SA2′,
then aiσi(bj)=0 for all i≥0 and j≥0. In particular, σ−i(ai)bj=0 for all i≥0 and j≥0. Hence g(x)∈rR[[x;σ]](Cf(x)).
On the other hand, leth(x)=∑k=0∞ckxk be an element
in R[[x;σ]] such that Cf(x)h(x)=0. It is clear
that aiσi(ck)=0 for all i≥0 and k≥0. So f(x)h(x)=(0). Since R satisfies SA2′
then rR[[x;σ]](T)=rR[[x;σ]](∪f(x)∈TCf(x)). Thus
(2.2)rR[[x;σ]](T)=⋂f(x)∈TrR[[x;σ]](f(x))=⋂f(x)∈TrR[[x;σ]](Cf(x))=(⋂f(x)∈TrR(Cf(x)))R[[x;σ]]=rR(CT)R[[x;σ]]. Therefore, ϕ is surjective.
(ii)→(i). Let f(x)=∑i=0∞aixi and g(x)=∑j=0∞bjxj be elements in R[[x;σ]] such that f(x)g(x)=0. By assumption, rR[[x,σ]](f(x))=BR[[x;σ]], for some right ideal B of R. Hence g(x)∈BR[[x;σ]] and we have
that bj∈B⊂rR[[x;σ]](f(x)) for all j≥0. So aiσi(bj)=0 for all i≥0 and j≥0.
(iii)→(i). Let f(x)=∑i≥0aixi and g(x)=∑j≥0bjxj be elements in R[[x;σ]] such that f(x)g(x)=0. By assumption, lR[[x;σ]](g(x))=R[[x;σ]]B for some left
ideal B of R. We can write f(x)=∑i≥0xiσ−i(ai)∈R[[x;σ]]B. By the equality of the polynomials with the
coefficients on the right side, we have that σ−i(ai)∈B⊆lR[[x;σ]](g(x)) for all i≥0. So aiσi(bj)=0 for all i≥0 and j≥0.
(i)→(iii). It is only
necessary to show that Ψ is surjective.
Let f(x)=∑i≥0aixi∈R[[x;σ]]. Define Cf(x)={ai,i≥0}, and for a subset T of R[[x;σ]], we denote the set ∪f(x)∈TCf(x) by CT. We show that
(2.3)lR[[x;σ]](f(x))=lR[[x;σ]](Cf(x)). In fact, given g(x)=∑j≥0bjxj∈lR[[x;σ]](f(x)), we have g(x)f(x)=0. Since R satisfies SA2′,
then bjσj(ai)=0 for all i≥0 and j≥0. Hence g(x)=∑j≥0xjσ−j(bj)∈lR[[x;σ]](Cf(x)).
On the other hand, letg(x)∈R[[x;σ]] such that g(x)Cf(x)=0. Thus g(x)ai=0 for all i≥0. So g(x)∑i≥0aixi=g(x)f(x)=0, and we have that g(x)∈lR[[x;σ]](f(x)).
We easily have that for each subsetT of R[[x;σ]],
(2.4)lR[[x;σ]](T)=lR[[x;σ]](⋃f(x)∈TCf(x)). We claim that lR[[x;σ]](Cf(x))=R[[x;σ]]lR(Cf(x)). In fact, let g(x)=∑j≥0bjxj such that g(x)Cf(x)=0. Then we have that 0=g(x)ai=∑j≥0bjxjai=∑j≥0xjσ−j(bj)ai. Thus σ−j(bj)∈lR(Cf(x)), and it follows that
(2.5)∑j≥0xjσ−j(bj)∈R[[x;σ]]lR(Cf(x)). The other
inclusion is trivial. So
(2.6)lR[[x;σ]](T)=⋂f(x)∈TlR[[x;σ]](Cf(x))=⋂f(x)∈TlR[[x;σ]](Cf(x))=R[[x;σ]](⋂f(x)∈TlR(Cf(x)))=R[[x;σ]]lR(CT).
Therefore,Ψ is surjective.Now we are able to prove the main results of this
paper.Theorem 2.8.
Letσ be an
automorphism of R.
(i)
Suppose thatR satisfies SA1′.
The following conditions are equivalent:
(a)
R is a right
(left) zip ring;(b)
R[x;σ] is a right
(left) zip ring;(c)
R[x,x−1,σ] is a right
(left) zip ring.(ii)
Suppose thatR satisfies SA2′.
The following conditions are equivalent:
(a)
R is right (left)
zip ring;(b)
R[[x;σ]] is right (left)
zip ring;(c)
R[[x,x−1;σ]] is right (left)
zip ring.Proof.
(i) We
will show the right case because the left case is similar.
Suppose thatR[x;σ] is right zip.
Let X be a subset of R such that rR(X)=0, and f(x)=∑i=0naixi∈R[x;σ] such that Xf(x)=0. Thus ai∈rR(X)=0 and it follows
that f(x)=0. By assumption, there exists X1={x0,…,xn} such that rR[x;σ](X1)=0. Hence rR(X1)=rR[x;σ](X1)∩R=(0).
Conversely, letY⊆R[x;σ] such that rR[x;σ](Y)=0. By Lemma 2.7, rR[x;σ](Y)=rR(T)R[x;σ], where T=CY=∪f(x)∈YCf(x) such that Cf(x)={σ−i(ai):0≤i≤n} with f(x)=∑i=0naixi∈Y. We have that rR(T)=0 and, by
assumption, there exists T1={σ−i1(ai1),…,σ−in(ain)} such that rR(T1)=0. For each σ−ij(aij)∈T1, there exists gaij(x)∈Y such that some
of the coefficients of gaij(x) are aij for each 1≤j≤n. Let Y0 be a minimal
subset of Y such that gaij(x)∈Y0 for each 1≤j≤n. Then Y0 is nonempty
finite subset of Y. Set T0=∪f(x)∈Y0(Cf(x)) and we have that T1⊆T0. Hence rR(T0)⊆rR(T1)=0. By Lemma 2.7, rR[x;σ](Y0)=rR(T0)R[x;σ] and it follows
that rR[x;σ](Y0)=0.
The proofs of(a)⇔(c) and of item
(ii) follow similarly.Letσ be an
endomorphism of R and δ:R→R an additive map
of R. The application δ is said to be a σ-derivation if δ(ab)=δ(a)b+σ(a)δ(b). The Ore extension R[x;σ,δ] is the set of
polynomials ∑i=0naixi with the usual
sum, and the multiplication rule is xa=σ(a)x+δ(a).Following [16], R is said to be (σ,δ)-compatible,
where σ is an
endomorphism of R and δ is a σ-derivation of R if ab=0⇔aσ(b)=0 and ab=0 implies that aδ(b)=0.In the next result we
obtain a necessary and sufficient condition forR[x;σ,δ] to be left zip,
when σ is an
endomorphism of R using the skew
version of Armendariz rings of [19].Theorem 2.9.
Letσ be an
endomorphism of R and δ a σ-derivation of R. Suppose that if f(x)g(x)=0 for f(x)=∑i=0naixi and g(x)=∑j=0mbjxj∈R[x;σ,δ], then aibj=0 for all 0≤i≤n and 0≤j≤m. Then R is left zip if
and only if R[x;σ,δ] is left zip.Proof.
LetX be any subset
of R[x;σ,δ] and CX=∪f(x)∈XCf(x), where Cf(x)={ai,0≤i≤n} with f(x)=∑i=0naixi. Suppose that lR[x;σ,δ](X)=0. We clearly have lR(CX)=0. By assumption, there exists {b0,…,bt}⊆CX such that lR(Y)=0. Let fbi(x)∈X be an element
of X with some of
its coefficients are equal to bi for all 1≤i≤t. Take X0 be a minimal
subset of X with this
property. We clearly have that X0 is a finite
set. We claim that lR[x;σ,δ](X0)=0. In fact, we easily have lR(CX0)=0, where CX0=∪f(x)∈X0Cf(x) with Cf(x) being defined as
before. Next, let g(x)=∑j=0mbjxj such that g(x)X0=0. Hence for any f(x)=∑i=0naixi∈X0, g(x)f(x)=0, and we have, by assumption, bjai=0 for all 0≤j≤m and 0≤i≤n. Thus bjCX0=0 for all 0≤j≤m and it follows
that g(x)=0. So lR[x;σ,δ](X0)=0.
Using the methods of Theorem2.8, the converse
follows.Remark 2.10.
LetR   be a ring and  σ
an endomorphism
of  R. Suppose that  R   is  σ-power
Armendariz and left zip. Using similar methods of [20, Theorem 1.8], R satisfies SA2′
and with similar ideas of Theorem 2.9, we have that  R
is a left zip
ring if and only if  R[[x;σ]]
is a left zip
ring.


## 3. Examples

In this section, we present some examples of rings that
satisfy SA1′ and SA2′, and they are zip rings. Moreover, an example of a σ-rigid ring
that is a zip ring is given.Example 3.1.1.
LetF be any field
and σ:F→F any
automorphism of F. Following [14, page 113], we consider the ring T(F,F) with
automorphism σ¯(a,b)=(σ(a),σ(b)) and we denote
it by σ. Note that
(3.1)T(F,F)={(ab0a):a,b∈F}. By [14, Proposition 15], T(F,F) satisfies SA1′,
and using similar methods, we can prove that T(F,F) satisfies SA2′.
We claim that T(F,F) is a zip ring.
In fact, the unique one-sided ideals of T(F,F) are {(0000)},
(3.2)I={(0b00):b∈F}, and T(F,F). Note that rT(F,F)(I)≠{0} and lT(F,F)(I)≠0. So we easily have that T(F,F) is a zip ring.Example 3.2.2.
LetF be any field
and σ a monomorphism
of F, and let
(3.3)R={(abc0ad00a):a,b,c∈F} with usual
addition and multiplication of matrix. Note that the monomorphism σ is naturally
extended to R, and R has the
following one-sided ideals:
(3.4)I1={(00000a000):a∈F},I2={(00c000000):c∈F},R  and
the zero ideal. We easily have rR(I2)≠0, lR(I2)≠0, rR(I1)≠0, and lR(I1)≠0. Now we clearly have that R is a zip ring
and by [14,
Proposition 17], R satisfies SA1′,
and with similar methods of [14, Proposition 17], we can prove that R satisfies SA2′.Example 3.3.3.3.
LetD be any domain with identity, R=D[x], σ an endomorphism
of R defined by σ(f(x))=f(0). Since R is a domain,
then R is right and
left zip. Moreover, using similar methods of [14, Example 5], we have that R satisfies SA1′
and SA2′.Example 3.4.4.
LetD and D1 be any domains, σ an monomorphism
of D, and τ an monomorphism
of D1. Set R=D×D1 with usual
addition and multiplication, and we define an endomorphism γ of R by γ(a,b)=(σ(a),τ(b)). We easily have that γ is a
monomorphism of R. Since D is σ-rigid and D1 is τ-rigid, we
easily obtain that R is γ-rigid. We
claim that R is left and
right zip. In fact, let I be any left
ideal of R. It is well known that I=A×B, where A is a left ideal
of D and B is a left ideal
of D1. Suppose that rR(I)=0. Then A≠0 and B≠0. It is not difficult to show that rD(A)=0 and rD1(B)=0. Since D and D1 are left zip,
then there exists a left finitely generated ideal L of D contained in A such that rD(L)=0 and a left
finitely generated ideal L1 of D1 contained in B such that rD1(L1)=0. Thus rR(L×L1)=0 and L×L1 is a left
finitely generated ideal of R contained in A×B. Hence R is left zip.
Using similar methods, we have that R is right zip.Example 3.5.5.
LetF be a field, σ an automorphism
of F,
(3.5)R={(abc0ad00a):a,b,c∈F}, and D a domain with
automorphism τ. Set T=R×D and we define
an endomorphism γ of T by γ(a,b)=(σ(a),τ(t)). It is clear that γ is an
automorphism of T and it is not
difficult to show that T satisfies SA1′ and SA2' because R and D satisfy SA1′ by
[14,Proposition 17]
and [14, Proposition
10], respectively, and using similar methods of [14,Proposition 17] and
[14, Proposition
10], R and D satisfy SA2′,
respectively.
Using similar methods of Example3.4, we have that T is right and
left zip and note that T is not γ-rigid, since T is not a
reduced ring.



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*Source: 496720-2008-04-14.xml*