# Prime Ideals and Strongly Prime Ideals of  Skew Laurent Polynomial Rings

Authors: E. Hashemi
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

We first study connections betweenα-compatible ideals of R and related ideals of the skew Laurent polynomials ring R[x,x−1;α], where α is an automorphism of R. Also we investigate the relationship of P(R) and Nr(R) of R with the prime radical and the upper nil radical of the skew Laurent polynomial rings. Then by using
Jordan's ring, we extend above results to the case where α is not surjective.


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## Body

## 1. Introduction

Throughout  the
paper,R always denotes an
associative  ring with unity. We use P(R), Nr(R),
and N(R) to denote the prime radical, the upper nil
radical, and the set of all nilpotent elements of R,
respectively.Recall that for
a ringR with an injective ring endomorphism α:R→R, R[x;α] is the Ore extension of R.
The set {xj}j≥0 is easily seen to be a left Ore subset of R[x;α],
so that one can localize R[x;α] and form the skew Laurent polynomials ring R[x,x−1;α].
Elements of R[x,x−1;α] are finite sum of elements of the form x−jrxi,
where r∈R and i,j are nonnegative integers. Multiplication is
subject to xr=α(r)x and rx−1=x−1α(r) for all r∈R.Now we consider
Jordan's construction of the ringA(R,α) (see [1], for more details). Let A(R,α) or A be the subset {x−irxi∣r∈R,i≥0} of the skew Laurent polynomial ring R[x,x−1;α].
For each j≥0,  x−irxi=x−(i+j)αj(r)x(i+j).
It follows that the set of all such elements forms a subring of R[x,x−1;α] with x−irxi+x−jsxj=x−(i+j)(αj(r)+αi(s))x(i+j)  and
(x−irxi)(x−jsxj)=x−(i+j)αj(r)αi(s)x(i+j) for r,s∈R and i,j≥0.
Note that α is actually an automorphism of A(R,α),
given by x−irxi to x−iα(r)xi,
for each r∈R and i≥0.
We have R[x,x−1;α]≅A[x,x−1;α],
by way of an isomorphism which maps x−irxj to α−i(r)xj−i.
For an α-ideal I of R,
put Δ(I)=∪i≥0x−iIxi.
Hence Δ(I) is α-ideal of A.
The constructions I→Δ(I), J→J∩R are inverses, so there is an order-preserving
bijection between the sets of α-invariant ideals of R and α-invariant ideals of A.According to
Krempa [2], an
endomorphism α of a ring R is called rigid if aα(a)=0 implies a=0 for a∈R. R is called an α-rigid ring [3] if there exists a rigid
endomorphism α of R.
Note that any rigid endomorphism of a ring is a monomorphism and α-rigid rings are reduced (i.e., R has no nonzero nilpotent element) by Hong et
al. [3]. Properties of α-rigid rings have been studied in Krempa
[2], Hirano [4], and Hong et al. [3, 5].On the other
hand, a ringR is called 2-primal if P(R)=N(R) (see [6]). Every reduced ring is obviously a 2-primal ring.
Moreover, 2 primal rings have been extended to the class of rings which satisfy Nr(R)=N(R),
but the converse does not hold [7, Example 3.3]. Observe that R is a 2-primal ring if and only if P(R)=Nr(R)=N(R),
if and only if P(R) is a completely semiprime ideal (i.e., a2∈P(R) implies that
a∈P(R) for a∈R) of R.
We refer to [6–12] for more detail on 2 primal rings.Recall that a
ringR is called strongly prime if R is prime with no nonzero nil ideals. An ideal P of R is strongly prime if R/P is a strongly prime ring. All (strongly) prime
ideals are taken to be proper. We say an ideal P of a ring R is minimal (strongly) prime if P is minimal among (strongly) prime ideals of R.
Note that (see [13]) Nr(R)=∩{P∣P is a minimal strongly prime ideal of R}.Recall that an
idealP of R is completely prime if ab∈P implies a∈P or b∈P for a,b∈R.
Every completely prime ideal of R is strongly prime and every strongly prime
ideal is prime.According to
Hong et al. [5], for
an endomorphism α of a ring R,
an α-ideal I is called to be α-rigid ideal if aα(a)∈I implies that a∈I for a∈R.
Hong et al. [5]
studied connections between α-rigid ideals of R and related ideals of some ring extensions.
Also they studied relationship of P(R) and Nr(R) of R with the prime radical and the upper nil
radical of the Ore extension R[x;α,δ] of R in the cases when either P(R) or Nr(R) is an α-rigid ideal of R and obtaining the following result. Let P(R) (resp., Nr(R)) be an α-rigid δ-ideal of R.
Then P(R[x;α,δ])⊆P(R)[x;α,δ] (resp., Nr(R[x;α,δ])⊆Nr(R)[x;α,δ]).In [14], the authors defined α-compatible rings, which are a generalization
of α-rigid rings. A ring R is called α-compatible if for each a,b∈R, ab=0⇔aα(b)=0.
In this case, clearly the endomorphism α is injective. In [14, Lemma 2.2], the authors
showed that R is α-rigid if and only if R is α-compatible and reduced. Thus, the α-compatible ring is a generalization of α-rigid ring to the more general case where R is not assumed to be reduced.Motivated by
the above facts, for an endomorphismα of a ring R,
we define α-compatible ideals in R which are a generalization of α-rigid ideals. For an ideal I,
we say that I is an α-compatible ideal of R if for each a,b∈R, ab∈I⇔aα(b)∈I.
The definition is quite natural, in the light of its similarity with the notion
of α-rigid ideals, where in Proposition 2.3, we
will show that I is an α-rigid ideal if and only if I is an α-compatible ideal and completely semiprime.In this paper,
we first study connections betweenα-compatible ideals of R and related ideals of the skew Laurent
polynomial ring R[x,x−1;α],
where α is an automorphism of R.
Also we investigate the relationship of P(R) and Nr(R) of R with the prime radical and the upper nil
radical of the skew Laurent polynomials. Then by using Jordan's ring, we extend
above results to the case where α is not surjective.


## 2. Prime Ideals and Strongly Prime Ideals of Skew Laurent Polynomial Rings

Recall that an
idealI of R is called an α-ideal if α(I)⊆I; I is called α-invariant if α−1(I)=I.
If I is an α-ideal, then α¯:R/I→R/I defined by α¯(a+I)=α(a)+I is an endomorphism. Then we have the following
proposition.Proposition 2.1.
LetI be an ideal of a ring R.
Then the following statements are equivalent:
(1)
I is an α-compatible ideal;(2)
R/I is α¯-compatible.Proof.
It is clear.Proposition 2.2.
LetI be an α-compatible ideal of a ring R.
Then
(1)
I is α-invariant;(2)
ifab∈I,
then aαn(b)∈I,αn(a)b∈I for every positive integer n;
conversely, if aαk(b) or αk(a)b∈I for some positive integer k,
then ab∈I.Proof.
This follows from [15, Lemma 2.2 and
Proposition 2.3].Recall from [16] that a one-sided ideal I of a ring R has the insertion of factors property (or simply, IFP) if ab∈I implies aRb⊆I for a,b∈R (Bell in 1970 introduced this notion for I=0).Proposition 2.3 (see [15], Proposition 2.3 2.4).
LetR be a ring, I an ideal of R,
and α:R→R an endomorphism of R.
Then the following conditions are equivalent:
(1)
I is α-rigid ideal of R;(2)
I is α-compatible, semiprime and has the IFP;(3)
I is α-compatible and completely semiprime.For anα-ideal I of R,
put Δ(I)=∪i≥0x−iIxi.Proposition 2.4.
(1) IfI is an α-compatible ideal of R,
then Δ(I) is an α-compatible ideal of A.
(2) IfJ is an α-compatible ideal of A,
then J=Δ(J0) and J0 is an α-compatible ideal of R.
(3) IfP is a completely (semi)prime α-compatible ideal of R,
then Δ(P) is a completely (semi)prime α-compatible ideal of A.
(4) IfQ is a completely (semi)prime α-compatible ideal of A,
then Q=Δ(Q0) and Q0 is a completely (semi)prime α-compatible ideal of R.
(5) IfP is a prime α-compatible ideal of R,
then Δ(P) is a prime α-compatible ideal of A.Proof.
(1) SinceI is an α-ideal of R, Δ(I) is an ideal of A.
Now, let
(x−irxi)(x−jsxj)∈Δ(I).
Hence x−(i+j)αj(r)αi(s)xi+j∈Δ(I) and that
αj(r)αi(s)∈I.
Thus αj(r)αi+1(s)∈I,
since I is α-compatible. Consequently (x−irxi)α(x−jsxj)∈Δ(I).
Therefore Δ(I) is α-compatible.
(2) LetJ0=J∩J and r∈J0.
Then αn(r)∈J0 for each n≥0.
Hence αn(x−nrxn)=r∈J for each n≥0.
Thus x−nrxn∈J,
since J is α-compatible. Therefore Δ(J0)⊆J0.
Now, let x−mrxm∈J.
Then αm(x−mrxm)∈J and that r∈J,
since J is α-compatible. Thus J⊆Δ(J0).
Consequently, Δ(J0)=J.
(3) Let(x−irxi)(x−jsxj)∈Δ(P).
Then x−(i+j)αj(r)αi(s)xi+j∈Δ(P) and that αj(r)αi(s)∈P.
Hence rs∈P,
by Proposition 2.2. Thus r∈P or s∈P,
since P is completely prime. Consequently, x−irxi∈Δ(P) or x−jsxj∈Δ(P).
(4) By (2),Q=Δ(Q0) and Q0 is a α-compatible ideal of R.
Since Q is completely (semi)prime and Q=Δ(Q0),
hence Q0 is completely (semi)prime. Let (x−irxi)A(x−jsxj)⊆Δ(P).
Then rRs⊆P,
by Proposition 2.2. Hence r∈P or s∈P,
since P is prime. Consequently x−irxi∈Δ(P) or x−jsxj∈Δ(P).
Therefore Δ(P) is a prime ideal of A.Theorem 2.5.
LetP be a strongly prime α-compatible ideal of R.
Then Δ(P) is a strongly prime ideal of A.Proof.
SinceP is a prime α-compatible ideal of R,
hence Δ(P) is a prime ideal of A,
by Proposition 2.4. We show that Δ(P) is a strongly prime ideal of A.
Assume J=I/Δ(P) is a nil ideal of A/Δ(P).
Let a∈Ii.
Then x−iaxi∈I.
Since I/Δ(P) is a nil ideal, hence (x−iaxi)n∈Δ(P) for some n≥0.
Hence x−ianxi=x−jpxj for some p∈P and j≥0.
Thus αj(an)=αi(p)∈P.
Hence an∈P,
since P is α-compatible. Then (Ii+P)/P is a nil ideal of R/P for each i≥0.
Hence Ii⊆P,
for each i≥0.
Therefore I⊆Δ(P).
Consequently, Δ(P) is a strongly prime ideal of A.Note that ifI is an α-ideal of R,
then I[x,x−1;α] is an ideal of the skew Laurent polynomials
ring R[x,x−1;α].Theorem 2.6.
Letα be an automorphism of R.
Let I be a semiprime α-compatible ideal of R.
Assume f(x)=∑i=rnaixi and g(x)=∑j=smbjxj∈R[x,x−1;α].
Then the following statements are equivalent:
(1)
f(x)R[x,x−1;α]g(x)⊆I[x,x−1;α];(2)
aiRbj⊆I for each i,j.Proof.
(1)⇒(2). Assume f(x)R[x,x−1;α]g(x)⊆I[x,x−1;α].
Then(†)(arxr+⋯+anxn)c(bsxs+⋯+bmxm)∈I[x,x−1;α]foreachc∈R.Hence anαn(cbm)∈I.
Thus ancbm∈I,
since I is α-compatible. Next, replacing c by cbm−1dane,
where c,d,e∈R.
Then (arxr+⋯+anxn)cbm−1dane(bsxs+⋯+bm−1xm−1)∈I[x,x−1;α].
Hence anαn(cbm−1danebm−1)∈I and that
ancbm−1danebm−1∈I,
since I is α-compatible. Thus (RanRbm−1)2⊆I.
Hence RanRbm−1⊆I,
since I is semiprime. Continuing this process, we
obtain anRbk⊆I,
for k=s,…,m.
Hence from α-compatibility of I,
we get
(arxr+⋯+anxn)R[x,x−1;α](bsxs+⋯+bm−1xm−1)⊆I[x,x−1;α].
Using induction on n+m,
we obtain aiRbj⊆I for each i,j.
(2)⇒(1). It follows from Proposition 2.2.Corollary 2.7.
Letα be an automorphism on R.
If I is a (semi)prime α-compatible ideal of R,
then I[x,x−1;α] is a (semi)prime ideal of R[x,x−1;α].Proof.
Assume thatI is a prime α-compatible ideal of R.
Let f(x)=∑i=rnaixi and g(x)=∑j=smbjxj∈R[x,x−1;α] such that f(x)R[x,x−1;α]g(x)⊆I[x,x−1;α].
Then aiRbj⊆I for each i,j,
by Theorem 2.6. Assume g(x)∉I[x,x−1;α].
Hence bj∉I for some j.
Thus ai∈I for each i=r,…,n,
since I is prime. Therefore f(x)∈I[x,x−1;α].
Consequently, I[x,x−1;α] is a prime ideal of R[x,x−1;α].Theorem 2.8.
If each minimal prime ideal ofR is α-compatible, then
P(R[x,x−1;α])⊆Δ(P(R))[x,x−1;α].Proof.
LetP be a minimal prime ideal of R.
By Proposition 2.4, Δ(P) is a α-compatible ideal of A.
Assume (a−irxi)A(x−jsxj)⊆Δ(P).
Then rRs⊆P,
since Δ(P) is α-compatible. Hence r∈P or s∈P.
Thus a−irxi∈Δ(P) or x−jsxj∈Δ(P).
Therefore Δ(P) is a prime ideal of A.
Thus Δ(P)[x,x−1;α] is a prime ideal of A[x,x−1;α],
by Corollary 2.7. Consequently, P(R[x,x−1;α])⊆Δ(P(R))[x,x−1;α].In [14], the authors give some examples of α-compatible rings however they are not α-rigid. Note that there exists a ring R for which every nonzero proper
ideal is α-compatible but R is not α-compatible. For example, let R=(FF0F),
where F is a field, and the endomorphism α of R is defined by α((ab0c))(ab0c)=(a00c) for a,b,c∈F.The following
examples show that there existsα-compatible ideals which are not α-rigid.Example 2.9 (see [2.95], Example 2.5).
LetF be a field. Let R={(ff10f)∣f,f1∈F[x]},
where F[x] is the ring of polynomials over F.
Then R is a subring of the 2×2 matrix ring over the ring F[x].
Let α:R→R be an automorphism defined by α((ff10f))(ff10f)=(fuf10f),
where u is a fixed nonzero element of F.
Let p(x) be an irreducible polynomial in F[x].
Let I={(0f100)∣f1∈〈p(x)〉},
where 〈p(x)〉 is the principal ideal of F[x] generated by p(x).
Then I is an α-compatible ideal of R but is not α-rigid. Indeed, since
(0g(x)00)α((0g(x)00))(0g(x)00)=(0000)∈I,
but (0g(x)00)∉I for g(x)∉〈p(x)〉.
Thus I is not α-rigid.Example 2.10.10 (see [17], Example 2.10).
Letℤ2 be the field of integers  modulo 2 and
A=ℤ2[a0,a1,a2,b0,b1,b2,c] be the free algebra of polynomials with zero
constant term in noncommuting indeterminates a0,a1,a2,b0,b1,b2,c over ℤ2.
Note that A is a ring without unity. Consider an ideal of ℤ2+A,
say I,
generated by a0b0, a1b2+a2b1, a0b1+a1b0, a0b2+a1b1+a2b0, a2b2, a0rb0, a2rb2, (a0+a1+a2)r(b0+b1+b2) with r∈A and r1r2r3r4 with r1,r2,r3,r4∈A.
Then I has the IFP. Let α:R→R be an inner automorphism (i.e., there exists
an invertible element u∈R such that α(r)=u−1ru for each r∈R). Then I is α-compatible, since I has the IFP. But I is not α-rigid, since I is not completely semiprime.Theorem 2.11.
Letα be an automorphism of R.
If each minimal prime ideal of R is α-compatible, then P(R[x,x−1;α])⊆P(R)[x,x−1;α].Proof.
It follows from Corollary2.7.The following
example shows that there exists a ringR such that all minimal prime ideals are α-compatible, but are not α-rigid.Example 2.12.12.12 (see [15], Example 2.12.11).
LetR=Mat2(ℤ4) be the 2×2 matrix ring over the ring ℤ4.
Then P(R)={(a11a12a21a22)∣aij∈2¯ℤ} is the only prime ideal of R.
Let α:R→R be the endomorphism defined by α((a11a12a21a22))(a11a12a21a22)=(a11−a12−a21a22).
Then α is an automorphism of R and P(R) is α-compatible. However, P(R) is not α-rigid, since (0100)α((0100))(0100)∈P(R),
but (0100)∉P(R).Theorem 2.13.
Letα be an automorphism of R.
If P is a completely (semi)prime α-compatible ideal of R,
then P[x,x−1;α] is a completely (semi)prime ideal of R[x,x−1;α].Proof.
LetP be a completely prime ideal of R. R/P is domain, hence it is a reduced ring. R/P is an α¯-compatible ring, hence R/P is α¯-rigid, by [14, Lemma 2.2]. Let f(x)¯, g(x)¯∈R/P[x,x−1;α¯] such that
f(x)¯g(x)¯=0.
Then f(x)¯=0 or g(x)¯=0,
by a same way as used in [3, Proposition 6]. Thus R[x,x−1;α]/P[x,x−1;α]≅R/P[x,x−1;α¯] is domain and P[x,x−1;α] is a completely prime ideal of R[x,x−1;α].Corollary 2.14.
Letα be an automorphism on R.
If P(R) is an α-rigid ideal of R,
then P(R[x,x−1;α])⊆P(R)[x,x−1;α].Proof.
P(R) is α-rigid, hence P(R) is a completely semiprime α-compatible ideal of R,
by Proposition 2.3. Therefore P(R[x,x−1;α])⊆P(R)[x,x−1;α],
by Theorem 2.13.Theorem 2.15.
Letα be an automorphism of R.
If P is a strongly (semi)prime α-compatible ideal of R,
then P[x,x−1;α] is a strongly (semi)prime ideal of R[x,x−1;α].Proof.
By Corollary2.7, P[x,x−1;α] is a prime ideal of R[x,x−1;α].
Hence
R[x,x−1;α]/P[x,x−1;α]≃R/P[x,x−1;α¯] is a prime ring. We claim that zero is the
only nil ideal of R/P[x,x−1;α¯].
Let J be a nil ideal of R/P[x,x−1;α¯].
Assume I be the set of all leading coefficients of
elements of J.
First we show that I is an ideal of R/P.
Clearly, I is a left ideal of R/P.
Let a¯∈I and r¯∈R/P.
Then there exists f(x)¯=a0¯+⋯+an−1¯xn−1+a¯xn∈J.
Hence (f(x)¯r¯)m=0,
for some nonnegative integers m.
Thus
a¯α¯n(r¯a¯)⋯α¯(m−1)n(ra¯)α¯mn(r¯)=0,
since it is the leading coefficient of (f(x)¯r¯)m.
Therefore (ar¯)m=0,
since R/P is α¯-compatible. Consequently, I is an ideal of R/P.
Clearly I is a nil ideal of R/P.
Hence I=0 and so J=0.
Therefore P[x,x−1;α] is a strongly prime ideal of R[x,x−1;α].Theorem 2.16.
Letα be an automorphism of R.
If each minimal strongly prime ideal of R is α-compatible, then Nr(R[x,x−1;α])⊆Nr(R)[x,x−1;α].Corollary 2.17.
Letα be an automorphism of R.
If Nr(R) is an α-rigid ideal of R,
then Nr(R[x,x−1;α])⊆Nr(R)[x,x−1;α].Proof.
Nr(R) is α-rigid, hence Nr(R) is a completely semiprime α-compatible ideal of R,
by Proposition 2.3, and that Nr(R) is a strongly semiprime ideal of R.
Therefore Nr(R[x,x−1;α])⊆Nr(R)[x,x−1;α],
by Theorem 2.15.Example2.12
also shows that there exists a ring R such that all minimal strongly prime ideals
are α-compatible, but are not α-rigid.Theorem 2.18.
Assume each minimal prime ideal ofR is α-compatible. Then the following areequivalent:
(1)
P(R[x,x−1;α]) is completely semiprime;(2)
P(R[x,x−1;α])=Δ(P(R))[x,x−1;α] and P(R) is completely semiprime.Proof.
(1)⇒(2). Suppose that P(R[x,x−1;α]) is a completely semiprime ideal of R[x,x−1;α].
It is enough to show that Δ(P(R))[x,x−1;α]⊆P(R[x,x−1;α]),
by Theorem 2.8. Let Q be a minimal prime ideal of R[x,x−1;α] and P=A∩Q.
Since P(R[x,x−1;α]) is a completely semiprime ideal of R[x,x−1;α], P is a completely semiprime ideal of A.
Clearly P is an α-invariant ideal of A.
Hence P=Δ(P0).
We claim that P0 is a minimal prime ideal of R.
Since P is completely prime, P0 is a completely prime ideal of R.
Let I be a minimal prime ideal of R such that I⊆P0.
By assumption, I is α-compatible. Hence Δ(I) is a prime α-compatible ideal of A.
Thus Δ(I)[x,x−1;α] is a prime ideal of R[x,x−1;α] and Δ(I)[x,x−1;α]⊆Δ(P0)[x,x−1;α]⊆Q.
Since Q is a minimal prime ideal of R[x,x−1;α],
hence Δ(I)[x,x−1;α]=Δ(P0)[x,x−1;α]=Q.
Therefore Δ(I)=Δ(P0) and that I=P0.
Consequently Δ(P(R))[x,x−1;α]⊆P(R[x,x−1;α]) and that P(R[x,x−1;α])=Δ(P(R))[x,x−1;α].
Since P(R[x,x−1;α])=Δ(P(R))[x,x−1;α] and P(R[x,x−1;α]) is completely semiprime, hence P(R) is completely semiprime.
(2)⇒(1). Since P(R) is α-compatible and completely semiprime, Δ(P(R)) is an α-compatible completely semiprime ideal of A.
Hence Δ(P(R))[x,x−1;α] is a completely semiprime ideal of A[x,x−1;α]=R[x,x−1;α].
Thus P(R[x,x−1;α])=Δ(P(R))[x,x−1;α] is a completely semiprime ideal of R[x,x−1;α].Corollary 2.19.
Letα be an automorphism of R.
Let each minimal prime ideal of R be α-compatible. Then the following are
equivalent:
(1)
P(R[x,x−1;α]) is completely semiprime;(2)
P(R[x,x−1;α])=P(R)[x,x−1;α] and P(R) is completely semiprime.



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