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 2 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 2 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 ' > 1, where ' = d(x). 0 0 @' If ' is square-free (that is, if gcd('; ' ) = 1, where ' = @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 2 R of degree smaller than deg '. If ' 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 2 R with deg r < deg ', 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 2 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 2 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 2 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 2 k[x], then we denote by a0 the derivative of a. Note that (since char(k) = 0) for every b 2 k[x] there exists a 2 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 ' = d(x), then the image d(k[x]) is the principal ideal of k[x] generated by '. 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 ' = d(x). Let a; b 2 k[x], b 6= 0 and gcd(a; b) = 1. Then the following properties are equivalent: a (1) d( b ) 2 k[x]; (2) b divides , where = gcd('; '0). Proof. Put ' = g , '0 = f , where f; g 2 k[x], gcd(f; g) = 1. (2) ) (1). Assume that = cb with 0 6= c 2 k[x]. Then we have: a ac ac acg 1 1 0 2 0 d( b ) = d( cb ) = d( ) = d( ' ) = '2 (d(acg)' − acgd(')) = '2 ((acg) ' − acg' ')) 0 acg'0 0 ac'0 0 = (acg) − ' = (acg) − = (acg) − acf; a that is, d( b ) 2 k[x]. a 1 (1) ) (2). Assume that d( b ) = u 2 k[x]. Then b2 (d(a)b − ad(b)) = u, so a0b' − ab0' = b2u: If b 2 k r f0g, then of course b j . Assume that deg b > 1, and let p be an irreducible s t polynomial from k[x] dividing b. Let b = p b1, b1 2 k[x], s > 1, p - b1, and let ' = p '1, '1 2 k[x], t > 0, p - '1. Then, by the above equality, we have 0 0 t+s 0 s+t−1 2s 2 (a b1'1 − ab1'1) p − sap b1'1p = p b1u: 0 If t 6 s, then this equality implies that p divides ap b1'1. But it is a contradiction because 0 p - p , p - b1, p - '1 and p - a (since p j b and gcd(a; b) = 1). Hence, t > s, that is, t−1 > s. But pt−1 divides '0, so ps divides gcd('; '0) = . s1 sr Now let b = vp1 ··· pr be a decomposition of b into irreducible polynomials (here v 2 k r f0g; 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 , so b divides . Corollary 2.2. Let d : k[x] ! k[x] be a nonzero derivation and let ' = d(x). Assume 0 that gcd('; ' ) = 1 and let α 2 k(x). Then d(α) 2 k[x] () α 2 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 2 M(R; d), then δ1 + δ2 2 M(R; d). If δ 2 M(R; d) and a 2 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 2 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 2 R = k[x]. This means, that M(k[x]; d) is a free k[x]-module and the set fδn; n = 0; 1;::: g is its basis. Now we introduce a new basis of M(k[x]; d). Assume that ' := d(x) 6= 0. If f is a given element of R[t; d], then for every g 2 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] = 'h for some h 2 R[t; d]. This means that the mapping ' [f; ] (for any f 2 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 ' 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 ' ; 2 0 1 0 0 e2(x) = t + ' t + 3 (' ') ; 3 3 0 2 0 0 1 000 2 0 00 0 3 e3(x) = t + 2 ' t + (' ') t + 4 (' ' + 4' ' ' + (' ) ) : It is easy to check the following proposition. Proposition 3.1. For every n = 0; 1; 2;::: we have n e (x) = tn + '0tn−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 fe0; e1; e2;::: g is a basis of the k[x]-module M(k[x]; d). Proof. Suppose that a0e0 + ··· + anen = 0, where n > 0 and a0; : : : ; an 2 k[x]. n Then, by Proposition 3.1, 0 = a0e0(x) + ··· + anen(x) = ant + h, where h 2 R[t; d] and degt h 6 n − 1.