行政院國家科學委員會專題研究計畫 成果報告
常值具多項式恆等式之代數斜導算(3/3)
計畫類別： 個別型計畫 計畫編號： NSC91-2115-M-002-002- 執行期間： 91 年 08 月 01 日至 93 年 07 月 31 日 執行單位： 國立臺灣大學數學系暨研究所 計畫主持人： 李秋坤 報告類型： 完整報告 報告附件： 出席國際會議研究心得報告及發表論文 處理方式： 本計畫可公開查詢中 華 民 國 93 年 12 月 10 日
Chen-Lian Chuang and Tsiu-Kwen Lee
Department of Mathematics, National Taiwan University Taipei 106, Taiwan
E-mail: [email protected] E-mail: [email protected]
Abstract. Motivated by Grzeszczuk’s paper [9], we give a detailed analysis of nilpo-tent derivations of semiprime rings. With this, many known results can be either generalized or deduced.
1
† 2000 Mathematics Subject Classification. 16W20, 16W25, 16W55.
‡ Key words and phrases. Nilpotent derivation, (semi-)prime ring, annihilating
nilpo-tency.
Members of Mathematics Division, National Center for Theoretical Sciences at Taipei.
§0. Introduction
Throughout, R is always a semiprime ring and Q its symmetric Martindale quo-tient ring. The center C of Q is called the extended centroid of R. By a derivation of R we mean a map δ: R → R satisfying
(x + y)δ = xδ+ yδ and (xy)δ = xδy + xyδ for all x ∈ R.
For b ∈ R, the map ad(b): x ∈ R 7→ [x, b] = xb − bx defines a derivation of R. A derivation of this form is called inner. More generally, a derivation δ of R is called X-inner if it is the restriction of an inner derivation of Q, that is, if there exists
b ∈ Q such that xδ = [x, b] for x ∈ R. An ideal I of R satisfying Iδ ⊆ I is called a δ-ideal of R. This paper is motivated by Grzeszczuk’s paper [9] and the object of our investigation is the following:
Definition 1. ([9]) A derivation δ is called nilpotent if δn = 0 for some n ≥ 0. The least such n is called the nilpotency of δ, denoted by nδ(R). The least integer m such that, for each nonzero δ-ideal I of R there exists 0 6= c ∈ I with Rδmc = 0, is called the annihilating nilpotency of δ and is denoted by mδ(R).
1
For a nilpotent derivation δ, any ideal I 6= 0 of R contains a δ-ideal J 6= 0: We merely set J def.= Pnδ(R)−1
i=0 (In)δ
i
, where n ≥ nδ(R). Therefore, mδ(R) is also the least integer m such that, for each nonzero ideal I of R, there exists 0 6= c ∈ I such that Rδmc = 0. When R is a prime ring, mδ(R) is then the least integer m such that Rδm
c = 0 for some nonzero c ∈ R. The annihilating nilpotency of δ is first
investigated in [9]; we give it this name for brevity. Nilpotent derivations enjoy many interesting properties and have been studied extensively [3, 5, 6, 9, 15]. When R has a prime characteristic p ≥ 2, most results impose restrictions on p. Our aim here is to remove these restrictions. Let us briefly illustrate our results: For a prime ring R of char R = p ≥ 2, [10, Corollary 3] asserts that a C-algebraic derivation δ of R must satisfy an identity in the following form with s as small as possible:
xδps + β1xδ
ps−1
+ · · · + βsxδ = xb − bx,
where βi ∈ C and where b ∈ Q is C-algebraic. Algebraic properties of δ should be characterized by this identity and by the minimum polynomial of b over C, both of which assume very simple forms for nilpotent δ (Theorems 2.2 and 2.3 below). We justify this point of view by calculating the nilpotency and the annihilating nilpotency of δ in terms of the number ps and the degree of b over C (Theorems 2.4 and 2.5 below). We then extend these to the semiprime case, which merely says that Q can be decomposed into a finite direct sum ⊕iQisuch that the nilpotent derivation restricted to each Qi∩ R behaves in the same way as in the prime case (Theorems 3.1, 3.2 and 3.3 below). As applications, we give an interesting connection between nilpotency and the annihilating nilpotency (Theorems 3.4 and 3.5 below), which generalizes [3, 5, 9]. Finally, we deduce a result of [9] as a corollary.
We organize this article as follows: In §1 we recall some notions and facts, mainly formulated for semiprime rings, which are either well-known or implicit in the liter-ature. In §2 we consider the prime case which shows the main results of this paper. In §3 we extend our results to semiprime case.
§1. Preliminaries
The section is devoted to some notions and facts about semiprime rings (Theorems 1.1–1.7). They are used only in §3 and can be skipped if the reader is interested only in the prime case. Our ring R is always semiprime. It is well-known that a derivation of R can be uniquely extended to a derivation of Q. So all derivations will be assumed to be defined on Q. By a differential polynomial, we mean a generalized polynomial
ϕ with coefficients in Q and with variables acted by derivations of R. If ϕ vanishes
on R then we call ϕ a differential identity. The following is actually implicit in [11]. There is a proof [13] in the spirit of the theory of orthogonal completion [2]. We sketch below a short proof following the line of [4]:
Theorem 1.1. A differential identity ϕ of R also vanishes on Q.
Proof. We transform ϕ into its reduced form ϕ0by the basic identities on [11, p. 58]. By [11, Theorem 2], ϕ0 is a consequence of a generalized polynomial identity ϕ00 of
R in a specific way. By [1, Theorem 6.4.1], ϕ00 also vanishes on Q and hence so does
ϕ0. The basic identities also hold for Q by the uniqueness of extension of derivations from R to Q. Our assertion follows from transforming ϕ0 back into ϕ.
This implies that a nilpotent derivation of R can be extended to a unique nilpotent derivation of Q of the same nilpotency and of the same annihilating nilpotency.
It is shown in [1] that the commutative ring C is a von Neumann regular in the sense that for any α ∈ C, there exists β ∈ C such that α = α2β. It is shown in [1]
that central idempotents of Q form a complete Boolean algebra B. For any subset
T of Q, we define E[T ] to be the smallest central idempotent such that E[T ]t = t for
all t ∈ T . (See [1, Lemma 3.1.1]). If T = {a}, where a ∈ Q, we denote E[T ] by E[a] for brevity. For a derivation δ of Q we write E[δ] to stand for E[Qδ]. We need the following from [11]:
Theorem 1.2. For any maximal ideal A of B, the set AQ forms a minimal prime
ideal of Q. Conversely, any minimal prime ideal of Q admits this form.
We remark that the intersection of all minimal prime ideals is zero. For a semiprime ring R, C may not be a field and we must be careful in extending our notions: We call
ai ∈ Q (1 ≤ i ≤ n) C-independent if all ai 6= 0 and for any αi ∈ C, Pn
i=1αiai = 0 implies that all αiai = 0. Analogously, derivations δi (1 ≤ i ≤ n) of R are said to be mutually outer if all δi 6= 0 and if for any b ∈ Q and αi ∈ C, the identity Pn
i=1αixδi = xb − bx for all x ∈ R implies α1xδ1 = · · · = αnxδn = 0 for all x ∈ R. (This is called strongly independent in [11] but we follow [12].) The following is also implicit in [11]. For brevity, we refer its proof to pp. 65–68 of [11].
Theorem 1.3. Let P be a minimal prime ideal of Q. For a ∈ Q, let a denote the
natural image of a in Qdef.= Q/P . For a derivation δ of Q, let δ denote the derivation
of Q induced canonically by δ.
(2) If a1, a2, . . . are C-independent, then nonzero ai’s are C-independent.
(3) If derivations δ1, δ2, . . . of Q are mutually outer, then so are nonzero δi’s. Additive endomorphisms of the abelian group (Q, +) form a ring End(Q, +). Derivations preserve the addition and hence are elements of End(Q, +). Let b ∈ Q and
δ, a derivation. We compute: for x ∈ Q, xad(b)δ = [x, b]δ = [xδ, b]+[x, bδ] = (xδ)ad(b)+
xad(bδ)
= xδ ad(b)+ad(bδ)
. We hence have the equality ad(b)δ = δ ad(b) + ad(bδ) in End(Q, +). We also interpret α ∈ C as the map α: x ∈ Q 7→ xα ∈ Q and, in this sense, α is an additive endomorphism of (Q, +). For δα in this order, we have
δα: x ∈ Q 7→ αxδ ∈ Q. But the map αδ sends x ∈ R to xαδ def.= (xα)δ = xδα + xαδ =
xδα+ xαδ. So we have the equality αδ = δα + αδ in End(Q, +).
An element b ∈ Q is said to be integral over C or C-integral if bn+ α
1bn−1+ · · · +
αn−1b + αn = 0 for some integer n ≥ 1 and αi ∈ C. The minimal such n is called the integral degree of b. A derivation δ of R is said to be C-integral if there exist
α1, . . . , αn−1 ∈ C such that
δn+ δn−1α1+ · · · + δαn−1 = 0.
The least such n is called the integral degree of δ. We have the following
Theorem 1.4. For b ∈ Q, the derivation ad(b) is C-integral if and only if b is
C-integral.
Proof. Suppose first that b is C-integral. We define the left multiplication `b and the right multiplication rb by
`b : x ∈ Q 7→ bx, rb : x ∈ Q 7→ xb.
Then `b, rb ∈ End(Q, +). Since `brb = rb`b and b is C-integral, we see that the subalgebra over C generated by 1, `band rbis commutative and is a finitely generated module over C. Thus every element in this subalgebra is C-integral. In particular, the element ad(b)def.= `b− rb is C-integral, as asserted.
Suppose next that ad(b) is C-integral of degree n. Consider the set Σ = {e ∈ B | (bn+
n X i=1
αibn−i)e = 0 for some αi ∈ C}.
We must show 1 ∈ Σ. For g ∈ B, if g ≤ e for some e ∈ Σ, then obviously g ∈ Σ. If
B. Assume on the contrary that 1 /∈ Σ. We then extend Σ to a maximal ideal A of B. By (1) of Theorem 1.3, the extended centroid of Q/AQ is the natural image C
of C. In Q/AQ, ad(b) is also C-integral and has the integral degree ≤ n. Expanding this integral relation of degree ≤ n into a linear GPI and then applying [14, Theorem 2], we conclude that the elements 1, b, · · · , bn are C-dependent. So there exist αi∈ C such that bn+Pn
i=1αibn−i ∈ AQ. That is, (bn+ Pn
i=1αibn−i)(1 − e) = 0 for some
e ∈ A. This says 1 − e ∈ Σ ⊆ A. But e ∈ A. So 1 = e + (1 − e) ∈ A, a contradiction.
Let b ∈ Q be C-integral. By the minimum polynomial of b ∈ Q over C, we mean an identity relation bn + α
1bn−1+ · · · + αn−1b + αn = 0, where αi ∈ C and where 1, b, . . . , bn−1 are C-independent. Minimum polynomials of b ∈ Q, if they exist, are unique in the following sense: If bn + α
1bn−1+ · · · + αn−1b + αn = 0 and
bm + β
1bm−1+ · · · + βm−1b + βm = 0 are both minimum polynomials of b over C, then n = m and αibn−i= βibn−i for i = 1, . . . , n. It is not true that any C-integral
b ∈ Q has a minimum polynomial. We need the following more precise description:
Theorem 1.5. Let b ∈ Q be a C-integral element of degree n. Then there exist
finitely many integers ti with 1 ≤ ti ≤ n and orthogonal gi ∈ B (i = 1, . . . , k) with Pk
i=1gi= 1 such that the following hold:
(1) bgi has the minimal polynomial of degree ti.
(2) E[bjg
i] = gi for j = 0, 1, · · · , ti− 1.
Proof. We define Dt to be the set of e ∈ B such that E[bi] ≥ e for 0 ≤ i < t and such that bie (0 ≤ i < t) are C-independent. If 0 6= f ∈ B and f ≤ e for some e ∈ D
t, then obviously f ∈ Dt. Let et def.=
W
Dt. Note et = 0 if and only if Dt = ∅. We
claim that if et 6= 0, then et ∈ Dt: Obviously, E[bi] ≥ et and particularly biet 6= 0 for 0 ≤ i < t. Suppose Pt−1i=0αibiet = 0 for some αi ∈ C. Multiplying the equality by any given e ∈ Dt and using eet = e, we see that
Pt−1
i=0αibie = 0. These bie are
C-independent, since e ∈ Dt. So all αibie = 0. This is true for all e ∈ Dt and hence for et def.=
W
Dt. So αibiet = 0. This shows the C-independence of biet (0 ≤ i < t). So et∈ Dt, as claimed.
Let n be the C-integral degree of b. If t > n, then for any e ∈ Dt, e, eb, . . . , ebn are not C-independent. So Dt = ∅ for t > n. Set D0 def.= B and e0 def.= 1. Since
D0 ⊇ D1 ⊇ · · · ⊇ Dn+1 = ∅, we have e0 ≥ e1 ≥ · · · ≥ en ≥ en+1 = 0. Set ˆ
et def.= et−et+1for 0 ≤ t ≤ n. Define ˜Dtto be the set consisting of central idempotents
e ≤ ˆet such that (bt+ Pt
i=1αibt−i)e = 0 for some αi ∈ C. If f ∈ B and f ≤ e for some e ∈ ˜Dt, then obviously f ∈ ˜Dt. Set ˜et def.= W ˜Dt. Write ˜et =
W
{fν | ν ∈ I} is an orthogonal subset of ˜Dt. For each ν ∈ I, we have
btf
ν+ αν1bt−1fν + · · · + ανt = 0
for some ανi ∈ C. By the orthogonal completeness of C, for each i with 1 ≤ i ≤ t there exists αi∈ C such that αifν = ανifν for all ν ∈ I and αi(1 − ˜et) = 0. Applying the orthogonal completeness of Q, we obtain that
bt˜et+ α1bt−1˜et+ · · · + αt = 0. Therefore, ˜et ∈ ˜Dt follows. Of course, we have ˜et ≤ ˆet.
We claim next that ˜et = ˆet. Otherwise, f def.= ˆet− ˜et6= 0. Notice that 0 6= f = ˆet− ˜et = ˆet(1 − ˜et) = et(1 − et+1)(1 − ˜et).
If f ∈ Dt+1, then f ≤ et+1 and so f = 0, a contradiction. Therefore, f ∈ Dt\ Dt+1. Notice that E[bi] ≥ e
t ≥ f for 0 ≤ i ≤ t − 1.
If g def.= f (1 − E[bt]) 6= 0, then gbt = 0. Since g ≤ ˆe
t, this implies that g ∈ ˜Dt and so g ≤ ˜et. But then g = g˜et = gf ˜et = 0, a contradiction. Therefore, we have E[bt] ≥ f and, particularly, bif 6= 0 for 0 ≤ i ≤ t. Since f /∈ D
t+1, bif (0 ≤ i ≤ t) cannot be C-independent. Thus there exist αi ∈ C such that
α0btf + α1bt−1f + · · · + αt−1bf + αtf = 0
and such that αibt−if 6= 0 for some i ≤ t. We have α0btf 6= 0, for otherwise the
elements bt−1f, . . . , bf, f would not be C-independent, contradicting the fact that
f ∈ Dt. By abelian regularity of C ([1, Theorem 2.3.9]), there exists β ∈ C such that (f α0)2β = f α0. Since edef.= f α0β ∈ B and 0 < e ≤ f , we multiply the above equality
by βe and obtain ebt ∈ Pt
i=1Cebt−i. This implies that e ∈ ˜Dt and so e ≤ ˜et. But
e ≤ f def.= ˆet− ˜et, a contradiction. So f = 0 follows, that is, ˜et= ˆet. So b satisfies an identity bteˆ
t+ Pt
i=1αibt−iˆet= 0, where αi∈ C. This is the min-imum polynomial of bˆet, since the elements bieˆt, where 0 ≤ i < t, are C-independent by the fact that ˆet ∈ Dt. Note that E[bieˆt] = ˆet for 0 ≤ i < t. We set gt = ˆet for t = 1, 2, · · · , n. Then the central idempotents gi, where 1 ≤ i ≤ n, are pairwise orthogonal andPni=1gi = 1. This completes our proof.
For the case char R = 0, it is proved in [11] that every C-integral derivation of R is X-inner. For the case char R = p ≥ 2, the corresponding result is [10, Corollary 3], which is for prime rings only. We need the following more detailed information:
Theorem 1.6. Assume char R = 0. If δ is a C-integral derivation of R, then there
exist a C-integral element b ∈ Q, integers ti≥ 1 and orthogonal ei∈ B (i = 1, . . . , k)
with Pki=1ei= 1 such that the following holds:
(1) δ = ad(b).
(2) bei has the minimal polynomial of degree ti.
(3) E[bje
i] = ei for j = 0, 1, · · · , ti− 1.
Proof. By assumption, δ is C-integral. Since charR = 0, δ is X-inner by [11].
Write δ = ad(b) for some b ∈ Q. In view of Theorem 1.4, b is C-integral. Now, we conclude by applying Theorem 1.5 to our b. This proves the theorem.
Theorem 1.7. Assume char R = p ≥ 2. If δ is a C-integral derivation of R with
degree n, then there exist integers si ≥ 0, ti with 1 ≤ ti ≤ n and orthogonal ei ∈ B
(i = 1, . . . , k) with Pki=1ei= 1 such that the following holds for δi def.= δei:
(1) δpij (0 ≤ j < si) are mutually outer and E[δp
j
i ] = ei for 0 ≤ j < si.
(2) δpisi =Psi−1
j=0 δp
j
i αj+ ad(bi) holds for some αj ∈ eiC and bi ∈ eiQ.
(3) bi has the minimum polynomial of degree ti and E[bji] = ei for 0 ≤ j < ti.
Proof. For s ≥ 0, we define Bs to be the set of e ∈ B such that δp
j
e (0 ≤ j < s) are
mutually outer and such that E[δpj
] ≥ e for 0 ≤ j < s. Obviously, if 0 6= f ∈ B and
f ≤ e for some e ∈ Bs, then f ∈ Bs. Set es def.= W
Bs. Note that es = 0 if and only if
Bs = ∅.
Claim 1. If es 6= 0, then es ∈ Bs: Pick orthogonal central idempotents fν ∈ Bs with Wνfν = es. For 0 ≤ j < s, we have E[δp
j
] ≥ fν for all fν and hence E[δp
j ] ≥ W fν = es. Suppose that Ps−1 j=0δp j
esαj− ad(b) = 0 holds for some αj ∈ C and b ∈ Q. Multiplying the equality by fν, we have
s−1 X j=0 δpjfναj− ad(bfν) = ¡s−1X j=0 δpjesαj− ad(b) ¢ fν = 0. Since fν ∈ Bs, the derivations δp
j
fν, where 0 ≤ j < s, are mutually outer and hence all δpjf
ναj = 0. This is true for all fν. Hence, δp
j
esαj = δp
j
(Wfν)αj = 0 for all 0 ≤ j < s. So δpj
es (0 ≤ j < s) are mutually outer. This proves the claim. By the C-integrality of δ, we have a relation δn+Pn−1
i=1 δn−iαi= 0, where αi∈ C. Let e ∈ Bs, where ps > n. If δp
i
e (0 ≤ i < s) are mutually outer, then δie (0 ≤ i ≤ n) are regular words in mutually outer derivations δe, δpe, . . . , δps−1
e ordered in this
order. By [11, Theorem 2], we have the identity e(zn + Pn−1i=1 αizn−i) = 0 of Q, where zi are distinct indeterminates. This is absurd. So Bs = ∅.
Set B0 def.= B and e0 def.= 1. Then B0 ⊇ B1 ⊇ B2 ⊇ · · · and hence e0 ≥ e1 ≥
e2 ≥ · · · . Set ˆes def.= es− es+1 for 0 ≤ ps ≤ n. For 0 ≤ ps ≤ n, let ˜Bs be the set of central idempotents e ≤ ˆes such that δp
s
e +Ps−1i=0 δpi
eαi− ad(b) = 0 for some
αi ∈ C and b ∈ Q. For f ∈ B, if f ≤ e for some e ∈ ˜Bs, then obviously f ∈ ˜Bs. Set ˜
es def.= W ˜Bs. Applying the same argument as that in Claim 1, we see ˜es ∈ ˜Bs. Notice that E[δpi
] ≥ es for 0 ≤ i < s.
Claim 2. ˜es = ˆes: Otherwise, f def.= ˆes − ˜es 6= 0. Note that f = ˆes(1 − ˜es) =
es(1 − es+1)(1 − ˜es). Therefore, f /∈ Bs+1. Suppose for the moment that g def.=
f (1 − E[δps)] 6= 0. Then δδpsg = 0, implying that g ∈ ˜B
s. and so g ≤ ˜es. Then
g = gf ˜es = 0, a contradiction. Therefore, we have g = 0 and so E[δp
s
] ≥ f . Since f /∈ Bs+1, δp
i
f (0 ≤ i ≤ s) are not mutually outer. Since E[δps
] ≥ f , we have δpif = 0 for 0 ≤ i ≤ s. So for some α
i ∈ C and b ∈ Q, Ps
i=0δp
i
f αi− ad(b) = 0 and yet some δpi
f αi 6= 0. Since f ∈ Bs, therefore δp
i
f (0 ≤ i < s) are mutually
outer. We must have δps
f αs 6= 0. By abelian regularity of C [1, Theorem 2.3.9], there exists β ∈ C such that (f αs)2β = f αs. Since edef.= f αsβ ∈ B and 0 < e ≤ f , we have δpse + s−1 X i=0 δpieα iβ − ad(beβ) = ¡Xs i=0 δpif α i− ad(b) ¢ eβ = 0.
This says e ∈ ˜Bs and so 0 < e ≤W ˜Bs def.= ˜es. But e ≤ f def.= ˆes− ˜es, a contradiction. It suffices to show each ˆesQ has the desired property. Replacing Q by ˆesQ and δ by δˆes, we may assume, with Claims 1 and 2, that E[δp
i
] = 1 for 0 ≤ i < s, that δpi
(0 ≤ i < s) are mutually outer and that
δps = s−1 X j=0 δpjα j+ ad(b)
for some αi ∈ C and b ∈ Q. Multiply the above equality by δ from the left and the right hand sides respectively, and then take their difference. We see that
s−1 X j=0
δpjαδj + ad(bδ) = 0.
In view of [11, Theorem 2], we have αδ
j = 0 for all j and bδ ∈ C. Therefore, the
C-integrality of δ implies that ad(b) is C-integral. By Theorem 1.4, b is C-integral.
§2. Prime Case
Throughout this section, R is prime and C is hence a field. The case of char R = 0 is well-known. But we include it here and sketch a proof for easy reference.
Theorem 2.1. Let R be a prime ring of char R = 0 and δ, a nilpotent derivation of
R. Then δ = ad(b) for a nilpotent element b ∈ Q with nilpotency, say, l. Moreover, mδ(R) = l and nδ(R) = 2l − 1.
Proof. By [10, Corollary 2], δ = ad(b) for some b ∈ Q. It follows from Theorem
1.1 that ad(b) is also a nilpotent derivation of Q with the same nilpotency. By [15, Theorem 2], b may be chosen to be a nilpotent element of the nilpotency, say l, and then nδ(R) = 2l − 1. Let c 6= 0 be such that Rδ
m
c = 0, where m = mδ(R). For all
x ∈ R, (−1)mxδmc = bmxc − µ m 1 ¶ bm−1xbc + · · · + (−1)m µ m m ¶ xbmc = 0.
If m < l, then the elements 1, b, . . . , bm are C-independent and, by [14, Theorem 2], we have c = 0, absurd. So m ≥ l. But xδl
bl−1 = 0 and bl−1 6= 0. So m = l by the minimality of m.
The case of char R = p ≥ 2 is analogous to Theorem 2.1 but is more complicated. Theorem 2.2 Assume that R is a prime ring of char R = p ≥ 2 and δ, a nilpotent
derivation of R. Then there exists an integer s ≥ 0 such that δps
= ad(b) for some
b ∈ Q and such that the derivations δpi are mutually outer for i = 0, . . . , s − 1.
Proof. By [10, Corollary 3], there exists an integer s ≥ 0 such that the derivations δpi
, where 0 ≤ i < s, are mutually outer and such that δps
=Ps−1i=0δpi
βi+ ad(b) for some βi∈ C and b ∈ Q. We must show that all βi= 0. Assume on the contrary that
βt6= 0 for some 0 ≤ t < s. Let t be the least such integer and rewrite
(1) δps =
s−1 X i=t
δpiβi+ ad(b).
Claim: For any n ≥ 0, there exist αi ∈ C (0 ≤ i < s − 1), not all vanishing, and
a ∈ Q such that (2) δps+n = s−1 X i=t δpiαi+ ad(a).
Granted the claim, we take n so large that δps+n
= 0. Then 0 = δsp+n
=Ps−1i=t δpi
αi+ ad(a), where αi ∈ C are not all vanishing. This contradicts the mutual outerness of
δpi
(0 ≤ i < s) and proves our assertion.
The claim is proved by induction on n: If n = 0, then the expression (1), in which
βt 6= 0, is already in the expected form. As the induction hypothesis, assume that the claim holds for n ≥ 0. Multiply (2) by δ from the right hand side:
δps+n+1= s−1 X i=t δpi+1α i+ s−1 X i=t δpiαδ i + δ ad(a) + ad(aδ).
But(2) multiplied by δ from the left hand side gives δps+n+1 = Ps−1
i=t δp
i+1
αi +
δ ad(a). The difference between these two expressions of δ1+ps+n
gives the identity Ps−1
i=t δp
i
αδ
i + ad(aδ) = 0. The mutual outerness of δp
i
(0 ≤ i < s) implies αδ i = 0 (t ≤ i < s) and hence ad(aδ) = 0 also. So α
iδ = δαi + αδi = δαi and ad(a)δ =
δ ad(a) + ad(aδ) = δ ad(a). That is, all α
i and ad(a) commute with δ. Let u be the greatest integer with αu 6= 0 in (2) and rewrite (2) as δp
s+n
= Pui=tδpi
αi+ ad(a). Raise both sides of this equality to the p-th power using the commutativity of ad(a) and αi: (3) δps+n+1 = u X i=t δpi+1αp i + ad(ap).
If u + 1 < s, then (3) is in the claimed form since the coefficient of δpu+1 is αp u 6= 0. If u + 1 = s, we replace δps
of (3) by the right hand side of (1) and obtain
δps+n+1 = δps−1(·) + · · · + δpt+1(·) + δpt(β
tαpu) + ad(αpub + ap), where the coefficient of δpt is β
tαpu6= 0, as claimed.
In Theorem 2.1 above, ad(b) is nilpotent, since δ is also nilpotent. We now char-acterize nilpotent ad(b) in terms of the minimum polynomial of b:
Theorem 2.3. Let R be a prime ring of char R = p ≥ 2 and let b ∈ Q define a
nilpotent derivation ad(b) of R. Then the minimum polynomial of b over C admits the form (bpt
− α)l = 0, where α ∈ C, t ≥ 0, l ≥ 1 are integers and (p, l) = 1.
Proof. Let n ≥ 0 be the nilpotency of ad(b) on R. The identity xad(b)n
= 0 of R also holds for x ∈ Q by [1, Theorem 6.4.1]. Let C be the algebraic closure of C. The tensor product Qdef.= Q⊗
The map r ∈ Q 7→ r ⊗ 1 ∈ Q gives a ring embedding. The linear identity xad(b)n
= 0 also extends linearly to x ∈ Q. The nilpotency of ad(b) acting on Q is thus also n. Consider the polynomial ring C[λ] in the commuting indeterminate λ. Let µ(λ) be the minimum polynomial of b over C. We claim that µ(λ) ∈ C[λ]. Let ν be the degree of µ(λ) and write µ(λ) = λν+ α
1λν−1+ · · · + αν, where α1, · · · , αν ∈ C. We must show all αi ∈ C. Note that C is a subfield of C. Pick a C-basis β0 def.= 1, β1, β2, . . .
of the C-space C and express each αi as a C-linear combination of the basis βj ∈ C:
αi= γi01 + γi1β1+ γi2β2+ · · · , where γji ∈ C. Substitute these expressions of αi into the minimum polynomial of b over C and collect terms according to βi:
0 = bν+ α1bν−1+ · · · + αν
= bν+ (γ10+ γ11β1+ · · · )bν−1+ · · · + (γν0+ γν1β1+ · · · )
= (bν+ γ
10bν−1+ · · · + γν0) + (·)β1+ (·)β2+ · · · .
In this expression, the left coefficients of β0def.= 1, β1, β2, . . . are all in Q, since b ∈ Q
and γi
j∈ C. By the defining property of the tensor product Q ⊗
CC, these coefficients of βi must all be vanishing, since β0= 1, β1, β2, . . . are C-independent. Particularly,
the coefficient of β0def.= 1 gives the equality bν+ γ10bν−1+ · · · + γν0 = 0. So b satisfies the polynomial λν+γ
10λν−1+· · ·+γν0 ∈ C[λ]. This polynomial also has the minimal degree ν of µ(λ) and hence must be equal to µ(λ). So µ(λ) ∈ C[λ] as claimed.
We pick a root λ0 ∈ C of µ(λ). Then ad(b) = ad(b − λ0). For x ∈ Q, we have
xad(b−λ0)n= x(b − λ 0)n+ n X i=1 (−1)i µ n i ¶ (b − λ0)ix(b − λ0)n−i = 0.
Set µ0(λ)def.= µ(λ)/(λ − λ0). Multiplying the above by µ0(b) from the left and using
µ0(b)(b − λ0) = µ(b) = 0, we obtain 0 = µ0(b)x(b − λ0)n for x ∈ Q. Since µ0(b) 6= 0,
(b − λ0)n = 0 by the primeness of Q [7]. So µ(λ) divides (λ − λ0)n. We hence have
µ(λ) = (λ − λ0)ν for some ν. Write ν = ptl with (l, p) = 1 and compute:
µ(λ) =(λ − λ0)ν = (λ − λ0)p tl = (λpt− λp0t)l =λptl− µ l 1 ¶ λpt(l−1)λpt 0 + · · · .
Since µ(λ) ∈ C[λ], ¡1l¢λp0t ∈ C. But l ∈ C is invertible, since (l, p) = 1. So λp0t ∈ C.
Set αdef.= λp0t ∈ C. We have µ(λ) = (λpt
− λp0t)l = (λpt
By the p-power expansion of an integer n ≥ 0, we mean an expression of the form
n =Ps≥0nsps = n0+ n1p + n2p2+ . . . , where 0 ≤ ns < p for each s.
Theorem 2.4. Assume that R is a prime ring of char R = p ≥ 2, and δ a nilpotent
derivation of R. Let δps
= ad(b) be as described in Theorem 2.2 and let (bpt
− α)l = 0
be as described in Theorem 2.3. Then mδ(R) = lps+t.
Proof. Let m = mδ(R) be the annihilating nilpotency of δ and let m = P
i≥0mipi be the p-power expansion of m. We have
δm =¡δm0+m1p+···+ms−1ps−1¢¡δmsps+ms+1ps+1+···¢
=¡δm0(δp)m1· · · (δps−1)ms−1¢¡δps¢ms+ms+1p+···
= ∆ ad(b)n,
where we have set ∆def.= δm0(δp)m1· · · (δps−1)ms−1 and ndef.= m
s+ ms+1p + · · · . Note that ∆ is a regular derivation word in the mutually outer derivations δ, . . . , δps−1
linearly ordered by δ < δp < · · · < δps−1
. Suppose that Rδm
c = 0 with c 6= 0. We
apply [11, Theorem 2] to the identity 0 = xδm
c = (x∆)ad(b)n
c and obtain the identity xad(b)nc = 0 for R. But 0 = xad(b)nc = xδpsnc. So m = psn by the minimality of m. We then have 0 =xδmc = x(δps)nc = xad(b)nc =(−1)n¡bnx − µ n 1 ¶ bn−1xb + µ n 2 ¶ bn−2xb2− · · ·¢c.
Note that lpt is the algebraic degree of b over C. If n < ptl, then left coefficients 1, b, . . . , bn of the above are C-independent and right coefficients must be all vanishing by [14, Theorem 2], contradicting c 6= 0. So n ≥ ptl. But ad(b)ptl = ad(bpt)l = ad(bpt
− α)l = ad(˜b)l, where ˜bdef.= bpt
− α has nilpotency l. We have the expansion xad(b)ptl = xad(˜b)l = l X i=0 (−1)i µ n i ¶ ˜bix˜bl−i= l−1 X i=1 (−1)i µ n i ¶ ˜bix˜bl−i.
We see that xad(b)ptl˜bl−1 = 0 using ˜bl = 0. So n = ptl and m = psptl = ps+tl, as asserted.
We are now ready to describe completely the nilpotency of a nilpotent derivation. Theorem 2.5. Assume that R is a prime ring of char R = p ≥ 2, and δ a nilpotent
derivation of R. Let δps
= ad(b) be as described in Theorem 2.2 and let (bpt
− α)l = 0
(1) If 2l0− 1 < p and 2li< p for all i > 0, then nδ(R) = ¡ (2l0− 1) + X i>0 2lipi ¢ ps+t = (2l − 1)ps+t.
(2) If 2l0− 1 ≥ p or if 2li ≥ p for some i > 0, then
nδ(R) =¡(2lu+ 1)pu+X i>u
2lipi¢ps+t,
where u ≥ 0 is the least integer such that 2lu+ 1 < p and 2li < p for all i > u. We need to tell whether ¡ni¢ ≡ 0 (mod p) or not. This is solved neatly by Lucas
in his Theorie des Nombres (pp. 417–420) and his solution is reproduced in [8]: Lemma 2.6. (Lucas) Given a prime p ≥ 2, let n = Ps≥0nsps and i =
P
s≥0isps
be the p-power expansions of integers n, i ≥ 0 respectively. Then ¡ni¢ ≡ Qs≥0¡ns
is
¢
(mod p), where we postulate ¡k0¢= 1 for k ≥ 0 and¡kj¢= 0 for j > k ≥ 0.
We recall two notations in number theory: For a real number r ≥ 0, let [r] be the greatest integer n such that n ≤ r and let {r} be the smallest integer n such that
n ≥ r. Obviously, {n
2} +
£n
2
¤
= n for any integer n ≥ 0. With this we compute the
central nonvanishing binomial coefficients (mod p):
Lemma 2.7. Given a prime p ≥ 2, let n be a given nonnegative integer with the
p-power expansion n = Ps≥0nsps. Let v be the least nonnegative integer such that
ns is even for s > v. Define
`(n)def.= © nv 2 ª pv+X s>v ns 2 p s and `0(n)def.= n − `(n). Then `0(n) ≤ `(n) and ¡n i ¢
≡ 0 (mod p) for `0(n) < i < `(n) but ¡ n `(n)
¢
6≡ 0 (mod p). Proof. Write `(n), `0(n) as `, `0 for brevity. In view of the equality {nv
2 } + [n2v] = nv,
the p-power expansion of `0 is given by
`0 =X s<v nsps+ £ nv 2 ¤ pv+X s>v ns 2 p s. If {nv
2 } > [n2v], then ` > `0. If {n2v} = [n2v], then nv is even. This happens only when
v = 0 by the definition of v. In this case, ` = n0
2 +
P
s>0 n2sps = `0. So always ` ≥ `0,
Since 0 ≤ {nv 2 } ≤ nv < p, ¡ n v {ns/2} ¢
6≡ 0 (mod p). For s > v, ns is even by the definition of v and, since 0 ≤ ns/2 ≤ ns < p, we also have
¡ n s ns/2 ¢ 6≡ 0 (mod p). By Lemma 2.6, we have¡n`¢≡¡ nv {nv/2} ¢ Q s>v ¡ n s ns/2 ¢ 6≡ 0 (mod p), as asserted. Given `0 < i < `, let i =P
s≥0isps be its p-power expansion. For brevity, let `s, `0s denote the coefficients of ps in the p-power expansions of `, `0 respectively. Since
`0 < i, there exists s such that i
s > `0s and is = `0s for all s > s. If s > v, then also i > `, since `0
s
def.
= ns/2 def.= `s for s > v. This contradicts the assumption
`0 < i < `. If s = v, then i v > `0v
def.
= [nv/2] and hence iv ≥ {nv/2} def.= `v. This implies i ≥ `, since is = `0s = ns/2 = `s for s > v. Again, this contradicts the assumption `0 < i < `. So we must have s < v. But then i
s > `0s def.= ns and, by our convention on binomial coefficients,
µ ns is ¶ def. = 0. By Lemma 2.6, we have ¡n i ¢ ≡¡n0 i0 ¢¡n 1 i1 ¢ · · ·¡ns is ¢ · · · ≡ 0 (mod p), as asserted.
Lemma 2.8. Let R be a prime ring of char R = p ≥ 2. Let b ∈ Q be nilpotent and
have the nilpotency m. Then for integer n ≥ 0, ad(b)n = 0 if and only if `(n) ≥ m,
where `(n) is as defined in Lemma 2.7.
Proof. Write `(n) as ` for short. Set `0= n − `. We have the expansion (4) xad(b)n= n X i=0 (−1)i µ n i ¶ bixbn−i.
For necessity (⇐), assume that ` ≥ m. Consider a typical term ¡ni¢bixbn−i, where 0 ≤ i ≤ n. If `0< i < ` then¡n
i ¢
≡ 0 mod p by Lemma 2.7. If i ≥ ` then bi= 0, since
i ≥ ` ≥ m. If i ≤ `0 then bn−i = 0, since n − i ≥ n − `0= ` ≥ m. So ¡n i ¢
bixbn−i = 0 always and ad(b)n = 0 follows. For sufficiency (⇒), assume that xad(b)n
= 0 for all x ∈ R. Since bi = 0 for i ≥ m, the sum of the terms with i < m in (4) gives a linear identity with right coefficients 1, b, . . . , bm−1, which are C-independent. By [14, Theorem 2], their corresponding left coefficients must vanish. That is, ¡ni¢bn−i = 0 for 0 ≤ i < m. So, if ` < m then ¡n`¢bn−` = 0. But ¡n
` ¢
6= 0 by Lemma 2.7 and bn−` 6= 0, since m > ` ≥ `0 def.= n − `. This is a contradiction.
Proof of Theorem 2.5. Let n = nδ(R) be the nilpotency of δ. Write n = psq + r, where q, r are nonnegative integers and 0 ≤ r < ps. With δps = ad(b), we write
0 = xδn = xδpsq+r = (xδr)(δps)q = (xδr)ad(b)q,
which is a linear generalized identity in xδr. Let r have the p-power expansion
r0+ r1p + . . . + rs−1ps−1. Then
is a regular word in the mutually outer derivations δ, δp, . . . , δps−1
linearly ordered by
δ < δp < · · · < δps−1. We apply [11, Theorem 2] to the differential identity (xδr)ad(b)q
and obtain the identity zad(b)q
= 0 for R. But 0 = zad(b)q
= zδpsq
. By the minimality of the nilpotency n, we have n = psq. So 0 = δn = δpsq = ad(b)q. It suffices to find q, which is obviously the nilpotency q of ad(b). Replacing δ by ad(b), we may assume s = 0 to start with. Set Qdef.= Q⊗
CC, where C is the algebraic closure of C. As explained in the proof of Theorem 2.3, ad(b) has the same nilpotency on Q as on
R and the minimal polynomial of b over C is also (bpt
− α)l = 0. Replacing R by Q, we may assume that the extended centroid C is algebraically closed. The minimum polynomial of b over C is thus of the form (b − α1/pt
)ptl
= 0. But b and b − α1/pt
define the same inner derivation. Replacing b by b − α1/pt
, we may further assume that b is nilpotent and has the nilpotency lpt.
Firstly, assume that 2l0− 1 < p and 2li< p for all i > 0. Set
ν def.= (2l0− 1)pt+ X i>0 2lipi+t = ¡ (2l0− 1) + X i>0 2lipi ¢ pt = (2l − 1)pt. We compute `(ν) = l0pt+ X i>0 lipi+t = (l0+ X i>0 lipi)pt = lpt.
Since lpt is the nilpotency of b, we have ad(b)ν = 0 by Lemma 2.8. Note that l
0≥ 1,
since (l, p) = 1. The p-power expansion of ν − 1 is thus given by
ν − 1 = X 0≤j<t (p − 1)pj+ (2l0− 2)pt+ X i>0 2lipi+t. If p is odd, then p − 1 is even and
`(ν − 1) = X 0≤j<t (p − 1) 2 p j+ (l 0− 1)pt+ X i>0 lipi+t < lpt. By Lemma 2.8, ad(b)ν−1 6= 0. If p = 2, then l
i = 0 for i ≥ 1 and ν = 2t. Thus
`(ν − 1) is equal to 0 or 2t−1depending on whether t = 0 or t > 0. But `(ν − 1) < l2t in either case. By Lemma 2.8, ad(b)ν−1 6= 0. So the nilpotency of ad(b) is ν, as asserted.
Now, we assume that 2l0− 1 ≥ p or 2li≥ p for some i > 0. Let u ≥ 0 be the least integer such that 2lu+ 1 < p and 2li < p for all i > u. Set
ν def.= (2lu+ 1)pu+t+ X i>u 2lipi+t = ¡ (2lu+ 1)pu+ X i>u 2lipi ¢ pt.
We compute `(ν) = (lu+ 1)pu+t+ X i>u lipi+t = ¡ (lu+ 1)pu+ X i>u lipi ¢ pt > lpt. So ad(b)ν = 0 by Lemma 2.8. The p-power expansion of ν − 1 is given by
ν − 1 = X 0≤j<u+t (p − 1)pj+ 2lupu+t+ X i>u 2lipi+t. If p is odd, then p − 1 is even and
`(ν − 1) = X 0≤j<u+t (p − 1) 2 p j+ lupu+t+X i>u lipi+t.
By our case assumption, either 2l0− 1 ≥ p or 2li ≥ p for some i > 0. Let j be the greatest j such that 2lj ≥ p. Note that j < u by the definition of u. For j < j < u, we have 2lj < p by the maximality of j and 2lj+ 1 ≥ p by the minimality of u. That is, lj = p−12 for j < j < u. Also, lj ≥ p2 > p−12 by the definition of j. So `(ν −1) < lpt and hence ad(b)ν−1 6= 0 by Lemma 2.8. For p = 2, the condition that 2l
u+ 1 < p and 2li< p for all i > u implies li= 0 for i ≥ u and u is hence the least such integer. So
ν = 2u+tand l
u−1= 1 by the minimality of u. So `(ν − 1) = 2u+t−1. Since (l, p) = 1,
l0 = 1 and the condition 2l0 − 1 < p = 2 surely holds. By our case assumption,
2lj ≥ p = 2 for some j > 0. So u − 1 > 0, implying l > 2u−1 and hence `(ν − 1) < l2t. By Lemma 2.8, ad(b)ν−1 6= 0. So the nilpotency of ad(b) is ν, as asserted.
§3. Semiprime Case
Our aim here is to extend the results in §2 to semiprime rings. So our R is a semiprime ring with extended centroid C and with symmetric Martindale quotient ring Q throughout. Inspired by the m, n-homogeneity of [9], we give a name for those rings on which nilpotent derivations under consideration behave in the same way as in the prime case:
Definition 2. Let R be a semiprime ring of characteristic 0 or a prime p ≥ 2. Let δ be a nilpotent derivation of R. We call R δ-homogeneous if one of the following two conditions holds:
(1) If char R = 0, then there exists b ∈ Q satisfying the following: (i) δ = ad(b). (ii) The minimum polynomial of this b over C admits the form bl = 0 for some integer
(2) If char R = p ≥ 2, then there exist b ∈ Q and an integer s ≥ 0 satisfying the following: (i) δps = ad(b). (ii) The derivations δ, δp, ..., δps−1 are mutually outer. (iii) The minimum polynomial of this b over C admits the form (bpt
− α)l = 0, where
α ∈ C, t ≥ 0, l ≥ 1 are integers and (p, l) = 1. (iv) E[δpi] = 1 for i = 0, 1, · · · , s − 1 and E[bn] = 1 for n = 1, 2, · · · , ptl − 1.
This definition is justified by the following two theorems:
Theorem 3.1. Let R be a semiprime δ-homogeneous ring of char R = 0, where δ
is a nilpotent derivation of R. Let δ, l be as described in (1) of Definition 2. Then mδ(R) = l and nδ(R) = 2l − 1.
The proof of Theorem 3.1 is similar to that of Theorem 3.2 below but is much simpler. We omit it for brevity.
Theorem 3.2. Let R be a semiprime δ-homogeneous ring of prime characteristic
p ≥ 2, where δ is a nilpotent derivation of R. Let δ and s, t, l be as described in (2) of Definition 2. Then mδ(R) = ps+tl and nδ is given as follows: Let l =
P
i≥0lipi
be the p-power expansion of l.
(1) If 2l0− 1 < p and 2li< p for all i > 0, then
nδ(R) = ¡ (2l0− 1) + X i>0 2lipi ¢ ps+t = (2l − 1)ps+t. (2) If 2l0− 1 ≥ p or if 2li ≥ p for some i > 0, then
nδ(R) = ((2lu+ 1)pu+ X i>u
2lipi)ps+t,
where u ≥ 0 is the least integer such that 2lu+ 1 < p and such that 2li < p for all
i > u.
Proof. Set m = ps+tl. We also let n = (2l − 1)ps+t for Case (1) and let n = ((2lu+ 1)pu+
P
i>u2lipi)ps+t for Case (2). We extend δ to Q. By Theorem 1.1, the nilpotency and the annihilating nilpotency of δ remain the same for δ thus extended, and the identity δps
= ad(b) also holds for Q. Obviously, the pi-th powers δpi
, where 0 ≤ i < s, of δ thus extended remain mutually outer. We may thus replace R by
Q. Let P be a fixed but arbitrary minimal prime ideal of Q. By Theorem 1.2, we
have Pδ ⊆ P . Consider Qdef.= Q/P . Let b be the natural image of b in Q, and δ the derivation of Q induced by δ. Obviously, δps = ad(b). By Theorem 1.3, the minimum polynomial of b over C is (bpt− α)l = 0 and δpi (0 ≤ i < s) remain mutually outer.
By Theorems 2.4 and 2.5 applied to Q, we see that mδ(Q/P ) = m and nδ(Q/P ) = n. This implies immediately that nδ(R) = n.
Let m0 = m
δ(R). There exists 0 6= c ∈ R such that Rδ
m0
c = 0 and so Qδm0c = 0. Choose a minimal prime ideal P of Q such that c /∈ P . Then Qδ
m0
c = 0, where Q def.= Q/P and c 6= 0. Thus m0 ≥ m
δ(Q) = m. On the other hand, let I be a nonzero ideal of R. Choose a minimal prime ideal P of Q such that I 6⊆ P . Then
I 6= 0. Since mδ(Q) = m, there exists 0 6= c ∈ I, that is, there exists c ∈ I \ P , such that Qδ
m
c = 0. That is, Qδm
c ⊆ P . But Qδm
c is orthogonally complete. There
exists a central idempotent e /∈ P such that Qδm
ce = 0 by [1, Proposition 3.1.11].
Choose a dense ideal J of R such that ceJ ∪ eJ ⊆ R. Since ce 6= 0, we choose a nonzero element c0 ∈ ceJ. Then c0 ∈ I and Qδm
c0 = 0. In particular, Rδm
c0 = 0 follows. Thus m ≥ m0 and so m = m0. This proves our assertion on m.
We need some more notions to state our main theorem: Central idempotents are always constants of derivations. If e is a central idempotent of Q, then eQ is the symmetric Martindale quotient ring of the semiprime ring eQ ∩ R by [1, Proposition 2.3.14]. For any derivation δ of R, we have (eQ ∩ R)δ ⊆ eQ ∩ R. That is, the restriction of δ to eQ ∩ R gives rise to a derivation of eQ ∩ R. A family of central idempotents e1, e2, . . . is called orthogonal if eiej = 0 for i 6= j. Orthogonal central idempotents e1, . . . , ek with e1+· · ·+ek = 1 give rise to the direct sum decomposition
Q = ⊕k
i=1eiQ. In this case, the ideal ⊕ki=1(eiQ ∩ R) is essential in R. Any direct sum decomposition of Q arises in this way. We are now ready to state our main result in semiprime rings:
Theorem 3.3. Let R be a semiprime ring of characteristic 0 or a prime p and
δ, a nilpotent derivation of R. Then there exist finitely many orthogonal central idempotents e1, . . . , ek in Q with e1 + · · · + ek = 1 such that each eiQ ∩ R is δi
-homogeneous, where δi is the restriction of δ to eiQ ∩ R.
Proof. Let e1, . . . , ek be central idempotents given in Theorems 1.4 and 1.7, respec-tively, according to char R = 0 or a prime p. Let δi be the restriction of δ to eiQ. It suffices to show the δi-homogeneity of eiQ. Let us consider the case char R = p ≥ 2 only. The case char R = 0 is similar and simpler. Replacing R, δ by R ∩ eiQ, δi respectively, we may assume the following:
(i) δpi
(0 ≤ i < s) are mutually outer and E[δpi
] = 1 for 0 ≤ i < s. (ii) δ satisfies the identity δps
+ δps−1
where b ∈ Q.
(iii) E[bi] = 1 for 1 ≤ i < m and b has the minimum polynomial over C:
bm+ β
m−1bm−1+ · · · + β0 = 0, βi∈ C.
Consider Q def.= Q/P , where P is a fixed but arbitrary minimal prime ideal of Q. Let αi, b, βj be the natural images of αi, b, βj in Q respectively, and δ the derivation of Q induced by δ. (We have Pδ ⊆ P by Theorem 1.2.) Obviously, we have the identities δps+ δps−1α1+ · · · + δαs = ad(b) and b
m
+ βm−1bm−1+ · · · + β0= 0. By (2) and (3) of Theorem 1.3, the derivations δpi, where 0 ≤ i < s, are mutually outer and the elements 1, b, . . . , bm−1 are C-independent. Write m = ptl with (l, p) = 1. Applying Theorems 2.2 and 2.3 respectively to the nilpotent derivations δ and ad(b) of the prime ring Q, we have all αi = 0 and (bpt− α)l = 0, where αdef.= −β
pt(l−1)/l.
This is true for any minimal prime ideal P of Q. It follows that all αi = 0 and (bpt
− α)l = 0. So δps
= ad(b) and b has the minimum polynomial (bpt
− α)l = 0, as asserted.
It is proved in [3] that, for a semiprime ring of prime characteristic p ≥ 2, the nilpo-tency of a derivation has the p-power expansion in the form n = nνpν+P
i>νnipi, where nν is odd and all ni (i > ν) are even. We sharpen this as follows:
Theorem 3.4. Let R be a semiprime ring of prime characteristic p ≥ 2, and δ a
nilpotent derivation with mδ(R) = lpv, where (l, p) = 1. Then nδ(R) is the greatest
integer ν ≤ (2l − 1)pv with the p-power expansion in the form ν = ν kpk+
P
i>kνipi,
where νk is odd and all νi (i > k) are even.
Proof. By Theorem 3.3, Q is a direct sum of finitely many rings Qi, 1 ≤ i ≤ q such that each Qi∩R is δi-homogeneous, where δiis the restriction of δ to Qi∩R. Observe that mδ(R) = mδ(Q) = max{mδi(Qi) | 1 ≤ i ≤ k}, nδ(R) = nδ(Q) = max{nδi(Qi) |
1 ≤ i ≤ k}, mδi(R ∩ Qi) = mδi(Qi) and nδi(R ∩ Qi) = nδi(Qi). Thus we may
assume that R is δ-homogeneous. We see easily that l here is the l in Theorem 3.2 and v here is equal to s + t, where s, t are as given in Theorem 3.2. Write l in the p-power expansion l = Pi≥0lipi. The nilpotency n given in Theorem 3.2 has the form ν described here. If (1) of Theorem 3.2 holds, then the nilpotency is just (2l − 1)pv and there is nothing to prove. So suppose (2) of Theorem 3.2 holds. Then
nδ(R) = ((2lu + 1)pu+ P
i>u2lipi)ps+t, where u ≥ 0 is the least integer such that 2lu+ 1 < p and such that 2li < p for all i > u. Write nδ in its p-power expansion
nδ(R) = P
the form described here and assume ν > nδ(R). It suffices to show ν > (2l − 1)pv: Let j be the greatest j such that νj > nj and νi = ni for i > j. If j < u + v, then
u + v > k and so νu+v is even. But nu+v = 2lu+ 1 is odd, contradicting νu+v = nu+v. If j ≥ u + v, then ν > u−1 X j=0 (p − 1)pj+v+ (2lu+ 1)pu+v + X i>u 2lipi+v =¡− 1 + 2(lu+ 1)pu+ X i>u 2lipi ¢ pv > (2l − 1)pv, where the last inequality follows from (lu+ 1)pu+
P
i>ulipi > l. This finishes our proof.
For a semiprime ring R of char R = 2, Chung and Luh [6] proved that the nilpo-tency of a nilpotent derivation must be a power of 2. We also have a sharper version for this:
Theorem 3.5. Let R be a semiprime ring of characteristic 2, and δ a nilpotent
derivation of R with mδ(R) = l2v, where (l, 2) = 1. Then nδ(R) is the unique 2-power integer between l2v and (2l − 1)2v.
Proof. By Theorem 3.4, nδ(R) is the greatest integer ≤ (2l − 1)pv. Therefore, it suffices to prove that nδ(R) is the least 2-power ≥ mδ(R) = l2v. As in the proof of Theorem 3.4, we may assume that R is δ-homogeneous. In the notation of Theorem 3.2, we observe that 2li < p = 2 implies li= 0. If (1) of Theorem 3.2 holds, then li = 0 for i > 0 and hence l = l0= 1. The nilpotency is thus (2l−1)2s+t = 2s+t= mδ(R), as
asserted. If (2) of Theorem 3.2 holds, then u is the greatest integer with lu−1 6= 0 and the nilpotency of δ is 2u+s+t. By the case assumption of (2), we also have 2l
j ≥ p = 2 for some j > 0. So u − 1 > 0 and the nilpotency 2u+s+t is thus the least 2-power
≥ l2s+t.
As an application of Theorem 3.3, we deduce [9, Theorem 5] in a slightly general-ized form:
Theorem 3.6. (Grzeszczuk [9]) Let R be a semiprime ring and δ, a nilpotent
deriva-tion of R. Suppose that, for any δ-ideal J 6= 0 of R, if the characteristic of J is a prime p ≥ 2, then p does not divide mδ(J) or nδ(J). Then δ = ad(b) for some b ∈ Q
with bmδ(R) = 0 and, moreover, bI + Ib ⊆ I for an essential ideal I of R.
Proof. For a prime p ≥ 2, let epbe the central idempotent of Q defined by the ideal Qp of Q generated by all additive p-torsion elements in Q. That is, Qp = epQ. We then
set e0= 1 −
W
pep. Then e0Q has characteristic zero. Replacing R by either e0Q ∩ R
or epQ ∩ R, we may assume that R is a semiprime ring of characteristic either 0 or a prime p. By Theorem 3.3, there exist finitely many orthogonal central idempotents
e1, . . . , ek in Q with e1+ · · · + ek = 1 such that each eiQ ∩ R is δi-homogeneous,
where δiis the restriction of δ to eiQ ∩ R. Note that eiQ is the symmetric Martindale quotient ring of eiQ ∩ R by [1, Proposition 2.3.14]. It suffices to prove our assertions for each δi. Replacing R by eiQ ∩ R, we may assume that R is δ-homogeneous as described in Definition 2. For the case char R = 0, let b, l be as described there. By Theorem 3.1, mδ(R) = l and nδ(R) = 2l − 1. So bmδ(R) = 0. For the case char R = p ≥ 2, let s, t, b, l, α be as described there. By Theorem 3.2, mδ(R) = lps+t and nδ(R) is as described there. Since p does not divide mδ(R) or nδ(R), we see that
s + t = 0. So b satisfies (b − α)l = 0. Noting that ad(b) = ad(b − α) and replacing b by b − α, we also have bl = bmδ(R) = 0. Now, we aim to find an essential ideal I such
that bI + Ib ⊆ I: We let
I def.= {y ∈ R | biy ∈ R for all i ≥ 1}.
Let y ∈ I. By induction on j, we show biybj ∈ R for all i, j ≥ 0: This is trivial if
j = 0. For j ≥ 1,
biybj = [biybj−1, b] + bi+1ybj−1 ∈ Rδ+ R = R.
This implies I = {y ∈ R | ybj ∈ R for all j ≥ 1}. So I is a two-sided ideal of R. Since
b ∈ Q, there is an essential ideal J such that biJ ⊆ R for 1 ≤ i < l and hence for all
i ≥ 0, since bl = 0. This shows that I is essential, as asserted.
Acknowledgements The authors are thankful to the referee for her/his useful suggestions and comments. This research was supported by National Science Council of Taiwan.
References
[1] K.I. Beidar, W.S. Martindale 3rd and A.V. Mikhalev, “Rings with Generalized Identities”, Marcel Dekker, Inc., New York–Basel–Hong Kong, 1996.
[2] K.I. Beidar and A.V. Mikhal¨ev, Orthogonal completeness and algebraic system, Uspekhi Mat. Nauk 40 (1985) 6, 79-115. English translation: Russian Math. Surveys, 40 (1985) 6, 51-95.
[3] L.O. Chung, Y. Kobayashi and J. Luh, Remark on nilpotency of derivations, Proc. Japan Acad. Ser. A 60(1984), 329-330.
[4] C.-L. Chuang, On compositions of derivations of prime rings, Proc. Amer. Math. Soc. 108 (1990), 647-652.
[5] L.O. Chung and J. Luh, Nilpotency of derivations, Canad. Math. Bull. 26 (1983), 341-346.
[6] L.O. Chung and J. Luh, Nilpotency of derivations. II, Proc. Amer. Math. Soc. 91(3) (1984), 357–358.
[7] T.S. Erickson, W.S. Martindale 3rd and J.M. Osborn, Prime nonassociative
alge-bras, Pacific J. Math. 60 (1975), 49-53.
[8] N.J. Fine, Binomial coefficients modulo a prime, Amer. Math. Monthly 54 (1947), 589-592.
[9] P. Grzeszczuk, On nilpotent derivations of semiprime rings, J. Algebra 149 (1992), 313-321.
[10] V.K. Kharchenko, Differential identities of prime rings, Algebra i Logika 17 (1978) 2, 220-238. English translation: Algebra and Logic, 17 (1978) 2, 154-168.
[11] V.K. Kharchenko, Differential identities of semiprime rings, Algebra i Logika 18 (1979) 1, 86-119. English translation: Algebra and Logic, 18 (1979) 1, 58-80. [12] V.K. Kharchenko and A.Z. Popov, Skew derivations of prime rings, Comm.
Alge-bra 20 (1992) 11, 3321-3345.
[13] T.-K. Lee, Semiprime rings with differential identities, Bull. Inst. Math. Acad. Sinica 20 (1992), 27-38.
[14] W.S. Martindale, 3rd, Prime rings satisfying a generalized polynomial identity, J. Algebra 12 (1969), 576-584.
[15] W.S. Martindale, 3rd and C. R. Miers, On the iterates of derivations of prime
rings, Pacific J. Math. 104 (1983) 1, 179-190.
IDENTITIES WITH A SINGLE SKEW DERIVATION Chen–Lian Chuang and Tsiu–Kwen Lee
Department of Mathematics National Taiwan University
Taipei 106, TAIWAN E–mail: [email protected]
E–mail: [email protected]
Abstract. Let R be a prime ring with extended centroid C and δ, a continuous skew derivation of R. We define the notion of K–polynomials which, in the case that δ is an ordinary derivation, reduces to polynomials of the form xδpn
+α1xδ
pn−1
where αi ∈ C. It is shown that all generalized identities with δ are consequences of GPIs of R and an identity in the form ψ(x) = xσm
b−bx, where ψ(x) is a K–polynomial
of minimal possible order m.
† 2000 Mathematics Subject Classification. 16W20, 16W25, 16W55.
‡ Key words and phrases. Automorphism, prime ring, skew derivation, GPI, K–
polynomial.
1. Results
An associative ring R is called prime if for any a, b ∈ R, aRb = 0 implies a = 0 or b = 0. Throughout here, R will always be an associative prime ring. Let Q and
RF stand for respectively the symmetric and the left Martindale quotient rings of R. (See [2] for definitions.) The center C of Q is the same as that of RF and is called the extended centroid of R. A map δ: R → R is called additive if (x + y)δ = xδ+ yδ for x, y ∈ R. By a derivation of R, we mean an additive map δ: R → R satisfying (xy)δ = xδy + xyδ for x, y ∈ R. Given b ∈ R, the map ad
b: x ∈ R 7→ xb − bx obviously defines a derivation, called the inner derivation defined by b. We call a derivation
outer, if it is not of this form. To analyze derivations of R, as shown in Kharchenko’s
theory [7, 8], we have to work in the larger ring Q. We introduce a topology on Q, which we may call the ideal topology, by endowing x ∈ Q with the neighborhood system consisting of x plus a nonzero two-sided ideal of R. Let g: Q → Q be an additive map. We call g a continuous map of R if it is a continuous map with respect to the ideal topology. It is easy to check that a derivation δ of Q is continuous if and only if Iδ ⊆ R for some nonzero ideal I of R. Inner derivations of Q are obviously continuous. Any derivation of R can be uniquely extended to a continuous derivation of R. A continuous derivation of R is called X–inner or X–outer according as it is equal to an inner derivation of Q or not. A continuous derivation δ of R is said to be C–algebraic modulo inner derivations if there exist b ∈ Q and β1, · · · , βn−1 ∈ C such that xδn+ β1xδ n−1 + β2xδ n−2 + · · · + βn−1xδ = xb − bx
for all x ∈ R. We call n the δ–order or simply the order of the above algebraic depen-dence. Let σ be an automorphism of Q. We call σ a bi–continuous automorphism of
R if both σ and σ−1 are continuous. Equivalently, there exist nonzero ideals I and J of R such that J ⊆ Iσ ⊆ R. Any automorphism of R can be uniquely extended to an automorphism of Q and gives rise to a bi–continuous automorphism of Q. Let A(R) be the set of bi–continuous automorphisms of R. It is shown in [9] that A(R) forms a group. We say that g1, g2, . . . ∈ A(R) are mutually outer if g−1i gj is not X-inner for i 6= j. All generalized polynomials considered in the paper have coefficients in
RF. Kharchenko’s powerful theory of differential identities [7, 8] yields the following thorough analysis of GPIs with a single continuous derivation:
Theorem. (Kharchenko) Let R be a prime ring and δ, an X–outer continuous
derivation of R. If δ is not C–algebraic modulo inner derivations, then any GPI of the form ϕ(xδjgk
i ), where gk ∈ A(R) are mutually outer, yields the GPI ϕ(zijk),
where zijk are distinct indeterminates. If δ is C–algebraic modulo inner derivations,
the form
xδpm + α1xδ
pm−1
+ · · · + αm−1xδ = xb − bx
for some α1, . . . , αm−1 ∈ C, and (3) any GPI of the form ϕ(xδ
jg k
i )j<pm, where
gk ∈ A(R) are mutually outer, yields the GPI ϕ(zijk)j<pm, where zijk are distinct
indeterminates.
This theorem is actually a combined consequence of several theorems, all variously formulated for various purposes. For the sake of completeness, we provide a proof below. Let us recall some notions: A set S of continuous derivations of R is called outer independent if for any δ1, . . . , δn, any α1, . . . , αn ∈ C and any b ∈ Q, the condition α1xδ1 + · · · + αnxδn = xb − bx for all x ∈ R implies α1 = · · · = αn = 0. We say the set S is ordered if S is endowed with a linear order <. By a regular word in an ordered independent set S of X–outer continuous derivations, we mean an expression of the form:
δs1
1 δs22· · · δnsn,
where (i) δi∈ S are such that δ1 < δ2 < · · · < δn and where (ii) each si< char R in case of char R = p ≥ 2.
Proof of the Theorem: We first show that all δi occurring in ϕ are regular words in an ordered outer independent set: If char R = 0, then all δn are regular words in the ordered outer independent singleton set {δ}. Assume char R = p ≥ 2. If
δ, δp, δp2, . . . are outer independent, then we order them by δ < δp< δp2 < · · · . Any
k ≥ 0 can be written as k = k0 + k1p + k2p2+ · · · , where each 0 ≤ ki < p. So
δk = δk0(δp)k1(δp2)k2· · · is a regular word in this ordered outer independent set. If
δ, δp, δp2
, . . . are dependent modulo inner derivations, let n be the largest such that δ, δp, . . . , δpn−1
are independent modulo inner derivations. We order δ, . . . , δpn−1
by δ < · · · < δpn−1 . For 0 ≤ k < pn = m, we write k = k 0+ k1p + · · · + kn−1pn−1, where each 0 ≤ ki < p. So δk = δk0(δp)k1· · · (δp n−1
)kn−1 is a regular word in the ordered
outer independent set {δ, . . . , δpn−1
}. We recall two more notions: An automorphism h of Q is called Frobenius if, in the case of char R = 0, αh = α for α ∈ C and if, in the case of char R = p ≥ 2, αh = αpn
for α ∈ C, where n is an fixed integer, which may be zero, positive or negative. Automorphisms h1, h2, . . . are said to be
strongly independent if hih−1j is not Frobenius for i 6= j. We then show that the mutually outer automorphisms gk are strongly independent: Assume on the contrary that hdef.= gigj−1 is Frobenius for some i 6= j. If char R = p ≥ 2 and there exists n 6= 0 such that αh = αpn
for all α ∈ C, then C = Cp and hence Cδ = (Cp)δ = 0.
If ϕ is trivial, then there is nothing to prove. Thus we may assume that ϕ is nontrivial. In view of Corollary p.371 [3], R is a GPI–ring and so δ is X–inner by a result in p.68 [8]. This contradicts our assumption. If αh = α for all α ∈ C, then h is X–inner by a result in p.140 [6]. This is absurd again, since g1, g2, . . . are mutually
outer. So gk occurring in ϕ are indeed strongly independent. The desired assertion of our Theorem now follows immediately from Theorem 2 of [4] (cited as Main Theorem in the abstract).
