Derivations of Ore extensions of the polynomial ring in one variable
Andrzej Nowicki
Dedicated to Professor Kazuo Kishimoto on his 70th birthday
Abstract
We present a description of all derivations of Ore extensions of the form R[t, d], where R is a polynomial ring in one variable over a field of characteristic zero.
1 Introduction
Throughout this paper k is a field of characteristic zero and R is a commutative k- algebra. If A and B are k-algebras such that A ⊆ B, then a mapping d : A → B is called a derivation if D is k-linear and D(uv) = D(u)v + uD(v) for all u, v ∈ A. If d : R → R is a derivation, then we denote by R[t, d] the Ore extension of R ([11]). Recall that R[t, d] is a noncommutative ring (often called a skew polynomial ring of derivation type [7]) of polynomials over R in an indeterminate t with multiplication subject to the relation tr = rt + d(r) for all r ∈ R. Basic properties and applications of R[t, d] we may find in many papers (for example: [9], [1], [6]).
In this paper we study derivations of Ore extensions in the case when R is the polyno- mial ring k[x] in one variable over k. If R = k[x] and d is the ordinary derivative ∂x∂, then R[t, d] coincides with the Weyl algebra A1, with one pair of generators ([2], [3]). In this case it is well known ([4]) that every derivation of R[t, d] is inner. We will show (Theorem 9.1) that every derivation of A = k[x][t, d] is inner if and only if A is isomorphic to A1. Thus, the structure of all derivations of k[x][t, d] is clear if d(x) is a nonzero element of k.
Now let d be a nonzero derivation of R = k[x] such that deg ϕ > 1, where ϕ = d(x).
If ϕ is square-free (that is, if gcd(ϕ, ϕ0) = 1, where ϕ0 = ∂ϕ∂x), then we show (Theorem 10.1) that every derivation D of R[t, d] has a unique presentation D = W + ∆r, where W is an inner derivation and ∆r is the unique derivation of R[t, d] such that ∆r(t) = r and
∆r(x) = 0 for some polynomial r ∈ R of degree smaller than deg ϕ.
If ϕ is not square-free, then descriptions of all derivations of R[t, d] are more compli- cated. A full description in this case we present in the last section of this paper. We prove (Theorem 11.2) that, in this case, every derivation D of R[t, d] has a unique decomposi- tion D = W + EH + ∆r, where W is inner, r ∈ R with deg r < deg ϕ, and where EH is a derivation introduced thanks to many preparatory facts and observations presented in this paper.
2 Preliminary
Let R be a commutative k-algebra and let R[t, d] be an Ore extension. Every nonzero element f from R[t, d] is of the form f = antn+· · ·+a1t1+a0, where n > 0, a0, . . . , an ∈ R, an6= 0. In this case the number n is called the degree of f and denoted by deg f or degtf . If f = 0, then we put deg f = −∞. If f, g ∈ R[t, d], then [f, g] denotes the difference f g − gf . In particular, [tn, r] =
n
P
i=1 n
idi(r)tn−i, for all r ∈ R and n > 1. Moreover, if f = antn+ · · · + a1t + a0, then [t, f ] = d(an)tn+ · · · + d(a1)t1+ d(a0).
If a ∈ k[x], then we denote by a0 the derivative of a. Note that (since char(k) = 0) for every b ∈ k[x] there exists a ∈ k[x] such that b = a0. This means that the usual derivative is a surjective derivation of k[x]. This fact implies that if d : k[x] → k[x] is a derivation and ϕ = d(x), then the image d(k[x]) is the principal ideal of k[x] generated by ϕ.
If d is a derivation of k[x], then we denote also by d the unique extension of d to the field k(x) of rational functions.
Proposition 2.1. Let d : k[x] → k[x] be a nonzero derivation and let ϕ = d(x). Let a, b ∈ k[x], b 6= 0 and gcd(a, b) = 1. Then the following properties are equivalent:
(1) d(ab) ∈ k[x];
(2) b divides ψ, where ψ = gcd(ϕ, ϕ0).
Proof. Put ϕ = gψ, ϕ0 = f ψ, where f, g ∈ k[x], gcd(f, g) = 1.
(2) ⇒ (1). Assume that ψ = cb with 0 6= c ∈ k[x]. Then we have:
d(ab) = d(accb) = d(acψ) = d(acgϕ ) = ϕ12 (d(acg)ϕ − acgd(ϕ)) = ϕ12 ((acg)0ϕ2− acgϕ0ϕ))
= (acg)0− acgϕϕ 0 = (acg)0− acϕψ0 = (acg)0 − acf, that is, d(ab) ∈ k[x].
(1) ⇒ (2). Assume that d(ab) = u ∈ k[x]. Then b12(d(a)b − ad(b)) = u, so a0bϕ − ab0ϕ = b2u.
If b ∈ k r {0}, then of course b | ψ. Assume that deg b > 1, and let p be an irreducible polynomial from k[x] dividing b. Let b = psb1, b1 ∈ k[x], s > 1, p - b1, and let ϕ = ptϕ1, ϕ1 ∈ k[x], t > 0, p - ϕ1. Then, by the above equality, we have
(a0b1ϕ1− ab01ϕ1) pt+s− sap0b1ϕ1ps+t−1= p2sb21u.
If t 6 s, then this equality implies that p divides ap0b1ϕ1. But it is a contradiction because p - p0, p - b1, p - ϕ1 and p - a (since p | b and gcd(a, b) = 1). Hence, t > s, that is, t − 1 > s.
But pt−1 divides ϕ0, so ps divides gcd(ϕ, ϕ0) = ψ.
Now let b = vps11· · · psrr be a decomposition of b into irreducible polynomials (here v ∈ k r {0}, s1 > 1, . . . , sr > 1, and p1, . . . , pr are pairwise nonassociated irreducible polynomials from k[x]). Then, by the above observation, all the polynomials ps11, . . . , psrr divide ψ, so b divides ψ.
Corollary 2.2. Let d : k[x] → k[x] be a nonzero derivation and let ϕ = d(x). Assume that gcd(ϕ, ϕ0) = 1 and let α ∈ k(x). Then d(α) ∈ k[x] ⇐⇒ α ∈ k[x].
3 Derivations from k[x] to k[x][t,d]
Let R be a commutative k-algebra and let R[t, d] be an Ore extension. We denote by M(R, d) the set of all derivations from R to R[t, d]. Observe that if δ1, δ2 ∈ M(R, d), then δ1 + δ2 ∈ M(R, d). If δ ∈ M(R, d) and a ∈ R, then (since R is commutative) the mapping rδ also belongs to M(R, d). Thus, M(R, d) is an R-module.
Now let R = k[x]. It is clear that every derivation δ : R → R[t, d] is uniquely determined by the value δ(x). Moreover, if w is an arbitrary element of R[t, d], then there exists a unique derivation δ ∈ M(R, d) such that δ(x) = w. For every n > 0 denote by δn the unique derivation from R to R[t, d] such that δn(x) = tn. Then every derivation δ : R → R[t, d] has a unique decomposition of the form δ = a0δ0+ · · · + anδn, where n > 0 and a0, . . . , an ∈ R = k[x]. This means, that M(k[x], d) is a free k[x]-module and the set {δn; n = 0, 1, . . . } is its basis. Now we introduce a new basis of M(k[x], d).
Assume that ϕ := d(x) 6= 0. If f is a given element of R[t, d], then for every g ∈ R[t, d]
the element [f, g] = f g − gf is of the form rntn+ · · · + r1t + r0, where r0, . . . , rn belong to d(R), so [f, g] = ϕh for some h ∈ R[t, d]. This means that the mapping ϕ1[f, ] (for any f ∈ R[t, d]) restricted to R is a derivation from R to R[t, d], that is, this mapping belongs to M(k[x], d).
If n > 0, then we denote by en the restriction of the mapping ϕ1 1
n+1tn+1, to R.
Thus, we have the derivations e0, e1, e2, . . . belonging to M(k[x], d). In particular:
e0(x) = 1, e1(x) = t + 12ϕ0,
e2(x) = t2+ ϕ0t + 13(ϕ0ϕ)0,
e3(x) = t3+32ϕ0t2+ (ϕ0ϕ)0t + 14(ϕ000ϕ2+ 4ϕ0ϕ00ϕ + (ϕ0)3) . It is easy to check the following proposition.
Proposition 3.1. For every n = 0, 1, 2, . . . we have en(x) = tn+ n
2ϕ0tn−1+ hn−2,
where hn−2 is an element of k[x][t, d] of the degree (with respect to t) smaller then n − 1.
Proposition 3.2. The set {e0, e1, e2, . . . } is a basis of the k[x]-module M(k[x], d).
Proof. Suppose that a0e0 + · · · + anen = 0, where n > 0 and a0, . . . , an ∈ k[x].
Then, by Proposition 3.1, 0 = a0e0(x) + · · · + anen(x) = antn+ h, where h ∈ R[t, d] and degth 6 n − 1. So an = 0 and, repeating the same argument, an−1 = · · · = a0 = 0. This means that the derivations e0, e1, . . . are linearly independent over k[x].
Assume that δ : k[x] → k[x][t, d] is a derivation. Let δ(x) = antn + · · · + a1t1+ a0, where n > 0 and a0, . . . , an ∈ k[x]. We will show, using an induction with respect to n, that δ is a linear combination of e0, e1, . . . over k[x]. For n = 0 this is clear. Let n > 0 and assume that this is true for all derivations of the form δ with degtδ(x) < n.
Consider the derivation δ1 = δ − anen. Observe that degtδ1(x) < n (Proposition 3.1).
So, by induction, δ1 = b0e0 + · · · + bn−1en−1 for some b0, . . . , bn−1 ∈ k[x]. Therefore, δ = δ1+ anen = b0e0 + · · · + bn−1en−1+ anen is a linear combination of e0, . . . , en over k[x].
In the next theorem, which is true for arbitrary k-domain R (even if k is of positive characteristic), we show when a derivation δ : k[x] → k[x][t, d] has an extension to a derivation of k[x][t, d].
Theorem 3.3. Let d be a derivation of a commutative k-domain R, and let A = R[t, d].
Assume that δ : R → A is a derivation and f ∈ A. Then the following conditions are equivalent.
(1) There exists a unique derivation D : A → A such that D|R = δ and D(t) = f . (2) [f, r] + [t, δ(r)] = δ(d(r)), for every r ∈ R.
Proof. (1) ⇒ (2). Let D : A → A be a derivation such that D|R = δ and D(t) = f . Let r ∈ R. Since [t, r] = d(r), we have
δ(d(r)) = D(d(r)) = D([t, r]) = [D(t), r] + [t, D(r)] = [f, r] + [t, δ(r)].
(2) ⇒ (1). We will define a function D : R[t, d] → R[t, d]. If w ∈ R[t, d], then we define the element D(w), using an induction with respect to deg w, in the following way.
If w = 0, then we put D(w) = 0. If deg w = 0, then w ∈ R and we put D(w) = δ(w).
Let deg w = n + 1, where n > 0, and assume that we know D(u) for all u ∈ R[t, d] with deg u 6 n. Then w is of the form w = ut + a, where a ∈ R, u ∈ R[t, d], deg u = n. We define
D(w) := δ(a) + D(u)t + uf.
So, we have a function D : R[t, d] → R[t, d]. It is clear that this function is k-linear. Now we will show that D(uv) = D(u)v + uD(v) for all u, v ∈ R[t, d]. We use an induction with respect to deg(uv). Note that, since R is a domain, deg(uv) = deg(u) + deg(v).
If u = 0 or v = 0, then of course D(uv) = D(u)v + uD(v). Assume now that u 6= 0 and v 6= 0. If deg(uv) = 0, then uv ∈ R and then u, v ∈ R, so D(uv) = δ(uv) = δ(u)v + uδ(v) = D(u)v + uD(v).
Let deg(uv) = 1. Then (deg(u) = 1 and deg v = 0) or (deg(u) = 0 and deg v = 1).
Assume that u = a + bt, v = c, where a, b, c ∈ R. Then we have:
D(uv) − D(u)v − uD(v) = D((a + bt)c) − D(a + bt)c − (a + bt)D(c)
= D(ac + bct + bd(c)) − D(a + bt)c − (a + bt)D(c)
= δ(ac + bd(c)) + δ(bc)t + bcf − (δ(a) + δ(b)t + bf )c − (a + bt)δ(c)
= δ(a)c + aδ(c) + δ(b)d(c) + bδ(d(c)) + δ(b)ct + bδ(c)t + bcf
−δ(a)c − δ(b)ct − δ(b)d(c) − bf c − aδ(c) − bδ(c)t − bd(δ(c))
= b · (δ(d(c)) + cf − f c − d(δ(c))) = b · (δ(d(c)) − [f, c] − [t, δ(c)])
= b · 0 = 0,
so in this case, D(uv) = D(u)v + uD(v). Now let u = c and v = a + bt, where a, b, c ∈ R.
Then
D(uv) − D(u)v − uD(v) = D(ca + cbt) − δ(c)(a + bt) − cD(a + bt)
= δ(ca) + δ(cb)t + cbf − δ(c)a − δ(c)bt − cδ(a) − cδ(b)t − cbf
= (δ(ca) − δ(c)a − cδ(a)) + (δ(cb) − δ(c)b − cδ(b))t
= (δ(ca) − δ(ca)) + (δ(cb) − δ(cb))t
= 0 + 0t = 0.
Hence, we proved that if deg(uv) 6 1, then D(uv) = D(u)v + uD(v).
Let deg(uv) = n + 1, where n > 1, and assume that our statement is true for all products of degrees 6 n.
Case 1. Assume that deg v = 0. Let v = b ∈ R. Then deg u = n + 1. Let u = a + wt, where a ∈ R, w ∈ R[t, d], deg w 6 n. Then uv = (a + wt)b = ab + wd(b) + wbt and hence
D(uv) = δ(ab) + D(wd(b)) + D(wb)t + wbf, D(u)v + uD(v) = (δ(a) + D(w)t + wf )b + (a + wt)δ(b).
Observe that deg wd(b) 6 n and deg wb 6 n. Hence, by induction, D(uv)−D(u)v−uD(v)
= δ(ab) + D(wd(b)) + D(wb)t + wbf − δ(a)b − D(w)tb − wf b − aδ(b) − wtδ(b)
= D(w)d(b) + wD(d(b)) + D(w)bt + wD(b)t + wbf − D(w)tb − wf b − wtδ(b)
= wD(d(b)) + wD(b)t + wbf − wf b − wtδ(b)
= w(D(d(b)) + D(b)t + bD(t) − D(t)b − tδ(b))
= w(D(d(b)) + D(bt) − D(tb)) = wD(d(b) + bt − tb) = wD(0) = 0.
Case 2. Assume that deg v > 1. Then deg u 6 n and then v = b + wt, where b ∈ R, w ∈ R[t, d], w 6= 0, deg w < deg v. So we have uv = ub + uwt, and deg(ub) 6 n, deg(uw) 6 n. Hence, by induction, we have:
D(uv) − D(u)v − uD(v) = D(ub + uwt) − D(u)(b + wt) − uD(b + wt)
= D(ub) + D(uw)t + uwD(t) − D(u)b − D(u)wt − uD(b) − uD(w)t − uwD(t)
= D(u)wt + uD(w)t − D(u)wt − uD(w)t = 0.
Therefore, we proved by induction that the mapping D is really a derivation of R[t, d].
Note that D|R = δ and D(t) = f and it is clear that such D is unique.
Lemma 3.4. Let d : R → R be a derivation of a commutative k-algebra R, and let A = R[t, d] be the Ore extension. Assume that δ : R → A is a derivation and f ∈ A.
Then the set M := {r ∈ R; [f, r] + [t, δ(r)] = δ(d(r))} is a k-subalgebra of R.
Proof. It is clear that 1 ∈ M and that if a, b ∈ M , then a + b ∈ M . Now let a, b ∈ M . We will show that ab ∈ M . We have:
[f, ab] + [t, δ(ab)] = [f, a]b + a[f, b] + [t, δ(a)b + aδ(b)]
= [f, a]b + a[f, b] + [t, δ(a)]b + δ(a)[t, b] + [t, a]δ(b) + a[t, δ(b)]
= ([f, a] + [t, δ(a)])b + a([f, b] + [t, δ(b)]) + δ(a)[t, b] + [t, a]δ(b)
= δ(d(a))b + aδ(d(b)) + δ(a)d(b) + d(a)δ(b) = δ(d(a)b + ad(b))
= δ(d(ab)).
Hence, ab ∈ M .
As a consequence of Lemma 3.4 and Theorem 3.3 we obtain
Corollary 3.5. Let R be a commutative k-domain generated over k by a subset S ⊆ R.
Let d : R → R be a derivation and let A = R[t, d] be the Ore extension. Assume that δ : R → A is a derivation and f ∈ A. Then the following conditions are equivalent.
(1) There exists a unique derivation D : A → A such that D|R = δ and D(t) = f . (2) [f, s] + [t, δ(s)] = δ(d(s)), for every s ∈ S.
Corollary 3.6. Let R = k[x] and let A = R[t, d] be an Ore extension. Assume that δ : R → A is a derivation and f ∈ A. Then there exists a unique derivation D : A → A such that D|R = δ and D(t) = f if and only if
(∗) [f, x] + [t, δ(x)] = δ(d(x)).
4 Scalar inner derivations
If f ∈ R[t, d], then we will denote by Wf the inner derivation [f, ]. We will say that Wf is a scalar inner derivation if all the coefficients of f belong to k, that is, if f ∈ k[t].
Since Wf = Wf +c for any c ∈ k, we may always assume that f is without constant term (i.e., f (0) = 0). Note that if Wf is a scalar inner derivation, then Wf(t) = 0. The following proposition says that if R = k[x], then the converse of this fact is also true.
Proposition 4.1. Let d be a derivation of R = k[x] with ϕ = d(x) 6= 0, and let D be a derivation of R[t, d]. If D(t) = 0, then there exists a unique polynomial f ∈ k[t] such that D = Wf and f (0) = 0.
Proof. Assume that D(t) = 0 and let δ : R → R[t, d] be the restriction of D to R. Using Proposition 3.2 we have an equality of the form δ = anen+ · · · + a1e1+ a0e0, where n > 0 and an, . . . , a0 ∈ k[x]. We will show, by an induction on n, that there exists a polynomial f ∈ k[t] such that D = Wf and f (0) = 0. Since the case D = 0 is obvious, we may assume that an 6= 0.
Let n = 0. Then δ(x) = a0 and, by (∗), [0, x] + [t, δ(x)] = δ(ϕ), so a0ϕ0 = d(a0) = [t, a0] = [t, δ(x)] = δ(ϕ) = ϕ0a0, that is, a0ϕ0 = ϕ0a0. This implies that (aϕ0)0 = 0, so a0 = c1ϕ for some c1 ∈ k, and we have δ = a0e0 = c1ϕe0 = c1ϕ1f[t, ] = [c1t, ].
Let f = c1t. Then f ∈ k[t], f (0) = 0 and D = Wf (because D(t) = 0 = Wf(t) and D(x) = δ(x) = [c1t, x] = Wf(x)).
Assume now that n > 1 and that for all numbers smaller then n our statement in true. Let δ(x) = anen + · · · + a0e0, with an 6= 0 and a0, . . . , an ∈ k[x]. Then, by (∗), [0, x] + [t, δ(x)] = δ(ϕ). Comparing in this equality the coefficients with respect to tn we get the equality a0nϕ = anϕ0, which implies that an = pn+1ϕ for some nonzero pn+1 ∈ k.
Now we have:
anen= pn+1ϕen= pn+1ϕ 1
(n + 1)ϕ[tn+1, ] = [cn+1tn=1, ],
where cn+1= n+11 pn+1 ∈ k r {0}.
Consider the derivation D1 = D − [cn+1tn+1, ]. Observe that D1(t) = 0 and D1|R = an−1en−1+ · · · + a0e0. So, by induction, D1 = [h, ] for some h ∈ k[t] with h(0) = 0.
Hence D = D1+ [cn+1tn+1, ] = Wh+ [cn+1tn+1, ] = Wf, where f = cn+1tn+1+ h ∈ k[t].
This completes the proof of existence of f . We will show yet, that such a polynomial f is unique. Suppose that f, g ∈ k[t], Wf = Wg and f (0) = g(0) = 0. Then Wf −g = 0.
Put h := f − g = cptp+ · · · + c1t1 and suppose that h 6= 0. Then ap 6= 0, p > 1, and we have
0 = Wh(x) = [cptp+ · · · + c1t + c0, x] = pcpϕtp−1+ hp−2,
for some hp−2 ∈ k[x][x, d] with degthp−26 p − 2. We have a contradiction: 0 = pcpϕ 6= 0.
Hence, h = 0, that is, f = g.
5 Derivations ∆
aLet R be a k-domain, and let d : R → R be a derivation. If a ∈ R, then Theorem 3.3 implies that there exists a unique derivation ∆a : R[t, d] → R[t, d] such that
∆a(t) = a, ∆a(r) = 0 for every r ∈ R.
In particular, ∆a(t) = a, ∆a(t2) = 2at + d(a), ∆a(t3) = 3at2 + 3d(a)t + d2(a) and
∆a(t4) = 4at3+ 6d(a)t2+ 4d2(a)t + d3(a).
Proposition 5.1. The derivation ∆a is inner if and only if a ∈ d(R).
Proof. Assume that ∆a = [f, ], for some f = antn+ · · · + a1t1+ a0 ∈ R[t, d]. Then a = ∆a(t) = [f, t] = −d(an)tn− · · · − d(a1)t − d(a0). In particular, a = d(−a0) ∈ d(R).
Assume now that a = d(b) for some b ∈ R and consider the inner derivation D = [−b, ]. We have here: ∆a(r) = D(a) = 0 for all r ∈ R, and D(t) = [−b, t] = [t, b] = d(b) = a = ∆a(t). Hence, ∆a= [−b, ] is an inner derivation of R[t, d].
As a consequence of this proposition we obtain
Corollary 5.2. Let R be a k-domain. If every derivation of R[t, d] is inner, then d is surjective.
Proposition 5.3. If D is a derivation of k[x][t, d] such that D(t) = a ∈ k[x], then D = W + ∆a, where W is a scalar inner derivation.
Proof. Let W = D − ∆a. Since W (t) = 0, Proposition 4.1 implies that W is a scalar inner derivation.
Proposition 5.4. Let d be a derivation of R = k[x] with ϕ = d(x) 6= 0. If a ∈ R, then
∆a = W + ∆r, where W is an inner derivation of R[t, d], and r ∈ k[x] is the remainder of a in the divisibility by ϕ.
Proof. Let a = bϕ + r, b, r ∈ k[x], deg r < deg ϕ. Then ∆a = ∆bϕ+r = ∆bϕ+ ∆r and, by Proposition 5.1, the derivation ∆bϕ is inner.
Proposition 5.5. Let R = k[x], d : R → R be a derivation with ϕ = d(x) 6= 0. Then for every a ∈ R there exists a unique derivation Da of R[t, d] such that Da(t) = a and Da(x) = ϕ. The derivation Da is equal to [t, ] + ∆a. The derivation Da is inner if and only if a is divisible by ϕ.
Proof. Let D = [t, ] + ∆a. Then D(x) = [t, x] + ∆a(x) = d(x) + 0 = ϕ and D(t) = [t, t] + ∆a(t) = 0 + a = a. This implies that the derivation Da exists and that Da = D. Since [t, ] is inner, the derivation Da is inner if and only if ∆a is inner. The last statement is equivalent (by Proposition 5.1) to the divisibility of a by ϕ.
Let R be a k-domain. Note that the inner derivation D0 = [t, ] : R[t, d] → R[t, d]
is an extension of the derivation d such that D0(t) = 0. If d 6= 0, then every derivation D : R[t, d] → R[t, d], such that D|R = d, satisfies the condition: D(t) ∈ R. The same is true if D|R = δ, where δ : R → R is a derivation such that δd = dδ.
Consider the derivation ∆ = ∆1. This is a unique derivation of R[t, d] such that
∆(t) = 1 and ∆(r) = 0 for all r ∈ R. In this case we have:
∆(antn+ · · · + a1t1+ a0) = nantn−1+ (n − 1)an−1tn−2+ · · · + 2a2t + a1.
So ∆ is equal to usual partial derivative ∂t∂ of R[t, d]. Note that (by Proposition 5.1) ∆ is inner if and only if 1 ∈ d(R).
6 Polynomials of the form D(t)
Proposition 6.1. Let d be a nonzero derivation of a k-domain R. Assume that D : R[t, d] → R[t, d] is a derivation such that D(t) = untn+ · · · + u1t1+ u0, where n > 0 and u0, . . . , un∈ R. If D is inner, then u0, . . . , un∈ d(R).
Proof. Assume that D = [f, ], where −f = amtm+ · · · + a1t + a0 ∈ R[t, d]. Then untn+ · · · + u1t1+ u0 = D(t) = [f, t] = [t, −f ] = d(am)tm+ · · · + d(a1)t + d(a0), Thus, we have u0 = d(a0) ∈ d(R), u1 = d(a1) ∈ d(R) and so on.
Proposition 6.2. Let R = k[x] and let d : R → R be a nonzero derivation with d(x) = ϕ 6= 0. Let D : R[t, d] → R[t, d] be a derivation such that D(t) = untn+ · · · + u1t + u0, where n > 0 and u0, . . . , un∈ R. Then D is inner if and only if u0, . . . , un∈ d(R).
Proof. Assume that u0, . . . , un ∈ d(R). Let u0 = d(b0), . . . , un = d(bn), for some b0, . . . , bn∈ R. Let f = −bntn− bn−1tn−1− · · · − bat − b0 and let D1 = D − [f, ]. Then D1 is a derivation of R[t, d] such that D(t) = 0. This derivation is inner (by Proposition 4.1). Hence the derivation D = D1 + [f, ] is inner. The opposite implication follows from Proposition 6.1.
Corollary 6.3. Let R = k[x] and let d : R → R be a nonzero derivation with d(x) = ϕ 6=
0. Assume that D : R[t, d] → R[t, d] is a derivation such that D(t) = untn+ · · · + u1t + u0, where n > 0 and u0, . . . , un∈ R. Then D = W + D1, where W is an inner derivation of R[t, d] and D1 is a derivation of R[t, d] such that D1(t) = rntn+ · · · + r1t1+ r0, where each ri (for i = 0, 1, . . . , n) is the remainder in the divisibility of ui by ϕ.
Proof. Let ui = aiϕ + ri, ai, ri ∈ k[x], deg ri < deg ϕ, for i = 0, 1, . . . , n. Note that each aiϕ belongs to d(R). Put aiϕ = d(bi) for i = 0, . . . , n, and let g = −bntn − · · · − b1t1 − b0. Consider the derivation D1 = D − [g, ]. Then D1(t) = rntn+ · · · + r1t1+ r0 and we have D = [g, ] + D1.
Note also the following
Proposition 6.4. Let d be a nonzero derivation of R = k[x], and let D be a derivation of R[t, d]. If D(x) = 0, then D(t) ∈ R. If D(x) 6= 0, then degtD(t) 6 1 + degtD(x), and moreover, there exists an inner derivation W of R[t, d] such that the D1(t) = D(t) and degtD1(t) = 1 + degtD1(x), where D1 := D + W .
Proof. Let D(t) = untn+ · · · + u0, D(x) = amtm+ · · · + a0, where all the coefficients u0, . . . , un, a0, . . . , am belong to R.
Assume that D(x) = 0 and suppose that n > 1 and un 6= 0. Then (∗) implies that [D(t), x] = 0. But [D(t), x] = nund(x)tn−1+ h, where h ∈ R[t, d] and degth < n − 1. So we have a contradiction: 0 = nund(x) 6= 0. Thus, if D(x) = 0, then D(t) = u0 ∈ R.
Assume now that D(x) 6= 0. Then m > 0, am 6= 0 and, by (∗), nund(x)tn−1+ h + d(am)tm + · · · d(a0) = ϕ0amtm + v, where ϕ = d(x), h, v ∈ R[t, d], degth < n − 2 and degtv < m. If n − 1 > m, then nund(x) = 0, but this is a contradiction. So, n − 1 6 m, that is, degtD(t) 6 1 + degtD(x).
Assume now that n − 1 < m. Then, by the above equality, d(am) = ϕ0am. Hence, a0mϕ = ϕ0am so,
am
ϕ
0
= 0 and this implies that am = cϕ, for some c ∈ k. Let W = [cm+11 tm+1, ], and let D1 = D − W . Then D1(t) = D(t) and degtD1(x) = m1 < m. If again n−1 < m1, then we repeat the same procedure for the derivation D1. Repeating this argument several times we obtain an inner derivation W of R[t, d] and a new derivation D1 = W + d having the required properties.
7 On some differential equation
In this section we consider a differential equation of the form uϕ = aϕ0− a0ϕ,
where ϕ is a nonzero polynomial from k[x], and u, a ∈ k[x].
Lemma 7.1. Let ϕ, u, a be as above. If deg u < deg ϕ and gcd(ϕ, ϕ0) = 1, then u = 0.
Proof. The equality uϕ = aϕ0− a0ϕ and the assumption gcd(ϕ, ϕ0) = 1 imply that ϕ divides a. Let a = bϕ with b ∈ k[x]. Then uϕ = bϕϕ0 − b0ϕ2− bϕϕ0 = −bϕ2, that is, u = −bϕ. But deg u < deg ϕ, so u = 0.
Lemma 7.2. Let u, ϕ ∈ k[x], u 6= 0, ϕ 6= 0, m = deg ϕ > 2, deg u < m. Assume that there exists a polynomial a ∈ k[x] such that uϕ = aϕ0− a0ϕ. Then deg u 6 m − 2.
Proof. Let ϕ = ϕmxm + · · · + ϕ1x + ϕ0, ϕ0, . . . , ϕm ∈ k, ϕm 6= 0. Suppose that deg u = m − 1. Let u = um−1xm−1 + · · · + u0, u0, . . . , um−1 ∈ k, um−1 6= 0. Let uϕ = aϕ0 − a0ϕ, a ∈ k[x]. Then a 6= 0 (because uϕ 6= 0). Put a = anxn+ · · · + a0, a0, . . . , an ∈ k, n > 0, an6= 0. Now we have the equality: (um−1xm−1+ · · · + u0)(ϕmxm+
· · · + ϕ0) = (anxn+ · · · + a0)(mϕmxm−1+ · · · + ϕ1) − (nan+ · · · + a1)(ϕmxm+ · · · + ϕ0).
Suppose that n > m. Comparing in this equality the coefficients of monomials of degree n + m − 1, we obtain that manϕm − nanϕm = 0, that is, m = n. But it is a contradiction. Hence, n 6 m.
Suppose that n = m. Then, comparing the coefficients of monomials of degree n + m − 1, we have a contradiction: 0 6= um−1ϕm = manϕm− manϕm = 0. Hence, n < m.
But in this case there is also a contradiction: 0 6= um−1ϕm = 0. Therefore, deg u 6 m − 2.
Lemma 7.3. Let u, ϕ ∈ k[x], u 6= 0, ϕ 6= 0, deg ϕ > 2. Assume that there exists a polynomial a ∈ k[x] such that uϕ = aϕ0− a0ϕ. Then deg u > s − 1, where s is the number of all pairwise different roots of the polynomial ϕ belonging to an algebraic closure k of k.
Proof. Put m := deg ϕ > 2. We may assume that ϕ is monic. Let ϕ = (x − a1)n1· · · (x − as)ns,
where a1, . . . , as are pairwise different elements of k, and where n1 > 1, . . . , ns > 1, n1+ · · · + ns= m. We will use the following notations:
g := (x − a1) · · · (x − as), f :=
s
P
i=1
ni(x − a1) · · · \(x − ai) · · · (x − as) =
s
P
i=1
nix−ag
i, ψ := (x − a1)n1−1· · · (x − as)ns−1.
Observe that ϕ = gψ, ϕ0 = f ψ, ψ = gcd(ϕ, ϕ0) and gcd(g, f ) = 1. Moreover, g0 =
s
P
i=1
(x − a1) · · · \(x − ai) · · · (x − as) =
s
P
i=1 g
x−ai, so we have:
f − g0 =
s
P
i=1
(ni− 1)(x − a1) · · · \(x − ai) · · · (x − as) =
s
P
i=1
(ni− 1)x−ag
i.
If h is a nonzero polynomial from k[x], then we denote by I(h) the initial nonzero monomial of h. In our case we have:
I(ϕ) = xm, I(ϕ0) = mxm−1, I(ψ) = xm−s, I(g) = xs, I(g0) = sxs−1,
and, if m > s, then I(f − g0) = (m − s)xs−1.
Now let a ∈ k[x] be such that uϕ = aϕ0 − a0ϕ. Then of course a 6= 0 and ugψ = af ψ − a0gψ, so ug = af − a0g. This implies that g divides a (because gcd(f, g) = 1). Let a = gb for some b ∈ k[x] r {0}. Then ug = bgf − b0g2− bg0g, so u = bf − b0g − bg0, that is,
(1) u = b(f − g0) − b0g.
Observe that ψ · (f − g0) = ψ0g. In fact: ψ · (f − g0) − ψ0g = ψf − (ψg0+ ψ0g) = ϕ0− (ψg)0 = ϕ0 − ϕ0 = 0. This means, that if there exists a polynomial b satisfying the equality (1), then every polynomial of the form b + cψ, where c ∈ k, also satisfies this equality. As a consequence of this fact, we may assume that b ∈ k[x] is a nonzero polynomial satisfying (1) and the coefficient of b with respect to xm−s is equal to zero.
We must prove that deg u > s − 1. Suppose that deg u < s − 1. Let u = us−2xs−2+
· · · + u0, u0, . . . , us−2 ∈ k, ui 6= 0 for some i ∈ {0, 1, . . . , s − 2}. Let b = bpxp+ · · · + b0, b0, . . . , bp ∈ k, p > 0 and bp 6= 0. Then the equality (1) is of the form:
us−2xs−2+ · · · + u0 = (bpxp+ · · · )((m − s)xs−1+ · · · ) − (pbpxp−1+ · · · )(xs+ · · · ).
Since p + s − 1 > s − 1 > s − 2, we have ((m − s) − p)bp = 0. But bp 6= 0, so p = (m − s).
However, by our assumption, the coefficient bm−s is equal to zero. Thus, we have a contradiction: 0 6= bp = bm−s = 0. Therefore, deg u > s − 1.
Remark 7.4. If f, g, ϕ, ψ are as in the proof of Lemma 7.3, then ψ2 divides ϕψ0 and
ϕψ0
ψ2 = f − g0. In fact, f − g0 = ϕψ0 −
ϕ ψ
0
= ϕψ0 − ψ12(ϕ0ψ − ϕψ0) = ϕ0ψ−ϕψ0ψ+ϕψ2 0 = ϕψψ20. Proposition 7.5. Let 0 6= ϕ ∈ k[x], deg ϕ > 1. The following two properties are equiva- lent.
(1) There exists a polynomial a ∈ k[x] such that ϕ = aϕ0− a0ϕ.
(2) ϕ = c(x − λ)m, for some λ, c ∈ k, c 6= 0 and m > 2.
Proof. (1) ⇒ (2). Assume that ϕ = aϕ0− a0ϕ for some a ∈ k[x]. Let m = deg ϕ and let s be as in Lemma 7.3. Using Lemma 7.1 for u = 1, we obtain that m > 2. Moreover, Lemma 7.3 implies that s = 1. Hence, ϕ = c(x − λ)m, 0 6= c ∈ k, λ ∈ k and m > 2.
(1) ⇒ (2) Assume that ϕ is of the form (2), and take a = m−11 (x − λ). Then:
ϕ − aϕ0+ a0ϕ = c(x − λ)m− mc(x − λ)m−1m−11 (x − λ) + m−11 c(x − λ)m
= c(x − λ)m 1 −m−1m +m−11 = c(x − λ)m0 = 0.
Note that every polynomial a ∈ k[x] of the form a = m−11 (x − λ) + αϕ, with α ∈ k, also satisfies the equality ϕ = aϕ0− a0ϕ.
Lemma 7.6. Let d be a nonzero derivation of R = k[x] and let ϕ = d(x) 6= 0. Let u ∈ k[x]r{0} and assume that there exists a polynomial a ∈ k[x] such that uϕ = aϕ0−a0ϕ.
For any n > 1 consider the inner derivation Dn= [aϕtn, ] of the ring k(x)[t, d]. Then:
(1) Dn(R[t, d]) ⊆ R[t, d].
(2) Denote by Dn the restriction of Dn to R[t, d]. Then Dn is a unique derivation of R[t, d] such that Dn(t) = utn, Dn(x) = naen−1(x).
Proof. We have: Dn(t) = [aϕtn, t] = −d(ϕa)tn = −a0ϕ−aϕϕ2 0ϕtn = uϕϕ2ϕtn = utn and Dn(x) = [ϕatn, x] = ϕa[tn, x] = nanϕ1 [tn, x] = naen−1(x). This implies (1) and (2). Theorem 7.7. Let R = k[x] and let d : R → R be a nonzero derivation with d(x) = ϕ 6= 0.
Let u ∈ k[x] r {0}. The following properties are equivalent.
(1) There exists a ∈ k[X] such that uϕ = aϕ0 − a0ϕ.
(2) There exists a derivation D1 of R[t, d] such that D1(u) = ut.
(3) There exist n > 1 and a derivation Dn of R[t, d] such that Dn(t) = utn. (4) For every n > 1 there exists a derivation Dn of R[t, d] such that Dn(t) = utn. Proof. The implication (1) ⇒ (4) follows from Lemma 7.6. The implications (4) ⇒ (2) and (2) ⇒ (3) are obvious. Now we will prove the implication (3) ⇒ (1).
Assume that Dn : R[t, d] → R[t, d] is a derivation such that Dn(t) = utn, where n > 1 is a fixed natural number. Let δ : R → R[t, d] be the restriction of Dn to R. We know, by Proposition 3.2, that δ = apep+ · · · + a1e1+ a0e0 for some a0, . . . , ap ∈ R. If δ = 0, then by (∗), 0 = [utn, x] = nud(x)tn−1+ · · · and we have a contradiction: 0 = nuϕ 6= 0. Hence δ 6= 0, p > 0 and ap 6= 0. We may assume (by Proposition 6.4) that n − 1 = p. Now comparing in (∗) the coefficients of tn−1, we obtain the equality nud(x) + d(ap) = ϕ0ap, that is, nuϕ + a0pϕ = ϕ0ap, so uϕ = aϕ0 − a0ϕ for a = 1nap. This completes the proof. Theorem 7.8. Let R = k[x] and let d : R → R be a nonzero derivation with d(x) = ϕ 6= 0.
Let f = untn+ · · · + u1t + u0 ∈ R[t, d], where n > 1, un 6= 0, u0, . . . , un ∈ R. Then the following conditions are equivalent.
(1) There exists a derivation D of R[t, d] such that D(t) = f .
(2) For every i ∈ {0, 1, . . . , n} there exists a derivation Mi of R[t, d] such that Mi(t) = uiti.
(3) For every i ∈ {1, . . . , n} there exists a polynomial ai ∈ k[x] such that uiϕ = aiϕ0 − a0iϕ.
Proof. (2) ⇒ (1). Let D := Mn+ · · · + M1+ M0. Then D is a derivation of R[t, d]
and D(t) = f .
(2) ⇒ (3). This follows from Theorem 7.7.
(3) ⇒ (2). Let i ∈ {0, 1, . . . , n}. If i 6= 0, then a derivation Mi exists by Theorem 7.7.
For i = 0 we may put M0 := ∆u0.
(1) ⇒ (2). Let D be a derivation of R[t, d] such that D(t) = f , and let δ : R → R[t, d]
be the restriction of D to R. Then, by Proposition 3.2, δ = apep+ · · · + a1e1+ a0e0, for some ap, . . . , a0 ∈ R. We may assume (by Proposition 6.4) that n − 1 = p. Moreover, comparing in the equality (∗) the coefficients of tn−1, we have nunϕ + apϕ = ϕ0ap, that is, unϕ = aϕ0− a0ϕ for a = n1ap. This implies, by Theorem 7.7, that there exists a derivation Mn: R[t, d] → R[t, d] such that Mn(t) = untn.
Consider now the new derivation D0 := D − Mn. Observe that D0(t) = un−1tn−1+
· · · + u1t1+ u0. So, doing the same for D0 we obtain the derivation Mn−1 and, repeating, we have derivations Mn, . . . , M1, and M0 = D − Mn− · · · − M1. This completes the proof.
8 Inner derivations of k(x)[t,d]
Let k(x) be, as usually, the field of rational functions in one variable over k (that is, k(x) is the field of fractions of k[x]), and let d : k[x] → k[x] be a nonzero derivation. Let A := k[x][t, d]. It is obvious that if D is a derivation of k(x)[t, d], then D(A) ⊆ A if and only if D(x) ∈ A and D(t) ∈ A.
Lemma 8.1. Let k(x), d, A be as above. Let λ ∈ k(x), n > 0. Consider the inner derivation W of k(x)[t, d] defined by W = [λtn, ]. Then W (A) ⊆ A ⇐⇒ d(λ) ∈ k[t].
Proof. Put ϕ := d(x). Observe that W (t) = [λtn, t] = −d(λ)tn and
W (x) = [λtn, x] = λ
n
X
i=1
n i
di(x)tn−i = λϕ · (rn−1tn−1+ · · · + r0),
for some r0, . . . , rn−1∈ k[x]. Hence, if W (A) ⊆ A, then it is clear that d(λ) ∈ k[x].
Assume now that d(λ) ∈ k[x]. Then W (t) = −d(λ)tn ∈ A. We will show that W (x) also belongs to A.
Let λ = ab, a, b ∈ k[x], b 6= 0, gcd(a, b) = 1. We know, by Proposition 2.1, that b | ψ = gcd(ϕ, ϕ0). Let ψ = bc, ϕ = gψ, where c, g ∈ k[x]. Then
λϕ = a
bϕ = ac
bcϕ = ac
ψϕ = ac
ψ gψ = acg ∈ k[x], and this implies that W (x) ∈ A. Hence W (A) ⊆ A.
Proposition 8.2. Let d : k[x] → k[x] be a nonzero derivation, d(x) = ϕ 6= 0, and let A = k[x][t, d]. Let
f := λntn+ · · · + λ1t1+ λ0 ∈ k(x)[t, d],
n > 0. Consider the inner derivations W , W0, . . . , Wn of the ring k(x)[t, d] defined as:
W = [f, ], W0 = [λ0t0, ], . . . , Wn = [λntn, ].
Then the following three conditions are equivalent.
(1) W (A) ⊆ A.
(2) W0(A) ⊆ A, · · · , Wn(A) ⊆ A.
(3) d(λ0), . . . , d(λn) ∈ k[x].
Moreover, if W (A) ⊆ A and D : A → A is the restriction of W to A, then D is an inner derivation of A if and only if λ0, . . . , λn∈ k[x].
Proof. The implication (2) ⇒ (1) follows from the fact that W = W0 + · · · + Wn. The equivalence (2) ⇐⇒ (3) follows from Lemma 8.1. We will prove the implication (1) ⇒ (3). Assume that W (A) ⊆ A. Then W (t) ∈ A. But
W (t) = [f, t] = [λntn+ · · · + λ0, t] = −d(λn)tn− · · · − d(λ0),
so d(λ0), . . . , d(λn) ∈ k[x]. Hence, the conditions (1), (2) and (3) are equivalent.
Let W (A) ⊆ A and let D := W |A. If λ0, . . . , λn ∈ k[x], then f ∈ A and it is clear that D is inner. Assume now that D = [g, ], for some g = rmtm+ · · · + r1t1+ r0 ∈ A.
Then r0, . . . , rm ∈ k[x] and we have:
d(λn)tn+ · · · + d(λ0) = [t, f ] = −W (t) = −D(t) = −[g, t] = d(rm)tm+ · · · + d(r0).
This means that d(λi) = d(ri) for i = 0, 1, . . . , n. Hence, each difference λi − ri belongs to Ker(d) = k, so λi − ri = ci ∈ k for i = 0, . . . , n. Therefore, λi = ri+ ci ∈ k[x], for all i = 0, . . . , n.
9 When all derivations of k[x][t,d] are inner?
It is well known that if R = k[x] and d = ∂x∂ , then the Ore extension R[t, d] coincides with the Weyl algebra with one pair of generators (see for example, [2], [3]). In this case it is also well known ([4]) that every derivation of R[t, d] is inner. Now we present a new proof of this fact and we show that if in an Ore extension of the form A = k[x][t, d] every derivation is inner, then A is isomorphic to the Weyl algebra over k with one pair of generators.
Theorem 9.1. If d is a nonzero derivation of R = k[x], then the following two properties are equivalent.
(1) Every derivation of R[t, d] is inner.
(2) d(x) ∈ k r {0}.
Proof. (1) ⇒ (2). Assume that every derivation of R[t, d] is inner. Then, in particular, the derivation ∆ = ∆1is inner, which implies, by Proposition 5.1, that 1 = d(b) for some b ∈ k[x]. Hence, 1 = b0d(x) so, d(x) is invertible in k[x], that is, d(x) ∈ k r {0}.
(2) ⇒ (1). Assume that d(x) = c ∈ k r {0}, and let D : R[t, d] → R[t, d] be an arbitrary derivation. If D(t) = 0, then we know (by Proposition 4.1) that D is inner.
Now let D(t) = untn+ · · · + u1t1 + u0, u0, . . . , un ∈ k[x], n > 0 and un 6= 0. Observe that, since 0 6= d(x) ∈ k, the image d(R) is equal to R. Hence, the coefficients u0, . . . , un belong to d(R) and hence, by Proposition 6.2, the derivation D is inner.
In the above theorem k is a field of characteristic zero. This assumption is important.
For positive characteristic we have:
Proposition 9.2. If char(k) = p > 0, then for any derivation d of R = k[x] there exists a derivation of R[t, d] which is not inner.
Proof. Suppose that there exists a derivation d of R = k[x] such that every derivation of R[t, d] is inner. Then, in particular, the derivation ∆ = ∆1 is inner (note that such derivation ∆1 exists even in positive characteristic), which implies that 1 = d(b) = b0d(x) for some b ∈ k[x]. Hence, d(x) = c ∈ k r {0}. It is easy to check that, in this case, the monomial xp−1 is not in the set d(R) and consequently, the derivation ∆xp−1 is not inner.
So we have a contradiction.
Now let k be again a field of characteristic zero. Note that Proposition 5.1 is true for any k-domain R. This implies that if any derivation of R[t, d] is inner, then the derivation d is surjective. We know, by [8] or [10] (Theorem 2.6.1), that if R is a field such that the transcendence degree of R over k is finite, then every derivation of R is not surjective. As a consequence of this fact, we obtain that if R is such a field, then there exists a derivation of R[t, d] which is not inner. In particular:
Proposition 9.3. If R is the field k(x) of rational functions over k, then for every deriva- tion d of R there exists a non-inner derivation of R[t, d].
Assume now that A = R[t, d] is the Weyl algebra with one pair of generators, that is, R = k[x] and d = ∂x∂ , where k is a field of characteristic zero. We know, by Theorem 9.1, that every derivation of A is inner. Thus, we have:
Proposition 9.4. Let R = k[x], d = ∂x∂ , A = R[t, d]. If f, g are such elements from A that [x, f ] = [t, g], then there exists w ∈ A such that [w, t] = f and [w, x] = g.
Proof. Consider the derivation δ : R → A such that δ(x) = g. The condition [x, f ] = [t, g] means, that [f, x] + [t, δ(x)] = 0 = δ(1) = δ(d(x)); it is exactly the condition (∗). Hence there exists a derivation D of A such that D(x) = g and D(y) = f . But, by Theorem 9.1, D = [w, ] for some w ∈ A. So, there exists w ∈ A such that [w, x] = D(x) = g and [w, t] = D(t) = f .
Note that in the Weyl algebra A = k[x][t,∂x∂ ] we have tnx = xtn+ ntn−1, for all n > 1.
In particular, if f = antn+ · · · + a0 ∈ A, then
[f, x] = nantn−1+ (n − 1)an−1tm−2+ · · · + 2a2t + a1.
Hence, in this case, the inner derivation [ , x] coincides with the derivation ∆ = ∆1. Let g = bmtm+ · · · + b0 ∈ A and f = antn+ · · · + a0 ∈ A. Using the above equality we see that, in this case, the condition (∗) (that is, [f, x] + [t, g] = 0) is equivalent to the equalities:
d(bi) + (i + 1)ai+1= 0, for i = 0, 1, . . . , n − 1,
and d(bj) = 0 for j > n. As a consequence of this equivalence and the fact that the derivation d = ∂x∂ is surjective, we obtain:
Proposition 9.5. Let A = k[x][t,∂x∂ ], (char(k) = 0). For any f ∈ A there exists a derivation D of A such that D(t) = f . For any g ∈ A there exists a derivation D of A such that D(x) = g.
Now let R be the ring k[x, y], of polynomials over k in two variables, and consider the Ore extension A = R[t, d], where d = ∂x∂ . Let δ : R → R[t, d] be the derivation ∂y∂ . It is clear, by Theorem 3.3, that there exists a unique derivation D : R[t, d] → R[t, d] such that D|R = δ and D(t) = 0. Observe that D is not inner. Indeed, suppose that D = [f, ] for some f = aptp+ · · · + a1t + a0 ∈ R[t, d]. Then a0, . . . , ap ∈ R = k[x, y] and, since D(t) = 0, d(a0) = · · · = d(ap) = 0, that is, a0, . . . , ap ∈ k[y]. But D(x) = 0, so ap = · · · , a1 = 0, and this implies that f = a0 ∈ k[y]. So we have a contradiction: 1 = D(y) = [a0, y] = 0.
Thus, we have:
Proposition 9.6. If A = k[x, y][t,∂x∂], then there exists a non-inner derivation of A. The next proposition is a consequence of the above fact.
Proposition 9.7. Let R = k[x, y] be the polynomial ring in two variables over a field k of characteristic zero, and let A = R[t, d] be an arbitrary Ore extension of R. Then there exists a non-inner derivation of A.
Proof. If d = 0, then it is obvious. Assume that d is a nonzero derivation of R = k[x, y] and suppose that every derivation of A = R[t, d] is inner. Then, by Proposition 5.1, d is surjective and this means, by [13], that the derivation d is locally nilpotent with a slice ([5], [10]). Now, by a theorem of Rentschler [12], there exists a k-automorphism σ : R → R such that σdσ−1 = ∂x∂ . This automorphism induces a natural isomorphism from A to k[x, y][t,∂x∂ ]. But, by Proposition 9.6, the algebra k[x, y][t,∂x∂ ] has a non-inner derivation. Thus, A has a non-inner derivation, and we have a contradiction.
10 The square-free case
Theorem 10.1. Let d be a nonzero derivation of R = k[x], and let ϕ = d(x) 6= 0 with deg ϕ > 1. Assume that gcd(ϕ, ϕ0) = 1. Then every derivation D of R[t, d] has a unique decomposition
D = W + ∆r,
where W is an inner derivation of R[t, d], and r ∈ R with deg r < deg ϕ. Moreover, if D is as above, then D is inner if and only if r = 0.
Proof. Let D be a derivation of R[t, d]. If D(t) = 0, then, by Proposition 4.1, D = Wf for some f ∈ k[t], so in this case: D = Wf + ∆0.
Assume now that D(t) 6= 0. By Corollary 6.3 we know that D = W + D1, where W is an inner derivation of R[t, d] and D1 is a derivation of R[t, d] such that D1(t) = untn+ · · · + u1t1+ u0, where u0, . . . , un ∈ R and deg ui < deg ϕ for all i = 0, 1, . . . , n.
By Theorem 7.8, for every i ∈ {1, . . . , n} there exists ai ∈ R such that uiϕ = aiϕ0− a0iϕ.
Since gcd(ϕ, ϕ0) = 1, Lemma 7.1 implies that ui = 0. Hence, we know that all the coefficients u1, . . . , unare equal to zero. This means, that D1 = ∆u0 with deg u0 < deg ϕ.
So we already proved that every derivation od R[t, d] is of the form W + ∆r with r ∈ R, deg r < deg ϕ. Moreover, by Proposition 5.1, such a derivation W + ∆r is inner if and only if r = 0.
Now we will show that the decomposition is unique. Suppose that W + ∆r = D = W1+ ∆r1, where W , W1are inner derivations and r, r1 ∈ R, deg r < deg ϕ, deg r1 < deg ϕ.
Then the derivation ∆r−r1 = W1− W is inner. But deg(r − r1) < deg ϕ so, by Proposition 5.1, r − r1 = 0. Therefore r = r1 and W1− W = ∆0 = 0. This completes the proof.
Corollary 10.2. Let R = k[x], let d : R → R be a nonzero derivation, and let ϕ = d(x) 6=
0 with deg ϕ > 1. Assume that gcd(ϕ, ϕ0) = 1. Let D be a derivation of R[t, d]. Then D(x) = ϕ · G, D(t) = ϕF t + r, for some F, G ∈ R[t, d] and r ∈ R. Moreover, D is inner if and only if ϕ | r.
Example 10.3. Let R = k[x] and let d : R → R be the derivation such that d(x) = x2+1.
There is no derivation D of R[t, d] such that D(t) = t. This fact is a consequence of Corollary 10.2.
Note the following consequence of Theorem 10.1.
Corollary 10.4. Let R = k[x], char(k) = 0 and let d := (ax + b)∂x∂ , with a, b ∈ k, a 6= 0.
For every derivation D of R[t, d] there exists a unique c = cD ∈ k such that D = W + c∆1, where W is an inner derivation of R[t, d]. Moreover, a derivation D of R[t, d] is inner if and only if cD = 0.
11 Derivations E
Hand a final description
Let d be a nonzero derivation of R = k[x] and let ϕ = d(x) 6= 0. Let m = deg ϕ and let s be the number of all pairwise different roots of the polynomial ϕ belonging to an algebraic closure k of k. Of course m > s. If m = s, then gcd(ϕ, ϕ0) = 1 and in this case we know, by Theorem 10.1, a description of all derivations of R[t, d].
In this section we assume that m − s > 1, and we denote by ψ the greatest common divisor of ϕ and ϕ0. Note that deg ψ = m − s > 1 and m > 2.
We will say that a polynomial H ∈ R[t, d] is special if it is of the form H = hntn+ · · · + h1t1,
where n > 1, hi ∈ R, deg hi < m − s for all i = 1, . . . , n. In particular, 0 from R[t, d] is a special polynomial.
If H ∈ R[t, d] is a special polynomial, then we denote by EH the derivation of R[t, d]
defined as:
EH(f ) =h
1 ψH, fi
, for all f ∈ R[t, d].
Observe that, by Propositions 2.1 and 8.2, the mapping EH is really a derivation from R[t, d] to R[t, d]. Moreover, it follows from Proposition 8.2 that the derivation EH is inner if and only if H = 0.
Lemma 11.1. Let n > 1 and let p ∈ {0, 1, . . . , m−s−1}. If H = xptn, then EH(t) = vptn, where vp ∈ k[x] r {0} and deg vp = p + s − 1. In particular, s − 1 6 deg vp 6 m − 2.
Proof. We use the same notations as in the proof of Lemma 7.3. Recall that ϕ = gψ, ϕ0 = f ψ, deg g = s and I(f − g0) = (m − s)xs−1.
