NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p. Contents. 1. Introduction Preliminaries 3. References

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1 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p ANNA MEDVEDOVSKY Abstract. We prove that the killing rate of certain egree-lowering recursion operators on a polynomial algebra over a finite fiel grows slower than linearly in the egree of the polynomial attacke. We also explain the motivating application: obtaining a lower boun for the Krull imension of a local component of a big mo-p Hecke algebra in the genus-zero case. We sketch the application for p = 2 an p = 3 in level one. The case p = 2 was first establishe in by Nicolas an Serre in 2012 using ifferent methos. Contents 1. Introuction 1 2. Preliminaries 3 3. The Nilpotence Growth Theorem (NGT) 5 4. A toy case of the NGT 7 5. Applications to mo-p Hecke algebras for p = 2, The proof of the NGT begins The content function an its properties Content of some proper fractions The nilgrowth witness Complements 26 References Introuction The main goal of this ocument is to prove the following Nilpotence Growth Theorem, about the killing rate of a recursion operator on a polynomial algebra over a finite fiel uner repeate application: Key wors an phrases. linear recurrences in characteristic p, mo-p moular forms, congruences between moular forms, Hecke algebras, p-regular sequences. 1

2 2 ANNA MEDVEDOVSKY Theorem A (Nilpotence Growth Theorem; see also Theorem 1). Let F be a finite fiel, an suppose that T : F[y] F[y] is a egree-lowering F-linear operator satisfying the following conition: The sequence {T (y n )} n of polynomials in F[y] satisfies a linear recursion over F[y] whose companion polynomial (i) X + a 1 X a F[y][X] has both total egree an y-egree. Then there exists a constant α < 1 so that the minimum power of T that kills y n is O(n α ). To prove this theorem, we reuce to the case where the companion polynomial of the recursion has an empty mile in its egree- homogeneous part: that is, when it has the form X + ay + (lower-orer terms) for some a F. Then we prove this empty-mile case (see Theorem 4 below) by constructing a function c : F[y] N { } that grows like (eg f) α an whose value is lowere by every application of T. In the special case where is a power of p, the function c takes y n to the integer obtaine by writing n base an then reaing the expansion in some smallers base, so that the sequence {c(y n )} n is p-regular in the sense of Allouche an Shallit [2]. The proof that c(t (y n )) < c(y n ), by strong inuction, uses higher-orer recurrences epening on n, so that n is compare to numbers whose base- expansion is not too ifferent. It is the author s hope that ieas from p-automata theory can eventually be use to sharpen an generalize the Nilpotence Growth Theorem. Motivating application of the Nilpotence Growth Theorem: The motivating application for the Nilpotence Growth Theorem (Theorem A above) is the so-calle nilpotence metho for establishing lower bouns on imensions of local components of Hecke algebras acting on mo-p moular forms of tame level N. These Hecke algebra components were first stuie by Jochnowitz in the 1970s [11], but the first full structure theorem, for p = 2 an N = 1, ue to Nicolas an Serre, appeare only in 2012 [15]. The Nicolas-Serre metho uses the recurrence satisfie by Hecke operators (see equation (5.1) below) to escribe the action of Hecke operators on moular forms moulo 2 completely explicitly but unfortunately these explicit formulas o not appear to generalize beyon p = 2. The case p 5 was then establishe by very ifferent techniques by Bellaïche an Khare for N = 1 [4], an later by Deo for general level [7]. The Bellaïche- Khare metho euces information about mo-p Hecke algebra components from corresponing characteristic-zero Hecke algebra components, which are known to be big by the Gouvêa-Mazur infinite fern construction ([10]; see also [8, Corollary 2.28]). The nilpotence metho is yet a thir technique, coming out of an iea of Bellaïche for tackling the case p = 3 an N = 1 as outline in [4, appenix], an implemente an evelope in level one for p = 2, 3, 5, 7, 13 in the present author s Ph.D. issertation [13]. Like Nicolas-Serre, the nilpotence metho stays entirely in characteristic p an makes use of the Hecke recursion; but instea of explicit Hecke action formulas, the Nilpotence Growth Theorem (Theorem A above) now plays the crucial imension-bouning role. See section 5 below for a taste of this metho for p = 2, 3, which completes the etermination of the structure of the Hecke algebra for p = 3 begun in [4, appenix] an recovers the Nicolas-Serre result for p = 2. (i) See section 2.4 for efinitions.

3 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p 3 The nilpotence metho using the Nilpotence Growth Theorem can be generalize for all (p, N) if the genus of X 0 (Np) is zero; see the forthcoming [12] for etails. Structure of this ocument: After a few preliminary efinitions in section 2, we state the Nilpotence Growth Theorem (restate as Theorem 1) in section 3 an iscuss the various conitions of the theorem. In section 4, we prove a toy version of the theorem (Theorem 2). In section 5, we use the toy version of NGT (Theorem 2) from the previous sections to prove that the mo-p level-one Hecke algebra for p = 2, 3 has the form F p x, y. This section illustrates the motivating application of the Nilpotence Growth Theorem, an is not require for the rest of the ocument. This woul be a reasonable stopping point for a first reaing. At this point, in section 6 the proof begins in earnest. There is a short overview of the structure of the proof in subsection 6.1. In subsection 6.2, we reuce to working over a finite fiel. In subsection 6.3, we reuce to the so-calle empty-mile NGT (Theorem 4). In subsection 6.4, we give the inuctive argument that reuces the proof of the empty-mile NGT to fining a nilgrowth witness function that satisfies certain properties. In section 7, we iscuss base-b representation of numbers an introuce the content function. In section 8, we prove a number of technical inequalities about the content function. In section 9 we efine the nilgrowth witness, finishing the proof of the empty-mile NGT, an hence of the NGT in full. Finally in section 10, we state a more precise version of the toy NGT an speculate on the optimality of some bouns. Acknowlegements: I am greatly inebte to my Ph.D. avisor Joël Bellaïche for sparking an supporting this investigation into moular forms moulo p. The initial iea for the nilpotence metho is his a most generous gift. I woul also like to thank Paul Monsky for many illuminating iscussions on the topic of mo-p moular forms an Hecke algebras, an Kiran Kelaya for helpful comments on an earlier version of Theorem 1. Part of the writing of this ocument was carrie out at the Max Planck Institute for Mathematics in Bonn, an I am grateful for their hospitality. Many extensive computations relate to this project were performe using SAGE [19]. 2. Preliminaries This section contains a brief review of a few unconnecte algebraic notions. All rings an algebras are assume to be commutative, with unity. We use the convention that the set of natural numbers starts with zero: N = {0, 1, 2,...}. Below, R is always a ring Structure of finite rings. If R finite, then R is artinian, hence a finite prouct of finite local rings. If R is a finite local ring with maximal ieal m, then the resiue fiel R/m is a finite fiel of characteristic p. Moreover, the grae pieces m n /m n+1 are finite R/m-vector spaces, so that that R has carinality a power of p. Basic examples of finite local rings are F p [t]/(t k ) an Z/p k Z Degree filtration on a polynomial algebra. If 0 f = n 0 c ny n is a polynomial in R[y] (so only finitely many of the c n are nonzero), then its y-egree, or just egree, is as usual efine to be eg f := max{n : c n 0}. Also let eg 0 :=.

4 4 ANNA MEDVEDOVSKY The egree function gives R[y] the structure of a filtere algebra: Let R[y] n := {f R[y] : eg f n}, an then R[y] = n 0 R[y] n an multiplication preserves the filtration as require Local nilpotence an the nilpotence inex. Let M be any R-moule (for example, M = R[y]) an T En R (M) an R-linear enomorphism. (In applications to Hecke algebras, R will be a finite fiel, M an infinite-imensional space of moular forms, an T a Hecke operator.) The operator T : M M is locally nilpotent on M if every element of M is annihilate by some power of T. If T is locally nilpotent, an f in M is nonzero, we efine the nilpotence inex of f with respect to T : N T (f) := max{k 0 : T k f 0}. Also set N T (0) :=. Suppose R = K is a fiel an M = K[y] an T : M M preserves the egree filtration: that is, T (K[y] n ) K[y] n. Then T is locally nilpotent if an only if T strictly lowers egrees, in which case we also have N T (f) eg f. For example, T = y is locally nilpotent on K[y]. If K has characteristic zero, then N T (f) = eg f; otherwise N T (f) char K Linear recurrences an companion polynomials. Now suppose that M is an R-algebra, an M is an M-moule (we will normally take M = M ). A sequence s = {s n } n M N satisfies a (monic) M-linear recurrence of orer if there exist elements a 1,..., a M so that (2.1) s n = a 1 s n a s n for all n. We o not a priori assume that a 0, but we o insist that the recursion alreay hol for n =. The companion polynomial of this linear recurrence is P (X) = X a 1 X 1 a M[X]. Example 2.1. The sequence s = {0, 1, y, y 2, y 3, y 4,...} R[y] N satisfies an R[y]-linear recursion of minimal orer 2: we have s n = ys n 1 for all n 2, but not for n = 1. The companion polynomial of the recurrence is therefore X 2 yx. Given any sequence s in M N, the set of companion polynomials of M-linear recurrences satisfie by s forms an ieal of M[X]. We recor this observation in the following form: Fact 2.1. If a sequence s M N satisfies the recurrence efine by some monic P M[X], then it also satisfies the recurrence efine by P Q for any other monic Q M[X]. In characteristic p, then, we get the following corollary, of which we will make crucial use: Corollary 2.1. If R has characteristic p, an s M N satisfies the orer- recurrence s n = a 1 s n 1 + a 2 s n a s n for all n, then for every k 0 the sequence s also satisfies the orer-p k recurrence (2.2) s n = a pk 1 s n p k + apk 2 s n 2p k + + apk s n p k for all n pk.

5 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p 5 Proof. Let P = X a 1 X 1 a be the companion polynomial of a recursion satisfie by s. By Fact 2.1, the sequence s also satisfies the recurrence whose companion polynomial is P pk = X pk a pk 1 Xpk p k a pk 2 Xpk 2p k a pk, which is exactly what is expresse in equation (2.2). If M can be embee into a fiel K, we have the following well-known characterization of power sequences in K N satisfying a fixe M-linear recurrence: Fact 2.2. An element α in K is a root of monic P M[X] if an only if the sequence {α n } n = {1, α, α 2,...} satisfies the linear recurrence with companion polynomial P. If the companion polynomial of such an M-linear recurrence has no repeate roots in K, it follows from the proposition that every solution to the recurrence is a linear combination of such power sequences on the roots of the companion polynomial. One can further escribe all K-sequences satisfying a general M-linear recursion see, for example, [6] an the historical references therein but we will not nee this below. 3. The Nilpotence Growth Theorem (NGT) 3.1. Statement of the NGT. We are now reay to state the most general version of the Nilpotence Growth Theorem (NGT). From now on, we will assume that R will be a finite ring, an M = R[y]. The eventual Hecke examples will come from the case where R is a finite fiel. Theorem 1 (Nilpotence Growth Theorem). Let R be a finite ring, an suppose that T : R[y] R[y] is an R-linear operator satisfying the following two conitions: (1) T lowers egrees: eg T (f) < eg f for every nonzero f in R[y]. (2) The sequence {T (y n )} n satisfies a filtere linear recursion over R[y]: that is, there exist a 1,..., a R[y], with eg a i i for each i, so that for all n, Suppose further that T (y n ) = a 1 T (y n 1 ) + + a T (y n ). (3) the coefficient of y in a is invertible in R. Then there exists a constant α < 1 so that N T (y n ) n α. In other wors, Theorem 1 implies that, uner a mil technical assumption (conition (3)), the nilpotence inex of a egree-lowering operator efine by a filtere linear recursion grows slower than linearly in the egree. The mil technical assumption is necessary in the theorem as state: see the iscussion in (4) in subsection 3.2 below Discussion of the NGT.

6 6 ANNA MEDVEDOVSKY (1) Connection with Theorem A: If T : R[y] R[y] satisfies the conitions of Theorem 1, then the companion polynomial of the recursion satisfie by the sequence {T (y n )} n is P T = X a 1 X 1... a R[y, X]. The conition eg a i i from (2) guarantees that the total egree of P T is exactly. In particular, in the case where R = F is a finite fiel, conition (3) implies that eg y P T = eg a =. In other wors, Theorem 1 over a finite fiel reuces to Theorem A, as implie. (2) Conition (1) guarantees that T is locally nilpotent. Moreover, N T (y n ) n, so that the function n N T (y n ) a priori grows no faster than linearly. (3) Conition (2) an connection to recursion operators: The conition that the sequence {T (y n )} n satisfies a linear recurrence is the efinition of a recursion operator, a notion that will be explore in a future paper. A natural source of filtere recursion operators (that is, satisfying aitional egree bouns as in conition (2) above) comes from the action of Hecke operators on algebras of moular forms of a fixe level. Namely, if f is a moular form of weight k an level N an T is a Hecke operator acting on the algebra M of forms of level N, then the sequence {T (f n )} n satisfies an M-linear recursion with companion polynomial X + a 1 X a, where a i comes from weight ki. See equations (5.1) an (5.2) below for examples over F p, [13, chapter 6] for a proof in level one when T is a prime Hecke operator, or the forthcoming [12] for more etails. This kin of recurrence in characteristic 2 was crucially use in Nicolas-Serre [14, section 3] to obtain the structure of the mo-2 Hecke algebra. Earlier appearances of the Hecke recurrence inclue [1] (two-imensional recurrence for {T l (E n 4 Em 6 )} n,m in characteristic zero) an [5, p. 594] (recurrence for U 2 acting on a power basis of overconvergent 2-aic moular functions). (4) Conition (3) is necessary as state: Consier the operator T : R[y] R[y] efine by T (y n ) = s n, where s n is the sequence {0, 1, y, y 2,...} with companion polynomial X 2 yx from Example 2.1. All conitions except (3) are satisfie, an it is easy to see that N T (y n ) = n in this case. For an example with a 0, consier the operator T with efining companion polynomial P T = X 2 + yx + y with initial values [T (1), T (y)] = [0, 1]. By inuction, eg T (y n ) = n 1. Therefore N T (y n ) = n. Computationally, it appears that if R = F p an eg a < but there exists an i with 0 < i < so that eg a i = i, then either N T grows logarithmically or else it grows linearly. In that sense, it appears that fullness of egree at the en of P T (that is, the presence of a y term) appears to be, at least generically, necessary to compensate for fullness of egree in the mile (that is, the presence of a y i X i term for some 0 < i < ), if one wants the growth of N T to be sublinear but not egenerate. Further explanation is necessary to unerstan this behavior well. (5) The constant α: The power α epens on R an only, an tens to 1 as. More precisely, the epenence on R is only through its maximal resiue characteristic; the length of R as a moule over itself affects only the implicit constant of the growth conition N T (y n ) n α. In the special case empty-mile case where the inequality eg a i < i is strict for every i <, we can take α to be log p k(p k 1) for k satisfying p k. See Theorem 2 or Theorem 4 below.

7 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p 7 (6) Finite characteristic is necessary: A counterexample in characteristic zero: Consier the operator T on Q[y] with P T = X 2 yx y 2 an egree-lowering initial terms [T (1), T (y)] = [0, 1]. This satisfies all three conitions of the NGT. It is easy to see that T (y n ) = F n y n 1, where F n is the n th Fibonacci number: the recursion is s n = ys n 1 + y 2 s n 2. Therefore T k (y n ) = F n F n 1 F n k+1 y n k, so that N T (y n ) = n. (Compare to characteristic p, where the operator efine by T (y n ) = F n y n 1 on F p [y] satisfies T p+1 0.) See also Proposition 10.1 for a family of examples in any egree. Computationally, it appears that generic characteristic-zero examples that o not egenerate (to log n growth) all exhibit linear growth. Over a finite fiel, computationally one sees a spectrum of O(n α ) growth for various α < 1. (7) Finiteness of R is necessary as state: A counterexample over F p (t): Let P T = X 2 tyx y 2 an start with [0, 1] again. Then T (y n ) = F n (t)y n 1 with F n (t) F p [t] monic of egree n 1, so that N T (y n ) = n again. However, see the empty-mile case (Theorem 4) for a special case that oes hol for infinite rings of characteristic p. 4. A toy case of the NGT Fix a prime p an take R = F p (ii), = p k, an assume that the recursion has an empty mile : eg a i < i for 0 < i <. Theorem 2 (Toy case of NGT). Let q = p k for some k 1. Suppose T : F p [y] F p [y] is a egree-lowering linear operator so that the sequence {T (y n )} n satisfies an F p [y]-linear recursion with companion polynomial P = X q + (terms of total egree < q) + ay q for some a F p. Then N T (y n ) n log(q 1)/ log q. F p [y][x] Most of the main features of the proof of Theorem 1 are alreay present in the proof of Theorem 2. We give it here because the proof is technically much simpler; unerstaning it may suffice for all but the most curious reaers The content function. Following Bellaïche in [4, appenix] for q = 3, we efine a function c : N N epening on q as follows. Given an integer n, we write it base q as n = i n iq i with 0 n i < q, only finitely many of which are nonzero, an efine the q-content of n as c(n) := c q (n) := i n i(q 1) i. For example, since 71 = [2 4 1] 5 in base 5, the 5-content of 71 is = 49. The following properties of the content function are easy to check. See also section 7.1, where the content function an variations are iscusse in etail. Proposition 4.1. (1) c(n) n log q (q 1) (2) c(q k n) = (q 1) k c(n) for all k 0 (ii) In fact R may be any ring of characteristic p. We work with Fp in this section for simplicity.

8 8 ANNA MEDVEDOVSKY (3) If 0 n < q, then c(n) = n. (4) If i is a igit base q an n i has no more than 2 igits base q, then c(n) c(n i) is either i or i 1. (5) If q n < q 2, then c(n q) = c(n) q Setup of the proof. We now efine the q-content of a polynomial f F p [y] through the q-content of its egree. More precisely, if 0 f = a n y n, let c(f) := max{c(n) : a n 0}. Also set c(0) :=. For example, the 3-content of 2y 9 + y 7 + y 2 is max{c(9), c(7), c(2)} = 5. Now let T : F p [y] F p [y] be a egree-lowering recursion operator whose companion polynomial P = X q + a 1 (y)x q a q (y) F p [y][x] satisfies eg a i (y) < i for 1 i < q an eg a q (y) q. To prove Theorem 2, we will show that T lowers the q-content of any f F p [y]: that is, that c(t f) < c(f). Since c(f) < 0 only if f = 0, the fact that T lowers q-content implies that N T (f) c(f). Proposition 4.1(1) will then imply N T (f) (eg f) log(q 1)/ log q, as esire. It suffices to prove that c(t f) < c(f) for f = y n. We will procee by strong inuction on n, each time using recursion of orer q k+1 corresponing to P qk, with k chosen so that q k+1 n < q k+2. This will allow us to compare the q-contents of n an n iq k for i small, so that the last k igits base q are unchange, rather than comparing q-contents of n an n i for i small, which can be very estructive to igits base q. (iii) The base case is n < q, in which case being q-content-lowering is the same thing as being egree-lowering (Proposition 4.1(3)) The inuction. For n q, we must show that c(t (y n )) < c(n) assuming that c(t (y m )) < c(m) for all m < n. As above, choose k 0 with q k+1 n < q k+2. By Corollary 2.1, the sequence {T (y n )} satisfies the orer-q k+1 recurrence T (y n ) = a 1 (y) qk T (y n qk ) + a 2 (y) qk T (y n 2qk ) + + a q (y) qk T (y n qk+1 ). Pick a term y m appearing in T (y n ) with nonzero coefficient; we want to show that c(m) < c(n). From the recursion, y m appears with nonzero coefficient in a i (y) qk T (y n iqk ) for some i. More precisely, y m appears in y jqk T (y n iqk ) for some y j appearing in a i (y), so that either j < i or i = j = q. Then y m jqk appears in T (y n iqk ), an by inuction we know that c(m jq k ) < c(n iq k ). To conclue that c(m) < c(n), it woul suffice to show that c(n) c(m) c(n iq k ) c(m jq k ), or, equivalently, that c(n) c(n iq k ) c(m) c(m jq k ). Since subtracting multiples of q k leaves the last k igits of n base q untouche, we may replace n an m by q k n q k an q k m q k, respectively, an then use Proposition 4.1(2) to cancel out a factor of (q 1) k. In other wors, we must show that c(n) c(n i) c(m) c(m j) (iii) I learne this technique from Gerbelli-Gauthier s proof [9] of the key technical lemmas of Nicolas-Serre [14].

9 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p 9 for n, m, i, j satisfying i n < q 2 an j m < n an either j < i or i = j = q. But this is an easy consequence Proposition 4.1(4) (5): For j < i, we know that c(n) c(n i) is at least i 1 an c(m) c(m j) is at most j i 1. An for i = j = q both sies equal q 1. This completes the proof of Theorem Toy case vs. general case. The proof of the full NGT (Theorem 1) procees by first reucing to the empty-mile case over a finite fiel (Theorem 4 below), of which Theorem 2 is a special case. Apart from the reuction step, most of the ifficulty in generalizing from Theorem 2 to Theorem 4 comes from working with more general versions of the content function. In particular, we must exten the notion of content to rational numbers an prove sufficiently strong analogues of Proposition 4.1 for the proof to procee: see sections Applications to mo-p Hecke algebras for p = 2, 3 This section gives an inication of how the NGT can give information about lower bouns of mop Hecke algebras, the author s main motivation for proving the theorem. More precisely, in this section we will use Theorem 2 to complete the proof of Theorem 24 (iv) of [4, appenix], which establishes the structure of the mo-3 Hecke algebra of level one. Simultaneously an using the same methos, we will give an alternate proof of the main result of [15], the structure of the mo-2 Hecke algebra in level one. See Theorem 3 below. More generally, the NGT can be use to obtain lower bouns on Krull imensions of local components of big mo-p Hecke algebras acting on forms of level N in the case where X 0 (Np) has genus zero, for this is precisely the conition for the algebra of moular forms of level N mo p to be a polynomial algebra over F p. For more etails, see [13] (for level one) or [12], to appear in the future. To generalize the nilpotence metho to all (p, N), one must generalize the NGT to all rings of S-integers in characteristic-p global fiels. We work in level one, an let p {2, 3}. Let M = M(1, F p ) F p q be the space of moular forms of level one moulo p in the sense of Swinnterton-Dyer an Serre (that is, reuctions of integral q-expansions). For p = 2, 3, Swinnterton- Dyer observes [17] that M = F p [ ], where = n i=1 (1 qn ) 24 F p q ; stanar imension formulas show that M k := F p [ ] k, the polynomials in of egree boune by k, coincies with the space of mo-p reuctions of q-expansions of forms of weight 12k, an hence Hecke-invariant. Further, one can show that K = n : p n F p M is the kernel of the operator T p. In particular, K, an hence every finite-imensional subspace K k := M k K, is Hecke-invariant. Let A k En Fp (K k ) be the algebra generate by the action of the Hecke operators T l with l prime an l p. Since K k K k+1, we have A k+1 A k. Let A := lim A k k. Then A is a profinite ring embeing into En(K): it is the shallow Hecke algebra acting on forms of level one mo p. (iv) We prove a weaker version of Proposition 35 loc. cit. In the notation of subsection 7.2 here, we show that, for f F 3[ ], we have c 3,2(T 2f) c 3,2(f) 1 an that c 9,6(T 7f) c 9,6(f) 3, which suffice to establish that the nilpotence inex grows slower than linearly. We o not show that c 3,2(T 7f) c 3,2(f) 2.

10 10 ANNA MEDVEDOVSKY The stanar pairing A K F p given by T, f a 1 (T f) is nonegenerate on both sies an continuous in the profinite topology on A. Therefore A is in continuous uality with K. By work of Tate an Serre [18, 16] we know that is the only Hecke eigenform in K F p. (v) This implies that A is a local F p -algebra with maximal ieal m an resiue fiel F p generate by the moifie Hecke operators T l := T l a l ( ), acting locally nilpotently on M = F p [ ]. Using eformation theory, we euce: x T 3, y T 5 if p = 2 Proposition 5.1. There is a surjection F p x, y A given by x T 2, y T 7 if p = 3. For p = 2, the fact that A is generate by T 3 an T 5 was first prove without eformation theory in [14]. For p = 3, Proposition 5.1 is state [4, appenix], using eformation theory of reucible pseuocharacters, as in [3]. See also [13, Chapter 7] for etaile eformation theory arguments for reucible Galois pseuocharacters in level one. The main result of this section is the following theorem: Theorem 3. The surjection F p x, y A of Proposition 5.1 is an isomorphism. The key input will be Theorem 2, as well as the following observation: if T is any Hecke operator an f is a moular form in a Hecke-invariant algebra M, then the sequence {T (f n )} n satisfies an M-linear recursion. For more etails on the Hecke recursion, see [13, Chapter 6] or the forthcoming [12], but here we will only nee some special cases for f = alreay given in [14] an [4, appenix]. For p = 2, we have (see [14, equations (13) (14)]) (5.1) T 3 ( n ) = T 3 ( n 3 ) + 4 T 3 ( n 4 ), n 3 T 5 ( n ) = 2 T 5 ( n 2 ) + 4 T 5 ( n 4 ) + T 5 ( n 5 ) + 6 T 5 ( n 6 ), n 6 with companion polynomials P 3 = X 4 + X + 4 an P 5 = X X X 2 + X + 6. Note that {T 5 ( n )} n also satisfies the recursion efine by P 5 = P 5(X ) = X 8 + X X + 8. An for p = 3, the recursions satisfie by {T 2 ( n )} an {T 7 ( n )} have companion polynomials (5.2) P 2 = X 3 X + 3, See Lemma 33 (vi) in [4, appenix]. P 7 = X 9 X 5 2 X 4 + ( 4 )X 2 + ( )X 9. Proof of Theorem 3. Let T an S be the generators of A from Proposition 5.1. Then T, S are filtere an egree-lowering recursion operators on F p [ ], each satisfying the conitions of Theorem 2. In other wors, there exists an α < 1 so that N( n ) := N T ( n ) + N S ( n ) n α. (v) Alternatively, one can use an observation of Serre to conclue that any Hecke eigenform in K Fp is in fact efine over F p, reucing the eigenform search to a finite computation. See [4, section 1.2 footnote]. (vi) Lemma 33 loc. cit. gives the egree-8 recurrence satisfie by {T7( n )} n; the recurrence satisfie by { n + T 7( n )} n has an extra factor of X.

11 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p 11 We now claim that the Hilbert-Samuel function of A grows faster than linearly, so that the Krull imension of A is at least 2. Inee, the Hilbert-Samuel function of A sens a positive integer k to im Fp A/m k = im Fp K[m k ] #{n : n K, m k n = 0} = #{n prime to p : N( n ) k} k 1/α, which is certainly faster than linear, since 1 α > 1. Therefore it grows at least quaratically, an the Krull imension of A is at least 2. By the Hauptiealsatz, the kernel of the surjection F p x, y A from Proposition 5.1 is trivial. Using the more precise bouns on α from Theorem 4, we can conclue that, for p = 2 we have α = max{log 4 2, log 8 4} = 2 3 ; an for p = 3 we have α = max{log 3 2, log 9 6} Compare to α = 1 2 obtaine for p = 2 by Nicolas an Serre in [14, 4.1]. Computations suggest that α = 1 2 also hols for p = 3, but we have not been able to prove this. We now begin the proof of Theorem The proof of the NGT begins 6.1. Overview of the proof. The proof procees as follows. (1) Reuce the NGT to the case where R is a finite fiel: See subsection 6.2 below. (2) Reuce to the empty-mile case: The NGT over a finite fiel is implie by a special empty-mile case (Theorem 4), where the companion polynomial has no terms of maximal total egree except for X an y (i.e., the highest-egree homogeneous part has an empty mile). Note that Theorem 4 hols over any ring of characteristic p. See subsection 6.3 below for the statement of Theorem 4 an the reuction step. (3) Prove Theorem 4: The main iea of the proof is as follows. Given the operator T satisfying the conitions of Theorem 4, we efine a function c T : N N that grows like n α for some α < 1. We exten this function to polynomials in R[y] via the egree. Finally, we use strong inuction to prove that applying T strictly lowers the c T -value of any polynomial in R[y]. Therefore, N T (y n ) is boune by c T (n) n α. The key features of this kin of proof are alreay present in the proof of Theorem 2 in section Reuction to the case where R is a finite fiel. Proposition 6.1. If Theorem 1 is true whenever R is a finite fiel, then Theorem 1 is true. Proof. First, suppose R is a finite artinian local ring with maximal ieal m an finite resiue fiel F. Let l be the orer of nilpotence of m (that is, l is the least positive integer so that m l = 0). Let T : R[y] R[y] be the operator in the statement of the theorem, an write T : F[y] F[y] for the operator obtaine by tensoring with with the quotient map R F. Theorem 1 for F guarantees

12 12 ANNA MEDVEDOVSKY that N T (y n ) n α for some α < 1. Let g(n) := max n n {N T (yn ) + 1} n α, so that g is nonecreasing, integer-value, an satisfies T g(eg f) f = 0 for every f in F[y]. Lifting back to R, we get that T g(eg f) f is in m[y] for all f R[y]. (vii) More generally, if f is in m i [y], then T g(eg f) sens f to m i+1 [y]. Since m l = 0, we have T l g(eg f) f = 0 for every f R[y], so that N T (y n ) l g(n) 1 n α. In the general case, R is a finite prouct of finite artinian local rings R i, an an R-linear operator T : R[y] R[y] ecomposes as T i where T i : R i [y] R i [y] is the (R i -linear) restriction of T to R i. From the paragraph above, we can choose α i < 1 so that N Ti (y n ) n α i for all n. Then α = max i {α i } works for T Reuction to the empty-mile NGT. From now on, we fix a prime p. Theorem 1 over a finite fiel of characteristic p is implie by the following special case in which the shape of the recursion satisfie by T is restricte. However, note that the statement below has no finiteness restrictions on the base ring, an no restriction on the coefficient on y. Theorem 4 (Empty-mile NGT). Let R be a ring of characteristic p, an suppose that T is a egree-lowering linear operator on R[y] so that the sequence {T (y n )} n satisfies a linear recursion whose companion polynomial has the shape X + ay + (terms of total egree D) for some D 1 an some constant a R. Let b be a power of p, an suppose that either b 1 or that D b 2. Then N T (y n ) n α for α = log(b D). log b The case = b an D = 1 has alreay been establishe in Theorem 2 (in fact, the same argument extens to = b an any D with 1 D b 1). Proposition 6.2 (Empty-mile NGT implies NGT). Theorem 4 implies Theorem 1. Proof. By Proposition 6.1, we may assume that we are working over a finite fiel F. Let P = X + a 1 X a F[y][X] be the filtere recursion satisfie by the sequence {T (y n )} n as in the setup of Theorem 1; recall that we insist that eg a =. We will show that P ivies a polynomial of the form X e y e + (terms of total egree < e) for e = q m (q 1), where q is a power of p an m 0. Then the sequence {T (y n )} n will also satisfy the recursion associate to a polynomial whose shape fits the requirements of Theorem 4. in a. (vii) If R is any ring an a R is an ieal, then a[y] R[y] is the ieal of polynomials all of whose coefficients are

13 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p 13 Let H be the egree- homogeneous part of P, so that P = H + (terms of total egree < ). We claim that there exists a homogeneous polynomial S F[y, X] so that H S = X e y e for some positive integer e of require form. Once we fin such an S, we know that P S will have the esire shape X e y e + (terms of total egree < e). To fin S, we ehomogenize the problem by setting y = 1: let h(x) := H(1, X) F[X], a monic polynomial of egree an nonzero constant coefficient. Let F be the splitting fiel of h(x); uner our assumptions on a, all the roots of h(x) are nonzero. Let q be the carinality of F. (Recall that we are assuming that F, an hence its finite extension F, is a finite fiel.) Every nonzero element α F, an hence every root of h(x), satisfies α q 1 = 1. Finally, let q m be a power of q not less than any multiplicity of any root of h(x). Since every root of h satisfies the polynomial X q 1 1, we know that h(x) ivies the polynomial ( X q 1 1 ) q m = X qm (q 1) 1. Set e = q m (q 1), an let s(x) be the polynomial in F[X] satisfying h(x)s(x) = X e 1. Now we finally rehomogenize again: if S F[y, X] is the homogenization of s(x), then Q S = X e y e, so that S is the homogeneous scaling factor for P that we seek The main inuction for the proof the empty-mile NGT. From now on, having alreay fixe p, we will always assume that R is a ring of characteristic p, not necessarily finite. Definition. If T : R[y] R[y] an R-linear operator, we will call T a (, D)-NRO, for nilpotent recursion operator, if T satisfies the conitions of Theorem 4: that is, T lowers egrees, an {T (y n )} satisfies an R[y]-linear recursion with companion polynomial X + ay + (terms of total egree D) for some 1 an some D 1. Note that any (, D)-NRO is a (, D )-NRO for any 1 D D. The proof of Proposition 6.2 shows that, if R is a finite fiel, then any T satisfying the conitions of Theorem 1 is in fact a (, D)-NRO for some an D. On the other han, we make the following efinition, for any triple (b,, D) with b D 1: Definition. A function c : Q 0 Q 0 is a (b,, D)-nilgrowth witness if it satisfies the following properties: (1) Discreteness: c(n) is containe in a lattice of Q (that is, M N with Mc(N) N). (2) Growth: c(n) n log(b D) log b as n. (3) Base property: 0 = c(0) < c(1) < < c( 1) an c( D) < c(). (4) Step property: For any k 0, an any pair (i, j) {0, 1,..., } 2 with either (i, j) = (, ) or i j D, an any integers n, m satisfying b k n < b k+1 an jb k m we have c(n) c(n ib k ) c(m) c(m jb k ). In this section, we prove, using strong inuction, that if b is a power of p, then the growth of the nilpotence inex of a (, D)-NRO is boune by the growth of a (b,, D)-nilgrowth witness:

14 14 ANNA MEDVEDOVSKY Proposition 6.3. Let T be a (, D)-NRO, an b a power of p not less than. If c is a (b,, D)- witness, then N T (y n ) c(n). Using the Growth Property above, we obtain an immeiate corollary: Corollary 6.4. Suppose that T is a (, D)-NRO, an let b = p log p. If there exists a (b,, D)- witness, then N T (y n ) n log(b D) log b. In other wors, in this section we reuce the proof of Theorem 4 to establishing the existence of a (p log p,, D)-witness for the (, D)-NRO T, provie that D is not too big. Proof of Proposition 6.3. Given a (b,, D)-witness c, we efine a new function c : R[y] N { } via c ( an y n) := max{c(n) : a n 0} an c(0) :=. We will show that T lowers the c-value of polynomials in R[y]: that is, that for any nonzero f R[y], we have c ( T (f) ) < c(f). It suffices to show this for f = y n. Write x n for T (y n ). We will use strong inuction to show that c(x n ) < c(n). The base case is all n with 0 n <. Since eg x n < n, the statement c(x n ) < c(n) for n < is implie by the statement that c is strictly increasing on {0, 1,..., 1}. This is the base property above. For n >, let k 0 be the integer so that b k n < b k+1. Let P (X) F[y][X] be the companion polynomial of the given recursion satisfie by the sequence {x n }. Let By assumption, P has the form I := {(i, j) : 0 j < j + D i } {(, )}. P = X + (i,j) I a i,j y j X i for some a i,j R. By Corollary 2.1, the sequence {x n } also satisfies the orer-b k recursion corresponing to P bk : namely, for all n b k, we have x n = (i,j) I a i,j y jbk x n ib k. We will show that, if y m appears with nonzero coefficient in one of the terms on the right-han sie above, then c(m) < c(n). Since c(x n ) is equal to one of these c(m)s, this will imply our claim. So suppose that y m appears with nonzero coefficient in the (i, j)-term on the right-han sie. That is, y m appears in y jbk x n ib k for some (i, j) I. That means that y m jbk appears in x n ib k. Note that i D, so that n ib k < n, an the inuction assumption applies: since y m jbk appears in x n ib k, we can assume that c(m jb k ) < c(n ib k ).

15 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p 15 To show that c(m) < c(n), it therefore suffices to show that c(n) c(m)? c(n ib k ) c(m jb k ), since the latter is assume to be strictly positive. But this, slightly rearrange, is just the step property from the efinition of a (b,, D)-witness above. We now aim to construct a (b,, D)-witness if b 1 or if D b 2. This will occupy the next three sections. In section 7 we investigate the properties of a content function, which writes numbers in one base an reas them in another. In section 8, we establish some inequalities about the content of proper fractions. In section 9, we use the content function to construct a (b,, D)-nilgrowth witness, completing the proof of Theorem The content function an its properties In this section we will introuce a function c : Q 0 Q 0 that will serve as a nilgrowth witnesses in the proof of Theorem 4. This type of function was first introuce by Bellaïche in the appenix to [4], inspire in a very loose way by the Nicolas-Serre coe in [14] Base-b representation of numbers. We fix an integer b 2 to be the base. Let D(b) = {0,..., b 1} be the alphabet of igits base b, an D(b) the set of wors on D(b), incluing the empty wor ɛ. The number-of-letters function for a wor x D(b) will be enote by l(x), for length. Let R(b) be the set of all bi-infinite pointe wors x =... x 2 x 1 x 0.x 1 x 2... on D(b) that start with 0 (the igit 0 repeate infinitely to the left). The set of finite wors D(b) naturally embes into R(b) via x ( 0)x.(0 ), where 0 is the igit 0 repeate infinitely to the right. More generally, any pointe right-infinite wor will be viewe as an element of R(b) by appening 0 on the left. For x R(b), we can efine the real number π b (x) R 0 by reaing it as a sequence of igits base b, via π b (x) := i x ib i. Since x i = 0 for i 0, this sum converges. The map π b is not injective: inee, i<k (b 1)bi = b k, so that that for any finite wor w an igit x b 1 (an any raix point placement), we have π b (w x (b 1) ) = π b (w (x + 1) 0 ). But we can choose a section of π b by restricting the omain: let R (b) R(b) the subset of those that o not en with (b 1). Then the reaing-base-b function π b : R (b) R 0 is a bijection, an the inverse map ρ b : R 0 R (b) takes a nonnegative real number q to its normal (that is, not ening in (b 1) ) base-b representation x = ρ b (q) satisfying π b (x) = q. The base-b representation ρ b (q) is eventually perioic (that is, ens with z for some finite wor z) if an only if q Q 0. For q Q 0, then, we know that ρ b (q) = x.yz,

16 16 ANNA MEDVEDOVSKY where x, y, z are in D(b). If we insist that x oes not start with 0, that first y an then z have minimal length among such representations, an finally that z (b 1), then x, y, an z are efine uniquely. We will assume this minimality from now on. Note that by construction x an y may be empty wors, but z has length at least 1. We efine, then, three constants associate associate to q Q 0 : l(q) = l b (q) := l(x) = max{0, log b q + 1}, s(q) = s b (q) := l(y) = min{k 0 : enominator of b k q is prime to b}, t(q) = t b (q) := l(z) = min{k 1 : enominator of b s(q) q ivies b k 1}. In particular, we know that, for q Q 0, we have (7.1) q = n + u b s(q) + m b s(q) (b t(q) 1), where n, u, an m are all integers with n = q = π b (x), u = π b (y), an m = π b (z). We will nee the following very simple lemma. Lemma 7.1. Given q Q 0 an a base b, if q qn, then (1) s b (q ) s b (q), (2) t b (q ) ivies t b (q). The proof follows from the fact that the enominator of q is a ivisor of the enominator of q. Alternatively, one can consier the effect of multiplication by integers on base-b expansions The content function. Now let b, β 2 be bases. Define the (b, β)-content of q R 0 (we will soon restrict to rational q) to be the result of reaing the base-b representation of q in base β: c b,β (q) := π β ( ρb (q) ). Note that π β makes sense as a function R(b) R 0 : the series i k x ib i always converges if the x i are boune. Example 7.1. (1) Since ρ 5 (196) = 1241, we have c 5,3 (196) = = 58. (2) We have ρ 7 ( 1 3 ) = 0.(2). Therefore c 7,5 (n) = 2 i 1 5 i = 1 2. (3) c 8,3 ( 1 6 ) = π 3(0.1(25) ) = ( ) i 0 3 2i = = The following lemma, which will be use frequently, is a irect computation. Lemma 7.2. For q Q 0, let s = s b (q) an t = t b (q). Then if q = n + u b + s (7.1) above, we have c b,β (q) = c b,β (n) + c b,β(u) β s + c b,β(m) β s (β t 1). We will also use the following growth estimate: Lemma 7.3. We have c b,β (n) n log b β. More precisely, for n 1, we have m b s (b t 1) as in equation

17 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p 17 β 1 n log b β < c b,β (n) < β(b 1) β 1 nlog b β. Proof. Let l = l b (n), so that for n 1 we have l = 1 + log b n, or log b n < l 1 + log b n. Then, on one han, the (b, β)-content of n is boune above by π β of the infinite pointe wor (b 1) l.(b 1) : c b,β (n) < k<l (b 1)β k = b 1 β(b 1) β 1 βl β 1 βlog b n β(b 1) = β 1 nlog b β. On the other han, the (b, β)-content of n is at least π β of the pointe wor 1(0 l 1 ).0 : c b,β (n) β l 1 > β 1 n log b β. In particular, if β < b, then c b,β grows slower than linearly in n. Remark. (1) If β < b, then the (b, β)-content function is never monotonically increasing, even on integers. Inee, c b,β (b k 1) = b 1 β 1 (βk 1), which is greater than c b,β (b k ) = β k as soon as β k > β 1 b 2. (2) The sequence {c b,β (n)} n N is b-regular in the sense of Allouche an Shallit [2]. They efine a sequence {s n } Q N to be b-regular if there exists a Q-vector space V, enomorphisms M 0,..., M b 1 of V, a vector v V, an a functional λ : V Q so that s πb (n l...n 0 ) = λm nl M n0 v. For {c b,β (n)}, we can take V = Q 2, M i = ( ) 1 i 0 β, v = ( 0 1 ) an λ = ( 1 0 ). Inee, b-regularity appears to be lurking in many places in this theory (viii), but we have not yet been able to make use of it. Finally, we recor a trivial property of c b,β : Lemma 7.4. For n R 0, an any k Z, we have c b,β (b k n) = β k c b,β (n) Content an the carry-igit wor. Even though the content function is not monotonic, it exhibits a certain amount of regularity uner aition. We quantify this regularity in this section. For m, n R 0, we efine r b (m, n) R(2) to be the wor of carry igits when the sum s = m+n is compute in base b. More precisely, let ρ b (m) = m, ρ b (n) = n, an ρ b (s) = s, an let r b (m, n) := r satisfying (7.2) m i + n i + r i 1 = s i + br i for all i in Z. Since m i, n i, an s i are all 0 for i > l b (s), an since r i {0, 1}, the set of equations above efines r i uniquely, inuctively own from i = l b (s). Example 7.2. (1) We compute r 3 (77, 11). We have ρ 3 (77) = 2212 an ρ 3 (11) = 102. Since the expansions are finite, the carry igits are compute simply in performing the aition (viii) For example, the Nicolas-Serre coe sequence h(n) efine in [14, 4.1] is 2-regular, as are its constituent parts n 3(n) an n 5(n).

18 18 ANNA MEDVEDOVSKY algorithm base 3: From this we rea off the carry-igit wor: r 3 (77, 5) = Note the shift one space to the right in our inexing from how one conventionally keeps track of carry igits in long aition. (2) We compute r 5 ( 53 60, ). We have ρ 5 ( 53 ) 60 = 0.4(20) ( an ρ ) = 0.10(3). Their sum is = π 5(1.02(40) ). Comparing the base-5 expansions of the two aens with the expansion of the sum allows us to compute the carry igits left to right Therefore r 5 ( 53 60, ) = 0.10(01). In this case, we will get the same infinite carry-igit wor if we take the limit of the finite carry-igit wors obtaine by truncating the expansions of the two aens. (3) We insist that r 10 ( 1 3, 2 3 ) = 0.1, even though any finite truncation of the ecimal expansions of 1 3 an 2 3 woul yiel no carry igits in the sum. In full generality (i.e., if the aens are not in Z[ 1 b ] but the sum is), one nees to know the expansion of the sum before one can compute the carry-igit wor. The carry igit wor exactly keeps track of the ifference between values of c b,β : Lemma 7.5. For m, n be in R 0, we have c b,β (m) + c b,β (n) = c b,β (m + n) + (b β) π β ( rb (m, n) ). Proof. Let s = m + n, an let m, n, s be the corresponing base-b expansions an r the carry-igit wor. Scaling equation (7.2) by β i an summing up over all i gives us mi β i + n i β i + r i 1 β i = s i β i + b r i β i or, equivalently, c b,β (m) + c b,β (n) + βπ β (r) = c b,β (m + n) + bπ β (r). We will typically use Lemma 7.5 when comparing c b,β (m) an c b,β (n) by analyzing c b,β (m n) an r b (m n, n). Example 7.3. (1) Consier (b, β) = (3, 2) an the = 88 example from above. We have c 3,2 (77) = π 2 (2212) = 28, an c 3,2 (88) = π 2 (10021) = 21. The ifference is accounte for by c 3,2 (11) = π 2 (102) = 6 an the carry igit wor evaluation in base 2. Namely, we have ( π 2 r3 (77, 11) ) = π 2 (1101) = 13, an then = , as in Lemma 7.5.

19 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p 19 (2) Let (b, β) = (5, 3), an consier the fin that example from above. Using Lemma 7.2, we c 5,3 ( ) = π 3(0.4(20) ) = π 3(4) + π 3(20) 3 3(3 2 1) = c 5,3 ( ) = π 3(0.10(3) ) = π 3(10) π 3(3) 3 2 (3 1) = 1 2 c 5,3 ( ) = π 3(1.02(40) ) = π 3 (1) + π 3(02) π 3(40) 3 2 (3 2 1) = π 3 ( r5 ( 53 60, )) = π 3 ( 0.10(01) ) = π 3(10) π 3(01) 3 2 (3 2 1) = As expecte from Lemma 7.5, we have = = (3) Finally, let b = 10 an return to the aition equation = 1. For a D(9), we have π β (0.a ) = a β 1. Therefore the two sies of the Lemma 7.5 equation agree: (LHS) c 10,β ( 1 3 ) + c 10,β( 2 3 ) = π β(0.3 ) + π β (0.6 ) = 9 β 1 ( (RHS) c 10,β (1) + (10 β) π β r10 ( 1 3, 2 3 )) =1 + (10 β) π β (0.1 ) = 9 β Content of some proper fractions In this section, we continue notation from the previous section, but further assume that > 1. We prove some inequalities about c b,β ( 1 ) an c b,β( D b ) that we will use in section 9 to prouce a (b,, D)-nilgrowth witness Unit fractions base b. To motivate the iscussion, we note that 1 1 = b 1 b = ) k 1(b k 1 b k. b Therefore, the base-b expansion of 1 wants to be 0.1 (b ) (b )2 (b ) Of course, unless b 1, this is not possible: (b ) k is not a igit base b for k large enough. However, letting ρ b ( 1 ) = 0.a 1a 2 a 3..., we can say the following: Lemma 8.1. For k 1, we have a i = (b ) i 1 for i = 1,..., k if an only if (b ) k <. Proof. We will establish this claim by inuction on k. The a k can be efine recursively via b k k 1 (8.1) a k = b a i b k i k k 1 = a i b k i. i=1 i=1 For k = 1, we have a 1 = b 1. Therefore a 1 = 1 if an only if b (b ) 1 <. So the claim for k = 1 is true. < 2, which is equivalent to

20 20 ANNA MEDVEDOVSKY Now suppose we alreay know that a i = (b ) i 1 for i < k. Then a k = (b ) k 1 if an only if ( ) b k k 1 1 > a i b k i (b ) k 1 i=1 = bk ( b k 1 + (b )b k (b ) k 2 b + (b ) k 1) as esire. = bk bk (b ) k b (b ) = (b )k, The same argument also implies the immeiate Corollary 8.2. If a i = (b ) i 1 for i < k, then a k = (b ) k 1 + (b ) k. We can now elineate what 1 must look like base b. Lemma 8.3. For b, the base-b expansion of 1 falls into one of five mutually exclusive cases. (1) ρ b ( 1 ) = (in other wors, a 1 2) iff b 2. (2) ρ b ( 1 ) = (i.e., a 1 = 1 an a 2 3) iff b 2 < b2 b+3 = b b+3. (3) ρ b ( 1 ) = (i.e., a 1 = 1, a 2 = 2, an a 3 4) iff b > 6 an = b 2. (4) ρ b ( 1 ) = 0.1 iff = b 1. (5) ρ b ( 1 ) = 0.10 iff = b. Proof. If b = 0 or b = 1, then a i = (b ) i 1 for all i (Lemma 8.1). Assume b 2. If b 2, then 1 2 b, so that a 1 2, as claime. Otherwise, we must have (b ) 1 <, so that a 1 = 1 an a 2 b. This means that a 2 3 unless both b = 2 an (b ) 2 <, in which case we have > 4 (an hence b > 6), an a 2 (b ) 2 = The carry-igit wor for a proper fraction base b. Now we aitionally fix a D with 1 D < an investigate ρ b ( D ) =: 0.e 1e 2 e The e k satisfy the same type of recursion as the a k, namely, b k D k 1 e k = e i b k i. In particular, e 1 = Db i=1, an the following generalization of Lemma 8.1 an Corollary 8.2 hols: Lemma 8.4. For k 0, we have e i = D(b ) i 1 for i = 1,..., k if an only if D(b ) k <. Moreover, if D(b ) k <, then e k+1 = D(b ) k + D(b ) k+1. To unerstan the relationship between the a k an the e k we efine the carry-igit wor r = 0.r 1 r 2 r 3... for the aition problem = D. (See section 7.3 for efinitions.) In other wors, set }{{} D times r (i) := 0.r (i) 1 r(i) 2 r(i) 3... := r b( i, 1 ) for 1 i D 1, an then set r j := 1 i=1 r(i) j.

21 NILPOTENCE ORDER GROWTH OF RECURSION OPERATORS IN CHARACTERISTIC p 21 Then 0 r j D 1 for every i; since D < 1, we have r 1 = 0. Putting together equations (7.2) applie to i + 1 for 1 i < D gives us the precise relationship between the a k, e k, an r k : (8.2) e k = Da k + r k+1 br k. In fact, we have a close formula for r k : Lemma 8.5. For k 1 we have Db k 1 b k 1 r k = D = D(b ) k 1 D (b ) k 1. Proof. The formula is true for k = 1 (in our case, all quantities are 0). For k 1, we will establish the formula for k + 1, starting with equation 8.2: Db k k 1 b r k+1 = e k Da k + br k = e i b k i k k 1 D + D a i b k i + br k i=1 Db k b k k 1 Db = D + b k i k b k (br i r i+1 ) + br k = D Here we use equation (8.2) in the form e i + Da i = br i r i+1 for 1 i < k to pass from the first line to the secon, an cancelation of a telescoping sum to pass from the secon to the thir. Finally, we note that Db k D i=1 b k D(b ) k = because the intervening terms D k i=1 ( k i)b k i ( ) i greatest-integer function to cancel. Corollary 8.6. D i=1 (b ) k are integers, an hence can pass through the (1) If (b ) k < for some k 0, then for every i k + 1, we have r i = (2) If D(b ) k < for some k 0, then for every i k + 1, we have r i = 0. D(b ) i (b, β)-content of proper fractions. As before, we have a triple (b,, D) with 1 D b subject to the conitions that b 2 an β := b D 2. Recall the content function c b,β from section 7.2, an let c b,β : Q 0 Q 0 be the function efine via c b,β (n) := c b,β( n ). Whenever the triple (b,, D) is unerstoo, we write c = c b,β, an let r = 0.r 1r 2... be the carryigit wor for = D as in section 8.2. Lemma 7.5 implies that (8.3) c(d) = Dc(1) D π β (r). In this section, we will establish some lower bouns on c(1) an c(d). First, we ispatch the cases = b an = b 1, which yiel easy explicit formulas. Lemma 8.7. Suppose that = b or = b 1, an 0 i <. Then c(i) = i D.