Prime Ideals and Strongly Prime Ideals of Skew Laurent Polynomial Rings

Abstract

We first study connections between 𝛼-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 𝑁_𝑟(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.

1. Introduction

Throughout the paper, 𝑅 always denotes an associative ring with unity. 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.

Recall that for a ring 𝑅 with an injective ring endomorphism 𝛼∶𝑅→𝑅, 𝑅[𝑥;𝛼] is the Ore extension of 𝑅. The set {𝑥^𝑗}_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^−𝑗𝑟𝑥^𝑖, 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 ring 𝐴(𝑅,𝛼) (see [1], for more details). Let 𝐴(𝑅,𝛼) or 𝐴 be the subset {𝑥^−𝑖𝑟𝑥^𝑖∣𝑟∈𝑅,𝑖≥0} of the skew Laurent polynomial ring 𝑅[𝑥,𝑥⁻¹;𝛼]. For each 𝑗≥0, 𝑥^−𝑖𝑟𝑥^𝑖=𝑥^{−(𝑖+𝑗)}𝛼^𝑗(𝑟)𝑥^(𝑖+𝑗). It follows that the set of all such elements forms a subring of 𝑅[𝑥,𝑥⁻¹;𝛼] with 𝑥^−𝑖𝑟𝑥^𝑖+𝑥^−𝑗𝑠𝑥^𝑗=𝑥^{−(𝑖+𝑗)}(𝛼^𝑗(𝑟)+𝛼^𝑖(𝑠))𝑥^(𝑖+𝑗) and (𝑥^−𝑖𝑟𝑥^𝑖)(𝑥^−𝑗𝑠𝑥^𝑗)=𝑥^{−(𝑖+𝑗)}𝛼^𝑗(𝑟)𝛼^𝑖(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^−𝑖𝑟𝑥^𝑖 to 𝑥^−𝑖𝛼(𝑟)𝑥^𝑖, 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^−𝑖𝑟𝑥^𝑗 to 𝛼^−𝑖(𝑟)𝑥^𝑗−𝑖. For an 𝛼-ideal 𝐼 of 𝑅, put Δ(𝐼)=∪_𝑖≥0𝑥^−𝑖𝐼𝑥^𝑖. 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 ring 𝑅 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 𝑁_𝑟(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)=𝑁_𝑟(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 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 [13]) 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. [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 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 and 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]).

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 endomorphism 𝛼 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 between 𝛼-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 𝑁_𝑟(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 ideal 𝐼 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. Let 𝐼 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. Let 𝐼 be an 𝛼-compatible ideal of a ring 𝑅. Then
(1)𝐼 is 𝛼-invariant;(2) if 𝑎𝑏∈𝐼, then 𝑎𝛼^𝑛(𝑏)∈𝐼,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.

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). Let 𝑅 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 an 𝛼-ideal 𝐼 of 𝑅, put Δ(𝐼)=∪_𝑖≥0𝑥^−𝑖𝐼𝑥^𝑖.

Proposition 2.4. (1) If 𝐼 is an 𝛼-compatible ideal of 𝑅, then Δ(𝐼) is an 𝛼-compatible ideal of 𝐴.
(2) If 𝐽 is an 𝛼-compatible ideal of 𝐴, then 𝐽=Δ(𝐽₀) and 𝐽₀ is an 𝛼-compatible ideal of 𝑅.
(3) If 𝑃 is a completely (semi)prime 𝛼-compatible ideal of 𝑅, then Δ(𝑃) is a completely (semi)prime 𝛼-compatible ideal of 𝐴.
(4) If 𝑄 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) If 𝑃 is a prime 𝛼-compatible ideal of 𝑅, then Δ(𝑃) is a prime 𝛼-compatible ideal of 𝐴.

Proof. (1) Since 𝐼 is an 𝛼-ideal of 𝑅, Δ(𝐼) is an ideal of 𝐴. Now, let (𝑥^−𝑖𝑟𝑥^𝑖)(𝑥^−𝑗𝑠𝑥^𝑗)∈Δ(𝐼). Hence 𝑥^{−(𝑖+𝑗)}𝛼^𝑗(𝑟)𝛼^𝑖(𝑠)𝑥^𝑖+𝑗∈Δ(𝐼) and that 𝛼^𝑗(𝑟)𝛼^𝑖(𝑠)∈𝐼. Thus 𝛼^𝑗(𝑟)𝛼^𝑖+1(𝑠)∈𝐼, since 𝐼 is 𝛼-compatible. Consequently (𝑥^−𝑖𝑟𝑥^𝑖)𝛼(𝑥^−𝑗𝑠𝑥^𝑗)∈Δ(𝐼). Therefore Δ(𝐼) is 𝛼-compatible.
(2) Let 𝐽₀=𝐽∩𝐽 and 𝑟∈𝐽₀. Then 𝛼^𝑛(𝑟)∈𝐽₀ for each 𝑛≥0. Hence 𝛼^𝑛(𝑥^−𝑛𝑟𝑥^𝑛)=𝑟∈𝐽 for each 𝑛≥0. Thus 𝑥^−𝑛𝑟𝑥^𝑛∈𝐽, since 𝐽 is 𝛼-compatible. Therefore Δ(𝐽₀)⊆𝐽₀. Now, let 𝑥^−𝑚𝑟𝑥^𝑚∈𝐽. Then 𝛼^𝑚(𝑥^−𝑚𝑟𝑥^𝑚)∈𝐽 and that 𝑟∈𝐽, since 𝐽 is 𝛼-compatible. Thus 𝐽⊆Δ(𝐽₀). Consequently, Δ(𝐽₀)=𝐽.
(3) Let (𝑥^−𝑖𝑟𝑥^𝑖)(𝑥^−𝑗𝑠𝑥^𝑗)∈Δ(𝑃). Then 𝑥^{−(𝑖+𝑗)}𝛼^𝑗(𝑟)𝛼^𝑖(𝑠)𝑥^𝑖+𝑗∈Δ(𝑃) and that 𝛼^𝑗(𝑟)𝛼^𝑖(𝑠)∈𝑃. Hence 𝑟𝑠∈𝑃, by Proposition 2.2. Thus 𝑟∈𝑃 or 𝑠∈𝑃, since 𝑃 is completely prime. Consequently, 𝑥^−𝑖𝑟𝑥^𝑖∈Δ(𝑃) or 𝑥^−𝑗𝑠𝑥^𝑗∈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^−𝑖𝑟𝑥^𝑖)𝐴(𝑥^−𝑗𝑠𝑥^𝑗)⊆Δ(𝑃). Then 𝑟𝑅𝑠⊆𝑃, by Proposition 2.2. Hence 𝑟∈𝑃 or 𝑠∈𝑃, since 𝑃 is prime. Consequently 𝑥^−𝑖𝑟𝑥^𝑖∈Δ(𝑃) or 𝑥^−𝑗𝑠𝑥^𝑗∈Δ(𝑃). Therefore Δ(𝑃) is a prime ideal of 𝐴.

Theorem 2.5. Let 𝑃 be a strongly prime 𝛼-compatible ideal of 𝑅. Then Δ(𝑃) is a strongly prime ideal of 𝐴.

Proof. Since 𝑃 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∈𝐼_𝑖. Then 𝑥^−𝑖𝑎𝑥^𝑖∈𝐼. Since 𝐼/Δ(𝑃) is a nil ideal, hence (𝑥^−𝑖𝑎𝑥^𝑖)^𝑛∈Δ(𝑃) for some 𝑛≥0. Hence 𝑥^−𝑖𝑎^𝑛𝑥^𝑖=𝑥^−𝑗𝑝𝑥^𝑗 for some 𝑝∈𝑃 and 𝑗≥0. Thus 𝛼^𝑗(𝑎^𝑛)=𝛼^𝑖(𝑝)∈𝑃. Hence 𝑎^𝑛∈𝑃, since 𝑃 is 𝛼-compatible. Then (𝐼_𝑖+𝑃)/𝑃 is a nil ideal of 𝑅/𝑃 for each 𝑖≥0. Hence 𝐼_𝑖⊆𝑃, for each 𝑖≥0. Therefore 𝐼⊆Δ(𝑃). Consequently, Δ(𝑃) is a strongly prime ideal of 𝐴.

Note that if 𝐼 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. Let 𝛼 be an automorphism of 𝑅. Let 𝐼 be a semiprime 𝛼-compatible ideal of 𝑅. Assume 𝑓(𝑥)=∑^𝑛_𝑖=𝑟𝑎_𝑖𝑥^𝑖 and 𝑔(𝑥)=∑^𝑚_𝑗=𝑠𝑏_𝑗𝑥^𝑗∈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)𝑎_𝑖𝑅𝑏_𝑗⊆𝐼 for each 𝑖,𝑗.

Proof. (1)⇒(2). Assume 𝑓(𝑥)𝑅[𝑥,𝑥⁻¹;𝛼]𝑔(𝑥)⊆𝐼[𝑥,𝑥⁻¹;𝛼]. Then(𝑎_𝑟𝑥^𝑟+⋯+𝑎_𝑛𝑥^𝑛)𝑐(𝑏_𝑠𝑥^𝑠+⋯+𝑏_𝑚𝑥^𝑚)∈𝐼[𝑥,𝑥⁻¹;𝛼]foreach𝑐∈𝑅. (†)Hence 𝑎_𝑛𝛼^𝑛(𝑐𝑏_𝑚)∈𝐼. Thus 𝑎_𝑛𝑐𝑏_𝑚∈𝐼, since 𝐼 is 𝛼-compatible. Next, replacing 𝑐 by 𝑐𝑏_𝑚−1𝑑𝑎_𝑛𝑒, where 𝑐,𝑑,𝑒∈𝑅. Then (𝑎_𝑟𝑥^𝑟+⋯+𝑎_𝑛𝑥^𝑛)𝑐𝑏_𝑚−1𝑑𝑎_𝑛𝑒(𝑏_𝑠𝑥^𝑠+⋯+𝑏_𝑚−1𝑥^𝑚−1)∈𝐼[𝑥,𝑥⁻¹;𝛼]. Hence 𝑎_𝑛𝛼^𝑛(𝑐𝑏_𝑚−1𝑑𝑎_𝑛𝑒𝑏_𝑚−1)∈𝐼 and that 𝑎_𝑛𝑐𝑏_𝑚−1𝑑𝑎_𝑛𝑒𝑏_𝑚−1∈𝐼, since 𝐼 is 𝛼-compatible. Thus (𝑅𝑎_𝑛𝑅𝑏_𝑚−1)²⊆𝐼. Hence 𝑅𝑎_𝑛𝑅𝑏_𝑚−1⊆𝐼, since 𝐼 is semiprime. Continuing this process, we obtain 𝑎_𝑛𝑅𝑏_𝑘⊆𝐼, for 𝑘=𝑠,…,𝑚. Hence from 𝛼-compatibility of 𝐼, we get (𝑎_𝑟𝑥^𝑟+⋯+𝑎_𝑛𝑥^𝑛)𝑅[𝑥,𝑥⁻¹;𝛼](𝑏_𝑠𝑥^𝑠+⋯+𝑏_𝑚−1𝑥^𝑚−1)⊆𝐼[𝑥,𝑥⁻¹;𝛼]. Using induction on 𝑛+𝑚, we obtain 𝑎_𝑖𝑅𝑏_𝑗⊆𝐼 for each 𝑖,𝑗.
(2)⇒(1). It follows from Proposition 2.2.

Corollary 2.7. Let 𝛼 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 that 𝐼 is a prime 𝛼-compatible ideal of 𝑅. Let 𝑓(𝑥)=∑^𝑛_𝑖=𝑟𝑎_𝑖𝑥^𝑖 and 𝑔(𝑥)=∑^𝑚_𝑗=𝑠𝑏_𝑗𝑥^𝑗∈𝑅[𝑥,𝑥⁻¹;𝛼] such that 𝑓(𝑥)𝑅[𝑥,𝑥⁻¹;𝛼]𝑔(𝑥)⊆𝐼[𝑥,𝑥⁻¹;𝛼]. Then 𝑎_𝑖𝑅𝑏_𝑗⊆𝐼 for each 𝑖,𝑗, by Theorem 2.6. Assume 𝑔(𝑥)∉𝐼[𝑥,𝑥⁻¹;𝛼]. Hence 𝑏_𝑗∉𝐼 for some 𝑗. Thus 𝑎_𝑖∈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 of 𝑅 is 𝛼-compatible, then 𝑃(𝑅[𝑥,𝑥⁻¹;𝛼])⊆Δ(𝑃(𝑅))[𝑥,𝑥⁻¹;𝛼].

Proof. Let 𝑃 be a minimal prime ideal of 𝑅. By Proposition 2.4, Δ(𝑃) is a 𝛼-compatible ideal of 𝐴. Assume (𝑎^−𝑖𝑟𝑥^𝑖)𝐴(𝑥^−𝑗𝑠𝑥^𝑗)⊆Δ(𝑃). Then 𝑟𝑅𝑠⊆𝑃, since Δ(𝑃) is 𝛼-compatible. Hence 𝑟∈𝑃 or 𝑠∈𝑃. Thus 𝑎^−𝑖𝑟𝑥^𝑖∈Δ(𝑃) or 𝑥^−𝑗𝑠𝑥^𝑗∈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=(^𝐹𝐹_0𝐹), where 𝐹 is a field, and the endomorphism 𝛼 of 𝑅 is defined by 𝛼((^𝑎𝑏_0𝑐))(^𝑎𝑏_0𝑐)=(^𝑎0_0𝑐) for 𝑎,𝑏,𝑐∈𝐹.

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

Example 2.9 (see [2.95], Example 2.5). Let 𝐹 be a field. Let 𝑅={(^𝑓𝑓1_0𝑓)∣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((^𝑓𝑓1_0𝑓))(^𝑓𝑓1_0𝑓)=(^{𝑓𝑢𝑓1}_0𝑓), 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). Let ℤ₂ 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. Let 𝛼 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 Corollary 2.7.

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 2.12.12.12 (see [15], Example 2.12.11). Let 𝑅=Mat₂(ℤ₄) be the 2×2 matrix ring over the ring ℤ₄. Then 𝑃(𝑅)={(^𝑎11𝑎12_𝑎21𝑎22)∣𝑎_𝑖𝑗∈2ℤ} is the only prime ideal of 𝑅. Let 𝛼∶𝑅→𝑅 be the endomorphism defined by 𝛼((^𝑎11𝑎12_𝑎21𝑎22))(^𝑎11𝑎12_𝑎21𝑎22)=(^{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. Let 𝛼 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. Let 𝑃 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. Let 𝛼 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. Let 𝛼 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 Corollary 2.7, 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+⋯+𝑎_𝑛−1𝑥^𝑛−1+𝑎𝑥^𝑛∈𝐽. Hence (𝑓(𝑥)𝑟)^𝑚=0, for some nonnegative integers 𝑚. Thus 𝑎𝛼^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,𝑥⁻¹;𝛼] is a strongly prime ideal of 𝑅[𝑥,𝑥⁻¹;𝛼].

Theorem 2.16. Let 𝛼 be an automorphism of 𝑅. If each minimal strongly prime ideal of 𝑅 is 𝛼-compatible, then 𝑁_𝑟(𝑅[𝑥,𝑥⁻¹;𝛼])⊆𝑁_𝑟(𝑅)[𝑥,𝑥⁻¹;𝛼].

Corollary 2.17. Let 𝛼 be an automorphism of 𝑅. If 𝑁_𝑟(𝑅) is an 𝛼-rigid ideal of 𝑅, then 𝑁_𝑟(𝑅[𝑥,𝑥⁻¹;𝛼])⊆𝑁_𝑟(𝑅)[𝑥,𝑥⁻¹;𝛼].

Proof. 𝑁_𝑟(𝑅) is 𝛼-rigid, hence 𝑁_𝑟(𝑅) is a completely semiprime 𝛼-compatible ideal of 𝑅, by Proposition 2.3, and that 𝑁_𝑟(𝑅) is a strongly semiprime ideal of 𝑅. Therefore 𝑁_𝑟(𝑅[𝑥,𝑥⁻¹;𝛼])⊆𝑁_𝑟(𝑅)[𝑥,𝑥⁻¹;𝛼], by Theorem 2.15.

Example 2.12 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 of 𝑅 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. Let 𝛼 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.

Acknowledgments

The author thanks the referee for his valuable comments and suggestions. This research is supported by the Shahrood University of Technology of Iran.