New York Journal of Mathematics

New York J. Math. 12 (2006) 349–356.

Compatible ideals and radicals of Ore extensions

E. Hashemi

Abstract. For a ring endomorphism α and an α-derivation δ, we introduce α-compatible ideals which are a generalization of α-rigid ideals and study the connections of the prime radical and the upper nil radical of R with the prime radical and the upper nil radical of the Ore extension R[x; α, δ] and the skew power series R[[x; α]]. As a consequence we obtain a generalization of Hong, Kwak and Rizvi, 2005.

Contents

1. Introduction 349 2. The prime and upper nil radicals of Ore extensions 351 References 355

1. Introduction

Throughout the paper R always denotes an associative ring with unity. R[x; α, δ] will stand for the Ore extension of R, where α is an endomorphism and δ an αderivation of R, that is, δ is an additive map such that δ(ab) = δ(a)b + α(a)δ(b), for all a, b ∈ R. We use P (R), N_{r}(R) and N(R) to denote the prime radical, the upper nil radical and the set of all nilpotent elements of R, respectively.

According to Krempa [15], 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 [9] 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 by Hong et al. [9]. Properties of αrigid rings have been studied in Krempa [15], Hirano [7] and Hong et al. [11] and [9].

On the other hand, a ring R is called 2-primal if P (R) = N(R) (see [2]). Every reduced ring is obviously a 2-primal ring. Moreover, 2-primal rings have been extended to the class of rings which satisfy N_{r}(R) = N(R), but the converse does not hold ([3], Example 3.3). Observe that R is a 2-primal ring if and only if P (R) = N_{r}(R) = N(R), if and only if P (R) is a completely semiprime ideal (i.e., a^{2} ∈ P (R) implies a ∈ P (R) for a ∈ R) of R. We refer to [2, 3, 7, 10, 14, 18, 19]

349^{ISSN 1076-9803/06}

Received May 29, 2006.

Mathematics Subject Classification. primary 16W20, 16N40, 16S36, secondary 16W60.

Key words and phrases. α-compatible rings, Ore extensions, skew power series, α-rigid rings, prime and upper nil radicals.

for more detail on 2-primal rings. Recall that a ring R 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 [17])

N_{r}(R) = ∩{P | P is a minimal strongly prime ideal of R}.

Recall that an ideal P 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. [11], for an endomorphism α of a ring R, an α-ideal I is called an α-rigid ideal if aα(a) ∈ I implies a ∈ I for a ∈ R. Hong et al.

[11] studied connections between the α-rigid ideals of R and the related ideals of some ring extensions. They also studied the relationship of P (R) and N_{r}(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 N_{r}(R) is an α-rigid ideal of R, obtaining the following result: Let P (R) (resp. N_{r}(R)) be an α-rigid δ-ideal of R. Then

P (R[x; α, δ]) ⊆ P (R)[x; α, δ]

(resp. N_{r}(R[x; α, δ]) ⊆ N_{r}(R)[x; α, δ]). Hong et al. [11] provided an example to show that the condition “P (R) is α-rigid” is not superfluous.

In [6], the authors defined α-compatible rings, which are a generalization of αrigid rings. A ring R is called α-compatible if for each a, b ∈ R, ab = 0 ⇔ aα(b) = 0. Moreover, R is said to be δ-compatible if for each a, b ∈ R, ab = 0 ⇒ aδ(b) = 0. If R is both α-compatible and δ-compatible, we say that R is (α, δ)-compatible. In this case, clearly the endomorphism α is injective. In ([6], 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 an α-rigid ring to the more general case where R is not assumed to be reduced.

Motivated by the above facts, for an endomorphism α of a ring R, we define αcompatible 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. Moreover, I is said to be a δ-compatible ideal if for each a, b ∈ R, ab ∈ I ⇒ aδ(b) ∈ I. If I is both α-compatible and δ-compatible, we say that I is an (α, δ)-compatible ideal. The definition is quite natural, in the light of its similarity with the notion of α-rigid ideals. Here, in Proposition 2.4, 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 give some examples of (α, δ)-compatible ideals which are not α-rigid. Then, we study connections between (α, δ)-compatible ideals of R and related ideals of some ring extensions. Also we investigate the relationship of P (R) and N_{r}(R) of R with the prime radical and the upper nil radical of the Ore extension R[x; α, δ] and the skew power series R[[x; α]].

Recall that an ideal I of R is called an α-ideal if α(I) ⊆ I; I is called α-invariant if α^{−1}(I) = I; I is called a δ-ideal if δ(I) ⊆ I; I is called an (α, δ)-ideal if it is both an α- and a δ-ideal. If I is an (α, δ)-ideal, then α : R/I → R/I defined by α(a + I) = α(a) + I is an endomorphism and δ : R/I → R/I defined by δ(a + I) = δ(a) + I is an α-derivation.

2. The prime and upper nil radicals of Ore extensions

In this section, our focus of study will be on the prime and the upper nil radicals of a ring R and those of the Ore extension R[x; α, δ] and the skew power series ring R[[x; α]].

Proposition 2.1. Let I be an ideal of a ring R. Then the following statements are equivalent:

(1) I is an (α, δ)-compatible ideal.

(2) R/I is (α, δ)-compatible.

Proof. (1)⇒(2) Clearly I is an (α, δ)-ideal. Let (a + I)(b + I) = 0 in R/I. Then ab ∈ I. Hence aα(b) ∈ I. Thus (a + I)α(b + I) = 0. Similarly, (a + I)α(b + I) = 0 implies (a + I)(b + I) = 0. If (a + I)(b + I) = 0, then ab ∈ I, so that aδ(b) ∈ I. Hence (a + I)δ(b + I) = 0.

(2)⇒(1) This is similar to the proof of (1)⇒(2). □

Lemma 2.2. Let I be an α-compatible ideal of a ring R. Then I is α-invariant.

Proof. Let α(a) ∈ I. Then 1α(a) ∈ I and so a ∈ I, since I is α-compatible. Thus I is an α-invariant ideal. □

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 (H.E. Bell in 1970 introduced this notion for I = 0).

The following proposition extends ([6], Lemma 2.1).

Proposition 2.3. Let I be an (α, δ)-compatible ideal of R and a, b ∈ R.

(1) If ab ∈ I, then aα^{n}(b) ∈ I and α^{n}(a)b ∈ I for every positive integer n. Conversely, if aα^{k}(b) or α^{k}(a)b ∈ I for some positive integer k, then ab ∈ I.

(2) If ab ∈ I, then α^{m}(a)δ^{n}(b), δ^{n}(a)α^{m}(b) ∈ I for any nonnegative integers m, n.

Proof. (1) If ab ∈ I, then α^{n}(a)α^{n}(b) ∈ I, since I is an α-ideal. Hence α^{n}(a)b ∈ I, since I is α-compatible. If α^{k}(a)b ∈ I, then α^{k}(a)α^{k}(b) ∈ I, since I is α-compatible. Hence α^{k}(ab) ∈ I and ab ∈ I, since I is α-invariant, by Lemma 2.2.

(2) It is enough to show that δ(a)α(b) ∈ I. If ab ∈ I, then by (1) and δcompatibility of I, α(a)δ(b) ∈ I. Hence δ(a)b = δ(ab) − α(a)δ(b) ∈ I. Thus δ(a)b ∈ I and δ(a)α(b) ∈ I, since I is α-compatible. □

Proposition 2.4. Let R be a ring, I be an ideal of R and α : R → R be an endomorphism of R. Then the following conditions are equivalent:

(1) I is an α-rigid ideal of R.

(2) I is α-compatible, semiprime and satisfies the IFP property.

(3) I is α-compatible and completely semiprime. If δ is an α-derivation of R, then the following are equivalent:

(4) I is an α-rigid δ-ideal of R.

(5) I is (α, δ)-compatible, semiprime and satisfies the IFP property.

(6) I is (α, δ)-compatible and completely semiprime.

Proof. (1)⇒(2) This follows from ([11], Propositions 2.2 and 2.4).

(2)⇒(1) Let aα(a) ∈ I. Then a^{2} ∈ I, since I is α-compatible. Hence aRa ⊆ I, since I satisfies the IFP property. Thus a ∈ I, since I is semiprime.

(4)⇒(6) By (1)⇒(3), I is α-compatible and completely semiprime. We show that aδ(b) ∈ I, when ab ∈ I. If ab ∈ I, then δ(ab) = α(a)δ(b) + δ(a)b ∈ δ(I) ⊆ I. Thus (α(a)δ(b))^{2} = δ(ab)α(a)δ(b) − δ(a)bα(a)δ(b) ∈ I, because δ(ab), bα(a) ∈ I. Since I is completely semiprime, we have α(a)δ(b) ∈ I and so aδ(b) ∈ I, by Proposition 2.3. □

In [6], the authors give some examples of (α, δ)-compatible rings which 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, consider the ring

R = ( \begin{matrix}F & F \\ 0 & F \end{matrix} ), where F is a field, and the endomorphism α of R is defined by

α (( \begin{matrix}a & b \\ 0 & c \end{matrix} )) = ( \begin{matrix}a & 0 \\ 0 & c \end{matrix} ) for a, b, c ∈ F.

The following examples show that there exist α-compatible ideals that are not α-rigid.

Example 2.5. Let F be a field. Let R = \{( \begin{matrix}f & f_{1} \\ 0 & f \end{matrix} ) | f, f_{1} ∈ 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 (( \begin{matrix}f & f_{1} \\ 0 & f \end{matrix} )) =

( \begin{matrix}f & uf_{1} \\ 0 & f \end{matrix} ), where u is a fixed nonzero element of F. Let p(x) be an irreducible

polynomial in F [x]. Let I = \{( \begin{matrix}0 & f_{1} \\ 0 & 0 \end{matrix} ) | f_{1} ∈ ⟨p(x)⟩\}, where ⟨p(x)⟩ is the prin-

cipal ideal of F [x] generated by p(x). Then I is an α-compatible ideal of R but it

is not α-rigid. Indeed, since ( \begin{matrix}0 & g(x) \\ 0 & 0 \end{matrix} ) α (( \begin{matrix}0 & g(x) \\ 0 & 0 \end{matrix} )) = ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) ∈ I, but

( \begin{matrix}0 & g(x) \\ 0 & 0 \end{matrix} ) ∉ I for g(x) ∉ ⟨p(x)⟩. Thus I is not α-rigid.

Example 2.6 ([12], Example 2). Let ℤ_{2} be the field of integers modulo 2 and A = ℤ_{2}[a_{0}, a_{1}, a_{2}, b_{0}, b_{1}, b_{2}, c] be the free algebra of polynomials with zero constant term in noncommuting indeterminates a_{0}, a_{1}, a_{2}, b_{0}, b_{1}, b_{2}, c over ℤ_{2}. Note that A is a ring without identity. Consider an ideal of ℤ_{2} +A, say I, generated by a_{0}b_{0}, a_{1}b_{2} +a_{2}b_{1}, a_{0}b_{1} + a_{1}b_{0}, a_{0}b_{2} + a_{1}b_{1} + a_{2}b_{0}, a_{2}b_{2}, a_{0}rb_{0}, a_{2}rb_{2}, (a_{0} + a_{1} + a_{2})r(b_{0} + b_{1} + b_{2}) with r ∈ A and r_{1}r_{2}r_{3}r_{4} with r_{1}, r_{2}, r_{3}, r_{4} ∈ A. Then I satisfies the IFP property. Let α : R → R be an inner automorphism (i.e., there exists an invertible element u ∈ R such that α(r) = u^{−1}ru for each r ∈ R). Then I is α-compatible, since I satisfies the IFP property. But I is not α-rigid, since I is not completely semiprime.

Definition 2.7. Given α and δ as above and integers j ≥ i ≥ 0, let us write f_{i}^{j} for the sum of all “words” in α and δ in which there are i factors of α and j − i factors of δ. For instance, f_{j}^{j} = α^{j}, f_{0}^{j} = δ^{j} and f_{j−1}^{j} = α^{j−1}δ + α^{j−2}δα + ⋯ + δα^{j−1}.

Note that if I is an (α, δ)-ideal of R, then I[x; α, δ] is an ideal of the Ore extension R[x; α, δ].

Theorem 2.8. Let I be an (α, δ)-compatible semiprime ideal of R. Assume f(x) = ∑_{i=0}^{n} a_{i}x^{i} and g(x) = ∑_{j=0}^{m} b_{j}x^{j} ∈ R[x; α, δ]. Then the following statements are equivalent:

(1) f(x)R[x; α, δ]g(x) ⊆ I[x; α, δ].

(2) a_{i}Rb_{j} ⊆ I for each i, j.

Proof. (1)⇒(2) Assume f(x)R[x; α, δ]g(x) ⊆ I[x; α, δ]. Then

(a_{0} + ⋯ + a_{n}x^{n})c(b_{0} + ⋯ + b_{m}x^{m}) ∈ I[x; α, δ] for each c ∈ R. (†)

Hence a_{n}α^{n}(cb_{m}) ∈ I. Thus a_{n}cb_{m} ∈ I, since I is α-compatible. Therefore a_{n}f_{i}^{j}(cb_{m}) ∈ I, by Proposition 2.3. Next, replace c by cb_{m−1}da_{n}e, where c, d, e ∈ R. Then (a_{0} + ⋯ + a_{n}x^{n})cb_{m−1}da_{n}e(b_{0} + ⋯ + b_{m−1}x^{m−1}) ∈ I[x; α, δ]. Hence a_{n}α^{n}(cb_{m−1}da_{n}eb_{m−1}) ∈ I and a_{n}cb_{m−1}da_{n}eb_{m−1} ∈ I, since I is α-compatible. Thus (Ra_{n}Rb_{m−1})^{2} ⊆ I. Hence Ra_{n}Rb_{m−1} ⊆ I, since I is semiprime. Continuing this process, we obtain a_{n}Rb_{k} ⊆ I, for k = 0, 1, ⋯, m. Hence by (α, δ)- compatibility of I, we get (a_{0}+⋯+a_{n}x^{n})R[x; α, δ](b_{0}+⋯+b_{m−1}x^{m−1}) ⊆ I[x; α, δ]. Using induction on n + m, we obtain a_{i}Rb_{j} ⊆ I for each i, j.

(2)⇒(1) This follows from Proposition 2.3. □

Corollary 2.9. If I is a (semi)prime (α, δ)-compatible ideal of R, then I[x; α, δ] is a (semi)prime ideal of R[x; α, δ].

Proof. Assume that I is a prime (α, δ)-compatible ideal of R. Let f(x) = ∑_{i=0}^{n} a_{i}x^{i} and g(x) = ∑_{j=0}^{m} b_{j}x^{j} ∈ R[x; α, δ] such that f(x)R[x; α, δ]g(x) ⊆ I[x; α, δ]. Then a_{i}Rb_{j} ⊆ I for each i, j, by Theorem 2.8. Assume g(x) ∉ I[x; α, δ]. Hence b_{j} ∉ I for some j. Thus a_{i} ∈ I for each i = 0, 1, ⋯, n, since I is prime. Therefore f(x) ∈ I[x; α, δ]. Consequently, I[x; α, δ] is a prime ideal of R[x; α, δ]. □

Theorem 2.10. If each minimal prime ideal of R is (α, δ)-compatible, then P (R[x; α, δ]) ⊆ P (R)[x; α, δ].

Proof. This follows from Corollary 2.9. □

The following example shows that there exists a ring R such that all minimal prime ideals are α-compatible, but are not α-rigid.

Example 2.11. Let R = Mat_{2}(ℤ_{4}) be the 2×2 matrix ring over the ring ℤ_{4}. Then

P (R) = \{( \begin{matrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} ) | a_{ij} ∈ \overline{2}ℤ\} is the only prime ideal of R. Let α : R → R

be the endomorphism defined by α (( \begin{matrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} )) = ( \begin{matrix}a_{11} & −a_{12} \\ −a_{21} & a_{22} \end{matrix} ). Then α

is an automorphism of R and P (R) is α-compatible. However, P (R) is not α-rigid,

since ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) α (( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} )) ∈ P (R), but ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) ∉ P (R).

Theorem 2.12. If P is a completely (semi)prime (α, δ)-compatible ideal of R, then P [x; α, δ] is a completely (semi)prime ideal of R[x; α, δ].

Proof. Let P be a completely prime ideal of R. R/P is a domain, hence it is a reduced ring. R/P is a (α, δ)-compatible ring, hence R/P is α-rigid, by ([6], Lemma 2.2). Let f(x), g(x) ∈ R/P [x; α, δ] such that f(x) g(x) = 0. Then f(x) = 0 or g(x) = 0, by ([9], Proposition 6). Thus R[x; α, δ]/P [x; α, δ] ≅ R/P [x; α, δ] is a domain and P [x; α, δ] is a completely prime ideal of R[x; α, δ]. □

Corollary 2.13 ([11], Proposition 3.8). If P (R) is an α-rigid δ-ideal of R, then P (R[x; α, δ]) ⊆ P (R)[x; α, δ].

Proof. P (R) is an α-rigid δ-ideal, hence P (R) is a completely semiprime (α, δ)- compatible ideal of R, by Proposition 2.4. Therefore P (R[x; α, δ]) ⊆ P (R)[x; α, δ], by Theorem 2.12. □

Theorem 2.14. If P is a strongly (semi)prime (α, δ)-compatible ideal of R, then P [x; α, δ] is a strongly (semi)prime ideal of R[x; α, δ].

Proof. By Corollary 2.9, P [x; α, δ] is a prime ideal of R[x; α, δ]. Hence

R[x; α, δ]/P [x; α, δ] ≅ R/P [x; α, δ]

is a prime ring. We claim that zero is the only nil ideal of R/P [x; α, δ]. Let J be a nil ideal of R/P [x; α, δ]. 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) = a_{0} + ⋯ + a_{n−1}x^{n−1} + ax^{n} ∈ J. Hence (f(x)r)^{m} = 0 for some nonnegative integer m. Thus a α^{n}(ra) ⋯ 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; α, δ] is a strongly prime ideal of R[x; α, δ]. □

Theorem 2.15. If each minimal strongly prime ideal of R is (α, δ)-compatible, then N_{r}(R[x; α, δ]) ⊆ N_{r}(R)[x; α, δ].

Corollary 2.16 ([11], Proposition 3.8). If N_{r}(R) is an α-rigid δ-ideal of R, then N_{r}(R[x; α, δ]) ⊆ N_{r}(R)[x; α, δ].

Proof. N_{r}(R) is an α-rigid δ-ideal, hence N_{r}(R) is a completely semiprime (α, δ)- compatible ideal of R, by Proposition 2.4, and N_{r}(R) is a strongly semiprime ideal of R. Therefore N_{r}(R[x; α, δ]) ⊆ N_{r}(R)[x; α, δ], by Theorem 2.15. □

As a parallel result to Theorems 2.8, 2.10, 2.12, 2.14 and 2.15, we have the following results for the skew power series ring R[[x; α]].

Proposition 2.17. Let I be an α-compatible semiprime ideal of R. Assume f(x) = ∑_{i=0}^{∞} a_{i}x^{i} and g(x) = ∑_{j=0}^{∞} b_{j}x^{j} ∈ R[[x; α]]. Then the following statements are equivalent:

(1) f(x)R[[x; α]]g(x) ⊆ I[[x; α]].

(2) a_{i}Rb_{j} ⊆ I for each i, j.

Proof. (1)⇒(2) Assume f(x)R[[x; α]]g(x) ⊆ I[[x; α]]. Let c be an arbitrary element of R. Then we have the following:

∑_{k=0}^{∞} ( ∑_{i+j=k} a_{i}x^{i}cb_{j}x^{j}) = ∑_{k=0}^{∞} ( ∑_{i+j=k} a_{i}α^{i}(cb_{j})) x^{k} ∈ I[[x; α]]. (§)

By induction on i + j, we show that a_{i}Rb_{j} ⊆ I for all i, j. From Equation (§), we obtain a_{0}Rb_{0} ⊆ I. This prove for i + j = 0. Now suppose that a_{i}Rb_{j} ⊆ I for i + j ≤ n − 1. From Equation (§), we have

∑_{i+j=k} a_{i}α^{i}(cb_{j}) ∈ I for all k ≥ 0. (§§)

Let c, d, e be arbitrary elements of R. For k = n replace c by cb_{n}da_{0}e. Then we obtain a_{0}cb_{n}da_{0}eb_{n} ∈ I, since a_{0}Rb_{j} ⊆ I for each j ≤ n − 1. Hence (Ra_{0}Rb_{n})^{2} ⊆ I and so Ra_{0}Rb_{n} ⊆ I, since I is semiprime. Continuing this process (replacing c by cb_{j}da_{n−j}e in Equation (§§) for j = 1, 2, ⋯, n − 1 and using α-compatibility of I), we obtain a_{i}Rb_{j} ⊆ I for i + j = n.

(2)⇒(1) This follows from Proposition 2.3. □

Corollary 2.18. If I is a (semi)prime α-compatible ideal of R, then I[[x; α]] is a (semi)prime ideal of R[[x; α]].

Theorem 2.19. If each minimal prime ideal of R is α-compatible, then P (R[[x; α]]) ⊆ P (R)[[x; α]].

Proof. This follows from Corollary 2.18. □

Theorem 2.20. If P is a completely (semi)prime α-compatible ideal of R, then P [[x; α]] is a completely (semi)prime ideal of R[[x; α]].

Proof. Let P be a completely prime ideal of R. R/P is a domain, hence it is a reduced ring. R/P is an α-compatible ring, hence R/P is α-rigid, by ([6], Lemma 2.2). Let f(x), g(x) ∈ R/P [[x; α]] such that f(x) g(x) = 0. Then f(x) = 0 or g(x) = 0, by ([9], Proposition 17). Thus R[[x; α]]/P [[x; α]] ≅ R/P [[x; α]] is a domain and P [[x; α]] is a completely prime ideal of R[[x; α]]. □

Corollary 2.21 ([11], Proposition 3.12). If P (R) is an α-rigid ideal of R, then P (R[[x; α]]) ⊆ P (R)[[x; α]].

Theorem 2.22. If P is a strongly prime α-compatible ideal of R, then P [[x; α]] is a strongly prime ideal of R[[x; α]].

Theorem 2.23. If each minimal strongly prime ideal of R is α-compatible, then N_{r}(R[[x; α]]) ⊆ N_{r}(R)[[x; α]].

Corollary 2.24 ([11], Proposition 3.12). If N_{r}(R) is an α-rigid ideal of R, then N_{r}(R[[x; α]]) ⊆ N_{r}(R)[[x; α]].

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Department of Mathematics, Shahrood University of Technology, Shahrood, Iran, P.O.Box: 316-3619995161 eb_{−}hashemi@yahoo.com or eb_{−}hashemi@shahrood.ac.ir

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