369 Condition (i) follows easily from the fact that (a + b)z = az + bz and that for a # 0 the order of az equals 1. Condition (ii,) follows from the associative law (ab)z = a(bz), see also 12, 51. This means, in particular, that 6, is a monomorphism and that 6,6;’ is a Jo-derivation of F. We will say that a sequence (a,, 6, ,..., 6, ,...) of mappings di from F’to F is admissible if (i) and (ii,) hold for all M. Such a sequence could be finite. We want to investigate the set of all admissible sequences for a given field F and we assume from now on that F is commutative. One known admissible sequence is (id, 6, 6*, 6’,..., a”,...) for an ordinary derivation 6 of F. It will be proved that (id, 0,O ,..., 0, 6 = 6,. 0, 0 ,..., 0, ((k + 1)/2) 6’ = 6,,, 0 ,..., 0, 0 ,..., 0, ((2k + I)(k + 1)/(2 x 3)) 6” = 6,, ,...) is admissible for a derivation 6 of F. This sequence is just a special case of a general type of sequence discussed in this paper in which the 6; = gi(S) for a derivation 6 of F and where the gi(x) are certain polynomials in one variable x with coef- ficients in K = {a E F; us = 0). Even though it seems likely that these sequences are admisssible we were not able to prove this in general. Further, it is shown that finite admissible sequences do exist which cannot form the beginning segment of a longer admissible sequence. Another obvious problem, not dealt with here, is the investigation of equivalence classes of admissible sequences under the equivalence relation given by the isomorphism of the corresponding rings F[ [z, a,, 6, ,..., d,,... 11 with multiplication defined by (M). 1.
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