代數斜導算及其常數

代數斜導算及其常數

中 華 民 國 93 年 2 月 3 日

Chen–Lian Chuang and Tsiu–Kwen Lee

Department of Mathematics, National Taiwan University Taipei 106, Taiwan

E–mail: chuang@math.ntu.edu.tw E–mail: tklee@math.ntu.edu.tw

Abstract. Let R be a prime ring with d a left R–algebraic derivation. All possible left R–algebraic relations of d are described by a specific ideal of the skew polynomial ring R[x; d]. Moreover, the prime radical and the minimal prime ideals over such ideal are also determined. As an application to the main theorem, the nilpotent case is completely obtained.

1

2000 Mathematics Subject Classification. 16W25, 16S36, 16N60.

Key words and phrases. Prime ring, Martindale quotient ring, skew polynomial ring,

algebraic derivation, central generator.

1. Results

Throughout this paper, R is always a prime ring with RF its left Martindale

quotient ring and Q its symmetric Martindale quotient ring. The center of RF (and

of Q also), denoted by C, is called the extended centroid of R. We refer these notation to [2] for details. Also, d is always a derivation of R. By this, we mean that d: R → R is an additive map such that d(ab) = d(a)b + ad(b) for all a, b ∈ R. Given b ∈ R, the map ad(b): r ∈ R 7→ [b, r]def.= br − rb obviously defines a derivation, called the inner derivation defined by b. We call a derivation outer if it is not of this form. It is well–known that every derivation d of R can be uniquely extended to a derivation of Q and also to a derivation of RF. For any subset S of RF, we define

S(d) def._{= {r ∈ S | d(r) = 0}, the set of constants of d on S. To analyze derivations}

of R, as shown in Kharchenko’s theory [9, 10], we have to work in the larger ring

Q. The derivation d is called X–inner if its extension to Q is inner. Otherwise, it is

called X–outer.

We denote R[x; d] the skew polynomial ring endowed with the multiplication rule:

xr = rx + d(r) for r ∈ R. Since the derivation d can be uniquely extended to a

derivation of RF, we can also construct RF[x; d] analogously and regard R[x; d] as

1

a subring of RF[x; d] in a natural way. For a0xn+ · · · + an−1x + an ∈ R[x; d] and

r ∈ R, we define

(a0xn+ · · · + an−1x + an) * r = a0dn(r) + · · · + an−1d(r) + anr ∈ R.

It is well–known that R forms a left R[x; d]–module under the above action. Our first aim is just to describe:

Adef.= {g(x) ∈ R[x; d] | g(x) * R = 0}.

Note that A is a two–sided ideal of R[x; d]. The ideal A describes all possible left

R–algebraic relations of d (its precise definition will be given below). Let P be the

ideal of R[x; d] containing A such that P/A is the prime radical of R[x; d]/A. Our next aim is to describe the ideal P and also the minimal prime ideals over A.

There is a natural interpretation of the ring R[x; d]/A: For a ∈ R, let aL: x ∈ R 7→

ax ∈ R denote the left multiplication by a. All aL are endomorphisms of the abelian

additive group (R, +). In the ring of endomorphisms of (R, +), let S be the subring generated by d and all aL, a ∈ R. Note that daL = d(a)L+ aLd for a ∈ R. The map

ϕ : a0xn+ · · · + an−1x + an 7→ (a0)Ldn+ · · · + (an−1)Ld + (an)L ∈ S

is a surjective ring homomorphism. We see easily that the kernel of ϕ is A. So

R[x; d]/A ∼= S. In this sense, our result amounts to describing the prime radical and minimal prime ideals of S.

To state our result, we need one more notation. If f (x) is a monic element of

RF[x; d], we define

hf (x)i def.= ¡RF[x; d]f (x)RF[x; d]

¢

∩ R[x; d].

In addition, if f (x) is central, then

hf (x)i = (f (x)RF[x; d]) ∩ R[x; d]

= {g(x) ∈ R[x; d] | g(x) is a multiple of f (x) in RF[x; d]}.

We will describe A, P in terms of ideals in the above form hf (x)i, where f (x) is central. For this purpose, we must investigate the center of RF[x; d]. Fortunately,

this has been completely done in [13]:

Proposition 1. (Matczuk [13]) (I) Assume char R = 0. If d =ad(−b) for some

b ∈ Q, then the center of RF[x; d] equals to C(d)[ζ], where ζ def.= x + b. If d is X–

outer, then the center of RF[x; d] is merely C(d). (II) Assume char R = p > 0. If

there exists b ∈ Q(d) _{and α}

1, · · · , αs−1∈ C(d) such that

(1) dps_{+ α}

1dp

s−1

then we let (1) be the one with s as mininmal as possible and set

ζ def.= xps + α1xp

s−1

+ · · · + αsx + b.

The center of RF[x; d] is equal to C(d)[ζ]. If there is no such expression (1), then the

center of RF[x; d] is merely C(d).

The derivation d is said to be left RF–algebraic (left R–algebraic) if there exist

b0 6= 0, b1, · · · , bn−1∈ RF (resp. R) such that

b0dn(r) + b1dn−1(r) + · · · + bn−1d(r) = 0

for all r ∈ R. We say that d is C–algebraic if all bi ∈ C. It is easy to see that the

left R–algebraicity, the left RF–algebraicity and the C–algebraicity of a derivation

d are all equivalent. If d is not left R–algebraic, then A = 0 and P = 0. Since the

ring R[x; d] is prime, the only minimal prime ideal is {0}. So there is nothing to do in this case. We hence assume that our d is left R–algebraic or, equivalently, left

RF–algebraic. Our main theorem is as follows:

Theorem 1. Let R be a prime ring and let d be a left RF–algebraic derivation of

R. Let ζ, b be as described in Proposition 1. Let µ(λ) be the minimal polynomial of b over C(d)_{. Then the following hold:}

(1) A = hµ(ζ)i.

(2) We factorize µ(λ) into the product of monic irreducible factors in C(d)_[λ]:

µ(λ) = π1(λ)n1π2(λ)n2· · · πk(λ)nk. Then all minimal prime ideals of R[x; d] over A

are hπs(ζ)i, s = 1, · · · , k, and P = hπ1(ζ)π2(ζ) · · · πk(ζ)i.

Before proceeding to the proof of Theorem 1, let us compute explicitly the ζ of Proposition 1 for a left RF–algebraic derivation d: We apply Kharchenko’s theorem

[10, Theorem 2]. If char R = 0, then d = ad(−b) for some b ∈ Q and we set ζ def.= x+b. If char R = p > 0, then d, dp_{, d}p2

, . . . are C–dependent modulo X–inner derivations.

Let s ≥ 0 be the minimal integer such that

dps, dps−1, · · · , dp, d

are C–dependent modulo X–inner derivations. By the minimality of s, there exist

αi∈ C and b ∈ Q such that

dps+ α1dp

s−1

+ · · · + αsd = ad(−b),

Case 1. d(b) ∈ d(C): Say, d(b) = d(α), where α ∈ C. Then d(b − α) = 0. Since b and b − α define the same X–inner derivation, we may replace b by b − α and assume that d(b) = 0. So we have

ζ = xps+ α1xp

s−1

+ · · · + αsx + b.

Case 2. d(b) /∈ d(C): Since d(αi) = 0, each αiconsidered as left multiplication

com-mutes with d. Since d(b) ∈ C, ad(b) also comcom-mutes with d. Using the commutativity, we raise both sides of (1) to the p–th power. This gives the equality:

dps+1+ αp₁dps+ · · · + αp_sdp= ad(−bp). Obviously, d(bp_{) = 0. So we have ζ = x}ps+1

+ αp₁xps

+ · · · + αp

sxp+ bp.

Moreover, we observe that the X–inner derivation ad(b) is RF–algebraic if and

only if b is C–algebraic. Note that a differential identity of R also vanishes on RF

[11, Theorem 2]. In particular, the restriction of d to C is C–algebraic. In view of [1, Theorem 1], C is finite–dimensional over C(d)_{. Thus this is also equivalent to say}

that b is C(d)_{–algebraic. We summarize what we have shown in the following:}

Lemma 1. Let d be a left RF–algebraic derivation and let ζ be as described in

Proposition 1. If char R = 0, then d = ad(−b) for some b ∈ Q and ζ = x + b. If

char R = p ≥ 2, then there exists the minimal integer s ≥ 0 such that

dps+ α1dp

s−1

+ · · · + αsd = ad(−b)

for some αi ∈ C(d) and b ∈ Q with d(b) ∈ C. In the case of d(b) ∈ d(C), we may

choose b ∈ Q(d) _{and ζ = x}ps

+ α1xp

s−1

+ · · · + αsx + b. In the case that d(b) /∈ d(C),

we have bp _{∈ Q}(d) _{and ζ = x}ps+1

+ αp₁xps

+ · · · + αp

sxp + bp. Moreover, b above is

always C(d)_{–algebraic.}

To prove Theorem 1, we need another important proposition from [13], which, unfortunately, is not explicitly formulated in [13]. So we include its proof here. See also [14, Theorem 3.3].

Proposition 2. (Matczuk) If I is an ideal of R[x; d], then there exists a unique

monic f (x) in the center of RF[x; d] such that any element of I is a multiple of f (x)

in RF[x; d], i.e., I ⊆ hf (x)i. Moreover, there exists a nonzero ideal I of R such that

If (x) ⊆ I.

Proof. If R ∩ I 6= 0, we simply take f (x) = 1 and set I def.= R ∩ I. Therefore, we assume that R ∩ I = 0. Let n be the minimal possible degree for nonzero elements of I. We define

Note that I is a nonzero ideal of R. Since R ∩ I = 0, it follows from the minimality of n that a1, · · · , an are uniquely determined by a. Write ai = βi(a) for a ∈ I. We

have βi(ra) = rβi(a) for r ∈ R. Each βi: I → R is thus a left R–module map. So

there exist qi∈ RF such that βi(a) = aqi. Set our desired monic polynomial to be

f (x)def.= xn_{+ q}

1xn−1+ · · · + qn ∈ RF[x; d].

We show that f (x) is a center element of RF[x; d]. Note that I2x ⊆ xI2+ d(I2) ⊆

xI2_{+ I. Thus I}2_{xf (x)R ⊆ I. It is also clear that I}2_{f (x)xR ⊆ I. We have}

I2(xf (x) − f (x)x)R = I2(d(q1)xn−1+ · · · + d(qn−1)x + d(qn))R ⊆ I.

By the minimality of n, this implies d(q1) = · · · = d(qn−1) = d(qn) = 0. Similarly,

for r ∈ R, I(rf (x) − f (x)r) falls in I and has degree < n. So I(rf (x) − f (x)r) = 0, implying that rf (x) = f (x)r. Given a ∈ RF, let J be a nonzero ideal of R such that

Ja ⊆ R. For r ∈ J we have

0 = [ra, f (x)] = [r, f (x)]a + r[a, f (x)] = r[a, f (x)],

and so J[a, f (x)] = 0. This implies af (x) = f (x)a for all a ∈ RF. All these together

say that f (x) falls in the center of RF[x; d], as asserted.

Finally, given any g(x) ∈ I, we write, using the division algorithm in RF[x; d],

g(x) = f (x)q(x) + r(x),

where q(x), r(x) ∈ RF[x; d] and the degree of r(x) is less than n. Pick a nonzero

ideal J of R such that Jq(x) ∪ Jr(x) ⊆ R[x; d]. Let a ∈ I and b ∈ J; then abg(x) =

af (x)bq(x) + abr(x). Since af (x) ∈ I, this implies that abr(x) ∈ I and has degree

less than n. Thus abr(x) = 0. That is, IJr(x) = 0 and so r(x) = 0. So any g(x) ∈ I is a multiple of f (x) in RF[x; d]. Also, it is clear that If (x) ⊆ I by the definition of

I. This completes the proof.

For convenience, we call f (x) described above the central generator of the ideal I. Proposition 2 says I ⊆ hf (x)i. We say I is stable if I = hf (x)i, that is, the ideal consists exactly of multiples of its central generator. Ideals of R[x; d] are somewhat decided by their central generators but still can involve complexities of the ideal structure of the ring R. But stable ideals are completely determined by their central generators. We now come to the proof of Theorem 1. We first note a simple but important property of central elements in RF[x; d]:

Lemma 2. If a0xn+ · · · + an−1x + an lies in the center of RF[x; d], then

Proof. Set f (x) = a0xn + · · · + an−1x + an. Let r ∈ RF. Note that a0dn(r) + · · · +

an−1d(r) + anr is the constant term of f (x)r and ran, the constant term of rf (x).

Since f (x)r and rf (x) are equal, so are their constant terms. Thus a0dn(r) + · · · +

an−1d(r) + anr = ran, proving the lemma.

Lemma 3. If π(λ) ∈ C(d)_{[λ] is monic and irreducible, then hπ(ζ)i is a prime ideal}

of R[x; d], where ζ is as given in Proposition 1.

Proof. Suppose that J1 and J2 are two ideals of R[x; d] such that Ji ⊇ hπ(ζ)i for

i = 1, 2 and such that J1J2 ⊆ hπ(ζ)i. Then, by Proposition 2, there exist central

generators g1(ζ) and g2(ζ) of J1 and J2, respectively. Choose a nonzero ideal J of

R such that Jπ(ζ) ⊆ R[x; d] and Jgi(ζ) ⊆ Ji for i = 1, 2. Thus

J2g1(ζ)g2(ζ) = Jg1(ζ)Jg2(ζ) ⊆ J1J2⊆ hπ(ζ)i.

If (g1(λ)g2(λ), π(λ)) = 1 in C(d)[λ], then there exist A(λ), B(λ) ∈ C(d)[λ] such that

A(λ)g1(λ)g2(λ) + B(λ)π(λ) = 1. In particular, we have

(2) A(ζ)g1(ζ)g2(ζ) + B(ζ)π(ζ) = 1.

Choose a nonzero ideal I of R such that IA(ζ) ∪ IB(ζ) ⊆ R[x; d]. By (2), 0 6= IJ2 ⊆ IA(ζ)(J2g1(ζ)g2(ζ)) + IB(ζ)J2π(ζ) ⊆ hπ(ζ)i,

a contradiction. Thus (g1(λ)g2(λ), π(λ)) 6= 1. The irreducibility of π(λ) implies that

either π(λ)|g1(λ) or π(λ)|g2(λ) in C(d)[λ]. Assume, without loss of generality, that

π(λ)|g1(λ). Therefore, there exists h(λ) ∈ C(d)[λ] such that g1(ζ) = h(ζ)π(ζ). This

implies that hg1(ζ)i ⊆ hπ(ζ)i. But J1 ⊆ hg1(ζ)i. So J1 ⊆ hπ(ζ)i, proving the lemma.

We are now ready to give the

Proof of Theorem 1. By Proposition 2, there exists the central generator f (x) of A and an ideal I 6= 0 of R such that If (x) ∈ A. Thus A ⊆ hf (x)i. Since If (x) ∈ A,

we have 0 = If (x) * R = I(f (x) * R) and hence f (x) * R = 0. This implies

hf (x)i * R = 0 and hence A = hf (x)i. By assumption, d is RF–algebraic. Assume

that d 6= 0. Let

ζ =n x + b, if char R = 0 xps+ α1xp

s−1

+ · · · + αsx + b, if char R = p > 0

be given as in Proposition 1. Applying Proposition 2, we may write (3) f (x) = ζn_{+ β}

where βi∈ C(d). We set eµ(λ) = λn+ β1λn−1+ · · · + βn ∈ C(d)[λ]. It follow from (3)

that eµ(b) is the constant term of f (x). By Lemma 2, we have 0 = f (x) * R = R eµ(b)

and so eµ(b) = 0. Thus µ(λ) divides eµ(λ) in C(d)_[λ].

On the other hand, µ(b) is the constant term of µ(ζ). Since µ(ζ) is central, we have

µ(ζ) * R = Rµ(b) = 0 by Lemma 2. It follows from the minimality of the degree

of f (x) that the degree of µ(λ) is equal to or greater than n and so µ(λ) = eµ(λ).

Hence, f (x) = µ(ζ) follows. We factorize µ(λ) into the product of monic irreducible factors in C(d)_{[λ]: µ(λ) = π}

1(λ)n1π2(λ)n2· · · πk(λ)nk, where ni are positive integers.

To show that P = hπ1(λ)π2(λ) · · · πk(λ)i and that minimal prime ideals of R[x; d]

over A are hπs(ζ)i, s = 1, · · · , k, we quote the following more general Theorem 2.

Theorem 2. Let µ(λ) ∈ C(d)_{[λ] be monic. We factorize µ(λ) into the product of}

monic irreducible factors in C(d)_{[λ]: µ(λ) = π}

1(λ)n1π2(λ)n2· · · πk(λ)nk, where ni are

positive integers. Then all minimal prime ideals of R[x; d] over hµ(ζ)i are hπs(ζ)i,

s = 1, · · · , k, where ζ is given as in Proposition 1. Moreover, the prime radical of R[x; d]/hµ(ζ)i is equal to hπ1(λ)π2(λ) · · · πk(λ)i/hµ(ζ)i.

The theorem above describes the prime radical and the minimal prime ideals over a stable ideal. For its proof, we need a lemma.

Lemma 4. If µ1(λ) and µ2(λ) are monic and coprime in C(d)[λ], then hµ1(ζ)i ∩

hµ2(ζ)i = hµ1(ζ)µ2(ζ)i, where ζ is given as in Proposition 1.

Proof. The inclusion hµ1(ζ)µ2(ζ)i ⊆ hµ1(ζ)i ∩ hµ2(ζ)i is obvious. For the reverse

inclusion, let f (x) ∈ hµ1(ζ)i ∩ hµ2(ζ)i. Write f (x) = g1(x)µ1(ζ) = g2(x)µ2(ζ),

where gi(x) ∈ RF[x; d]. Since µ1(λ) and µ2(λ) are coprime in C(d)[λ], there exist

A(λ), B(λ) ∈ C(d)_{[λ] such that}

A(ζ)µ1(ζ) + B(ζ)µ2(ζ) = 1. Thus g1(x) =A(ζ)g1(x)µ1(ζ) + g1(x)B(ζ)µ2(ζ) =A(ζ)g2(x)µ2(ζ) + g1(x)B(ζ)µ2(ζ) =¡A(ζ)g2(x) + g1(x)B(ζ) ¢ µ2(ζ), implying that f (x) =¡A(ζ)g2(x) + g1(x)B(ζ) ¢ µ1(ζ)µ2(ζ).

So f (x) ∈ hµ1(ζ)µ2(ζ)i. Thus hµ1(ζ)i ∩ hµ2(ζ)i ⊆ hµ1(ζ)µ2(ζ)i, proving the lemma.

Proof of Theorem 2. Let Q is an ideal of R[x; d], which is a minimal prime ideal

over A. Then

By the primeness of Q, we see that Q contains hπi(ζ)i for some i. By Lemma 3,

hπi(ζ)i is a prime ideal of R[x; d]. The minimality of Q implies that Q = hπi(ζ)i.

This proves that all possible minimal prime ideals of R[x; d] over hµ(ζ)i are hπs(ζ)i,

s = 1, · · · , k. Conversely, we show each hπi(ζ)i is a minimal prime ideal over A: Let

Q0 be a prime ideal of R[x; d] such that hπi(ζ)i ⊇ Q0 ⊇ hµ(ζ)i. Applying the same

argument above yields Q0 ⊇ hπj(ζ)i for some j and so hπi(ζ)i ⊇ hπj(ζ)i. Suppose

i 6= j. There exist A(ζ), B(ζ) ∈ C(d)_{[ζ] such that}

A(ζ)πi(ζ) + B(ζ)πj(ζ) = 1.

Choose a nonzero ideal I of R such that IA(ζ) ∪ Iπi(ζ) ∪ IB(ζ) ∪ Iπj(ζ) ⊆ R[x; d].

Then

0 6= I2⊆ IA(ζ)Iπi(ζ) + IB(ζ)Iπj(ζ) ⊆ IA(ζ)Iπi(ζ) + hπj(ζ)i ⊆ hπi(ζ)i,

a contradiction. Let H be the ideal of R[x; d] such that H/hµ(ζ)i is the prime radical of R[x; d]/hµ(ζ)i. Choose a positive integer m ≥ ni for all i. Then

hπ1(λ)π2(λ) · · · πk(λ)im ⊆ hµ(ζ)i ⊆ H.

But H is a semiprime ideal of R[x; d]. So hπ1(λ)π2(λ) · · · πk(λ)i ⊆ H follows. On the

other hand, by Lemma 3 each hπi(ζ)i is a prime ideal of R[x; d] and so

H ⊆ hπ1(λ)i ∩ · · · hπ1(λ)i ⊆ hπ1(λ)π2(λ) · · · πk(λ)i,

where the second inclusion is implied by Lemma 4. Thus H = hπ1(λ)π2(λ) · · · πk(λ)i.

The proof is now complete.

2. An Application to the Nilpotent Case Firstly, we need an important notion:

Definition. ([4, 3]) Let d be a nilpotent derivation of R. The least integer m such that dm_{(R)c = 0 for some nonzero c ∈ R is called the annihilating nilpotency of d}

and is denoted by md(R).

For a nilpotent derivation d of a prime ring R, there is another interesting con-struction, which has been employed extensively and fruitfully in the literature [4]–[8]: In the ring R[x; d], we consider the two–sided ideal hxm_{i, where m = m}

d(R). For r ∈ R, xmr = rxm+ µ m 1 ¶ d(r)xm−1+ · · · + µ m m − 1 ¶ dm−1(r)x + dm(r).

Therefore, if a0xn + a1xn−1 + · · · + an ∈ hxmi, then an ∈ Rdm(R). This implies

Rdm_{(R) contains hx}m_{i ∩ R, which is an ideal of R. Since Rd}m_{(R) has nonzero left}

annihilator, we have hxm_{i ∩ R = 0. Extend hx}m_{i to an ideal M of R[x; d] maximal}

with respect to the property that M ∩ hxm_{i = 0. Then we see easily that M is a}

prime ideal of R[x; d]. The quotient ring r[x; d]/M is called the d–extension of R [6]. Our aim is to prove that M is equal to the ideal P described in Theorem 1. For this purpose we need a structure result of nilpotent derivations. The following is given in [3, Theorems 1–4].

Theorem 3. Let d be a nilpotent derivation of a prime ring R.

(1) In the case of char R = 0, there exists a nilpotent b ∈ Q with the nilpotency l such that d = ad(−b) and md(R) = l.

(2) In the case of char R = p ≥ 2, let s be the least integer ≥ 1 such that dpi

,

0 ≤ i ≤ s, are C–dependent modulo X–inner ones. Then there exists b ∈ Q such that

dps

= ad(−b) and such that the minimal polynomial of b over C assumes the form (bpt

− α)l _{= 0, where α ∈ C}(d)_{, l, t ≥ 0 and (l, p) = 1. Moreover, m}

d(R) = ps+tl.

For a nilpotent derivation d, we have the following detailed description of M defined above:

Theorem 4. For a nilpotent derivation d of a prime ring R, let A, P be as described

in Theorem 1 and M, the ideal described above. In the notation of Theorem 3, we have the following

(1) In the case of char R = 0, A = hζl_{i and P = M = hζi, where ζ = x + b and}

where l is the nilpotency of b.

(2) In the case of char R = p ≥ 2, there are two subcases given as follows: (i) Suppose that b is chosen such that d(b) = 0. Set ζ = xps _{+ b. Then}

A = h(ζpt − α)li and M = P = hζpt−u − α1/pui, where u is the largest integer such that 0 ≤ u ≤ t and α1/pu

∈ C(d)_.

(ii) Suppose that d(b) /∈ d(C). Set ζ = xps+1

+ bp_{. Then}

A = h(ζpt−1 − α)li and M = P = hζpt−1−u − α1/pui, where u is the largest integer such that 0 ≤ u ≤ t − 1 and α1/pu _{∈ C}(d)_.

We need the following lemma. See [2, Theorem 2.3.3] for the proof.

Lemma 5. Let v1, v2, · · · , vn be C–independent elements in RF and let I be a

nonzero ideal of R. Then there exist finitely many ai, bi∈ I such that

P

iaivjbi = 0

Proof of Theorem 4. We will keep the notation of Theorem 3 in the following.

Firstly, assume that char R = 0: Then, by Theorem 3, d = ad(−b) for some nilpotent b ∈ Q with the nilpotency l, that is, bl _{= 0 but b}l−1_{6= 0. By Proposition 1,}

the center of RF[x; d] is equal to C(d)[ζ], where ζ = x + b. The minimal polynomial

of b over C(d) _{is obviously λ}l_{, where λ is a commuting indeterminate over C}(d)_{. It}

follows from Theorem 1 that A = hζl_{i and P = hζi. By Lemma 3, P is prime, as}

asserted. We compute the ideal M described above: By Theorem 3, md(R) = l.

Now, we look at the central generator of hxl_{i: Write x = ζ + b. Noting that b}l _{= 0}

and ζ is central, we have

xl = (ζ + b)l = ζl+ µ l 1 ¶ ζl−1b + · · · + µ l l − 1 ¶ ζbl−1,

implying that hxl_{i ⊆ hζi. Since 1, b, . . . , b}l−1 _{are C–independent, by Lemma 5}

there exist finitely many ri, r0i ∈ I such that

P iribjri0 = 0 for 0 ≤ j < l − 1 but P iribl−1r0i6= 0. We have hxli 3X i rixlri0 = µ l l − 1 ¶ ζ³ X i ribl−1r0i ´ 6= 0.

This shows that ζ is the central generator of the ideal hxl_{i. Since M extends hx}l_i,

the central generator of M divides the central generator ζ of hxl_{i. Since M ∩ R = 0,}

the central generator of M cannot be 1 and hence must be ζ. But hζi also extends

hxl_{i and intersects R trivially. By the maximality of M, it follows M = hζi = P.}

Next, we assume char R = p ≥ 2. Let s be the least integer≥ 1 such that dpi , 0 ≤ i ≤ s, are C–independent modulo X–inner derivations. Then, by Theorem 3, there exists b ∈ Q such that dps

= ad(−b) and such that the minimal polynomial of

b over C assumes the form (bpt

− α)l _{= 0, where α ∈ C, l, t ≥ 0 and (l, p) = 1. By}

Theorem 3 again, md(R) = ps+tl.

Our next step is to find the central generator of the ideal hxm_{i, where m =}

md(R) = ps+tl. For this purpose, we must first decide ζ described in Proposition 1.

Analogous to Lemma 1, we divide our argument into two cases:

Case 1. d(b) ∈ d(C): By Lemma 1, we may assume that d(b) = 0 and so ζ = xps +b. Applying d to (bpt

− α)l _{= 0, we obtain}

−l(bpt− α)l−1d(α) = 0.

the center of RF[x; d]. With this, we compute xm_{= x}ps+tl _{= (ζ − b)}ptl _{= (ζ}pt _{− b}pt₎l ₌¡_(ζpt_{− α) − (b}pt_{− α)}¢l = (ζpt− α)l− µ l 1 ¶ (ζpt− α)l−1(bpt − α) + · · · + (−1)l−1 µ l l − 1 ¶ (ζpt − α)(bpt − α)l−1_{+ (−1)}l µ l l ¶ (bpt − α)l_.

Since (bpt_−alp)l _{= 0, we have x}m _{∈ hζ}pt_{−αi and hence x}m _{⊆ hζ}pt_{−αi. On the other}

hand, since l is the nilpotency of bpt

− α, the elements 1, bpt

− α, . . . ,¡bpt

− α¢l−1 are

C–independent. By Lemma 5, there exist ri, r0i ∈ I such that

P iri(bp t − α)j_r0 i = 0 for 0 ≤ j < l − 1 but P_iri(bp t − α)l−1_r0

i 6= 0. Multiplying the above displayed

expression of xm _{by r}

i, ri0 from the left and right respectively and adding them up,

we have 0 6=X i rixmr0i= l(ζp t − α)³ X i ri ¡ bpt− α¢l−1r_i0 ´ ∈ hxmi.

The central generator of the ideal hxm_{i is thus ζ}pt_{− α.}

Let u be the largest integer such that 0 ≤ u ≤ t and α1/pu

∈ C(d)_{. Set β = α}1/pu . Then ζpt _{− α = (ζ}pt−u _{− β)}pu_{. Note that λ}pt−u _{− β ∈ C}(d)_{[λ] is irreducible. Since}

M ⊇ hxm_{i, the central generator of M is a divisor of (λ}pt−u_{− β)}pu_{, say (λ}pt−u_{− β)}v_,

where 0 ≤ v ≤ pu_{. Also, since M ∩ R = 0, v must be > 0. We have}

M ⊆ h(ζpt−u − β)vi ⊆ hζpt−u − βi.

Since hζpt−u

− βi ∩ R = 0, it follows that M = hζpt−u

− βi by the maximality of M.

We now compute the ideals A and P by Theorem 1: Since the minimal polynomial of b over C(d) _{is (λ}pt _{− α)}l _{= (λ}pt−u_{− β)}pul _{∈ C}(d)_{[λ], the ideal A is thus equal to}

h(ζpt−u − β)puli.

The only irreducible factor of (λpt−u _{− β)}pul _{is λ}pt−u_{− β. So the ideal P is given by}

hζpt−u

− βi and is hence equal to M, as asserted.

Case 2. d(b) /∈ d(C): We have ζ = xps+1 _{+ b}p _{by Lemma 1. We claim t > 0:}

Assume otherwise t = 0. That is, (b − α)l _{= 0. Applying d, we obtain}

l(b − α)l−1(d(b) − d(α)) = 0.

Since d(b) /∈ d(C), d(b) − d(α) 6= 0. But l 6≡ 0 modulo p by our assumption and

Applying d to (bpt

− α)l _{= 0 and using d(b) ∈ C and t > 0, we obtain l(b}pt

− α)l−1_{d(α) = 0. It follows d(α) = 0. So α is also in the center of R}

F[x; d]. We compute analogously xm = xps+tl = (ζ − bp)pt−1l = (ζpt−1− bpt)l =¡(ζpt−1− α) − (bpt − α)¢l = (ζpt−1 − α)l− µ l 1 ¶ (ζpt−1 − α)l−1(bpt− α) + · · · + (−1)l−1 µ l l − 1 ¶ (ζpt−1− α)(bpt− α)l−1+ (−1)l µ l l ¶ (bpt− α)l.

As in Case 1, the central generator of hxm_{i is ζ}pt−1 _{− α. We let u be the largest}

integer such that 0 ≤ u ≤ t − 1 and α1/pu _{∈ C}(d)_{. Set β = α}1/pu_{. Then ζ}pt−1_{− α =}

(ζpt−1−u

− β)pu

. Arguing as in Case 1, we have

M = hζpt−1−u− βi.

The minimal polynomial of bp _{over C}(d) _{is (λ}pt−1

− α)l _{= (λ}pt−1−u − β)pu_l ∈ C(d)_[λ]. By Theorem 1, A is equal to h(ζpt−1−u − β)pu_l i.

Since the only irreducible factor of (λpt−1−u

−β)pu_l

is λpt−1−u

−β, the ideal P is given

by hζpt−1−u

− βi and is also equal to M, as asserted.

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