Derivations of Ore Extensions of the Polynomial Ring in One Variable

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). 0 0 ∂ϕ If ϕ is square-free (that is, if gcd(ϕ, ϕ ) = 1, where 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 decomposition 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.

1 2 Preliminary

Let R be a commutative k-algebra and let R[t, d] be an Ore extension. Every nonzero n 1 element f from R[t, d] is of the form f = ant +···+a1t +a0, where n > 0, a0, . . . , an ∈ R, an 6= 0. In this case the number n is called the degree of f and denoted by deg f or degt f. If f = 0, then we put deg f = −∞. If f, g ∈ R[t, d], then [f, g] denotes the diﬀerence n n P n i n−i fg − gf. In particular, [t , r] = i d (r)t , for all r ∈ R and n > 1. Moreover, if i=1 n n 1 f = ant + ··· + a1t + a0, then [t, f] = d(an)t + ··· + d(a1)t + 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: a (1) d( b ) ∈ k[x]; (2) b divides ψ, where ψ = gcd(ϕ, ϕ0). Proof. Put ϕ = gψ, ϕ0 = fψ, where f, g ∈ k[x], gcd(f, g) = 1. (2) ⇒ (1). Assume that ψ = cb with 0 6= c ∈ k[x]. Then we have:

a ac ac acg 1 1 0 2 0 d( b ) = d( cb ) = d( ψ ) = d( ϕ ) = ϕ2 (d(acg)ϕ − acgd(ϕ)) = ϕ2 ((acg) ϕ − acgϕ ϕ)) 0 acgϕ0 0 acϕ0 0 = (acg) − ϕ = (acg) − ψ = (acg) − acf, a that is, d( b ) ∈ k[x]. a 1 (1) ⇒ (2). Assume that d( b ) = u ∈ k[x]. Then b2 (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 s t polynomial from k[x] dividing b. Let b = p b1, b1 ∈ k[x], s > 1, p - b1, and let ϕ = p ϕ1, ϕ1 ∈ k[x], t > 0, p - ϕ1. Then, by the above equality, we have 0 0 t+s 0 s+t−1 2s 2 (a b1ϕ1 − ab1ϕ1) p − sap b1ϕ1p = p b1u. 0 If t 6 s, then this equality implies that p divides ap b1ϕ1. But it is a contradiction because 0 p - p , 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. s1 sr Now let b = vp1 ··· pr be a decomposition of b into irreducible polynomials (here v ∈ k r {0}, s1 > 1, . . . , sr > 1, and p1, . . . , pr are pairwise nonassociated irreducible s1 sr polynomials from k[x]). Then, by the above observation, all the polynomials p1 , . . . , pr 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 0 that gcd(ϕ, ϕ ) = 1 and let α ∈ k(x). Then d(α) ∈ k[x] ⇐⇒ α ∈ k[x].

2 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 n δn the unique derivation from R to R[t, d] such that δn(x) = t . 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] n the element [f, g] = fg − gf is of the form rnt + ··· + r1t + r0, where r0, . . . , rn belong to 1 d(R), so [f, g] = ϕh for some h ∈ R[t, d]. This means that the mapping ϕ [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). 1 1 n+1 If n > 0, then we denote by en the restriction of the mapping ϕ n+1 t , to R. Thus, we have the derivations e0, e1, e2,... belonging to M(k[x], d). In particular:

e0(x) = 1, 1 0 e1(x) = t + 2 ϕ , 2 0 1 0 0 e2(x) = t + ϕ t + 3 (ϕ ϕ) , 3 3 0 2 0 0 1 000 2 0 00 0 3 e3(x) = t + 2 ϕ t + (ϕ ϕ) t + 4 (ϕ ϕ + 4ϕ ϕ ϕ + (ϕ ) ) . It is easy to check the following proposition.

Proposition 3.1. For every n = 0, 1, 2,... we have n e (x) = tn + ϕ0tn−1 + h , n 2 n−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]. n Then, by Proposition 3.1, 0 = a0e0(x) + ··· + anen(x) = ant + h, where h ∈ R[t, d] and degt h 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]. n 1 Assume that δ : k[x] → k[x][t, d] is a derivation. Let δ(x) = ant + ··· + a1t + 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 degt δ(x) < n.

3 Consider the derivation δ1 = δ − anen. Observe that degt δ1(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) − bfc − aδ(c) − bδ(c)t − bd(δ(c)) = b · (δ(d(c)) + cf − fc − d(δ(c))) = b · (δ(d(c)) − [f, c] − [t, δ(c)]) = b · 0 = 0,

4 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 − wfb − aδ(b) − wtδ(b) = D(w)d(b) + wD(d(b)) + D(w)bt + wD(b)t + wbf − D(w)tb − wfb − wtδ(b) = wD(d(b)) + wD(b)t + wbf − wfb − 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)).

5 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. 0 Let n = 0. Then δ(x) = a0 and, by (∗), [0, x] + [t, δ(x)] = δ(ϕ), so a0ϕ = d(a0) = 0 0 0 a0 0 [t, a0] = [t, δ(x)] = δ(ϕ) = ϕ a0, that is, a0ϕ = ϕ a0. This implies that ( ϕ ) = 0, so 1 a0 = c1ϕ for some c1 ∈ k, and we have δ = a0e0 = c1ϕe0 = c1ϕ f [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 0 0 get the equality anϕ = anϕ , which implies that an = pn+1ϕ for some nonzero pn+1 ∈ k. Now we have: 1 a e = p ϕe = p ϕ [tn+1, ] = [c tn=1, ], n n n+1 n n+1 (n + 1)ϕ n+1

6 1 where cn+1 = n+1 pn+1 ∈ k r {0}. n+1 Consider the derivation D1 = D − [cn+1t , ]. 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. n+1 n+1 n+1 Hence D = D1 + [cn+1t , ] = Wh + [cn+1t , ] = Wf , where f = cn+1t + 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. p 1 Put h := f − g = cpt + ··· + c1t and suppose that h 6= 0. Then ap 6= 0, p > 1, and we have p p−1 0 = Wh(x) = [cpt + ··· + c1t + c0, x] = pcpϕt + hp−2, for some hp−2 ∈ k[x][x, d] with degt hp−2 6 p − 2. We have a contradiction: 0 = pcpϕ 6= 0. Hence, h = 0, that is, f = g.

5 Derivations ∆a Let 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.

2 3 2 2 In particular, ∆a(t) = a, ∆a(t ) = 2at + d(a), ∆a(t ) = 3at + 3d(a)t + d (a) and 4 3 2 2 3 ∆a(t ) = 4at + 6d(a)t + 4d (a)t + d (a).

Proposition 5.1. The derivation ∆a is inner if and only if a ∈ d(R).

n 1 Proof. Assume that ∆a = [f, ], for some f = ant + ··· + a1t + a0 ∈ R[t, d]. Then n a = ∆a(t) = [f, t] = −d(an)t − · · · − 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.

7 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:

n 1 n−1 n−2 ∆(ant + ··· + a1t + a0) = nant + (n − 1)an−1t + ··· + 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 : n 1 R[t, d] → R[t, d] is a derivation such that D(t) = unt + ··· + u1t + u0, where n > 0 and u0, . . . , un ∈ R. If D is inner, then u0, . . . , un ∈ d(R).

m Proof. Assume that D = [f, ], where −f = amt + ··· + a1t + a0 ∈ R[t, d]. Then n 1 m unt + ··· + u1t + u0 = D(t) = [f, t] = [t, −f] = d(am)t + ··· + 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) = n ϕ 6= 0. Let D : R[t, d] → R[t, d] be a derivation such that D(t) = unt + ··· + 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 n n−1 b0, . . . , bn ∈ R. Let f = −bnt − bn−1t − · · · − 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.

8 Corollary 6.3. Let R = k[x] and let d : R → R be a nonzero derivation with d(x) = ϕ 6= n 0. Assume that D : R[t, d] → R[t, d] is a derivation such that D(t) = unt +···+u1t+u0, where n > 0 and u0, . . . , un ∈ R. Then D = W + D1, where W is an inner derivation of n 1 R[t, d] and D1 is a derivation of R[t, d] such that D1(t) = rnt + ··· + r1t + 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 n each aiϕ belongs to d(R). Put aiϕ = d(bi) for i = 0, . . . , n, and let g = −bnt − · · · − 1 n 1 b1t − b0. Consider the derivation D1 = D − [g, ]. Then D1(t) = rnt + ··· + r1t + 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 degt D(t) 6 1 + degt D(x), and moreover, there exists an inner derivation W of R[t, d] such that the D1(t) = D(t) and degt D1(t) = 1 + degt D1(x), where D1 := D + W .

n m Proof. Let D(t) = unt + ··· + u0, D(x) = amt + ··· + 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 n−1 [D(t), x] = 0. But [D(t), x] = nund(x)t + h, where h ∈ R[t, d] and degt h < n − 1. So we have a contradiction: 0 = nund(x) 6= 0. Thus, if D(x) = 0, then D(t) = u0 ∈ R. n−1 Assume now that D(x) 6= 0. Then m > 0, am 6= 0 and, by (∗), nund(x)t + h + m 0 m d(am)t + ··· d(a0) = ϕ amt + v, where ϕ = d(x), h, v ∈ R[t, d], degt h < n − 2 and degt v < m. If n − 1 > m, then nund(x) = 0, but this is a contradiction. So, n − 1 6 m, that is, degt D(t) 6 1 + degt D(x). 0 Assume now that n − 1 < m. Then, by the above equality, d(am) = ϕ am. Hence, 0 0 0 am amϕ = ϕ am so, ϕ = 0 and this implies that am = cϕ, for some c ∈ k. Let W = 1 m+1 [c m+1 t , ], and let D1 = D − W . Then D1(t) = D(t) and degt D1(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.

9 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 0 0 there exists a polynomial a ∈ k[x] such that uϕ = aϕ − a ϕ. Then deg u 6 m − 2.

m Proof. Let ϕ = ϕmx + ··· + ϕ1x + ϕ0, ϕ0, . . . , ϕm ∈ k, ϕm 6= 0. Suppose m−1 that deg u = m − 1. Let u = um−1x + ··· + u0, u0, . . . , um−1 ∈ k, um−1 6= 0. Let 0 0 n uϕ = aϕ − a ϕ, a ∈ k[x]. Then a 6= 0 (because uϕ 6= 0). Put a = anx + ··· + a0, m−1 m a0, . . . , an ∈ k, n > 0, an 6= 0. Now we have the equality: (um−1x + ··· + u0)(ϕmx + n m−1 m ··· + ϕ0) = (anx + ··· + a0)(mϕmx + ··· + ϕ1) − (nan + ··· + a1)(ϕmx + ··· + 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 0 0 polynomial a ∈ k[x] such that uϕ = aϕ − a ϕ. 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

n1 ns ϕ = (x − a1) ··· (x − as) , 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), s s P P g f := ni(x − a1) ··· (x\− ai) ··· (x − as) = ni , x−ai i=1 i=1 n1−1 ns−1 ψ := (x − a1) ··· (x − as) .

Observe that ϕ = gψ, ϕ0 = fψ, ψ = gcd(ϕ, ϕ0) and gcd(g, f) = 1. Moreover, g0 = s s P P g (x − a1) ··· (x\− ai) ··· (x − as) = , so we have: x−ai i=1 i=1

s s 0 P P g f − g = (ni − 1)(x − a1) ··· (x\− ai) ··· (x − as) = (ni − 1) . x−ai i=1 i=1 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,

10 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 = afψ − 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. s−2 We must prove that deg u > s − 1. Suppose that deg u < s − 1. Let u = us−2x + p ··· + u0, u0, . . . , us−2 ∈ k, ui 6= 0 for some i ∈ {0, 1, . . . , s − 2}. Let b = bpx + ··· + b0, b0, . . . , bp ∈ k, p > 0 and bp 6= 0. Then the equality (1) is of the form: s−2 p s−1 p−1 s us−2x + ··· + u0 = (bpx + ··· )((m − s)x + ··· ) − (pbpx + ··· )(x + ··· ).

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 0 ϕψ0 0 0 ϕ0 ϕ ϕ0 1 0 0 ϕ0ψ−ϕ0ψ+ϕψ0 ϕψ0 ψ2 = f − g . In fact, f − g = ψ − ψ = ψ − ψ2 (ϕ ψ − ϕψ ) = ψ2 = ψ2 .

Proposition 7.5. Let 0 6= ϕ ∈ k[x], deg ϕ > 1. The following two properties are equivalent. (1) There exists a polynomial a ∈ k[x] such that ϕ = aϕ0 − a0ϕ. m (2) ϕ = c(x − λ) , 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, m Lemma 7.3 implies that s = 1. Hence, ϕ = c(x − λ) , 0 6= c ∈ k, λ ∈ k and m > 2. 1 (1) ⇒ (2) Assume that ϕ is of the form (2), and take a = m−1 (x − λ). Then: 0 0 m m−1 1 1 m ϕ − aϕ + a ϕ = c(x − λ) − mc(x − λ) m−1 (x − λ) + m−1 c(x − λ) m m 1 m = c(x − λ) 1 − m−1 + m−1 = c(x − λ) 0 = 0. 1 Note that every polynomial a ∈ k[x] of the form a = m−1 (x − λ) + αϕ, with α ∈ k, also 0 0 satisﬁes the equality ϕ = aϕ − a ϕ.

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. a n For any n > 1 consider the inner derivation Dn = [ ϕ t , ] 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 n R[t, d] such that Dn(t) = ut , Dn(x) = naen−1(x).

11 a n a n a0ϕ−aϕ0 n uϕ n n Proof. We have: Dn(t) = [ ϕ t , t] = −d( ϕ )t = − ϕ2 ϕt = ϕ2 ϕt = ut and a n a n 1 n Dn(x) = [ ϕ t , x] = ϕ [t , x] = na nϕ [t , 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. n (3) There exist n > 1 and a derivation Dn of R[t, d] such that Dn(t) = ut . n (4) For every n > 1 there exists a derivation Dn of R[t, d] such that Dn(t) = ut . 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). n Assume that Dn : R[t, d] → R[t, d] is a derivation such that Dn(t) = ut , 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 n−1 0 comparing in (∗) the coeﬃcients of t , we obtain the equality nud(x) + d(ap) = ϕ ap, 0 0 0 0 1 that is, nuϕ + apϕ = ϕ ap, so uϕ = aϕ − a ϕ for a = n ap. 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. n Let f = unt + ··· + 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) = i uit .

(3) For every i ∈ {1, . . . , n} there exists a polynomial ai ∈ k[x] such that uiϕ = 0 0 aiϕ − aiϕ.

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, n−1 0 comparing in the equality (∗) the coeﬃcients of t , we have nunϕ + apϕ = ϕ ap, that is, 0 0 1 unϕ = aϕ − a ϕ for a = n ap. This implies, by Theorem 7.7, that there exists a derivation n Mn : R[t, d] → R[t, d] such that Mn(t) = unt . 0 0 n−1 Consider now the new derivation D := D − Mn. Observe that D (t) = un−1t + 1 0 ··· + u1t + u0. So, doing the same for D we obtain the derivation Mn−1 and, repeating, we have derivations Mn,...,M1, and M0 = D −Mn −· · ·−M1. This completes the proof.

12 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

n X n W (x) = [λtn, x] = λ di(x)tn−i = λϕ · (r tn−1 + ··· + r ), i n−1 0 i=1 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. a Let λ = b , 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 a ac ac ac λϕ = ϕ = ϕ = ϕ = gψ = acg ∈ k[x], b bc ψ ψ 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 n 1 f := λnt + ··· + λ1t + λ0 ∈ k(x)[t, d], n > 0. Consider the inner derivations W, W 0,..., W n of the ring k(x)[t, d] deﬁned as:

0 n W = [f, ], W 0 = [λ0t , ],..., W n = [λnt , ].

Then the following three conditions are equivalent. (1) W (A) ⊆ A.

(2) W 0(A) ⊆ A, ··· , W n(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 = W 0 + ··· + W n. 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

n n W (t) = [f, t] = [λnt + ··· + λ0, t] = −d(λn)t − · · · − d(λ0),

13 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 m 1 that D is inner. Assume now that D = [g, ], for some g = rmt + ··· + r1t + r0 ∈ A. Then r0, . . . , rm ∈ k[x] and we have:

n m d(λn)t + ··· + d(λ0) = [t, f] = −W (t) = −D(t) = −[g, t] = d(rm)t + ··· + 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 ∆ = ∆1 is 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. n 1 Now let D(t) = unt + ··· + u1t + 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 0 derivation ∆1 exists even in positive characteristic), which implies that 1 = d(b) = b d(x) for some b ∈ k[x]. Hence, d(x) = c ∈ k r {0}. It is easy to check that, in this case, the p−1 monomial x is not in the set d(R) and consequently, the derivation ∆xp−1 is not inner. So we have a contradiction.

14 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 derivation 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. ∂ n n n−1 Note that in the Weyl algebra A = k[x][t, ∂x ] we have t x = xt + nt , for all n > 1. n In particular, if f = ant + ··· + a0 ∈ A, then

n−1 m−2 [f, x] = nant + (n − 1)an−1t + ··· + 2a2t + a1.

Hence, in this case, the inner derivation [ , x] coincides with the derivation ∆ = ∆1. m n Let g = bmt + ··· + b0 ∈ A and f = ant + ··· + 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 p some f = apt +···+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:

15 ∂ 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 −1 ∂ σ : R → R such that σdσ = ∂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 0 deg ϕ > 1. Assume that gcd(ϕ, ϕ ) = 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) = n 1 unt + ··· + u1t + 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

0 0 uiϕ = aiϕ − aiϕ.

0 Since gcd(ϕ, ϕ ) = 1, Lemma 7.1 implies that ui = 0. Hence, we know that all the coeﬃcients u1, . . . , un are 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 , W1 are 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.

16 Corollary 10.2. Let R = k[x], let d : R → R be a nonzero derivation, and let ϕ = d(x) 6= 0 0 with deg ϕ > 1. Assume that gcd(ϕ, ϕ ) = 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 EH and 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 0 algebraic closure k of k. Of course m > s. If m = s, then gcd(ϕ, ϕ ) = 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 0 divisor of ϕ and ϕ . 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

n 1 H = hnt + ··· + h1t , 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: h 1 i EH (f) = ψ H, f , 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.

p n n Lemma 11.1. Let n > 1 and let p ∈ {0, 1, . . . , m−s−1}. If H = x t , then EH (t) = vpt , 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 = fψ, deg g = s and I(f − g0) = (m − s)xs−1.

17 xp n n xp Now we have: EH (t) = [ ψ t , t] = vpt , where vp = −d( ψ ). Let us calculate:

gxp 1 p p 1 p 0 0 p 2 p−1 2 vp = −d( ϕ ) = ϕ2 (gx d(ϕ) − d(gx )ϕ) = ϕ2 (gx ϕ ϕ − g x ϕ − px gϕ ) p ϕ0 0 p p−1 p ϕ0 0 p p−1 p 0 p p−1 = gx ϕ − g x − px g = x ψ − g x − px g = x f − g x − px g = (f − g0)xp − pxp−1g.

The assumption p ∈ {0, 1, . . . , m − s − 1} implies that m − s − p ∈ {1, 2, . . . , m − s}, so p+s−1 m − s − p 6= 0 and so, the initial monomial of vp is equal to (m − s − p)x . Hence, deg vp = p + s − 1. Now we are ready to prove the following main result of this paper.

Theorem 11.2. Let d be a nonzero derivation of R = k[x], and let ϕ = d(x). Assume that m = deg ϕ > 2 and ϕ is not square-free. Then every derivation D of R[t, d] has a unique decomposition D = W + EH + ∆r, where W is an inner derivation of R[t, d], H ∈ R[t, d] is a special polynomial, and r ∈ R with deg r < m. Moreover, if D is as above, then D is inner if and only if H = 0 and 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 + E0 + ∆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) = n 1 unt + ··· + u1t + 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

0 0 uiϕ = aiϕ − aiϕ.

Since the degree of each ui is smaller then m, Lemma 7.2 implies that deg ui 6 m − 2, for i = 1, . . . , n. n Consider the polynomial unt . Suppose that un 6= 0. Then, by Lemma 7.3, deg un > s − 1, where s is the number of all pairwise diﬀerent roots of the polynomial ϕ belonging to an algebraic closure k of k. So, s − 1 6 deg un 6 m − 2. Put deg un = p + s − 1, where p ∈ {0, 1, . . . , m − s − 1}. Let Mc := Ecxptn , where 0 6= c ∈ k. By Lemma 11.1, there exists a nonzero coeﬃcient n c ∈ k such that Mc(t) = vpt with 0 6= vp ∈ k[x], where the initial monomial of vp n coincides with the initial monomial of un. Let D2 = D1 − Ecxptn . Then D2(t) = unt + n−1 1 un−1t + ··· + u1t + u0, where the polynomials u0, . . . , un−1 are the same as in D1(t) and deg un < deg un. If un 6= 0, then we repeat the same procedure. Moreover, we repeat n n−1 1 this procedure for all the successive polynomials unt , un−1t , . . . , u1t . Thus, we see that D1 = EH +M, for some special H ∈ R[t, d], where M is a derivation 0 of R[t, d] such that M(t) = u0 ∈ R. We know, by Proposition 5.3, that M = ∆u0 + W , where W 0 is an inner derivation of R[t, d].

18 So we already proved that every derivation of R[t, d] is of the form W + EH + ∆r with a special H ∈ R[t, d] and r ∈ R, deg r < deg ϕ. Moreover, by Propositions 5.1 and 6.2, such a derivation W + EH + ∆r is inner if and only if H = 0 and r = 0. For a proof that such a decomposition is unique we use the same argument as in the proof of Theorem 10.1. 2 ∂ 2 ∂ 2 Consider the Ore extension A = k[x][t, x ∂x ]. Here d = x ∂x , ϕ = x and ψ = 0 1 n n gcd(ϕ, ϕ ) = x. For any n > 1, denote by En the derivation Et . Then En(t) = [ x t , t] = 1 n n n−1 −d( x )t = t and En(x) = nxt + hn−2, for some hn−2 ∈ A with degt hn−2 < n − 1. Put

n−1 Mn := En − [nt , ].

p p+1 Since d (x) = p!x for p = 0, 1,... , it is easy to check that Mn is such a derivation of n n−1 A that Mn(t) = t and Mn(x) = nxt . Every derivation of the form Mn is of course non-inner. As a consequence of Theorem 11.2 we obtain the following description of all derivations of A.

2 ∂ Example 11.3. Every derivation D of A = k[x][t, x ∂x ] has a unique decomposition

D = W + cnMm + ··· + c1M1 + c0∆1, where n > 1, W is an inner derivation of A, and c0, . . . , cn ∈ k. Moreover, if D is as above, then D is inner if and only if cn = ··· = c0 = 0.

Using Theorem 11.2 we may produce similar examples for any Ore extension k[x][t, d] 0 with deg ψ > 1 (where ψ = gcd(ϕ, ϕ ) and ϕ = d(x)).

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Faculty of Mathematics and Computer Science, N. Copernicus University, 87–100 Toru´n,Poland, (e-mail: [email protected])