American Journal of Mathematics, Vol. 109, No. 1. (Feb., 1987), pp

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A Simple Algorithm for Cyclic Vectors Nicholas M. Katz American Journal of Mathematics, Vol. 109, No. 1. (Feb., 1987), pp. 65-70. Stable URL: http://links.jstor.org/sici?

1 A Simple Algorithm for Cyclic Vectors Nicholas M. Katz American Journal of Mathematics, Vol. 109, No. 1. (Feb., 1987), pp Stable URL: American Journal of Mathematics is currently published by The Johns Hopkins University Press. Your use of the JSTOR archive indicates your acceptance of JSTOR's Terms and Conditions of Use, available at JSTOR's Terms and Conditions of Use provides, in part, that unless you have obtained prior permission, you may not download an entire issue of a journal or multiple copies of articles, and you may use content in the JSTOR archive only for your personal, non-commercial use. Please contact the publisher regarding any further use of this work. Publisher contact information may be obtained at Each copy of any part of a JSTOR transmission must contain the same copyright notice that appears on the screen or printed page of such transmission. The JSTOR Archive is a trusted digital repository providing for long-term preservation and access to leading academic journals and scholarly literature from around the world. The Archive is supported by libraries, scholarly societies, publishers, and foundations. It is an initiative of JSTOR, a not-for-profit organization with a mission to help the scholarly community take advantage of advances in technology. For more information regarding JSTOR, please contact support@jstor.org. Mon Oct 15 12:57:

2 A SIMPLE ALGORITHM FOR CYCLIC VECTORS Statement of Results. Let R be a commutative ring with unit, a :R -t R a derivation of R to itself, and t E R an element with a(t)= 1. We denote by Ra = Ker(8)the subring of "constants." For any constant a E Ra, the element t + a of R also satisfies a(t + a) = 1. Fix an integer iz r 1, and a triple (V, D, Z') consisting of a free R-module V of rank 12, an additive mapping D: V -t V satisfying for all f E R, v E V, and an R-basis Z' = (eo,..., el,- of V. An element v E V is said to be a cyclic vector for (V,D) if v, Dv,..., DUp1(v) is an R-basis of V; a basis of this form is called a cyclic basis. Suppose now that (n- I)! is invertible in R. For each constant a E Ra, we define an element c(z',t - a) in V by the formula THEOREM 1. Suppose that R is a local Z[l/(n - l)!]-algebra whose maximal ideal coiztaii?~ t - a. Then c(z, t - a) is a cyclic vector. THEOREM2. Let R be a ring iiz which (n - I)! is invertible, and let k be a sub-field of Ra. Suppose that #(k) > n(iz - I), and let 00,al,..., be 1 + n(iz - 1) distinct elenzeizts of k. Then Zariski locally on Spec(R), one of the vectors c(z, t - a;),0 5 i 5 n(n - I), is a cyclic vector. Manuscript received 30 November American Journul of Mathematics 109 (1987),

3 66 NICHOLAS M. KATZ Proofs. We first compute the derivatives of c(2, t - a). For this, we introduce the elements c(i,,j)e V, indexed by i,,iintegers 2 0, defined inductively as follows: By definition of c(2, t - a), we have Successively applying D, we easily verify by induction on i that for i r 0 we have A straightforward induction on i +j shows that for i +.j have 5 n - 1,we and so in particular we have (***) c(i, 0) = e; for i = 0, 1,..., n - 1. Therefore (**) yields the congruence D'c(2, t - a) ejmod (t - a)v, for 0 _i i ,

4 CYCLIC VECTORS from which Theorem 1follows, by Nakayama's lemma. To prove Theorem 2, we argue as follows. For 0 5 i 5 IZ variable X E R, we define elements ci(z, X) in V by 67-1, and In AU(V),we visibly have with P(X) the value at X of a polynomial P(T) E R[T]of degree 5 IZ(IZ - 1). By (***), we have ci(z, 0) = ei, so At X = t - a with a constant, (**) gives ci(z, t - a) = Dic(Z, t - a). Therefore c(z, t - a)is a cyclic vector if and only if P(t - a)lies in R '. We must show that the ideal I in R generated by the 1 + n(lz - 1) values P(t - a;)is the unit ideal. Let us write explicitly Then But for i #,j, the differences (t - a;)- (t - a;) = a, - ai lie in k ', so in R '; hence the van der Monde determinant lies in R '. Therefore the ideal I is equal to the ideal generated by the coefficients vo,..., Y,,(,,-~,of P(X).But vo = P(0) = 1. Q.E.D.

5 68 NICHOLAS M. KATZ Reinavks. (1) Suppose 2 = (eo,...,el,-,) is a cyclic basis to begin with, i.e., eo is a cyclic vector and ei = Dieo for 0 5 i 5 iz - 1. Then c(0, 0) = eo, and c(0, j) = 0 forj > 0. Therefore c(2, t - a) = eo is the cyclic vector we began with. (2) Suppose R is a field, and n 2 2. If (n - I)! is not invertible in R, (V, D) may admit no cyclic vector. For take a prime numberp, R = Fl,(t), a = d/dt, V = Ru, D(fl,...,f,,)= (af,,..., af,,). Because dl' = 0, we have Dl' = 0, so (V, D) admits no cyclic vector if p 5 iz - 1. (3) Suppose R is a field, iz 2 2, and (n - I)! invertible in R. For a suitably chosen basis d, c(d, t) can vanish. Indeed, if eo is a cyclic vector, and if ei = D1eo for 0 Ii 5 n - 2, then so we can solve for el,- to force c(2, t) = 0. (4) If R is a field in which (12 - I)! is invertible, and which is a finitely generated extension of an algebraically closed subfield k of Ra, then we can use Theorem 1 to produce cyclic vectors. Notice first that for any finite subset S of R, there exists a a-stable k-subalgebra Ro of R which is finitely generated as a k-algebra, and which contains S (in terms of generators xl,...,xn of R/k, write each a(xi) and each s E S as a ratio of k-polynomials in the xi's whose denominators are nonzero in R; then take for Ro the k-subalgebra of R generated by the xi and by the inverses of the denominators of both the a(x,) and the s E S). Given (V, D, d) over R, we apply this to the set S consisting of t and of the n2 coefficients a;; of the connection matrix, defined by Over the resulting Ro, we have a canonical descent (Vo, D, d) of the original (V, D, d) over R. For any k-valued point of X = Spec(Ro), we have ring inclusions Ro c Ox,,c R, and by Theorem 1 we know that c(d, t - t(x)) is a cyclic vector for Vo ORoOX,,,., SO a fortiori c(2, t - t(x)) is a cyclic vector for V = VoOK(, R itself. (5) The heuristic which leads to considering c(2, t) is the following. Suppose R = C[[t]], a = d/dt. If ho,..., h,,-, is a horizontal R-basis of V, i.e., an R-basis with Dh, = 0 for 0 5 i 5 n - 1, then

6 CYCLIC VECTORS is obviously a cyclic vector. Now given any v E V, the t-adically convergent series (cf. [2], proof of 8.9) is the unique solution of S v mod tv, D(S) = 0. Therefore if Z = (eo,..., el,-l) is any R-basis of V, then (do,..., c?,,-~) is, by Nakayama's lemma, a horizontal R-basis, and consequently is a cyclic vector. But if v E V is a cyclic vector, then so, by Nakayama's lemma, is v + t"vo for any vo E V,simply because, for 0 5 i 5 n - 1, Di(v + tnvo)= Div mod tlz-'v. Therefore in the above double sum, we may neglect all terms with.j -k k 2 n,to conclude that is a cyclic vector. But this last vector is easily seen to be c(z, t). (6) The proof of Theorem 2 also yields the following variant: If R is a ring in which (iz(iz - I))! is invertible, then Zariski locally on Spec(R), one of the vectors c(z,t - i), 0 5 i 5 iz(n - I), is a cyclic vector. PRINCETON UNIVERSITY

7 NICHOLAS M. KATZ REFERENCES I. F. T. Cope. Fornlal solutions of irregular linear differential equations, Part 11..4rir~r.J. Matlr, 58 (1936) P. Deligne, Eyrrtrtiorrs tli~;f~rt.rrtiel/rs ti poirzts sirrgtrliers rdg~rliers, Springer Lecture Notes it1 Mathematics. 163 (1970). 3. N. Kata. Nilpotent connections and the monodromy theorem: Applications of a result of Turrittin, Ptrb. Muth. I.H.E.S.,39 (1970),

Journal of the American Mathematical Society, Vol. 2, No. 2. (Apr., 1989), pp

An Estimate for Character Sums Nicholas M. Katz Journal of the American Mathematical Society, Vol. 2, No. 2. (Apr., 1989), pp. 197-200. Stable URL: http://links.jstor.org/sici?sici=0894-0347%28198904%292%3a2%3c197%3aaefcs%3e2.0.co%3b2-l