On a combinatorial method for counting smooth numbers in sets of integers

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1 On a combinatorial method for counting smooth numbers in sets of integers Ernie Croot February 2, 2007 Abstract In this paper we develop a method for determining the number of integers without large prime factors lying in a given set S. We will apply it to give an easy proof that certain sufficiently dense sets A and B always produce the expected number of smooth sums a + b, a A, b B. The proof of this result is completely combinatorial and elementary. 1 Introduction Given a set S, a common uestion one tries to answer is whether S contains the expected number of y-smooth integers, which are those integers n where P (n) y, where P (n) denotes the largest prime factor of n. We denote the number of integers in S with this property by Ψ(S, y); and for a number > 0, we denote the set of all y-smooths positive integers by Ψ(, y). So, Ψ([], y) = Ψ(, y), where here [] denotes the set of integers {1, 2,..., }. If S [] is typical, then one would expect that Ψ(S, y) S Ψ(, y). (1) For example, fix a real number 0 < θ 1 and an integer a 0, and let S be the set of numbers of the form p + a, where p is prime. S is often Supported by an SF grant. 1

2 called a set of shifted primes. It is conjectured that where Ψ(S, θ ) π()ψ(, θ ) ρ(u) = lim ρ(θ 1 )π(), (2) Ψ(, 1/u ). This function ρ is called Dickman s function, and it was proved in [7] that the limit exists. Unfortunately, proving (2) remains a difficult, open problem; however, in [9] J. B. Friedlander gave a beautiful proof that for θ > (2 e) 1 = , Ψ(S, θ ) π(), and in [1], R. Baker and G. Harman proved that for θ , Ψ(S, θ ) > log α, for some α > 1 and > 0 (a). There are several methods for attacking the general uestion of proving that (1) holds for a particular set S. One such method involves exponential sums and the circle method, and another uses a Buchstab or inclusionexclusion identity in combination with a sieve method (such as the Large Sieve). For example, one way that one could count the number of y-smooth integers in a set S is via the following inclusion-exclusion identity: Ψ(S, y) = S S p + S p1 p 2 S p1 p 2 p 3 +, y<p p prime y p 1 <p 2 p 1,p 2 prime y p 1 <p 2 <p 3 p 1,p 2,p 3 prime where S d denotes the set of elements of S divisible by d. One problem that one immediately faces is that these prime products d = p 1 p 2 p k can be very close to, and when that is the case, in many applications one does not have good estimates for the size of S d. That problem can be fixed if one knows that most of the elements of S have a divisor of size between, say, δ and 1 δ that is a product of primes < y, because no such integer could have a divisor d > 1 δ that is a product of primes y. One would also need asymptotic estimates for the sizes of S d for all d < 1 δ in order to make this approach work; in particular, something like S d = S (1 + o(1)), for all d < 1 δ. (3) d 2

3 Actually, all one needs is that this holds for most d < 1 δ, in some appropriate sense. Often, one only has such estimates for d < (as is the case for shifted primes). In the smooth sieve method that we will present, asymptotic estimates for S d will not be needed; all that will be needed are good lower bounds on the size of S d in a certain average sense. Furthermore, our method works when different positive weightings of the naturals are applied (besides the usual weighting, which assigns value 1 to every natural number). 1.1 Local-Global Sets and the Main Theorem The key structure that we use in the development of our smooth sieve method is a Local-Global Set, which we abbreviate as LG set, and define as follows: Definition. Let ɛ, c (0, 1) be parameters. Then, A [] is an LG-set with parameters ɛ, c if the following conditions hold: If 1, 2 A are distinct then [ 1, 2 ] > ; {n : n for some A with c } (1 ɛ). otes: This second condition is saying that all but an ɛ proportion of the integers n are divisible by some element of A of size at most c. The first condition tells us that A O( A /) (consider the number of n which are divisible by some A). The sets A that we construct in this paper will have A = o(), and so we will have o(1). A c A On the other hand, the second condition implies that ɛ (4) A c The last ineuality here follows from the fact that the second sum counts the number of positive integers n that are divisible by an element A 3

4 satisfying c ; and, such n can be divisible by at most one such by the first property of LG sets. So, the LG sets we will be considering will have that the sum of reciprocals of elements of A that are c are within ɛ of 1. A central result of the paper, proved in section 3, is the following existence theorem: Theorem 1 For every 0 < ɛ < ɛ 0, for every c 1 ɛ 2250, and for every > 0 (ɛ), there exists an LG set of integers A [] with parameters ɛ and c. The origin of our name Local-Global Set comes from the following observation: Suppose that A [] is an LG set with parameters ɛ and c. Suppose that S [], and let S(a; ) denote the number of integers in S that are congruent to a modulo. Fix a non-negative integer a, and suppose that we can prove that For every A, S(a; ) > (1 ɛ) S. (5) If S were euidistributed amongst the residue classes modulo, then S(a; ) = S / +O(1); so, we are assuming a lower bound that is off from the expected amount by a factor of 1 ɛ. ow, since A is an LG set we have that each element of S a can be divisible by at most one member of A; so, A ( S(a; ) S ) S 1 1 ɛ S. A If we let A 0 be those A such that S(a; ) < S /, then we have from (5) that for every A 0, S S(a; ) < ɛ S. Thus, S S(a; ) = ( S(a; ) S ) A A < 4ɛ S. + 2 S S(a; ) A 0 4

5 Thus, a good uniform lower bound (5) implies euidistribution of the elements of S in residue classes a (mod ) in some average sense. One can think of this as some type of local-to-global phenomenon. 1.2 The Local-Global Sieve for Smooth umbers Our smooth sieve result below will be stated in terms of weight functions, rather than sets, as they are more general and more flexible than sets. So, we suppose that w(n) 0 is defined for natural numbers, and then our problem is to give good estimates for the size of W (, θ) := w(n). n P (n) θ Letting Σ := n w(n), we note that since the sum W (, θ) has only a fraction ρ(θ 1 ) times as many terms as Σ, we expect that W (, θ) ρ(1/θ)σ. Our smooth sieve, which is proved in section 4, gives conditions for when this is the case: Theorem 2 Suppose 0 < ɛ < ɛ 0, and let A be the LG set with parameters ɛ and c = 1 ɛ 2250 given by Theorem 1. 1 Let A 1 = { A, c : P () θ }, and A 2 = { A, c : P () > θ }. Define the constants ρ 1 = A 1 1, and ρ 2 = A This theorem makes use of the structure of the LG set given by theorem 1, which is described in section 3; and so, the theorem doesn t hold for just any LG set with parameters ɛ and c. 5

6 Suppose A 1 A 2 w(n) > (ρ 1 ɛ)σ; and, (6) n n w(n) > (ρ 2 ɛ)σ. (7) n n and Then, we have that (ρ 1 ɛ)σ < W (, θ) < (ρ 1 + 2ɛ)Σ, (8) (ρ(1/θ) 3ɛ)Σ < W (, θ) < (ρ(1/θ) + 4ɛ)Σ. (9) Moreover, if one is only able to show (6), then one can still deduce the lower bound W (, θ) > (ρ 1 ɛ)σ > (ρ(1/θ) 3ɛ)Σ. (10) To see that the assumptions in this theorem are weaker than (3), let w(n) be the indicator function for a set S []. Then, (6) and (7) are just saying that we have a lower bound of the general form on average. S d > (1 ɛ) S d We will close this section by giving a simple application of the above theorem, and in the next section we will give a more sophisticated application: Suppose that S [] and S = 1 o(1). How many ordered pairs (s 1, s 2 ) S S have the property that s 1 s 2 is θ -smooth? We expect there to be about ρ(1/θ) S 2 such pairs. To prove this using the above theorem, define w(n) to be the number of ways of writing n = s 1 s 2, s 1, s 2 S. Then, the sum over all w(n) with n 1 divisible by is 1 ( ) S(a; ), 2 a=0 This expression is minimized if the elements of S are as euidistributed amongst the residue classes modulo as is possible; and so, this expression can be shown to be at least S 2 /2 in size for = o( S ). So, for any ɛ > 0 6

7 we have that for sufficiently large both (6) and (7) hold. It follows that the number of pairs (s 1, s 2 ), s 1 > s 2 such that P (s 1 s 2 ) θ is ρ(1/θ) S 2 /2; so, there are ρ(1/θ) S 2 pairs (s 1, s 2 ) with P (s 1 s 2 ) θ. The remainder of the paper is organized as follows. In the next section we will prove that under certain conditions a set S has about ρ(1/θ) S 2 ordered pairs (s 1, s 2 ) S S such that P (s 1 + s 2 ) θ ; and, in the process of proving this, we will present a large sieve ineuality that follows easily from properties of LG sets. In the final sections we will give proofs of Theorems 1 and 2. 2 An Application of Theorems 1 and 2 Given sets of integers A, B [] having elements each, it is an interesting and studied uestion to determine the number of ordered pairs (a, b) A B such that P (a + b) y. There are several ways of attacking this sort of problem, one of which is to use the circle method and exponential sums over smooth numbers, and another is to use the large sieve. We could also ask how τ(a + b) (the number of divisors of a + b) is distributed, or how large P (a + b) can be. These types of uestions were given a thorough treatment in a series of beautiful papers by A. Balog and A. Sárközy [2], [3], [4], and [5]; P. Erdős, H. Maier, and A. Sárközy [8]; A. Sárközy and C. L. Stewart [11], [12], [13], [14]; C. Pomerance, A. Sárközy, and C. L. Stewart [10]; and R. de la Bretèche [6]. The paper by de la Brèteche is more relevant to the main result of this section, and we give here one of his theorems: Theorem 3 Suppose that A, B []. For a given integer y, let u = (log )/ log y. Then, uniformly for 3, exp((log ) 2/3+ɛ ) < y we have #{a A, b B : P (a + b) y} ( ( )) log(u + 1) = A B ρ(u) 1 + O. A B log y R. de la Bretèche used estimates for exponential sums and the circle method to prove this result. otice that if A B (/ log ) 2, then his result fails to prove that there are the expected number of sums that are y-smooth for any y <, because in this case the big-oh term is 1. What 7

8 makes his theorem so powerful is the fact that the parameter y is allowed to go all the way down to exp((log ) 2/3+ɛ ). Let us now consider what happens in the case when y = θ (and so u = 1/θ): Is it possible to show that if A, B > c, for some 0 < c < 1, then we get the expected number of sums a + b being y-smooth? It is easy to see that the answer is no, no matter how close to 1 we take c to be. For example, if θ + c > 1 we could take A and B to both be the set of integers that are divisible by some prime number p 1 c. otice here that A c. The sums a + b, a, b A are numbers of the form pk, where k < 2 c, and such a sum is θ -smooth if and only if k is θ -smooth. Thus, one would expect (and can show) that #{a, b A : P (a + b) θ } A B Ψ(2 c, θ ) 2 c ρ(c/θ). On the other hand, the proportion of θ smooths is ρ(1/θ), which is not ρ(c/θ). So, the type of result we might try to prove is the following: Theorem 4 Given 0 < θ 1, and 0 < γ < γ 0 if A, B [] satisfy A, B > (8/γ) 1 (γ/8)2250, then for sufficiently large #{a A, b B : P (a+b) (2) θ } = (ρ(1/θ)+δ) A B, where δ < γ. The same result holds for differences a b. In section 2.2 we will give a short proof of this theorem using Theorem 2 and a version of the Large Sieve. 2.1 A Local-Global Large Sieve For our proof of Theorem 4 we will reuire a form of the Large Sieve, which can be proved via modifying the usual proofs of the large sieve; however, we will prove it here (perhaps surprisingly) through a brief and elementary application of LG sets. Theorem 5 Given ɛ > 0 and sufficiently large, let A [] be any LG set for parameters ɛ and c. Suppose that C [], and let C(a, ) denote the number of elements of C that are congruent to a modulo. Then, we have that A c 1 a=0 ( C(a, ) C ) 2 < C (ɛ C + c ). 8

9 Proof. We note that if b, c C, and b c, then if A divides b c, we must have that is uniue; otherwise, if A also divides b c, then [, ] > and [, ] (b c). Thus, we have that Thus, C 2 > #{(b, c) C 2 : b c, (b c)} A c = A c 1 ( C(a, ) 2 C(a, ) ) a=0 c C + A c C ( c + C ) < From this and (4) it follows that A c 1 a=0 A c 1 C(a, ) 2. a=0 1 C(a, ) 2. a=0 ( C(a, ) C ) 2 = 1 C(a, ) 2 C 2 A c a=0 A c 2.2 Proof of Theorem 4. < (1 (1 ɛ)) C 2 + c C = C (ɛ C + c ). We let ɛ = γ/8, and then for c = 1 ɛ 2250 we have from Theorem 1 that for sufficiently large, there is an LG set A [2] with parameters ɛ, c. We let A 1 and A 2 be as in Theorem 2 (with replaced with 2). Let α, β be the indicator functions for the sets A and B, respectively; let A(a, ) = α(n), and B(a, ) = β(n); n n a (mod ) and define the weight function w(n) = (α β)(n) = 9 a+b=n n n a (mod ) α(a)β(b). 1

10 Then, for A = A 1 or A 2 we get A n 2 n w(n) = 1 A(a, )B( a, ) A a=0 = 1 (A(a, ) A /)(B( a, ) B /) A a=0 + A B A 1. (11) Applying the Cauchy-Schwarz ineuality and Theorem 5 with C = A and C = B to this last double sum we deduce for A, B > ɛ 1 c that 1 A(a, ) A / B(a, ) B / < 2ɛ A B. (12) A a=0 Combining this with (11) we deduce that w(n) > A B γ A A n 2 n Thus, the conditions of Theorem 2 are met for ɛ replaced with γ/4, and we deduce that since Σ = A B in our case, then #{a A, b B : P (a + b) (2) θ } ρ(1/θ) A B < γ A B. 3 Proof of Theorem 1 In the course of our proof we will need to make use of the following conseuence of Brun s upper bound sieve: Theorem 6 Suppose that P is a subset of the primes. The number of integers not divisible by any prime in P is ( 1 1 ). p p P 10

11 Define A(k, δ) to be the set of all integers of the form p 1 p k, p 1 > p 2 > > p k > δ, where For i = 1, 2,..., k 1, p 1 p i p i ; while, p 1 p k < p k. (13) We will show that for every 0 < ɛ < ɛ 0, there exists δ > 0 so that A = k 1 A(k, δ) (14) is an LG set with parameters ɛ and c = 1 ɛ The Set A satisfies the First Property for an LG Set First, we show that the set A described in the previous subsection satisfies the first condition for being an LG set, namely that for any distinct pair of integers n 1, n 2 A, we have lcm(n 1, n 2 ) > : Say n 1 A(k, δ) and n 2 A(l, δ); so, we have the prime factorizations n 1 = p 1 p k, p 1 > p 2 > > p k ; and n 2 = 1 l, 1 > 2 > > l. Without loss of generality, we can assume that p k l. ow, if there is some prime i which is distinct from the primes p 1,..., p k, then we would have that the lcm of n 1 and n 2 is divisible by the product of primes i p 1 p k, and this product exceeds, because from (13) i p 1 p k > i p k l p k. So we are left to consider what happens when the i s are a subset of the p i s. We break this case into two sub-cases, with the first one where p k = l, and the second where p k < l. In the case p k = l, we must have that there exists one of the primes p i > l such that p i is distinct from 1,..., l, since otherwise we would have n 1 = n 2. But now our assumption gives p 1 p k p i 1 l < l p i 1, which is impossible. 11

12 So, we may assume p k < l. For this case, let j < k be the index where p j = l (which exists since i s are a subset of the p i s). Then, we have From (13) this gives p 1 p j 1 l. l = p j p 1 p j 1 l < l, which is impossible. So, we conclude that the set A given by (14) satisfies the first condition for being an LG set. 3.2 The Set A Satisfies the Second Property for an LG Set Define F to be the set of all integers n satisfying such that log < n <, If n = u 2 v, v suarefree, then u < log. We have that the number of n that are not contained in F is at most 1 u 2 log. So, u log F = O(/ log ). ow define G to be the integers n F that are not divisible by any element of A. Our goal will be to show that G is small. We first observe that one can deduce from (13) that every n G has the property that for all δ < x, p n, p prime p x p x. (15) For if is the largest x for which this ineuality fails (such exists), then is prime, and we would have p n, p prime p 12 p <,

13 which would imply p, p n p is an element of A dividing n G, contradiction. We conclude that for n G, p > p n p<x n (log 2 ) p n p > p x (log 3 ) p n p p x x log 3. (16) This brings us now to the conceptual heart of the proof of Theorem 1: Given x, the average of p n,p<x log p over all n is log x + O(1); however, for most n, as we vary x, this sum should fluctuate about log x, meaning that for most n there is an δ < x for which (16) fails to hold. The rest of the proof amounts to formalizing this intuition. Let R = , and let J = J(δ) = log(1/2δ). (17) log(r) ote that R J δ 1/2. Let G(h) be the set of all n G for which exactly h integers j among 1, 2,..., J satisfy Write where G 1 = p n, p prime p δrj 0 h J/14 log p > 3δRj log. 2 G = G 1 G 2, G(h), and G 2 = J/14<h J G(h). To show that G is small we will show that both G 1 and G 2 are small G 2 is small Define z(n) = z(n, δ, R) := J j=1 p n, p prime p δrj log p. 13

14 Since every n G(h) also lies in G, from (16) we have that for j = 1,..., J, log p > δr j log 3 log log. p n, p prime p δrj So, for n G(h) we will have z(n) > (1 + o(1))(δ log ) J R J(J+1)/2 (3/2) h. (18) Lemma 1 We have that z(n) (δ log ) J R J(J+1)/2 exp(2j/ R). n n Proof of the Lemma. We have that z(n) = (log p 1 ) (log p J ) < p 1,...,p J prime p j δrj p 1,...,p J prime p j δrj a 1,...,a J 0 a j J j+1 j a 1 + +a j J (log p 1 ) (log p J ) [p 1,..., p J ] 1 j J a j 0 (δ log ) J a 1,...,a J 0 a j J j+1 j a 1 + +a j J Here we have used the basic estimate for a 1 p δrj p prime 1 j J a j 0 [p 1,..., p J ] log a j p p R ja j a j. (19) p M p prime log a p p M 2 (log x) a 1 dx x loga M. a ow, suppose that for every j J we have j a a j J, where 0 a j J j + 1, 14

15 and that exactly k of the a i s are non-zero. Then, R ja j R J(J+1)/2 J+k. 1 j J Thus, the final expression of (19) is (δ log ) J R J(J+1)/2 J 1 k J = (δ log ) J R J(J+1)/2 J 1 k J ( ) J k R k ( J k b 1 + +b k =J b i 1 )( J 1 k 1 (δ log ) J R ( 2J 1 J(J+1)/2 J+1/2 R k 1/2 2k 1 1 k J ( < (δ log ) J R J(J+1)/ ) 2J 1 R (δ log ) J R J(J+1)/2 exp(2j/ R). R k ) ) With our choice of R = we will have from Lemma 1 that for δ > 0 sufficiently small, z(n) < 2 ( δ log )J R J(J+1)/2. n G Combining this with (18) we deduce G(h) < (1.0001) J (2/3) h. It follows that for sufficienty large, G 1 is small G 2 < 3(1.0001) J (2/3) J/14 < (0.98)J 3 To bound G 1 from above, we will bound G(h) from above for h J/14: Given h J/14 we split G(h) into smaller subsets as follows G(h) = G(h; B), B [J] B =h 15.

16 where G(h; B) is the set of all n G with the property that p n, p prime p δrj log p 3δRj log 2 if and only if j B. Given B, we let B = [J] \ B. We note that every n G(h; B) has no prime divisors p lying in any of the intervals n, prime <p [ 3δRj /4 log 2, δrj ], where j B. (20) For if some n G(h; B) had such a prime divisor p, then we would have that 1 < 3δRj /4 p log 2 p log 4, n, prime δrj which would contradict (16), and therefore we would have that n G. From Theorem 6, together with the fact that n G(h; B) has no prime divisors lying in the intervals (20), we deduce that G(h; B) ( 1 1 ) (3/4) B. p j B 3δRj /4 log 2 <p< δr j p prime Thus, for h J/14 and for δ > 0 sufficiently small, we will have from Stirling s formula that ( J G(h) )(3/4) J h < (0.991)J. h 3J Thus, for δ > 0 sufficiently small and sufficiently large, G 1 < (0.991)J G is small, and the Conclusion of the Proof of the Theorem From the upper bounds on G 1 and G 2 we deduce for R = and δ > 0 sufficiently small and sufficiently large, G G 1 + G 2 2(0.991)J 3 16 ;

17 and so, the number of integers n not divisible by any element from the set A is at most ( F ) + G < (0.991) J. Thus, we have that 1 a A a > 1 (0.991) J. ow, since the prime divisors of the elements of A exceed δ, we will have from Theorem 6 and Mertens Theorem that for 1/2 < M < there are M ( 1 1 ) M p δ log p δ p prime elements of A that are less than M. Thus, for 1/2 < c 1, 1 a log(1/c). δ a A a c a A c <a It follows that that the number of integers n divisible by some element of A that is c is > (1 (0.991) J Cδ 1 log(1/c)), a where C > 0 is some constant. This will exceed (1 ɛ) provided (0.991) J < ɛ 2, and Cδ 1 log(1/c) < ɛ 2. (21) From (17) we have that for ɛ < 1/2 the first ineuality holds provided δ < (ɛ/2) 1 log(r)/ log(0.991), The second ineuality of (21) holds provided log(c) > ɛδ 2C > (ɛ/2)2 log(r)/ log C So, for ɛ > 0 sufficiently small and (ɛ/2)2 (log R)/ log c > 1 ɛ 2250 > 1, 2C all but at most (1 ɛ) of the integers n will be divisible by some element of A. 17

18 4 Proof of Theorem 2. First, we remark that 1 ɛ n A n c 1 n = ρ 1 + ρ 2 < 1 + o(1). (22) From the properties of the set A defined at the beginning of section 3 we have that if n, n, and A 1, then n/ < 0 θ, where 0 is the smallest prime divisor of. Thus, n is θ -smooth. Also, if n where A 2, then n is obviously not θ -smooth, since has a prime factor exceeding θ in this case. We deduce from this and (6) that W (, θ) w(n) (ρ 1 ɛ)σ. Also, A 1 n n W (, θ) Σ A 2 which, together with (6), (7) and (22), implies w(n), n n W (, θ) (1 ρ 2 + ɛ)σ (ρ 1 + 2ɛ)Σ. Thus, we have proved (8) and the first ineuality in (10). To prove (9) and the second ineuality in (10) we must relate ρ(1/θ) and ρ 1 and ρ 2. To do this we observe that since ρ(1/θ) (1 + o(1))ψ(, θ ) (1 + o(1)) = ρ 1 + o(1), A 1 and 1 ρ(1/θ) 1 + o(1) Ψ(, θ ) (1 + o(1)) A 2 = ρ 2 + o(1), we deduce from (22) that o(1) < ρ(1/θ) ρ 1 < ɛ + o(1). Thus, (9) follows. 18

19 5 Acknowledgements I would like to thank the referee for the many excellent suggestions. References [1] R. C. Baker and G. Harman, Shifted Primes without Large Prime Factors Acta Arith. 83 (1998), [2] A. Balog and A. Sárközy, On sums of Integers Having Small Prime Factors. I, II, Studia Sci. Math. Hungar. 19 (1984), [3] -, On Sums of Seuences of Integers. I, Acta Arith. 44 (1984), [4] -, On Sums of Seuences of Integers. II, Acta Math. Hungar. 44 (1984), [5] -, On Sums of Seuences of Integers. III, Acta Math. Hungar. 44 (1984), [6] R. de la Bretèche, Sommes sans Grand Facteur Premier [Sums without Large Prime Factors] Acta Arith. 88 (1999), [7] Dickman, On the Freuency of umbers Containing Prime Factors of a Certain Relative Magnitude Ark. Mat. Astr. Fys 22 (1930), [8] P. Erdős, H. Maier, and A. Sárközy, On the Distribution of the umber of Prime Factors of Sums a + b Trans. Amer. Math. Soc. 302 (1987), [9] J. B. Friedlander, Shifted Primes without Large Prime Factors, ATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 265, Kluwer Acad. Publ., Dordrecht, [10] C. Pomerance, A. Sárközy, and C. L. Stewart, On Divisors of Sums of Integers. III. Pacific J. Math. 133 (1988), [11] A. Sárközy and C. L. Stewart, On Divisors of Sums of Integers. I Acta Math. Hungar. 48 (1986), [12], On Divisors of Sums of Integers. II., J. Reine Angew. Math. 365 (1986),

20 [13] A. Sárközy and C. L. Stewart, On Divisors of Sums of Integers. IV. Canad. J. Math. 40 (1988), [14], On Divisors of Sums of Integers. V. Pacific J. Math. 166 (1994),