INVARIANT POLYNOMIALS of ORE EXTENSIONS by Q-SKEW DERIVATIONS

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 140, Number 11, November 2012, Pages 3739–3747 S 0002-9939(2012)11268-7 Article electronically published on March 12, 2012

INVARIANT POLYNOMIALS OF ORE EXTENSIONS BY q-SKEW DERIVATIONS

CHEN-LIAN CHUANG, TSIU-KWEN LEE, AND CHENG-KAI LIU

(Communicated by Harm Derksen)

Abstract. Let R be a prime ring with the symmetric Martindale quotient ring Q. Suppose that δ is a quasi-algebraic q-skew σ-derivation of R.Fora minimal monic semi-invariant polynomial π(t)ofQ[t; σ, δ], we show that π(t) is also invariant if char R = 0 and that either π(t) − c for some c ∈ Q or π(t)p is a minimal monic invariant polynomial if char R = p ≥ 2. As an application, we prove that any R-disjoint prime ideal of R[t; σ, δ] is the principal ideal p(t) for an irreducible monic invariant polynomial p(t) unless σ or δ is X-inner.

1. Introduction It will be assumed throughout that R is an associative prime ring in the sense that for any a, b ∈ R, aRb = 0 implies a =0orb =0.Letσ be an automorphism of R.Byaσ-derivation of R,wemeanamapδ : R → R satisfying δ(x + y)=δ(x)+δ(y)andδ(xy)=δ(x)y + σ(x)δ(y)

for all x, y ∈ R.Givenb ∈ R,themapadb,σ : x ∈ R → bx − σ(x)b defines a σ-derivation of R, called the inner σ-derivation defined by b. Analogously, for a −1 unit u ∈ R,themapIu : x ∈ R → uxu defines an automorphism of R, called the inner automorphism defined by the unit element u. Let Q be the symmetric Martindale quotient ring of R. The center of Q, denoted by C, is called the extended centroid of R. (See [1, Chapter 2] for details.) The σ-derivation δ of R, together with its automorphism σ, can be uniquely extended to a σ-derivation of Q [15, Lemma 1]. A σ-derivation of R is called X-inner if its extension to Q is equal to ad b,σ for some b ∈ Q. An automorphism of R is called X-inner if its extension to Q is equal to I u for some unit u ∈ Q. Following [14], define for each integer j the C-space

Φ(j) def.= {u ∈ Q | ur = σj (r)u for all r ∈ Q}.

j By [22, Chapter 3, Lemma 12.1], any 0 = u ∈ Φ(j) is a unit such that σ =Iu.We recall a very useful property due to Kharchenko and Popov [15]. Lemma 1.1. If δ is an X-outer σ-derivation, then σ(α)=α and either δ(α)=0 or σ = I δ(α) for each α ∈ C.

Received by the editors June 29, 2010 and, in revised form, April 28, 2011. 2010 Mathematics Subject Classification. Primary 16S36, 16N60, 16W25, 16R50. Key words and phrases. Prime ring, (semi-)invariant polynomial, q-skew σ-derivation. The first two authors are members of the Mathematics Division, NCTS (Taipei Office).

c 2012 American Mathematical Society Reverts to public domain 28 years from publication 3739

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Proof. Given α ∈ C and x ∈ R, δ(αx)=δ(α)x + σ(α)δ(x)andδ(xα)=δ(x)α + σ(x)δ(α) yield (σ(α) − α)δ(x)=σ(x)δ(α) − δ(α)x.Soσ(α)=α and σ(x)δ(α) − δ(α)x = 0 by the outerness of δ.Sincex ∈ R is arbitrary, δ(α) = 0 implies σ =Iδ(α) by [22, Chapter 3, Lemma 12.1].

We shall need the following generalization of Lemma 1.1.

Lemma 1.2. Let δ be an X-outer σ-derivation. For a unit u ∈ Q,ifδ I u = γ I uδ, where γ ∈ C,thenσ(u)=γu and either δ(u)=0or σ = I u−1 δ(u). Proof. Expand 0 = δ(1) = δ(uu−1)=δ(u)u−1 + σ(u)δ(u−1). We see that δ(u−1)= −σ(u)−1δ(u)u−1. With this, we compute for x ∈ R −1 −1 −1 −1 δI u(x)=δ(uxu )=δ(u)xu + σ(u)δ(x)u + σ(u)σ(x)δ(u ) = δ(u)xu−1 + σ(u)δ(x)u−1 − σ(u)σ(x)σ(u)−1δ(u)u−1.

−1 The assumption δI u(x)=γI uδ(x)=γuδ(x)u gives the identity δ(u)xu−1 + σ(u) − γu δ(x)u−1 − σ(u)σ(x)σ(u)−1δ(u)u−1 =0. But δ is X-outer. By [15, Theorem 1], σ(u)=γu and for x ∈ R, 0=δ(u)xu−1 − σ(u)σ(x)σ(u)−1δ(u)u−1 = δ(u)xu−1 − (γu)σ(x)(γu)−1δ(u)u−1 = δ(u)xu−1 − uσ(x)u−1δ(u)u−1 = u(u−1δ(u)x − σ(x)u−1δ(u))u−1.

−1 So u δ(u) ∈ I σ.Ifδ(u) =0,then σ =Iu−1δ(u), as asserted. The Ore extension of R by a σ-derivation δ, denoted by R[t; σ, δ], is the set of n polynomial expressions a0 + a1t + ···+ ant ,wherea0,...,an ∈ R, with compo- nentwise addition and multiplication subjected to the rule tr = σ(r)t + δ(r)for all r ∈ R.WeformQ[t; σ, δ] analogously. Ore extensions have been extensively investigated in various directions. In the study of R[t; σ, δ], two crucially important notions are the notion of a right invariant polynomial and that of a semi-invariant polynomial. We omit “right” here for brevity and recall the definition below. Definition 1 ([16, 17, 18, 20]). We call f(t) ∈ Q[t; σ, δ] a cv-polynomial if it is associated to an automorphism τ of Q such that f(t)r − τ(r)f(t) ∈ Q for any r ∈ R. A cv-polynomial f(t)iscalledsemi-invariant if f(t)r − τ(r)f(t)=0 for all r ∈ R. A semi-invariant polynomial f(t) is called invariant if addition- ally f(t)t =(at + b)f(t)forsomea, b ∈ Q. A (semi-)invariant polynomial is called minimal if it is nonconstant and its degree is minimal among all nonconstant (semi-)invariant polynomials. The importance of these polynomials is based on the following: cv-polynomials determine R-stable homomorphisms from other Ore extensions of R into R[t; σ, δ] ([18] or [8, Theorem 12]). Semi-invariant polynomials determine the algebraic de- pendency of δ ([5]). Invariant polynomials determine the ideal structure of R[t; σ, δ] ([17, 20]). In particular, R[t; σ, δ] has nontrivial R-disjoint ideals iff R[t; σ, δ]has nonconstant (semi-)invariant polynomials ([20, Theorem 2.6]). For a simple ring R, R[t; σ, δ] is simple iff R[t; σ, δ] has no nonconstant (semi-)invariant polynomials. (Semi-)invariant polynomials for X-inner skew derivations can be found easily as follows.

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def. Lemma 1.3. If δ = ad b,σ,whereb ∈ Q,thenπ(t) = t−b is the minimal invariant polynomial. Proof. Clearly, π(t)r = σ(r)π(t)forr ∈ R.Also, (t − b)t =(t − b)(t − b + b)=(t − b)2 +(t − b)b =(t − b)2 + σ(b)(t − b)=(t − b + σ(b))(t − b). So t − b is invariant. Let π(t) be a minimal monic semi-invariant polynomial of R[t; σ, δ]. It is shown in [20] that a factor of π(t)deg π(t) forms a minimal monic invariant polynomial. But for an ordinary derivation δ (namely for δ with σ =1),π(t) is also invariant when char R = 0 and either π(t) − c for some c ∈ Q or π(t)p is a minimal invariant polynomial when char R = p ≥ 2. Our aim here is to show that this is also true if δ is q-skew in a sense we now make clear. For any subset S of Q, deﬁne S(σ) def.= {r ∈ S | σ(r)=r},S(δ) def.= {r ∈ S | δ(r)=0} and S(σ,δ) def.= S(σ) ∩ S(δ). Deﬁnition. A σ-derivation δ is called q-skew, where 0 = q ∈ C(σ,δ),ifσ−1δσ = qδ.

2. Results Throughout the sequel, let δ be a q-skew σ-derivation and let π(t) be a minimal monic semi-invariant polynomial. In order for π(t) to exist, we assume that δ is quasi-algebraic [20, p. 147]. We aim to describe the minimal monic invariant polynomial in terms of π(t). By Lemma 1.3, we may consider X-outer δ.For X-outer δ, π(t) is given as follows. Theorem 2.1 ([4, Theorem 1] and [9, Theorem 15]). Let δ be an X-outer quasi- algebraic q-skew σ-derivation of R. Then there exists the least integer ν ≥ 1 such that qν =1and any minimal monic semi-invariant polynomial π(t) ∈ Q[t; σ, δ] is given by ⎧ ⎪ tν + b if char R =0, ⎨⎪ s−1 π(t)= s j ⎪ νp νp ≥ ⎩⎪ t + bj t + b if char R = p 2, j=0 s j where b ∈ Q and bj ∈ Φ(ν(p − p )) for 0 ≤ j ≤ s − 1. We shall retain the notation above throughout. For brevity, we treat all char- acteristics simultaneously. If char R =0,thenallbj are interpreted as 0. For any u ∈ Φ(deg π(t)), π(t)+u is also semi-invariant. Clearly, all minimal monic semi-invariant polynomials of Q[t; σ, δ] are so obtained. Moreover, if σdeg π(t) is X- outer, viz. Φ(deg π(t)) = 0, then π(t) is the unique minimal monic semi-invariant polynomial of Q[t; σ, δ]. The following is important. Lemma 2.2. If 0 = u ∈ Φ(i),thenν|i2 and σ(u)=qiu.

i Proof. Since 0 = u ∈ Φ(i), we have σ =Iu.Sinceδ is q-skew, i i i i δI u = δσ = q σ δ = q I uδ. By Lemma 1.2, σ(u)=qiu. Inductively, σ(u)=qiu for any ≥ 0. Setting = i, i2 i i2 2 q u = σ (u)=Iu(u)=u.Soq = 1 and hence ν|i by the minimality of ν.

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As a direct application, if σ is X-inner, then q =1.Forifσ =Iv,thenσ(v)=qv by Lemma 2.2, implying q =1sinceσ(v)=Iv(v)=v. To ﬁnd minimal monic invariant polynomials, we need more information on σ(bj),δ(bj),σ(b)andδ(b).

(σ,δ) (σ,δ) Lemma 2.3. For 1 ≤ j1,thenb, b0 ∈ Q also. If ν =1, (σ) s s then char R = p ≥ 2, b0 ∈ Q , δ(b0)=b − σ(b) ∈ Φ(p ) and δ(b) ∈ Φ(p +1).

def. Proof. Set n =degπ(t) for simplicity. So n= ν if char R =0andn = νps if s j (σ) char R = p ≥ 2. For 0 ≤ j

s−1 νpj [t, π(t)] = δ(bj )t +(σ(b) − b)t + δ(b). j=0

Since deg[t, π(t)] < deg π(t), we have [t, π(t)] ∈ Φ(n + 1) by the minimality of π(t). If ν>1, then δ(bj )=0for0≤ j

s−1 pj s [t, π(t)] = δ(bj )t +(δ(b0)+σ(b) − b)t + δ(b) ∈ Φ(p +1). j=1

Comparing coeﬃcients of this yields δ(bj )=0for1≤ j

Lemma 2.4. [t, π(t)k]=kδ(b)π(t)k−1 for k ≥ 1 and [π(t),δ(b)] = 0. k k−1 i k−i−1 Proof. By the Leibniz rule, [t, π(t) ]= i=0 π(t) [t, π(t)]π(t) . By Lemma 2.3 (or its proof), [t, π(t)] = δ(b). It suﬃces to show [δ(b),π(t)] = 0. If ν>1, then δ(b) = 0 by Lemma 2.3 and there is nothing to prove. So assume ν =1and def. s ps char R = p ≥ 2. By Lemma 2.3, u = b−σ(b)=δ(b0) ∈ Φ(p ). So σ(u)=q u = u by Lemma 2.2. With this, we obtain inductively σk(b)=b − ku for k ≥ 0. In s s s particular, σp (b)=b − psu = b.Soδ(b)=δσp (b)=σp δ(b). By the semi- s invariance of π(t), we have π(t)δ(b)=σp (δ(b))π(t)=δ(b)π(t), as wanted.

Corollary 2.5. In the notation of Theorem 2.1, the following are equivalent: (1) π(t) is invariant; (2) π(t)t = tπ(t);(3)δ(b)=0.

Proof. (2) ⇒ (1) is trivial. For (1) ⇒ (3), let π(t) be invariant. So π(t)t =(t−c)π(t) for some c ∈ Q.Thencπ(t)=tπ(t) − π(t)t = δ(b) by Lemma 2.4. Comparing degrees gives c =0andδ(b)=0.(3)⇒ (2) holds since [t, π(t)] = δ(b)byLem- ma 2.4.

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We are ready for our main result. Theorem 2.6. Let δ be X-outer. (1) If char R =0or q =1 ,thenπ(t) is also invariant. (2) Suppose that char R = p ≥ 2 and q =1.Ifδ(b)=δ(c) for some c ∈ Φ(ps),thenπ(t)−c is invariant. Otherwise, π(t)p is a minimal monic invariant polynomial. Proof. If char R =0orq =1,then ν>1andsoδ(b) = 0 by Lemma 2.3. By Corollary 2.5, π(t) is invariant. So suppose that char R = p ≥ 2andq =1. Then ν =1anddegπ(t)=ps. For any c ∈ Φ(ps), π(t) − c is also monic and semi-invariant. If δ(b − c)=0,thenπ(t) − c is invariant by Corollary 2.5. So we suppose that δ(b − c) =0forany c ∈ Φ(ps). By Lemma 2.4, tπ(t)p − π(t)pt = [t, π(t)p]=pδ(b)π(t)p−1 =0.Soπ(t)p is invariant. We show its minimality as an invariant polynomial. Let f(t) ∈ Q[t; σ, δ] be a monic invariant polynomial. Then f(t)t =(t−c)f(t)forsomec ∈ Q.Socf(t)=tf(t)−f(t)t =[t, f(t)], implying c =0, since deg[t, f(t)] < deg f(t) by the monicity of f(t). So [t, f(t)] = 0. By (1) of [20, −i def. ∈ s Proposition 2.8], write f(t)= i=0 uiπ(t) ,whereu0 =1andui Φ(ip ). Since ips q =1,σ(ui)=q ui = ui by Lemma 2.2. So [t, ui]=σ(ui)t + δ(ui) − uit = δ(ui). With this and Lemma 2.4, we compute −i −i −i 0=[t, f(t)] = [t, uiπ(t) ]= [t, ui]π(t) + ui[t, π(t) ] i=0 i=0 i=0 −1 −i −i−1 = δ(ui)π(t) + ( − i)uiδ(b)π(t) i=1 i=0 −1 −1 −i−1 −i−1 = δ(ui+1)π(t) + ( − i)uiδ(b)π(t) i=0 i=0 −1 −i−1 = ( − i)uiδ(b)+δ(ui+1) π(t) . i=0 In the last expression above, the term of degree ≥ ( − 1)ps is contributed by −1 −1 u0δ(b)+δ(u1) π(t) = δ(b)+δ(u1) π(t) .

ps(−1) The coeﬃcient of t in [t, f(t)] yields the equality δ(b)+δ(u1)=0.Ifp , u1 − then is a unit and δ(b+ ) = 0, contradicting our assumption that δ(b c) =0for any c ∈ Φ(ps). So p| .Then ≥ p and deg f(t)=degπ(t) ≥ deg π(t)p.Sincef(t) is arbitrary, the minimality of π(t)p as a monic invariant polynomial is proved.

The second assertion of Theorem 2.6 can be simpliﬁed if σ is also X-outer. Theorem 2.7. Assume that δ, σ are X-outer, char R = p ≥ 2 and q =1.Thenπ(t) or π(t)p is a minimal monic invariant polynomial according as δ(b)=0or δ(b) =0 s − j respectively. In the former case, π(t)=tp + s 1 b tp + b ∈ Q(σ,δ)[t; σ, δ].Inthe j=0 j p ps+1 s−1 p pj+1 p ∈ (σ,δ) latter case, π(t) = t + j=0 bj t + b Q [t; σ, δ]. Proof. Since q =1,δσ = σδ.Soδσs = σsδ for any s. Given 0 = u ∈ Φ(s), s I u = σ and so δI u =Iuδ, implying δ(u) = 0 by Lemma 1.2, since σ is X-outer. So δ(Φ(s)) = 0. Now, if δ(b)=0,thenπ(t) is also invariant by Corollary 2.5. If

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δ(b) =0,then δ(b − c)=δ(b) =0forany c ∈ Φ(ps), since δ(Φ(ps)) = 0. By Theorem 2.6, π(t)p is a minimal monic invariant polynomial. For the rest, since σ is X-outer, we have δ(u) = 0 for any u ∈ Φ(j) by Lemma 1.2. s By Theorem 2.1, b0 ∈ Φ(p − 1), implying δ(b0)=0.Sob − σ(b)=δ(b0)=0by Lemma 2.3. Together with Lemma 2.3, we have thus shown (σ) (σ,δ) (∗) b ∈ Q and bj ∈ Q for 0 ≤ j

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An ideal of R[t; σ, δ]isR-disjoint if and only if its invariant generator is not 1. By definition, f(t) is maximal among ideals with the invariant generator f(t). We are interested in R-disjoint prime ideals of R[t; σ, δ]. It is not hard to see that p(t) is prime for any irreducible invariant p(t). Our aim here is the following converse. Theorem 3.2. Let δ be an X-outer σ-derivation. Unless char R = p ≥ 2 and σ is X-inner (necessarily q =1), any R-disjoint prime ideal of R[t; σ, δ] is of the form p(t) for an irreducible monic invariant polynomial p(t). The above says that R-disjoint prime ideals of R[t; σ, δ] are maximal. This problem has been considered for Ore extensions of derivation type in [2, 11, 23] and of automorphism type in [3, 10]. Leroy and Matczuk [19] proved the same result under the assumption σδ = δσ plus some mild conditions. We need the following generalization of [19, Lemma 1.8]. Theorem 3.3. Assume that δ is X-outer. Unless char R = p ≥ 2 and σ is X-inner (necessarily q =1), monic invariant polynomials have all their coefficients in Q(σ,δ) and hence commute with t. Proof. Let ⎧ ⎨⎪ π(t)ifcharR =0orq =1; M(t) def.= π(t)ifcharR = p ≥ 2,q =1andσ is X-outer with δ(b)=0; ⎩⎪ π(t)p if char R = p ≥ 2,q =1andσ is X-outer with δ(b) =0. By Theorems 2.6 and 2.7, M(t) is a minimal monic invariant polynomial. If char R =0orq =1,then M(t)=π(t) ∈ Q(σ,δ)[t; σ, δ] by Lemma 2.3. If char R = p ≥ 2, q =1andσ is X-outer, then M(t) ∈ Q(σ,δ)[t; σ, δ]byTheo- rem 2.7. So M(t) ∈ Q(σ,δ)[t; σ, δ] always. By [20, Proposition 3.4], any monic invariant polynomial can be written in the form uω(t)M(t)l,wherel ≥ 0isan integer, u ∈ Q is a unit and ω(t) is a nonzero central polynomial. By (iii) of [20, Theorem 3.7], there exist an integer >0 and a unit v ∈ Q such that ζ def.= vM(t) is a nonzero central polynomial of minimal degree and the central polynomial ω(t) canbewrittenintheform n n−1 ω(t)=αnζ + αn−1ζ + ··· (σ,δ) with αi ∈ C . Since [M(t),t]=0,wehave0=[ζ,t]=[v, t]M(t) , implying 0=[v, t]=(v − σ(v))t − δ(v). So σ(v)=v and δ(v) = 0, that is, v ∈ Q(σ,δ).So ζ ∈ Q(σ,δ)[t; σ, δ]andω(t) ∈ Q(σ,δ)[t; σ, δ]. Since uω(t)M(t)l and M(t)l are monic, so is uω(t). Hence u is the inverse of the leading coefficient of ω(t), implying u ∈ Q(σ,δ). Whence uω(t)M(t)l ∈ Q(σ,δ)[t; σ, δ], as asserted. We are now ready for Proof of Theorem 3.2. Let be the set of ideals I of R such that σ(I) ⊆ I and δ(I) ⊆ I.ForI ∈, R[t; σ, δ]I ⊆ I[t; σ, δ], where I[t; σ, δ]isthesetofpolynomials i ∈ ∈ ∈ i ait with ai I.IfI,J ,thenIJ . Define def. Q0 = {a ∈ Q | aI ⊆ R for some 0 = I ∈}. ∈ (σ,δ) Then Q0 forms a subring of Q.Givena Q arbitrarily, let I = 0 be an ideal ⊆ def. of R such that aI R.SetJ = I + θ θ(I), where the summation ranges over all products θ of σ and δ. Clearly, 0 = J ∈and aJ ⊆ R, implying a ∈ Q0.So

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(σ,δ) Q0 ⊇ Q . Clearly, Q0 ⊇ R.GivenanR-disjoint prime ideal P of R[t; σ, δ], let p(t)beitsmonicinvariantgenerator.Givenf(t) ∈ p(t),writef(t)=g(t)p(t), (σ,δ) where g(t) ∈ Q[t; σ, δ]. By Theorem 3.3, p(t) ∈ Q [t; σ, δ] ⊆ Q0[t; σ, δ]. With the left division algorithm in Q0[t; σ, δ], we see g(t) ∈ Q0[t; σ, δ]. Let 0 = I ∈be such that g(t)I ⊆ R[t; σ, δ]. Let J be a nonzero ideal of R such that p(t)J ⊆P.Since [p(t),t]=0,p(t)R[t; σ, δ]=R[t; σ, δ]p(t). With this, we compute f(t)R[t; σ, δ]IJ = g(t)p(t)R[t; σ, δ]IJ = g(t)R[t; σ, δ]p(t)IJ = g(t)R[t; σ, δ]σdeg p(I)p(t)J ⊆ g(t)R[t; σ, δ]IP⊆g(t)I[t; σ, δ]P⊆R[t; σ, δ]P⊆P. Since P is R-disjoint, we have IJ P. The primeness of P implies f(t) ∈P. But f(t) ∈ p(t) is arbitrary. This proves P = p(t). For the irreduciblility of p(t), suppose that p(t)=g(t)h(t), where g(t)andh(t) are monic invariant. Pick a nonzero ideal I of R such that Ig(t) ∪ Ih(t) ⊆ R[t; σ, δ]. Then Ig(t)R[t; σ, δ]Ih(t)= IR[t; σ, δ]Ideg g(t)g(t)h(t) ⊆ p(t), implying Ig(t) ⊆ p(t) or Ih(t) ⊆ p(t).That is, g(t)=1orh(t)=1.Sop(t) is irreducible.

InthecasecharR = p ≥ 2, our δ and σ above are both X-outer. The problem remains open if one of δ, σ is X-inner, as was raised in [19].

Acknowledgments The authors would like to thank the referees for the valuable suggestions which have served to simplify and clarify the paper greatly. This research was supported by the NSC and NCTS/TPE of Taiwan.

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Department of Mathematics, National Taiwan University, Taipei 106, Taiwan E-mail address: [email protected] Department of Mathematics, National Taiwan University, Taipei 106, Taiwan E-mail address: [email protected] Department of Mathematics, National Changhua University of Education, Changhua 500, Taiwan E-mail address: [email protected]

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