ISRAEL JOURNAL OF MATHEMATICS 175 (2010), 157–178 DOI: 10.1007/s11856-010-0007-z

HIGHER DERIVATIONS OF ORE EXTENSIONS

BY

Chen-Lian Chuang and Tsiu-Kwen Lee Department of Mathematics, National Taiwan University, Taipei 106, Taiwan e-mail: [email protected], [email protected] AND

Cheng-Kai Liu Department of Mathematics, National Changhua University of Education Changhua 500, Taiwan e-mail: [email protected] AND

Yuan-Tsung Tsai Department of Applied Mathematics, Tatung University, Taipei 104, Taiwan e-mail: [email protected] ABSTRACT def.

Let R be a prime ring and δ a derivation of R. Divided powers Dn = d 1 dn of ordinary differentiation dx form Hasse–Schmidt higher derivan! dxn tions of the Ore extension (skew polynomial ring) R[x; δ]. They have been used crucially but implicitly in the investigation of R[x; δ]. Our aim is to explore this notion. The following is proved among others: Let Q be the def.

left Martindale quotient ring of R. It is shown that S = Q[x; δ] is a quasiinjective (R, R)-module and that any (R, R)-bimodule endomorphism of S can be uniquely expressed in the form θ(f ) =

∞ X

ζn Dn (f )

for f ∈ Q[x; δ],

n=0

where ζn ∈ CS (R), the centralizer of R in S. As an application, we also use the Ore extension R[x; δ] to deduce Kharchenko’s theorem for a single derivation. These results are extended to the Ore extension R[X; D] of R by a sequence D of derivations of R.

Received December 5, 2007 and in revised form April 12, 2008

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1. Introduction Throughout, R is a prime ring and Q its left Martindale quotient ring. The center of Q, denoted by C, is called the extended centroid of R. We refer this notation to Beidar et al. [2] for details. By a derivation of R, we mean a map δ : R → R such that δ(a + b) = δ(a) + δ(b) and δ(ab) = δ(a)b + aδ(b) for all a, b ∈ R. Given b ∈ R, the map adb : r ∈ R 7→ [b, r] obviously defines a derivation, called the inner derivation defined by b. Let Der(R) denote the set of derivations of R and X be a family of non-commuting indeterminates. Given a map φ : X → Der(R), we associate each x ∈ X with the derivation δ = φ(x). We enumerate X as a (finite or transfinite) sequence x0 , x1 , . . . and let D be the corresponding sequence of derivations δ0 , δ1 , . . ., where δi = φ(xi ). The map φ is thus implicitly encoded in the two sequences X and D. Let R[X; D] denote the ring of polynomials in non-commuting indeterminates x ∈ X and with coefficients in R subjected to the commutation rule xa = ax + δ(a)

for a ∈ R, x ∈ X and δ = φ(x).

The algebraic structure of R[X; D] is clearly determined by the ring R and the sequence D of its derivations. We hence call R[X; D] the Ore extension of R by D. For basic properties of R[X; D], we refer the reader to Burkov [3]. We stress here that the indeterminates x ∈ X do not commute with each other. We also stress that the map φ may not be injective. So distinct x ∈ X can be associated with the same derivation via φ. If φ(X) = 0 then R[X; D] is traditionally denoted by RhXi, the ring freely generated by X over R. If R happens to be a field then RhXi is the free R-algebra generated by X. If X consists of a single element x, we denote R[X; D] by R[x; δ], where δ = φ(x). The ring R[x; δ] will frequently be our example. This Ore extension has been most extensively investigated in various directions. See, for example, [1], [4]–[8], [10], [11], [16]–[19]. Derivations of R can be uniquely extended to Q. We can hence form Q[X; D] as well. Algebraic properties of R[X; D] are closely related to those of Q[X; D]. Our aim is to investigate the following notions: def.

Definition: Set S = Q[X; D] for brevity. Let V be an (R, R)-subbimodule of S. An additive map θ : V → S is called an (R, R)-bimodule map if θ(rv) = rθ(v)

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and θ(vr) = θ(v)r for r ∈ R and v ∈ V . If V = S then θ is called an (R, R)-bimodule endomorphism of S. Let Λ be the set of all (R, R)-bimodule def. endomorphisms of S. Set CS (R) = {f ∈ S | f r = rf for all r ∈ R}. def.

Example 1: Let S = Q[x; δ]. For n ≥ 0, define Dn : S → S by ! X X i i Dn ai x = ai xi−n . n i≥0

i≥n

Clearly, Dn is a left R-module endomorphism. For a ∈ Q, xm a = Pm m
m−i (a)xi . So i=0 i δ X  X    m   m m m−i m i i−n Dn (xm a) = Dn δ (a)xi = δ m−i (a) x , i i n i=0 i=n    X  m  m m−n m m − n m−i m Dn (x )a = x a= δ (a)xi−n . n n i=n i − n Since

      m m−n m i m! = for i ≥ n, = (m − i)!(i − n)!n! n i−n i n

we have Dn (xm a) = Dn (xm )a. So Dn is also a right R-module endomorphism. P Hence Dn ∈ Λ. This amounts to saying that if f = j≥0 xj bj , where bj ∈ Q,
P then Dn (f ) = j≥n nj xj−n bj . Note that   m m−n 1 dn m (x ) for n, m ≥ 0, Dn (xm ) = x = n! dxn n
def. 1 dn = 0 if m < n. To be suggestive, we also denote Dn by n! where m n dxn and call it the divided n-th power of the differentiation d/dx. The importance of (R, R)-bimodule endomorphisms comes partially from the following simple observation: Lemma 1: If ψ ∈ Λ then ψ(CS (R)) ⊆ CS (R). Proof. Let f ∈ CS (R) and r ∈ R. By the R-linearity, we have ψ(f )r − rψ(f ) = ψ(f r) − ψ(rf ) = ψ(f r − rf ) = 0. This property was implicitly used in the crucial computation of CS (R) ([1, proof of Theorem 4, p. 97]) in the fundamental work due to Amitsur in the following manner: If f ∈ CS (R) has the minimal degree> 0 then Dn (f ) is a

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constant for n > 0. The same argument reappeared in some important later work, for example [15, proof of Lemma 1.7, p. 1255] and [17, proof of Proposition 2.4, p. 16]. We explain this in detail in the following: P i Example 2: In the notation of Example 1, assume that f = m i=0 ai x ∈ CS (R) has the minimal degree m > 0. By Lemma 1, m   X i Dn (f ) = ai xi−n ∈ CS (R). n i=n
By the minimality of f , all Dn (f ) are constants for n > 0. That is, ni ai = 0 for all i, n with m ≥ i > n > 0. In the case of charR = 0, this implies m = 1. So f = a1 x + a0 . The condition [f, r] = 0 for r ∈ R yields a1 ∈ C and the identity a1 δ(r) + ada0 (r) = 0 for r ∈ R. In the case of charR = p > 0, this Ps i implies ai = 0 for i not a p-power. We may thus write f = i=0 bi xp + a0 , def.

where m = ps and bi = api for 0 ≤ i ≤ s. For r ∈ R, 0 = [f, r] =

s X

i

[bi , r]xp + [a0 , r] +

i=0

=

s X

s X

i

bi [xp , r]

i=0

i

[bi , r]xp + [a0 , r] +

i=0

s X

i

bi δ p (r).

i=0

So [bi , R] = 0 and hence bi ∈ C. The above gives the identity s X

i

bi δ p (r) + ada0 (r) = 0

i=0

for r ∈ R. The leading coefficient of f may be assumed to be 1. The monic f is called the minimal semi-invariant polynomial and the corresponding identity  δ(r) + ad (r) = 0 if charR = 0 or a0 P s i s−1 p p δ (r) + bi δ (r) + ada (r) = 0 if charR = p > 0 i=0

0

is called the minimal quasi-algebraic relation of δ. (See [16].)

2. Higher derivations It turns out that (R, R)-bimodule endomorphisms are closely related to generalization of derivations. By a Hasse–Schmidt higher derivation of a ring R, we

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mean a finite or infinite sequence of maps {Dn }n≥0 from R into R with D0 = 1, the identity map, such that for a, b ∈ R and n ≥ 0,

Dn (a + b) = Dn (a) + Dn (b)

and Dn (ab) =

n X

Di (a)Dn−i (b).

i=0

By abuse of language, we also call each Dn a higher derivation. Example 3: The (R, R)-bimodule endomorphisms Dn defined in Example 1 form Pn a Hasse–Schmidt higher derivation. To show Dn (f g) = k=0 Dk (f )Dn−k (g) for P P f, g ∈ Q[x; δ], write f = i ai xi and g = j xj bj , where ai , bj ∈ Q. Then f g = P i+j bj . Since Dk are (R, R)-bimodule endomorphisms, it suffices to show ij ai x
j

Pn Pn i+j . Dn (x ) = k=0 Dk (xi )Dn−k (xj ). This is clear since i+j = k=0 ki n−k n This connection of higher derivations with (R, R)-bimodule endomorphisms is not accidental at all. To explore it, we have to generalize Hasse–Schmidt’s notion slightly. Definition: By a higher derivation of order n ≥ 1 of a ring R, we mean a map δ : R → R satisfying the following: For each 0 < i < n, there exist higher derivations δi and δi0 of orders i and n − i respectively such that for a, b ∈ R,

δ(a + b) = δ(a) + δ(b) and δ(ab) = δ(a)b +

n−1 X

δi (a)δi0 (b) + aδ(b).

i=1

If δ, δi above all vanish on a subring T of R then δ is clearly an (T, T )-bimodule endomorphism. This gives the connection between higher derivations and bimodule endomorphisms. We will see that the simplest (R, R)-bimodule endomorphisms of R[X; D] are higher derivations vanishing on R (Theorem 3). Clearly, higher derivations of order one are merely ordinary derivations. Let Ω be the set of words in X. Formally, Ω is the free monoid generated by X. The total number of letters in a word W ∈ Ω is called its length and denoted by lh(W ). Let W = y1 y2 · · · yn ∈ Ω, where the yi ∈ X, be a word of length n. Given a subset S = {i1 , . . . , is } of {1, . . . , n}, where 1 ≤ i1 < i2 < · · · < is ≤ n, we set def.

WS = yi1 yi2 · · · yis .

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def.

We call WS an expansion subword of W . Let S 0 = {1, . . . , n} \ S be the complement of S in {1, . . . , n}. For any word B ∈ Ω, we define X def. WS 0 , DB (W ) = S⊆{1,...,n}: WS =B

P

where the summation above ranges over all subsets S with WS = B. There may be more than one subset S with WS = B or none at all. In the latter case, we have DB (W ) = 0. The empty word is denoted as 1 and D1 is the identity P map. Given f = i ai Ai ∈ Q[X; D], where ai ∈ Q and Ai ∈ Ω, we define X def. DB (f ) = ai DB (Ai ). i


Example 4: For any B ∈ Ω, DB (B) = 1. For any x ∈ X, Dxk (xn ) = nk xn−k
for n ≥ k, since the number of k-element subsets of {1, . . . , n} is nk . So the 1 dn divided n-th power n! dxn of the differentiation d/dx is a special instance of our more general notion DB here. For x, y ∈ X, we have Dxy (x2 y 2 xy) = 4xyxy + 2xy 2 x + x2 y 2 and Dxyx (x2 y 2 xy) = 4xy 2 .
We consider the composition DA DB for A, B ∈ Ω. For W ∈ Ω, let A,WB be the number of subsets S of {1, 2, . . . , lh(W )} such that WS = A and WS 0 = B. Clearly, we have X  W  (◦) DA DB = DW . A, B W ∈Ω

This is a finite sum since there are only finitely many W ∈ Ω with Clearly, we also have (•)

W A, B

DA DB = DB DA




= 6 0.

for any A, B ∈ Ω.

n+m
n+m and xxn , xm = n+m Example 5: The only W ∈ Ω with xnW m . , xm 6= 0 is x
By (◦), Dxn Dxm = n+m m Dxn+m . This can be verified directly as follows:   1 dn 1 dm (n + m)! 1 dn+m n+m Dx n Dx m = = = Dxn+m . n! dxn m! dxm n!m! (n + m)! dxn+m n Similarly, we have Dvx Dyz = Dvxyz + Dvyzx + Dyzvx + Dvyxz + Dyvxz + Dyvzx , Dxy Dyx = 2Dyx2 y + Dyxyx + 2Dxy2 x + Dxyxy .

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The map DB : Q[X; D] → Q[X; D] is a left R-module endomorphism by definition. We will see that DB is also a right R-module endomorphism. But this seems rather difficult to prove directly as in Examples 1 and 3, mainly because of the complication of writing Aa, where A ∈ Ω and a ∈ Q, in the P standard form i ai Ai , where ai ∈ Q and Ai ∈ Ω. The following pleasant property is hence proved in a quite different way. Lemma 2: For B ∈ Ω and f, g ∈ Q[X; D], we have X DB (f g) = DB1 (f )DB2 (g), B1 B2 =B

where the summation ranges over all ordered pairs B1 , B2 ∈ Ω with B1 B2 = B. Proof. Let Q{X} be the unital ring freely generated by the ring Q and the ` noncommuting set X. Formally, Q{X} is the coproduct Q Z ZhXi, where Z is the ring of integers and ZhXi the unital free ring generated by the set X [2, p. 24]. Each element of Q{X} can be written uniquely as a finite sum of products in the form τ = a0 y1 a1 · · · yn an ∈ Q{X}, where ai ∈ Q and yi ∈ X. Consider the product τ above. Given a subset S of {1, . . . , n}, define τS to be the product obtained by substituting 1 for each yi with i ∈ S. Write S = {i1 , . . . , is }, where 1 ≤ i1 < i2 < · · · < is ≤ def. n and set τ (S) = yi1 yi2 · · · yis ∈ Ω. Given a word B ∈ Ω, define a map ∆B : Q{X} → Q{X} as follows: X def. τS , ∆B (τ ) = S⊆{1,...,n}: τ (S)=B

where the summation ranges over all subsets S of {1, . . . , n} with τ (S) = B. If there exist no such subsets S with τ (S) = B then we set ∆B (τ ) = 0. Given P f ∈ Q{X}, write f = i τi , where τi are distinct products and define X def. ∆B (f ) = ∆B (τi ). i

Given f, g ∈ Q{X}, we clearly have X ∆B (f g) =

B1 B2 =B

∆B1 (f )∆B2 (g),

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where the summation ranges over all ordered pairs B1 , B2 ∈ Ω with B1 B2 = B. Let I be the ideal of Q{X} generated by xr − rx − δ(r), where x ∈ X, r ∈ Q and δ = φ(x). Clearly, ∆B (I) ⊆ I. So ∆B induces a map from Q{X}/I to Q{X}/I. By definition Q{X}/I is isomorphic to Q[X; D]. Identify Q{X}/I with Q[X; D]. We see that the map induced by ∆B on Q{X}/I is equal to the map DB defined on Q[X; D]. The claimed equality for DB thus follows from the corresponding equality for ∆B . Remark: If lh(B) = n, then DB is a higher derivation of order n. 3. (R, R)-bimodule endomorphisms Recall that Ω denotes the set of words in X. Our first result is the following Theorem 3: Let R be a prime ring and let V be an (R, R)-bisubmodule of def.

S = Q[X; D]. Suppose that θ : V → S is an (R, R)-bimodule map. Then for each A ∈ Ω, there exist ζA ∈ CS (R) such that X θ(f ) = ζA DA (f ) for all f ∈ V. A∈Ω

Moreover, if V = S, this expression is unique.

For the proof of Theorem 3, we need some important notions due to Kharchenko [12, 14]. We recall them below and meanwhile fix our notation: Let Qop denote the ring opposite to Q and let Z be the ring of integers. EleP ments of the tensor product Q ⊗Z Qop admit the form i ri ⊗ ri0 , where ri ∈ Q P and ri0 ∈ Qop . For f ∈ Q[X; D] and β = i ri ⊗ ri0 ∈ Q ⊗Z Qop , we define X def. f ·β = ri0 f ri . i

For f ∈ Q[X; D] and β, γ ∈ Q ⊗Z Qop , we obviously have f · (βγ) = (f · β) · γ. Let L denote the subring of Q ⊗Z Qop generated by the elements of the form r ⊗ r0 for all r, r0 ∈ R ∪ {1}. Thus we can regard Q[X; D] as a right L-module. For a subset Y ⊆ Q[X; D], we define def.

Y ⊥ = {β ∈ L | f · β = 0 for all f ∈ Y }. Note that Y ⊥ is an (R, R)-submodule of L. On the other hand, for U ⊆ L, we define def. U ⊥ = {f ∈ Q[X; D] | f · β = 0 for all β ∈ U }.

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def.

Lemma 4: For Y ⊆ S = Q[X; D] and finitely many a1 , . . . , an ∈ Q, we have (Y + Ca1 + · · · + Can )⊥⊥ = Y ⊥⊥ + a1 CS (R) + · · · + an CS (R). def.

Proof. Set V = Ca1 + · · · + Can . For β ∈ (Y + V )⊥ , we have Y ⊥⊥ · β = 0 and V CS (R) · β = (V · β)CS (R) = 0. So we have the inclusion (Y + V )⊥⊥ ⊇ Y ⊥⊥ + V CS (R). For the reverse inclusion, we proceed by induction on n = dimC V . Assume n = 1. Write V = aC for some 0 6= a ∈ Q. Let f ∈ (Y + aC)⊥⊥ . For β ∈ Y ⊥ , we define the map θ : a · Y ⊥ → Q[X; D] by θ : a · β 7→ f · β

for β ∈ Y ⊥ .

This is well-defined: For β ∈ Y ⊥ , if a · β = 0 then β ∈ Y ⊥ ∩ a⊥ = (Y + aC)⊥ and so f · β = 0, since f ∈ (Y + aC)⊥⊥ . Enumerate all expansion subwords of words occurring in f as a finite sequence A0 , A1 , . . .. We may write X def. θ(a · β) = f · β = θi (a · β)Ai , i

def.

where all θi (a · β) ∈ Q. Clearly, θ(a · β) = f · β is an (R, R)-bimodule map. Particularly, for r ∈ R, X X θi (r(a · β))Ai = θ(r(a · β)) = rθ(a · β) = rθi (a · β)Ai . i

i

So θi (r(a·β)) = rθi (a·β). Each θi : a·Y ⊥ → Q is thus a left R-module map. Pick a two-sided ideal 0 6= I of R such that If ⊆ R[X; D]. For β ∈ Y ⊥ ∩ (R ⊗ I op ), def.

we have θ(a · β) = f · β ∈ R[X; D]. Hence each θi is a left R-module map from
the two-sided ideal a · Y ⊥ ∩ (R ⊗ I op ) of R into R. So there exist qi ∈ Q such that θi (a · β) = (a · β)qi for all β ∈ Y ⊥ ∩ (R ⊗ I op ) and hence for all β ∈ Y ⊥ . Define X def. h = qi Ai ∈ Q[X; D]. i

def.

We thus have θ(a · β) = f · β = (a · β)h for all β ∈ Y ⊥ . Since θ is an (R, R)-bimodule map, we have (a · β)rh = θ (a · β)r) = θ(a · β)r = (a · β)hr

for r ∈ R.

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So (a · β)(rh − hr) = 0 for β ∈ Y ⊥ and r ∈ R. This implies rh = hr for r ∈ R. That is, h ∈ CS (R). Then we have (a · β)h = (ah) · β. But for β ∈ Y ⊥ , def.

f · β = θ(a · β) = (a · β)h = (ah) · β. So (f − ah) · β = 0 for β ∈ Y ⊥ . That is, f − ah ∈ Y ⊥⊥ and so f ∈ ah + Y ⊥⊥ ⊆ aCS (R) + Y ⊥⊥ , as asserted. We have thus shown the case dimC V = 1. Let n > 1. Pick a C-subspace U of V with dimC U = n−1. Write V = U +aC def. for some 0 = 6 a ∈ V . Set Z = Y + U . Applying the case n = 1, we have (Z + aC)⊥⊥ = Z ⊥⊥ + aCS (R). By the induction hypothesis, Z ⊥⊥ = (Y + U )⊥⊥ = Y ⊥⊥ + U CS (R). It follows that (Y + V )⊥⊥ = (Z + aC)⊥⊥ = Z ⊥⊥ + aCS (R) = Y ⊥⊥ + U CS (R) + aCS (R) = Y ⊥⊥ + V CS (R). For f ∈ Q[X; D], lh(f ) is defined as the maximum among lh(A), where A ranges over all distinct words occurring in f . We are now ready for def.

Proof of Theorem 3. For n ≥ 0, let Sn = {f ∈ Q[X; D] | lh(f ) ≤ n}, which is clearly an (R, R)-subbimodule of S. We also set def. ˜ n def. = {A ∈ Ω | lh(A) = n}. Ωn = {A ∈ Ω | lh(A) ≤ n} and Ω

Elements of Sn are Q-linear combinations of words in Ωn . Given f ∈ Sn , write (†)

f = a1 A1 + · · · + ak Ak + terms with words of length < n,

˜ n , and define where A1 , . . . , Ak are distinct words in Ω f˜ = a1 A1 + · · · + ak Ak . def. We call f˜ the leading part of f . Set S˜n = {f˜ | f ∈ Sn }. We make S˜n an P ˜ n, (R, R)-bimodule as follows: For g = i bi Bi ∈ S˜n where bi ∈ Q and Bi ∈ Ω we define X def. g∗β = (bi · β)Bi . i

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(Actually, S˜n is naturally isomorphic to the quotient module Sn /Sn−1 .) Let f ∈ Sn be as given by (†). For β ∈ L, we see that f · β = a1 · βA1 + · · · + ak · βAk + terms with words of length < n. So we have fg · β = f˜ ∗ β.

(‡)

That is, the map f ∈ Sn 7→ f˜ ∈ S˜n is an (R, R)-bimodule map. def. Set Vn = Sn ∩ V . Clearly, Vn is an (R, R)-subbimodule of S. We proceed by induction on n to show the existence of ζA ∈ CS (R) for A ∈ Ωn such that X θ(f ) = ζA DA (f ) for all f ∈ Vn . A∈Ωn

For A ∈ Ω with lh(A) > n, we have DA (Vn ) = 0 and hence ζA DA does not P affect the value of A∈Ω ζA DA on Vn . The assertion then follows. For n = 0, let a ∈ V0 and β ∈ L. If a · β = 0, then θ(a) · β = θ(a · β) = 0. Thus θ(a) ∈ a⊥⊥ . By Lemma 4, a⊥⊥ = aCS (R), so θ(a) = aζa for some ζa ∈ CS (R). For b ∈ V0 , we write θ(b) = bζb , where ζb ∈ CS (R). Recall that a · L and b · L are both nonzero (R, R)-subbimodules of Q. Let β, β 0 ∈ L such that a · β = b · β 0 ∈ (a · L) ∩ (b · L). Then (a · β)ζa = θ(a · β) = θ(b · β 0 ) = (b · β 0 )ζb . So ((a · L) ∩ (b · L))(ζa − ζb ) = 0, implying that ζa = ζb . Hence there exists ζ1 ∈ CS (R) such that θ(a) = ζ1 a = ζ1 D1 (a) for all a ∈ V0 . This proves the case for n = 0. For n > 0, as the induction hypothesis, we assume that ζA ∈ CS (R) for A ∈ Ωn−1 have been found such that X θ(f ) = ζA DA (f ) for all f ∈ Vn−1 . A∈Ωn−1

Define θn : V → S by θn (f ) = θ(f ) −

X

ζA DA (f ).

A∈Ωn−1

Clearly, θn is also an (R, R)-bimodule map such that θn (f ) = 0 for all f ∈ Vn−1 . def. Set V˜n = {f˜ | f ∈ Vn , lh(f ) = n}, which is clearly an (R, R)-bisubmodule of S˜n . We define the map θ˜n : f˜ ∈ V˜n 7→ θn (f )

for f ∈ Vn .

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This is well-defined: If f˜ = g˜ for f, g ∈ Vn then f − g ∈ Vn−1 and hence θn (f − g) = 0. By (‡), θ˜n is an (R, R)-bimodule map. Pick an (R, R)-bimodule U of S˜n which is maximal with respect to the property U ∩ V˜n = 0. Then U ⊕ V˜n is essential in S˜n in the sense that any nonzero (R, R)-bimodule of S˜n intersects U ⊕ V˜n nontrivially. Extend the bimodule map θ˜n to U ⊕ V˜n by setting u + v 7→ θ˜n (v) for u ∈ U ˜ n , the bi-submodule RA of S˜n intersects U ⊕ V˜n and v ∈ V˜n . Given A ∈ Ω nontrivially. So there is a two-sided ideal I 6= 0 of R such that IA ⊆ U ⊕ V˜n . For a ∈ I and β ∈ a⊥ , we have
θ˜n (aA) · β = θ˜n (aA ∗ β) = θ˜n (a · β)A = 0.

By Lemma 4, θ˜n (aA) ∈ a⊥⊥ = aCS (R). Let ζ ∈ CS (R) be such that θ˜n (aA) = aζ. For β ∈ L, we have

(a · β)ζ = (aζ) · β = θ˜n (aA) · β = θ˜n (aA) ∗ β = θ˜n (a · β)A .

For another a0 ∈ I, let ζ 0 ∈ CS (R) be such that θ˜n (a0 A) = a0 ζ 0 . Analogously,
(a0 · β 0 )ζ 0 = θ˜n (a0 · β 0 )A for β 0 ∈ L. If a · β = a0 · β 0 then

(a · β)ζ = θ˜n (a · β)A = θ˜n (a0 · β 0 )A = (a0 · β)ζ 0 .
This implies (a · L) ∩ (a0 · L) (ζ − ζ 0 ) = 0. But (a · L) ∩ (a0 · L) is a nonzero two-sided ideal of R. So ζ = ζ 0 . We have thus shown that ζ is independent of the choice of a ∈ I. Write ζ, I respectively as ζA , IA to denote their dependence P on A. So θ˜n (aA) = aζA for all a ∈ IA . Let g = i ai Ai ∈ U ⊕ V˜n . Pick a T two-sided ideal J 6= 0 of R such that Jai ∈ i IAi . For r ∈ J, we have ! X X X ˜ ˜ ˜ rθn (g) = θn (rg) = θn rai Ai = rai ζA = r ai ζA . i

i

i

i

i

P That is, r θ˜n (g) − i ai ζAi ) = 0 for r ∈ J. Since J is a nonzero two-sided ideal P P P of R, it follows that θ˜n (g) = i ai ζAi . Note that i ai ζAi = A∈Ω˜ n ζA DA (g). For f ∈ Vn , we have X θn (f ) = θ˜n (f˜) = ζA DA (f˜). ˜n A∈Ω

But f − f˜ ∈ Sn−1 . So

P

ζA DA (f − f˜) = 0, that is, X X ζA DA (f ) = ζA DA (f˜). ˜n A∈Ω

˜n A∈Ω

˜n A∈Ω

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Combining all these, we have thus shown that for f ∈ Vn , X θn (f ) = θ˜n (f˜) = ζA DA (f ). ˜n A∈Ω

It follows that X X X θ(f ) = ζA DA (f ) + ζA DA (f ) = ζA DA (f ) for f ∈ Vn . A∈Ωn−1

˜n A∈Ω

A∈Ωn

This completes our induction. For the uniqueness, assume that there exists another ηA in CS (R) such that P θ(f ) = A∈Ω ηA DA (f ) for all f ∈ S. We prove that ηA = ζA for all A ∈ Ω by induction on m = lh(A). For m = 0, let f = 1. We have θ(f ) = ζ1 = η1 . P For lh(A) = m > 0, set f = A. We have θ(f ) = ζA + B∈Ωn−1 ζB DB (A) = P ηA + B∈Ωn−1 ηB DB (A). Since ζB = ηB for all B ∈ Ωn−1 , we have ζA = ηA . This proves the uniqueness. The proof is now complete. def.

Corollary 5: The (R, R)-bimodule S = Q[X; D] is quasi-injective in the sense that any (R, R)-bimodule map from an (R, R)-bisubmodule into S can be extended to an (R, R)-bimodule endomorphism of S. Lemma 4 is interesting in itself and we extend it to its full generality. Theorem 6: Let Y be a subset of Q[X; D] and f1 , . . . , fn ∈ Q[X; D]. Then (Y + Cf1 + · · · + Cfn )⊥⊥ = Y ⊥⊥ + Λf1 + · · · + Λfn , where Λ is the ring of all (R, R)-bimodule endomorphisms of Q[X; D]. Proof. For β ∈ (Y + Cf1 + · · · + Cfn )⊥ and θ ∈ Λ, we have Y ⊥⊥ · β = 0 and θ(fi ) · β = θ(fi · β) = 0 for i = 1, . . . , n. So we have (Y + Cf1 + · · · + Cfn )⊥⊥ ⊇ Y ⊥⊥ + Λf1 + · · · + Λfn . For the reverse inclusion, we prove it by induction on n. First n = 1, let g ∈ (Y + Cf1 )⊥⊥ . Then for β ∈ Y ⊥ , f1 ·β = 0 implies g ·β = 0. The set f1 ·Y ⊥ is an (R, R)-bisubmodule of Q[X; D]. We define θ : f1 ·Y ⊥ → g·Y ⊥ by θ(f1 ·β) = g·β. Thus θ is well-defined and is an (R, R)-bimodule endomorphism. By Corollary 5, θ can be extended to an (R, R)-bimodule endomorphism of Q[X; D], also denoted θ. So θ ∈ Λ. For β ∈ Y ⊥ , we have θ(f1 ) · β = θ(f1 · β) = g · β. Hence (θ(f1 ) − g) · Y ⊥ = 0. So θ(f1 ) − g ∈ Y ⊥⊥ and g ∈ Y ⊥⊥ + Λf1 . We have (Y + Cf1 )⊥⊥ ⊆ Y ⊥⊥ + Λf1 , which proves the case n = 1.

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Next, n ≥ 2; (Y +Cf1 +· · ·+Cfn )⊥⊥ = ((Y +Cf1 +· · ·+Cfn−1 )+Cfn )⊥⊥ ⊆ (Y + Cf1 + · · · + Cfn−1 )⊥⊥ + Λfn ⊆ Y ⊥⊥ + Λf1 + · · · + Λfn . So we have proved the reverse inclusion. If 1 6= B ∈ Ω then DB (Q) = 0. So Λa = CS (R)a by Theorem 3. With this, we see that Theorem 6 does generalize Lemma 4.

4. Invertibility We now apply Theorem 3 to analyze invertibility of (R, R)-bimodule endomorphisms. We recall some notions from [20, 21]: First, we fix an arbitrary complete order < on X (that satisfies descending chain condition) by Zorn’s Lemma. Then we extend < to a linear order on words in X, also denoted by <, as follows: We postulate xi1 · · · xil < xj1 · · · xjk if and only if (1) l < k, or (2) l = k and there exists 1 ≤ t ≤ l such that xit < xjt and such that xis = xjs for 1 ≤ s < t. That is, we order words first by length and then lexicographically for words of the same length. This is called Hall order [9]. Any nonzero f ∈ Q[X; D] may be written in the form f = a0 A0 + a1 A1 + · · · , where ai ∈ Q\{0} and Ai are distinct words in Ω. In this expression of f , the ≤-greatest word, denoted by lw(f ), is called the leading word of f . We start with left invertibility: Theorem 7: The following conditions are equivalent for θ ∈ Λ: (1) There exists ϕ ∈ Λ such that ϕθ = 1. (2) θ is injective. (3) θ |CS (R) , the restriction of θ to CS (R), is injective. Proof. The implications (1) ⇒ (2) and (2) ⇒ (3) are trivial. For the implication (3) ⇒ (2). Assume otherwise θ is not injective. Pick def.

0 = 6 f ∈ ker(θ) with A = lw(f ) being mininal. Write f = aA + · · · . For β ∈ a⊥ , f · β ∈ ker(θ) and lw(f · β) < lw(f ). By the minimality of lw(f ) among nonzero elements of ker(θ), f · β = 0. So f · β = 0 for β ∈ a⊥ . By

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Lemma 4, f ∈ a⊥⊥ = aCS (R). Write f = aζ, where ζ ∈ CS (R). By Lemma 1, θ(ζ) ∈ CS (R). Hence aRθ(ζ) = aθ(ζ)R = θ(aζ)R = θ(f )R = 0. By the primeness of R, θ(ζ) = 0, contradicting the injectivity of θ |CS (R) . def.

For the implication (2) ⇒ (1), we define ϕ : θ(S) → S by setting ϕ(θ(f )) = f for f ∈ S. Then ϕ is an (R, R)-bimodule map defined on the (R, R)-bisubmodule θ(S) of S. By Corollary 5, ϕ can be extended to be defined on S. Hence ϕ ∈ Λ satisfies ϕθ = 1. This complete the proof. Theorem 8: The following conditions are equivalent for θ ∈ Λ: (1) There exist ϕ ∈ Λ such that θϕ = 1. (2) θ |CS (R) , the restriction of θ to CS (R), is surjective. Proof. The implication (1) ⇒ (2) follows immediately from Lemma 1. For P the implication (2) ⇒ (1), we write θ = A∈Ω ζA DA , where ζA ∈ CS (R), by Theorem 3. We want to determine ηB ∈ CS (R) for each B ∈ Ω such that P ϕ = B∈Ω ηB DB satisfies θϕ = 1. Write ! X X
θϕ = θ ηB DB = θ ηB DB . B∈Ω

B∈Ω


Using Lemma 2 and (◦), we bring each θ ηB DB into the form granted by Theorem 3. In view of (◦), we have
(∗) θ ηB DB = θ(ηB )DB + terms with DA for A ∈ Ω with lh(A) > lh(B). We thus compute the desired ηB by induction on lh(B). By (∗), we have θϕ = θ(η1 )D1 + terms with DA for nonempty word A ∈ Ω. Since θ |CS (R) is surjective, we can find η1 ∈ CS (R) such that θ(η1 ) = 1. As the induction hypothesis, assume that ηB has been found for those words with length < lh(B). In view of (∗), only θ(ηB DB ) and those θ(ηB1 DB1 ) with lh(B1 ) < lh(B) can contribute to give terms with DB . Let ηDB , where η ∈ P CS (R), be the term with DB contributed by B1 ∈Ω: lh(B1 )
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existence of such ηB follows by the surjectivity of θ |CS (R) . This completes the induction. An immediate consequence of Theorems 7 and 8 is the following characterization of invertible (R, R)-bimodule endomorphisms: Theorem 9: An (R, R)-bimodule endomorphism is invertible if and only if its restriction to CS (R) is invertible. One might expect that the surjectivity of an (R, R)-bimodule endomorphism would imply the surjectivity of its restriction to CS (R). This is actually false in general.

5. The case R[x; δ] It is interesting to apply our results to the one-indeterminate case R[x, δ]. For this purpose, we need information of CS (R). This is almost provided by Example 2 and is implicit in [17]. Theorem 10: Consider R[x; δ], that is, X = {x} and D = {δ}. Set ζ to be the monic polynomial in CS (R) of the minimal nonzero degree if it exists. def.

Otherwise, set ζ = 1. Then CS (R) = C[ζ], the polynomial ring in ζ. Proof. Clearly, CS (R) ⊇ C[ζ]. For the reverse inclusion, if ζ = 1 then all elements of CS (R) have zero degree and hence CS (R) = C. Assume that ζ has degree> 0. Since ζ is monic, given f ∈ CS (R), we may write f = gζ + h, where g, h ∈ Q[x, δ] with deg(h) < deg(ζ). For r ∈ R, we have 0 = [f, r] = [g, r]ζ + [h, r]. If [g, r] 6= 0 then deg([g, r]ζ) ≥ deg(ζ) > deg([h, r]), absurd. So [g, r] = [h, r] = 0. That is, g, h ∈ CS (R). The assertion follows by a simple induction on deg(f ). Our first application is to differential identities. By this we mean an equality of the form X aij δ i (r)bij = 0 for all r ∈ R. ij

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We translate this as follows: δ(r) = [x, r] = xr − rx, δ 2 (r) = [x, [x, r]] = x2 r − 2xrx + rx2 and, in general,   i X i i−s s δ i (r) = (−1)s x rx . s s=0 With these, the above equality can then be expressed in Q[x, δ] as X fi (x)rgi (x) = 0 for all r ∈ R, i

where fi (x), gi (x) ∈ Q[x; δ]. Our interpretation of Kharchenko’s theory [13, 14] for such identity is the following: Pn Theorem 11: For fi , gi ∈ Q[x, δ], i=1 fi rgi = 0 for all r ∈ R if and only if Pn i=1 fi ⊗ gi = 0 in Q[x; δ] ⊗ Q[x; δ]. CS (R)

Proof. The implication ⇐ is trivial. For the implication ⇒, let m be the degree of ζ in Theorem 10. If m > 0 then any element of Q[x; δ] may be written in the P Pm−1 form i j=0 ai xj hij (ζ), where ai ∈ Q and hij (ζ) ∈ C[ζ] = CS (R). Express each fi in this form and move elements of C[ζ] to the right hand side of r. With this, we may assume that each fi has degree< m if m ≥ 1. If all fi = 0 then we P are done. Otherwise, we pick such an expression with i deg(fi ) as small as possible. Let 0 = 6 a ∈ Q be the leading coefficient of f1 . For β ∈ a⊥ and r ∈ R, X (fi (x) · β)rgi (x) = 0. P

P

i

Since i deg(fi · β) < i deg(fi ), we have fi (x) · β = 0 for all i. By Lemma 4, fi ∈ aCS (R). But CS (R) = C[ζ] and deg(fi ) < deg(ζ) if ζ = 6 1. So fi ∈ aC. Write fi = aαi for some αi ∈ C. We have n n X  X 0= fi (x)rgi (x) = ar αi gi (x) for all r ∈ R. P

i=1

i=1

 Pn P n So aR α g (x) = 0 and i i i=1 i=1 αi gi (x) = 0. So i fi ⊗ gi = 0 in Q[x; δ] ⊗ Q[x; δ] as asserted. CS (R)

We remark that Theorem 11 above can be extended to R[X; D] in general. The crucial step is a description of CS (R), which is more complicated in the case of |X| > 2. We resume this work elsewhere. Theorem 3 applied to R[x; δ] yields the following:

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Theorem 12: Let R be a prime ring and δ a derivation of R. Any (R, R)bimodule endomorphism θ : Q[x; δ] → Q[x; δ] can be written uniquely in the form ∞ X θ(f ) = ζn Dn (f ) for f ∈ Q[x; δ], n=0

def. 1 dn n! dxn .

where ζn ∈ CS (R) and Dn =

With this, we study the invertibility of (R, R)-bimodule endomorphisms of R[x; δ]. We start with the simplest case that CS (R) = C. This is equivalent to the condition that δ is not quasi-algebraic. Theorem 13: Assume that CS (R) = C. Let θ ∈ Λ. The following conditions are equivalent: (1) θ is invertible. (2) θ(1) 6= 0. (3) θ is injective. Proof. By Theorem 12, write θ = ζ0 + ζ1 D1 + . . . where ζ0 , ζ1 , . . . ∈ C. So θ(1) = ζ0 . By Theorem 9, the invertiblity of θ is equivalent to that of θ |CS (R) , that is, the invertiblity of θ |C . But θ(α) = ζ0 α for α ∈ CS (R) = C. So (1) ⇔ (2). By Theorem 7, the injectivity of θ is equivalent to that of θ |CS (R) . The latter is clearly equivalent to ζ0 = 6 0. So (2) ⇔ (3). We now study the surjectivity of (R, R)-bimodule endomorphisms. This is easy in the case of charR = 0. Theorem 14: If charR = 0 and CS (R) = C then any nonzero θ ∈ Λ is P surjective. Write θ = 6 0. Then ker θ = i≥k ζi Di , where ζi ∈ C and ζk = {f ∈ Q[x; δ] : deg(f ) < k}. Proof. By (◦) and (•), we may write θ = Dk (ζk D0 +

k!1! k!1! ζk+1 D1 + · · · ) = (ζk D0 + ζk+1 D1 + · · · )Dk . (k + 1)! (k + 1)!

k!1! By Theorem 13, ζk D0 + (k+1)! ζk+1 D1 + · · · is invertible. So θ(S) = Dk (S) = S and ker θ = ker Dk = {f ∈ Q[x; δ] : deg(f ) < k}.

So, contrary to our expectation, the surjectivity of θ ∈ Λ does not imply that of θ |CS (R) . For the case of charR = p > 0, we need two lemmas:

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HIGHER DERIVATIONS OF ORE EXTENSIONS

Lemma 15: Let p be a prime. For any integer n ≥ 1, p divides

175 (pn)! (n!)p .

Proof. Let n have the p-power expansion n = ns ps + ns+1 ps+1 + ns+2 ps+2 + · · · where ns = 6 0 and where 0 ≤ ni < p for all i. Then pn has the p-power expansion pn = ns ps+1 + ns+1 ps+2 + ns+2 ps+3 + · · · . By Lucas Lemma, we have      pn 0 ns ≡ · · · modulo p. n ns ns+1
Since ns ≥ 1, we have pn n ≡ 0 modulo p. With this, we see that       (pn)! pn (p − 1)n 2n n = ··· ≡ 0 modulo p, (n!)p n n n n as asserted. Lemma 16: If charR = p > 0 then Dnp = 0 for any integer n ≥ 1. Proof. By (•) or Example 5, we see that Dn1 Dn2 · · · Dnk =

(n1 + · · · + nk )! Dn1 +···+nk . n1 ! · · · nk !

Let k = p and n1 = · · · = nk = n. We have Dnp = 15.

(pn)! (n!)p Dpn

= 0 by Lemma

Theorem 17: If charR = p > 0 and CS (R) = C then, for any θ ∈ Λ, either θp = 0 or θ is invertible. Proof. By Theorem 12, write θ = ζ0 + ζ1 D1 + . . . where ζ0 , ζ1 , · · · ∈ C. If ζ0 = 0 then by Lemma 16, θp = ζ1p D1p + ζ2p D2p + · · · = 0. If ζ0 = 6 0 then θ is invertible by Theorem 13. If θp = 0 then θ cannot be surjective. So in the case of charR = p > 0 and CR (S) = C, if θ ∈ Λ is surjective then so is θ |CS (R) . Actually, the exception happens only when charR = 0 and CS (R) = C, as shown in following: Theorem 18: Unless charR = 0 and CS (R) = C, if θ ∈ Λ is surjective then so is θ |CS (R) and θ is hence right invertible. Proof. Let us divide our argument into two cases: Case 1. charR = 0: By assumption, CS (R) = 6 C. By Example 2, δ is Xinner, say δ = adb . Then CS (R) = C[ζ] where ζ = x − b. Replacing x by ζ + b,

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we see that S = Q[ζ], the ring generated by Q and ζ. Fix a C-basis {b0 , b1 , . . . } of Q. As vector spaces over C, we can write S = b0 C[ζ] ⊕ b1 C[ζ] ⊕ · · · . Then θ(S) = b0 θ(C[ζ]) ⊕ b1 θ(C[ζ]) ⊕ · · · . Hence if θ(S) = S then θ(C[ζ]) = C[ζ]. So the surjectivity of θ implies that of θ |CS (R) . Case 2. charR = p > 0. If CS (R) = C then either θp = 0 or θ is invertible by Theorem 17. If θp = 0 then θ cannot be surjective. If θ is invertible then θ |CS (R) is also invertible and hence surjective. We hence assume that CS (R) 6= C. By Example 2, there exist integer s ≥ 0, and β1 , . . . , βs ∈ C, b ∈ Q such that CS (R) = C[ζ] where s

s−1

ζ = xp + β1 xp

+ · · · + βs x + b.

Fix a C-basis {b0 , b1 , . . . } of Q; we can write S as a direct sum of vector spaces over C: s −1 X pX S= bi C[ζ]xj . i

j=0

Assume that θ |CS (R) is not surjective. So there is f (ζ) ∈ C[ζ] \ θ(C[ζ]). Since s θ is surjective, there is g(x) ∈ S such that θ(g(x)) = b0 f (ζ)xp −1 . Write P P s −1 g(x) = i pj=0 bi hij (ζ)xj , where hij (ζ) ∈ C[ζ]. By Theorem 12, write θ = P∞ k=0 ηk Dk , where ηk ∈ C[ζ]. Using Lemma 2, we compute θ(g(x)) =

∞ X

ηk Dk

i

k=0

=

X i

= b0

bi

∞ X

∞ X

bi hij (ζ)xj

j=0

s

ηk

k=0

∞ X k=0

= b0

s −1  X pX

ηk

pX −1 X k



Dl (hij (ζ))Dk−l (xj )

j=0 l=0

s pX −1 X k

j=0 l=0

     Dl (h0j (ζ))Dk−l (xj ) + b1 · + b2 · + · · · s

ηk Dk (h0,ps −1 (ζ))xp

−1

s

+ (·)xp

k=0

s

= b0 θ(h0,ps −1 (ζ))xp

−1

s

+ b0 (·)xp

−2

−2

   + · · · + b1 · + · · ·

  + · · · + b1 · + · · · .

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This gives θ(h0,ps −1 (ζ)) = f (ζ), contradicting that f (ζ) ∈ C[ζ] \ θ(C[ζ]). Acknowledgements. The authors would like to thank the referee for the valuable suggestions which help clarify the paper.

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