Derivations of Ore extensions of the polynomial ring in one variable

Consider the derivation δ 1 = δ − a n e n . Observe that deg t δ 1 (x) < n (Proposition 3.1).So, by induction, δ 1 = b 0 e 0 + · · · + b n−1 e n−1 for some b 0 , . . . , b n−1 ∈ k[x]. Therefore,δ = δ 1 + a n e n = b 0 e 0 + · · · + b n−1 e n−1 + a n e n is a linear combination of e 0 , . . . , e n overk[x]. □In the next theorem, which is true for arbitrary k-domain R (even if k is of positivecharacteristic), we show when a derivation δ : k[x] → k[x][t, d] has an extension to aderivation 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 areequivalent.(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 haveδ(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 wedefine 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] withdeg u n. Then w is of the form w = ut + a, where a ∈ R, u ∈ R[t, d], deg u = n. WedefineD(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. Nowwe will show that D(uv) = D(u)v + uD(v) for all u, v ∈ R[t, d]. We use an induction withrespect 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 ≠ 0and v ≠ 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.ThenD(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) 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 allproducts of degrees 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 n. Then uv = (a + wt)b = ab + wd(b) + wbt and henceD(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) n and deg wb 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 n and then v = b + wt, where b ∈ R,w ∈ R[t, d], w ≠ 0, deg w < deg v. So we have uv = ub + uwt, and deg(ub) n,deg(uw) 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 letA = 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 leta, 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 obtainCorollary 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 thatδ : 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 thatδ : R → A is a derivation and f ∈ A. Then there exists a unique derivation D : A → Asuch that D|R = δ and D(t) = f if and only if(∗) [f, x] + [t, δ(x)] = δ(d(x)). □4 Scalar inner derivationsIf f ∈ R[t, d], then we will denote by W f the inner derivation [f, ]. We will say thatW f is a scalar inner derivation if all the coefficients of f belong to k, that is, if f ∈ k[t].Since W f = W f+c for any c ∈ k, we may always assume that f is without constant term(i.e., f(0) = 0). Note that if W f is a scalar inner derivation, then W f (t) = 0. Thefollowing 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) ≠ 0, and let D be aderivation of R[t, d]. If D(t) = 0, then there exists a unique polynomial f ∈ k[t] such thatD = W f and f(0) = 0.Proof. Assume that D(t) = 0 and let δ : R → R[t, d] be the restriction of D toR. Using Proposition 3.2 we have an equality of the form δ = a n e n + · · · + a 1 e 1 + a 0 e 0 ,where n 0 and a n , . . . , a 0 ∈ k[x]. We will show, by an induction on n, that there existsa polynomial f ∈ k[t] such that D = W f and f(0) = 0. Since the case D = 0 is obvious,we may assume that a n ≠ 0.Let n = 0. Then δ(x) = a 0 and, by (∗), [0, x] + [t, δ(x)] = δ(ϕ), so a 0 ϕ ′ = d(a 0 ) =[t, a 0 ] = [t, δ(x)] = δ(ϕ) = ϕ ′ a 0 , that is, a 0 ϕ ′ = ϕ ′ a 0 . This implies that ( a 0ϕ) ′ = 0, soa 0 = c 1 ϕ for some c 1 ∈ k, and we have δ = a 0 e 0 = c 1 ϕe 0 = c 1 ϕ 1 [t, ] = [c f 1t, ].Let f = c 1 t. Then f ∈ k[t], f(0) = 0 and D = W f (because D(t) = 0 = W f (t) andD(x) = δ(x) = [c 1 t, x] = W f (x)).Assume now that n 1 and that for all numbers smaller then n our statement intrue. Let δ(x) = a n e n + · · · + a 0 e 0 , with a n ≠ 0 and a 0 , . . . , a n ∈ k[x]. Then, by (∗),[0, x] + [t, δ(x)] = δ(ϕ). Comparing in this equality the coefficients with respect to t n weget the equality a ′ nϕ = a n ϕ ′ , which implies that a n = p n+1 ϕ for some nonzero p n+1 ∈ k.Now we have:1a n e n = p n+1 ϕe n = p n+1 ϕ(n + 1)ϕ [tn+1 , ] = [c n+1 t n=1 , ],6

where c n+1 = 1n+1 p n+1 ∈ k {0}.Consider the derivation D 1 = D − [c n+1 t n+1 , ]. Observe that D 1 (t) = 0 and D 1 |R =a n−1 e n−1 + · · · + a 0 e 0 . So, by induction, D 1 = [h, ] for some h ∈ k[t] with h(0) = 0.Hence D = D 1 + [c n+1 t n+1 , ] = W h + [c n+1 t n+1 , ] = W f , where f = c n+1 t n+1 + h ∈ k[t].This completes the proof of existence of f. We will show yet, that such a polynomialf is unique. Suppose that f, g ∈ k[t], W f = W g and f(0) = g(0) = 0. Then W f−g = 0.Put h := f − g = c p t p + · · · + c 1 t 1 and suppose that h ≠ 0. Then a p ≠ 0, p 1, and wehave0 = W h (x) = [c p t p + · · · + c 1 t + c 0 , x] = pc p ϕt p−1 + h p−2 ,for some h p−2 ∈ k[x][x, d] with deg t h p−2 p − 2. We have a contradiction: 0 = pc p ϕ ≠ 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 3.3implies 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 (t 2 ) = 2at + d(a), ∆ a (t 3 ) = 3at 2 + 3d(a)t + d 2 (a) and∆ a (t 4 ) = 4at 3 + 6d(a)t 2 + 4d 2 (a)t + d 3 (a).Proposition 5.1. The derivation ∆ a is inner if and only if a ∈ d(R).Proof. Assume that ∆ a = [f, ], for some f = a n t n + · · · + a 1 t 1 + a 0 ∈ R[t, d]. Thena = ∆ a (t) = [f, t] = −d(a n )t n − · · · − d(a 1 )t − d(a 0 ). In particular, a = d(−a 0 ) ∈ 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 obtainCorollary 5.2. Let R be a k-domain. If every derivation of R[t, d] is inner, then d issurjective. □Proposition 5.3. If D is a derivation of k[x][t, d] such that D(t) = a ∈ k[x], thenD = 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 scalarinner derivation. □Proposition 5.4. Let d be a derivation of R = k[x] with ϕ = d(x) ≠ 0. If a ∈ R, then∆ a = W + ∆ r , where W is an inner derivation of R[t, d], and r ∈ k[x] is the remainderof 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 + ∆ rand, 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) ≠ 0. Thenfor every a ∈ R there exists a unique derivation D a of R[t, d] such that D a (t) = a andD a (x) = ϕ. The derivation D a is equal to [t, ] + ∆ a . The derivation D a is inner if andonly if a is divisible by ϕ.Proof. Let D = [t, ] + ∆ a . Then D(x) = [t, x] + ∆ a (x) = d(x) + 0 = ϕ andD(t) = [t, t] + ∆ a (t) = 0 + a = a. This implies that the derivation D a exists and thatD a = D. Since [t, ] is inner, the derivation D a is inner if and only if ∆ a is inner. Thelast 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 D 0 = [t, ] : R[t, d] → R[t, d]is an extension of the derivation d such that D 0 (t) = 0. If d ≠ 0, then every derivationD : R[t, d] → R[t, d], such that D|R = d, satisfies the condition: D(t) ∈ R. The same istrue 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:∆(a n t n + · · · + a 1 t 1 + a 0 ) = na n t n−1 + (n − 1)a n−1 t n−2 + · · · + 2a 2 t + a 1 .So ∆ is equal to usual partial derivative ∂ ∂tinner if and only if 1 ∈ d(R).of R[t, d]. Note that (by Proposition 5.1) ∆ is6 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) = u n t n + · · · + u 1 t 1 + u 0 , where n 0 andu 0 , . . . , u n ∈ R. If D is inner, then u 0 , . . . , u n ∈ d(R).Proof. Assume that D = [f, ], where −f = a m t m + · · · + a 1 t + a 0 ∈ R[t, d]. Thenu n t n + · · · + u 1 t 1 + u 0 = D(t) = [f, t] = [t, −f] = d(a m )t m + · · · + d(a 1 )t + d(a 0 ), Thus, wehave u 0 = d(a 0 ) ∈ d(R), u 1 = d(a 1 ) ∈ 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 ≠ 0. Let D : R[t, d] → R[t, d] be a derivation such that D(t) = u n t n + · · · + u 1 t + u 0 ,where n 0 and u 0 , . . . , u n ∈ R. Then D is inner if and only if u 0 , . . . , u n ∈ d(R).Proof. Assume that u 0 , . . . , u n ∈ d(R). Let u 0 = d(b 0 ), . . . , u n = d(b n ), for someb 0 , . . . , b n ∈ R. Let f = −b n t n − b n−1 t n−1 − · · · − b a t − b 0 and let D 1 = D − [f, ]. ThenD 1 is a derivation of R[t, d] such that D(t) = 0. This derivation is inner (by Proposition4.1). Hence the derivation D = D 1 + [f, ] is inner. The opposite implication followsfrom Proposition 6.1. □8

Corollary 6.3. Let R = k[x] and let d : R → R be a nonzero derivation with d(x) = ϕ ≠0. Assume that D : R[t, d] → R[t, d] is a derivation such that D(t) = u n t n +· · ·+u 1 t+u 0 ,where n 0 and u 0 , . . . , u n ∈ R. Then D = W + D 1 , where W is an inner derivation ofR[t, d] and D 1 is a derivation of R[t, d] such that D 1 (t) = r n t n + · · · + r 1 t 1 + r 0 , whereeach r i (for i = 0, 1, . . . , n) is the remainder in the divisibility of u i by ϕ. □Proof. Let u i = a i ϕ + r i , a i , r i ∈ k[x], deg r i < deg ϕ, for i = 0, 1, . . . , n. Note thateach a i ϕ belongs to d(R). Put a i ϕ = d(b i ) for i = 0, . . . , n, and let g = −b n t n − · · · −b 1 t 1 − b 0 . Consider the derivation D 1 = D − [g, ]. Then D 1 (t) = r n t n + · · · + r 1 t 1 + r 0and we have D = [g, ] + D 1 . □Note also the followingProposition 6.4. Let d be a nonzero derivation of R = k[x], and let D be a derivationof R[t, d]. If D(x) = 0, then D(t) ∈ R. If D(x) ≠ 0, then deg t D(t) 1 + deg t D(x), andmoreover, there exists an inner derivation W of R[t, d] such that the D 1 (t) = D(t) anddeg t D 1 (t) = 1 + deg t D 1 (x), where D 1 := D + W .Proof. Let D(t) = u n t n + · · · + u 0 , D(x) = a m t m + · · · + a 0 , where all the coefficientsu 0 , . . . , u n , a 0 , . . . , a m belong to R.Assume that D(x) = 0 and suppose that n 1 and u n ≠ 0. Then (∗) implies that[D(t), x] = 0. But [D(t), x] = nu n d(x)t n−1 + h, where h ∈ R[t, d] and deg t h < n − 1. Sowe have a contradiction: 0 = nu n d(x) ≠ 0. Thus, if D(x) = 0, then D(t) = u 0 ∈ R.Assume now that D(x) ≠ 0. Then m 0, a m ≠ 0 and, by (∗), nu n d(x)t n−1 + h +d(a m )t m + · · · d(a 0 ) = ϕ ′ a m t m + v, where ϕ = d(x), h, v ∈ R[t, d], deg t h < n − 2 anddeg t v < m. If n − 1 > m, then nu n d(x) = 0, but this is a contradiction. So, n − 1 m,that is, deg t D(t) 1 + deg t D(x).Assume now that( )n − 1 < m. Then, by the above equality, d(a m ) = ϕ ′ a m . Hence,′a ′ mϕ = ϕ ′ aa m so, mϕ= 0 and this implies that a m = cϕ, for some c ∈ k. Let W =[c 1m+1 tm+1 , ], and let D 1 = D − W . Then D 1 (t) = D(t) and deg t D 1 (x) = m 1 < m. Ifagain n−1 < m 1 , then we repeat the same procedure for the derivation D 1 . Repeating thisargument several times we obtain an inner derivation W of R[t, d] and a new derivationD 1 = W + d having the required properties. □7 On some differential equationIn this section we consider a differential equation of the formuϕ = aϕ ′ − a ′ ϕ,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, ϕ ′ ) = 1, then u = 0.9

and, if m > s, then I(f − g ′ ) = (m − s)x s−1 .Now let a ∈ k[x] be such that uϕ = aϕ ′ − a ′ ϕ. Then of course a ≠ 0 and ugψ =afψ − a ′ gψ, so ug = af − a ′ g. This implies that g divides a (because gcd(f, g) = 1). Leta = gb for some b ∈ k[x] {0}. Then ug = bgf − b ′ g 2 − bg ′ g, so u = bf − b ′ g − bg ′ , that is,(1) u = b(f − g ′ ) − b ′ g.Observe that ψ ·(f −g ′ ) = ψ ′ g. In fact: ψ ·(f −g ′ )−ψ ′ g = ψf −(ψg ′ +ψ ′ g) = ϕ ′ −(ψg) ′ =ϕ ′ − ϕ ′ = 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 aconsequence of this fact, we may assume that b ∈ k[x] is a nonzero polynomial satisfying(1) and the coefficient of b with respect to x m−s is equal to zero.We must prove that deg u s − 1. Suppose that deg u < s − 1. Let u = u s−2 x s−2 +· · · + u 0 , u 0 , . . . , u s−2 ∈ k, u i ≠ 0 for some i ∈ {0, 1, . . . , s − 2}. Let b = b p x p + · · · + b 0 ,b 0 , . . . , b p ∈ k, p 0 and b p ≠ 0. Then the equality (1) is of the form:u s−2 x s−2 + · · · + u 0 = (b p x p + · · · )((m − s)x s−1 + · · · ) − (pb p x p−1 + · · · )(x s + · · · ).Since p + s − 1 s − 1 > s − 2, we have ((m − s) − p)b p = 0. But b p ≠ 0, so p = (m − s).However, by our assumption, the coefficient b m−s is equal to zero. Thus, we have acontradiction: 0 ≠ b p = b m−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 ψ 2 divides ϕψ ′ and′ϕψ ′= f − g ′ . In fact, f − g ′ = ϕ′− ϕψ 2 ψ ψ =ϕ ′− 1 (ϕ ′ ψ − ϕψ ′ ) = ϕ′ ψ−ϕ ′ ψ+ϕψ ′= ϕψ′ . □ψ ψ 2 ψ 2ψ 2Proposition 7.5. Let 0 ≠ ϕ ∈ k[x], deg ϕ 1. The following two properties are equivalent.(1) There exists a polynomial a ∈ k[x] such that ϕ = aϕ ′ − a ′ ϕ.(2) ϕ = c(x − λ) m , for some λ, c ∈ k, c ≠ 0 and m 2.Proof. (1) ⇒ (2). Assume that ϕ = aϕ ′ − a ′ ϕ for some a ∈ k[x]. Let m = deg ϕ andlet 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 ≠ c ∈ k, λ ∈ k and m 2.(1) ⇒ (2) Assume that ϕ is of the form (2), and take a = 1 (x − λ). Then:m−1ϕ − aϕ ′ + a ′ ϕ = c(x − λ) m − mc(x − λ) m−1 11(x − λ) + c(x − λ)mm−1 m−1= c(x − λ) m ( 1 − mm−1 + 1m−1)= c(x − λ) m 0 = 0.Note that every polynomial a ∈ k[x] of the form a = 1satisfies the equality ϕ = aϕ ′ − a ′ ϕ. □m−1(x − λ) + αϕ, with α ∈ k, alsoLemma 7.6. Let d be a nonzero derivation of R = k[x] and let ϕ = d(x) ≠ 0. Letu ∈ k[x]{0} and assume that there exists a polynomial a ∈ k[x] such that uϕ = aϕ ′ −a ′ ϕ.For any n 1 consider the inner derivation D n = [ a ϕ tn , ] of the ring k(x)[t, d]. Then:(1) D n (R[t, d]) ⊆ R[t, d].(2) Denote by D n the restriction of D n to R[t, d]. Then D n is a unique derivation ofR[t, d] such that D n (t) = ut n , D n (x) = nae n−1 (x).11

Proof. We have: D n (t) = [ a ϕ tn , t] = −d( a ϕ )tn = − a′ ϕ−aϕ ′ϕt n = uϕ ϕt n = ut n andϕ 2ϕ 2D n (x) = [ a ϕ tn , x] = a ϕ [tn , x] = na 1nϕ [tn , x] = nae n−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 ≠ 0.Let u ∈ k[x] {0}. The following properties are equivalent.(1) There exists a ∈ k[X] such that uϕ = aϕ ′ − a ′ ϕ.(2) There exists a derivation D 1 of R[t, d] such that D 1 (u) = ut.(3) There exist n 1 and a derivation D n of R[t, d] such that D n (t) = ut n .(4) For every n 1 there exists a derivation D n of R[t, d] such that D n (t) = ut n .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 D n : R[t, d] → R[t, d] is a derivation such that D n (t) = ut n , where n 1is a fixed natural number. Let δ : R → R[t, d] be the restriction of D n to R. We know, byProposition 3.2, that δ = a p e p + · · · + a 1 e 1 + a 0 e 0 for some a 0 , . . . , a p ∈ R. If δ = 0, thenby (∗), 0 = [ut n , x] = nud(x)t n−1 + · · · and we have a contradiction: 0 = nuϕ ≠ 0. Henceδ ≠ 0, p 0 and a p ≠ 0. We may assume (by Proposition 6.4) that n − 1 = p. Nowcomparing in (∗) the coefficients of t n−1 , we obtain the equality nud(x) + d(a p ) = ϕ ′ a p ,that is, nuϕ + a ′ pϕ = ϕ ′ a p , so uϕ = aϕ ′ − a ′ ϕ for a = 1 a n p. This completes the proof. □Theorem 7.8. Let R = k[x] and let d : R → R be a nonzero derivation with d(x) = ϕ ≠ 0.Let f = u n t n + · · · + u 1 t + u 0 ∈ R[t, d], where n 1, u n ≠ 0, u 0 , . . . , u n ∈ R. Then thefollowing 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 M i of R[t, d] such that M i (t) =u i t i .(3) For every i ∈ {1, . . . , n} there exists a polynomial a i ∈ k[x] such that u i ϕ =a i ϕ ′ − a ′ iϕ.Proof. (2) ⇒ (1). Let D := M n + · · · + M 1 + M 0 . 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 ≠ 0, then a derivation M i exists by Theorem 7.7.For i = 0 we may put M 0 := ∆ 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 = a p e p + · · · + a 1 e 1 + a 0 e 0 , forsome a p , . . . , a 0 ∈ R. We may assume (by Proposition 6.4) that n − 1 = p. Moreover,comparing in the equality (∗) the coefficients of t n−1 , we have nu n ϕ + a p ϕ = ϕ ′ a p , that is,u n ϕ = aϕ ′ − a ′ ϕ for a = 1 n a p. This implies, by Theorem 7.7, that there exists a derivationM n : R[t, d] → R[t, d] such that M n (t) = u n t n .Consider now the new derivation D ′ := D − M n . Observe that D ′ (t) = u n−1 t n−1 +· · · + u 1 t 1 + u 0 . So, doing the same for D ′ we obtain the derivation M n−1 and, repeating,we have derivations M n , . . . , M 1 , and M 0 = D − M n − · · · − M 1 . 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. LetA := k[x][t, d]. It is obvious that if D is a derivation of k(x)[t, d], then D(A) ⊆ A if andonly 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 innerderivation W of k(x)[t, d] defined by W = [λt n , ]. Then W (A) ⊆ A ⇐⇒ d(λ) ∈ k[t].Proof.Put ϕ := d(x). Observe that W (t) = [λt n , t] = −d(λ)t n andW (x) = [λt n , x] = λn∑i=1( ni)d i (x)t n−i = λϕ · (r n−1 t n−1 + · · · + r 0 ),for some r 0 , . . . , r n−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)t n ∈ A. We will show that W (x)also belongs to A.Let λ = a , a, b ∈ k[x], b ≠ 0, gcd(a, b) = 1. We know, by Proposition 2.1, thatbb | ψ = gcd(ϕ, ϕ ′ ). Let ψ = bc, ϕ = gψ, where c, g ∈ k[x]. Thenλϕ = a b ϕ = acbc ϕ = acψ ϕ = ac gψ = acg ∈ k[x],ψand 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 ≠ 0, and letA = k[x][t, d]. Letf := λ n t n + · · · + λ 1 t 1 + λ 0 ∈ k(x)[t, d],n 0. Consider the inner derivations W , W 0 , . . . , W n of the ring k(x)[t, d] defined as:W = [f, ], W 0 = [λ 0 t 0 , ], . . . , W n = [λ n t n , ].Then the following three conditions are equivalent.(1) W (A) ⊆ A.(2) W 0 (A) ⊆ A, · · · , W n (A) ⊆ A.(3) d(λ 0 ), . . . , d(λ n ) ∈ k[x].Moreover, if W (A) ⊆ A and D : A → A is the restriction of W to A, then D is an innerderivation of A if and only if λ 0 , . . . , λ n ∈ 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. ButW (t) = [f, t] = [λ n t n + · · · + λ 0 , t] = −d(λ n )t n − · · · − d(λ 0 ),13

so d(λ 0 ), . . . , d(λ n ) ∈ k[x]. Hence, the conditions (1), (2) and (3) are equivalent.Let W (A) ⊆ A and let D := W |A. If λ 0 , . . . , λ n ∈ k[x], then f ∈ A and it is clearthat D is inner. Assume now that D = [g, ], for some g = r m t m + · · · + r 1 t 1 + r 0 ∈ A.Then r 0 , . . . , r m ∈ k[x] and we have:d(λ n )t n + · · · + d(λ 0 ) = [t, f] = −W (t) = −D(t) = −[g, t] = d(r m )t m + · · · + d(r 0 ).This means that d(λ i ) = d(r i ) for i = 0, 1, . . . , n. Hence, each difference λ i − r i belongsto Ker(d) = k, so λ i − r i = c i ∈ k for i = 0, . . . , n. Therefore, λ i = r i + c i ∈ k[x], for alli = 0, . . . , n. □9 When all derivations of k[x][t,d] are inner?It is well known that if R = k[x] and d = ∂ , then the Ore extension R[t, d] coincides∂xwith the Weyl algebra with one pair of generators (see for example, [2], [3]). In this caseit is also well known ([4]) that every derivation of R[t, d] is inner. Now we present a newproof of this fact and we show that if in an Ore extension of the form A = k[x][t, d] everyderivation is inner, then A is isomorphic to the Weyl algebra over k with one pair ofgenerators.Theorem 9.1. If d is a nonzero derivation of R = k[x], then the following two propertiesare equivalent.(1) Every derivation of R[t, d] is inner.(2) d(x) ∈ k {0}.Proof. (1) ⇒ (2). Assume that every derivation of R[t, d] is inner. Then, inparticular, the derivation ∆ = ∆ 1 is inner, which implies, by Proposition 5.1, that 1 = d(b)for some b ∈ k[x]. Hence, 1 = b ′ d(x) so, d(x) is invertible in k[x], that is, d(x) ∈ k {0}.(2) ⇒ (1). Assume that d(x) = c ∈ k {0}, and let D : R[t, d] → R[t, d] be anarbitrary derivation. If D(t) = 0, then we know (by Proposition 4.1) that D is inner.Now let D(t) = u n t n + · · · + u 1 t 1 + u 0 , u 0 , . . . , u n ∈ k[x], n 0 and u n ≠ 0. Observethat, since 0 ≠ d(x) ∈ k, the image d(R) is equal to R. Hence, the coefficients u 0 , . . . , u nbelong 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 existsa derivation of R[t, d] which is not inner.Proof. Suppose that there exists a derivation d of R = k[x] such that every derivationof R[t, d] is inner. Then, in particular, the derivation ∆ = ∆ 1 is inner (note that suchderivation ∆ 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 {0}. It is easy to check that, in this case, themonomial x p−1 is not in the set d(R) and consequently, the derivation ∆ x p−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 forany k-domain R. This implies that if any derivation of R[t, d] is inner, then the derivationd is surjective. We know, by [8] or [10] (Theorem 2.6.1), that if R is a field such that thetranscendence degree of R over k is finite, then every derivation of R is not surjective. Asa consequence of this fact, we obtain that if R is such a field, then there exists a derivationof 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 derivationd 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 = ∂ , where k is a field of characteristic zero. We know, by Theorem 9.1,∂xthat every derivation of A is inner. Thus, we have:Proposition 9.4. Let R = k[x], d = ∂ , A = R[t, d]. If f, g are such elements from A∂xthat [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 tn x = xt n + nt n−1 , for all n 1.In particular, if f = a n t n + · · · + a 0 ∈ A, then[f, x] = na n t n−1 + (n − 1)a n−1 t m−2 + · · · + 2a 2 t + a 1 .Hence, in this case, the inner derivation [ , x] coincides with the derivation ∆ = ∆ 1 .Let g = b m t m + · · · + b 0 ∈ A and f = a n t n + · · · + a 0 ∈ A. Using the above equalitywe see that, in this case, the condition (∗) (that is, [f, x] + [t, g] = 0) is equivalent to theequalities:d(b i ) + (i + 1)a i+1 = 0, for i = 0, 1, . . . , n − 1,and d(b j ) = 0 for j n. As a consequence of this equivalence and the fact that thederivation d = ∂ is surjective, we obtain:∂xProposition 9.5. Let A = k[x][t, ∂ ], (char(k) = 0). For any f ∈ A there exists a∂xderivation D of A such that D(t) = f. For any g ∈ A there exists a derivation D of Asuch that D(x) = g. □Now let R be the ring k[x, y], of polynomials over k in two variables, and consider theOre extension A = R[t, d], where d = ∂∂. Let δ : R → R[t, d] be the derivation . It is∂x ∂yclear, by Theorem 3.3, that there exists a unique derivation D : R[t, d] → R[t, d] such thatD|R = δ and D(t) = 0. Observe that D is not inner. Indeed, suppose that D = [f, ] forsome f = a p t p +· · ·+a 1 t+a 0 ∈ R[t, d]. Then a 0 , . . . , a p ∈ R = k[x, y] and, since D(t) = 0,d(a 0 ) = · · · = d(a p ) = 0, that is, a 0 , . . . , a p ∈ k[y]. But D(x) = 0, so a p = · · · , a 1 = 0,and this implies that f = a 0 ∈ k[y]. So we have a contradiction: 1 = D(y) = [a 0 , y] = 0.Thus, we have:15

Proposition 9.6. If A = k[x, y][t, ∂ ], then there exists a non-inner derivation of A. □∂xThe 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 kof characteristic zero, and let A = R[t, d] be an arbitrary Ore extension of R. Then thereexists a non-inner derivation of A.Proof. If d = 0, then it is obvious. Assume that d is a nonzero derivation ofR = k[x, y] and suppose that every derivation of A = R[t, d] is inner. Then, by Proposition5.1, d is surjective and this means, by [13], that the derivation d is locally nilpotent witha slice ([5], [10]). Now, by a theorem of Rentschler [12], there exists a k-automorphismσ : R → R such that σdσ −1 = ∂ . This automorphism induces a natural isomorphism∂xfrom A to k[x, y][t, ∂∂]. But, by Proposition 9.6, the algebra k[x, y][t, ] has a non-inner∂x ∂xderivation. Thus, A has a non-inner derivation, and we have a contradiction. □10 The square-free caseTheorem 10.1. Let d be a nonzero derivation of R = k[x], and let ϕ = d(x) ≠ 0 withdeg ϕ 1. Assume that gcd(ϕ, ϕ ′ ) = 1. Then every derivation D of R[t, d] has a uniquedecompositionD = W + ∆ r ,where W is an inner derivation of R[t, d], and r ∈ R with deg r < deg ϕ. Moreover, if Dis 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 = W f for some f ∈ k[t], so in this case: D = W f + ∆ 0 .Assume now that D(t) ≠ 0. By Corollary 6.3 we know that D = W + D 1 , whereW is an inner derivation of R[t, d] and D 1 is a derivation of R[t, d] such that D 1 (t) =u n t n + · · · + u 1 t 1 + u 0 , where u 0 , . . . , u n ∈ R and deg u i < deg ϕ for all i = 0, 1, . . . , n.By Theorem 7.8, for every i ∈ {1, . . . , n} there exists a i ∈ R such thatu i ϕ = a i ϕ ′ − a ′ iϕ.Since gcd(ϕ, ϕ ′ ) = 1, Lemma 7.1 implies that u i = 0. Hence, we know that all thecoefficients u 1 , . . . , u n are equal to zero. This means, that D 1 = ∆ u0 with deg u 0 < 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 andonly if r = 0.Now we will show that the decomposition is unique. Suppose that W + ∆ r = D =W 1 +∆ r1 , where W , W 1 are inner derivations and r, r 1 ∈ R, deg r < deg ϕ, deg r 1 < deg ϕ.Then the derivation ∆ r−r1 = W 1 −W is inner. But deg(r −r 1 ) < deg ϕ so, by Proposition5.1, r − r 1 = 0. Therefore r = r 1 and W 1 − 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) ≠0 with deg ϕ 1. Assume that gcd(ϕ, ϕ ′ ) = 1. Let D be a derivation of R[t, d]. ThenD(x) = ϕ · G, D(t) = ϕF t + r, for some F, G ∈ R[t, d] and r ∈ R. Moreover, D is innerif and only if ϕ | r. □Example 10.3. Let R = k[x] and let d : R → R be the derivation such that d(x) = x 2 +1.There is no derivation D of R[t, d] such that D(t) = t. This fact is a consequence ofCorollary 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) ∂ , with a, b ∈ k, a ≠ 0.∂xFor every derivation D of R[t, d] there exists a unique c = c D ∈ 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 ifand only if c D = 0. □11 Derivations E H and a final descriptionLet d be a nonzero derivation of R = k[x] and let ϕ = d(x) ≠ 0. Let m = deg ϕ andlet s be the number of all pairwise different roots of the polynomial ϕ belonging to analgebraic closure k of k. Of course m s. If m = s, then gcd(ϕ, ϕ ′ ) = 1 and in this casewe 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 commondivisor 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 formH = h n t n + · · · + h 1 t 1 ,where n 1, h i ∈ R, deg h i < m − s for all i = 1, . . . , n. In particular, 0 from R[t, d] is aspecial polynomial.If H ∈ R[t, d] is a special polynomial, then we denote by E H the derivation of R[t, d]defined as:[ ]1E H (f) = H, f , for all f ∈ R[t, d].ψObserve that, by Propositions 2.1 and 8.2, the mapping E H is really a derivation fromR[t, d] to R[t, d]. Moreover, it follows from Proposition 8.2 that the derivation E H is innerif and only if H = 0.Lemma 11.1. Let n 1 and let p ∈ {0, 1, . . . , m−s−1}. If H = x p t n , then E H (t) = v p t n ,where v p ∈ k[x] {0} and deg v p = p + s − 1. In particular, s − 1 deg v p m − 2.Proof. We use the same notations as in the proof of Lemma 7.3. Recall that ϕ = gψ,ϕ ′ = fψ, deg g = s and I(f − g ′ ) = (m − s)x s−1 .17

Now we have: E H (t) = [ xpψ tn , t] = v p t n , where v p = −d( xp ). Let us calculate:ψv p = −d( gxpϕ ) = 1 ϕ 2 (gx p d(ϕ) − d(gx p )ϕ) = 1 ϕ 2 (gx p ϕ ′ ϕ − g ′ x p ϕ 2 − px p−1 gϕ 2 )= gx p ϕ′ϕ − g′ x p − px p−1 g = x p ϕ′ψ − g′ x p − px p−1 g = x p f − g ′ x p − px p−1 g= (f − g ′ )x p − px p−1 g.The assumption p ∈ {0, 1, . . . , m − s − 1} implies that m − s − p ∈ {1, 2, . . . , m − s}, som − s − p ≠ 0 and so, the initial monomial of v p is equal to (m − s − p)x p+s−1 . Hence,deg v p = 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). Assumethat m = deg ϕ 2 and ϕ is not square-free. Then every derivation D of R[t, d] has aunique decompositionD = W + E H + ∆ r ,where W is an inner derivation of R[t, d], H ∈ R[t, d] is a special polynomial, and r ∈ Rwith deg r < m. Moreover, if D is as above, then D is inner if and only if H = 0 andr = 0.Proof. Let D be a derivation of R[t, d]. If D(t) = 0, then, by Proposition 4.1,D = W f for some f ∈ k[t], so in this case: D = W f + E 0 + ∆ 0 .Assume now that D(t) ≠ 0. By Corollary 6.3 we know that D = W + D 1 , whereW is an inner derivation of R[t, d] and D 1 is a derivation of R[t, d] such that D 1 (t) =u n t n + · · · + u 1 t 1 + u 0 , where u 0 , . . . , u n ∈ R and deg u i < deg ϕ for all i = 0, 1, . . . , n.By Theorem 7.8, for every i ∈ {1, . . . , n} there exists a i ∈ R such thatu i ϕ = a i ϕ ′ − a ′ iϕ.Since the degree of each u i is smaller then m, Lemma 7.2 implies that deg u i m − 2, fori = 1, . . . , n.Consider the polynomial u n t n . Suppose that u n ≠ 0. Then, by Lemma 7.3, deg u n s − 1, where s is the number of all pairwise different roots of the polynomial ϕ belongingto an algebraic closure k of k. So, s − 1 deg u n m − 2. Put deg u n = p + s − 1, wherep ∈ {0, 1, . . . , m − s − 1}.Let M c := E cx p tn, where 0 ≠ c ∈ k. By Lemma 11.1, there exists a nonzero coefficientc ∈ k such that M c (t) = v p t n with 0 ≠ v p ∈ k[x], where the initial monomial of v pcoincides with the initial monomial of u n . Let D 2 = D 1 − E cx p t n. Then D 2(t) = u n t n +u n−1 t n−1 + · · · + u 1 t 1 + u 0 , where the polynomials u 0 , . . . , u n−1 are the same as in D 1 (t)and deg u n < deg u n . If u n ≠ 0, then we repeat the same procedure. Moreover, we repeatthis procedure for all the successive polynomials u n t n , u n−1 t n−1 , . . . , u 1 t 1 .Thus, we see that D 1 = E H +M, for some special H ∈ R[t, d], where M is a derivationof R[t, d] such that M(t) = u 0 ∈ R. We know, by Proposition 5.3, that M = ∆ u0 + W ′ ,where W ′ 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 + E H + ∆ r witha special H ∈ R[t, d] and r ∈ R, deg r < deg ϕ. Moreover, by Propositions 5.1 and 6.2,such a derivation W + E H + ∆ 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 theproof of Theorem 10.1. □Consider the Ore extension A = k[x][t, x 2 ∂ ].∂Here d = x2 , ϕ = ∂x ∂x x2 and ψ =gcd(ϕ, ϕ ′ ) = x. For any n 1, denote by E n the derivation E t n. Then E n (t) = [ 1 x tn , t] =−d( 1 x )tn = t n and E n (x) = nxt n−1 + h n−2 , for some h n−2 ∈ A with deg t h n−2 < n − 1. PutM n := E n − [nt n−1 , ].Since d p (x) = p!x p+1 for p = 0, 1, . . . , it is easy to check that M n is such a derivation ofA that M n (t) = t n and M n (x) = nxt n−1 . Every derivation of the form M n is of coursenon-inner. As a consequence of Theorem 11.2 we obtain the following description of allderivations of A.Example 11.3. Every derivation D of A = k[x][t, x 2 ∂ ] has a unique decomposition∂xD = W + c n M m + · · · + c 1 M 1 + c 0 ∆ 1 ,where n 1, W is an inner derivation of A, and c 0 , . . . , c n ∈ k. Moreover, if D is asabove, then D is inner if and only if c n = · · · = c 0 = 0. □Using Theorem 11.2 we may produce similar examples for any Ore extension k[x][t, d]with deg ψ 1 (where ψ = gcd(ϕ, ϕ ′ ) and ϕ = d(x)).References[1] J. Cozzens, C. Faith, Simple Noetherian Rings, Cambridge Tracts in Mathematics,69, Cambridge University Press, 1975.[2] J. Dixmier, Sur les algèbres de Weyl, Bull. Soc. Math. France, 96(1968), 209 - 242.[3] J. Dixmier, Sur les algèbres de Weyl. II, Bull. Soc. Math. France 2 e sèrie, 94(1970),289 - 301.[4] J. Dixmier, Enveloping Algebras, Academie - Verlag, Berlin, 1977.[5] A. van den Essen, Polynomial automorphisms and the Jacobian Conjecture, Progressin Mathematics 190, 2000.[6] D. A. Jordan, Noetherian Ore extensions and Jacobson rings, J. London Math. Soc.,10(1975), 281 - 291.[7] K. Kishimoto, On abelian extensions of rings, I, Mathematical Journal of OkayamaUniversity, 14(1970), 159 - 174.19

[8] K. Kishimoto, A. Nowicki, On the image of derivations for fields of characteristiczero, Communications in Algebra, 23(12)(1995), 4557 - 4562.[9] J. Kovacic, An Eisenstein criterion for noncommutative polynomials, Proc. of Amer.Math. Soc., 34(1972), 25 - 29.[10] A. Nowicki, Polynomial derivations and their rings of constants, N. Copernicus UniversityPress, Toruń, 1994.[11] O. Ore, Theory of non-commutative polynomials, Ann. of Math., 34(1933), 480 - 508.[12] R. Rentschler, Opérations du groupe additif sur le plan affine, C. R. Acad. Sc. Paris267(1968), 384 - 387.[13] Y. Stein, On the density of image of differential operators generated by polynomials,J. Analyse Math., 52(1989), 291 - 300.Faculty of Mathematics and Computer Science,N. Copernicus University,87–100 Toruń, Poland,(e-mail: anow@mat.uni.torun.pl)20