Witt#5e: Generalizing integrality theorems for ghost-witt vectors [not completed, not proofread]

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1 Witt vectors. Part 1 Michiel Hazewinkel Sienotes by Darij Grinberg Witt#5e: Generalizing integrality theorems for ghost-witt vectors [not complete, not proofrea In this note, we will generalize most of the results in [4, replacing the Witt polynomials w n by the more general polynomials w F,n efine for any pseuo-monotonous map F : P N N the meaning of pseuo-monotonous will soon be explaine below. Whenever possible, the proofs will be one by simply copypasting the corresponing proofs from [4 an oing the necessary changes - which often will be trivial, though sometimes new thinking will be require. I will even try to keep the numbering of the results in this note consistent with the numbering of the results in [4, so that for instance Theorem i in this note will be the generalization of Theorem i in [4 for as many i as possible. This explains why there are gaps in the numbering: e. g., there is no numbere result between Theorem 17 an Lemma 19 in this note, because Lemma 18 of [4 was just an auxiliary result an nees not be generalize to the w F,n. First, let us introuce some notation 1 : Definition 1. Let P enote the set of all primes. A prime means an integer n > 1 such that the only ivisors of n are n an 1. The wor ivisor means positive ivisor. Definition 2. We enote the set {0, 1, 2,...} by N, an we enote the set {1, 2, 3,...} by N +. Note that our notations conflict with the notations use by Hazewinkel in [1; in fact, Hazewinkel uses the letter N for the set {1, 2, 3,...}, which we enote by N +. Definition 3. Let Ξ be a family of symbols. We consier the polynomial ring Q [Ξ this is the polynomial ring over Q in the ineterminates Ξ; in other wors, we use the symbols from Ξ as variables for the polynomials an its subring Z [Ξ this is the polynomial ring over Z in the ineterminates Ξ. 2. For any n N, let Ξ n mean the family of the n-th powers of all elements of our family Ξ consiere as elements of Z [Ξ 3. Therefore, whenever P Q [Ξ is a polynomial, then P Ξ n is the polynomial obtaine from P after replacing every ineterminate by its n-th power. 4 Note that if Ξ is the empty family, then Q [Ξ simply is the ring Q, an Z [Ξ simply is the ring Z. 1 The first 6 of the following 10 efinitions are the same as the corresponing efinitions in [4. 2 For instance, Ξ can be X 0, X 1, X 2,..., in which case Z [Ξ means Z [X 0, X 1, X 2,... Or, Ξ can be X 0, X 1, X 2,...; Y 0, Y 1, Y 2,...; Z 0, Z 1, Z 2,..., in which case Z [Ξ means Z [X 0, X 1, X 2,...; Y 0, Y 1, Y 2,...; Z 0, Z 1, Z 2,... 3 In other wors, if Ξ = ξ i i I, then we efine Ξ n as ξi n i I. For instance, if Ξ = X 0, X 1, X 2,..., then Ξ n = X0 n, X1 n, X2 n,... If Ξ = X 0, X 1, X 2,...; Y 0, Y 1, Y 2,...; Z 0, Z 1, Z 2,..., then Ξ n = X0 n, X1 n, X2 n,...; Y0 n, Y1 n, Y2 n,...; Z0 n, Z1 n, Z2 n,... 4 For instance, if Ξ = X 0, X 1, X 2,... an P Ξ = X 0 + X 1 2 2X 3 + 1, then P Ξ n = X n 0 + X n 1 2 2X n

2 Definition 4. If m an n are two integers, then we write m n if an only if m is coprime to n. If m is an integer an S is a set, then we write m S if an only if m n for every n S. Definition 5. A nest means a nonempty subset N of N + such that for every element N, every ivisor of lies in N. Here are some examples of nests: For instance, N + itself is a nest. For every prime p, the set {1, p, p 2, p 3,...} is a nest; we enote this nest by p N. For any integer m, the set {n N + n m} is a nest; we enote this nest by N m. For any positive integer m, the set {n N + n m} is a nest; we enote this nest by N m. For any integer m, the set {n N + n m} is a nest; we enote this nest by N m. Another example of a nest is the set {1, 2, 3, 5, 6, 10}. Clearly, every nest N contains the element 1 5. Definition 6. If N is a set 6, we shall enote by X N the family X n n N of istinct symbols. Hence, Z [X N is the ring Z [ X n n N this is the polynomial ring over Z in N ineterminates, where the ineterminates are labelle X n, where n runs through the elements of the set N. For instance, Z [ X N+ is the polynomial ring Z [X1, X 2, X 3,... since N + = {1, 2, 3,...}, an Z [ X {1,2,3,5,6,10} is the polynomial ring Z [X1, X 2, X 3, X 5, X 6, X 10. If A is a commutative ring with unity, if N is a set, if x N A N is a family of elements of A inexe by elements of N, an if P Z [X N, then we enote by P x N the element of A that we obtain if we substitute x for X for every N into the polynomial P. For instance, if N = {1, 2, 5} an P = X X 2 X 5 X 5, an if x 1 = 13, x 2 = 37 an x 5 = 666, then P x N = We notice that whenever N an M are two sets satisfying N M, then we canonically ientify Z [X N with a subring of Z [X M. In particular, when P Z [X N is a polynomial, an A is a commutative ring with unity, an x m m M A M is a family of elements of A, then P x m m M means P x m m N. Thus, the elements xm for m M \ N are simply ignore when evaluating P x m m M. In particular, if N N+, an x 1, x 2, x 3,... A N +, then P x 1, x 2, x 3,... means P x m m N. Definition 7. Let n Z \ {0}. Let p P. We enote by v p n the largest nonnegative integer m satisfying p m n. Clearly, p vpn n an v p n 0. Besies, v p n = 0 if an only if p n. We also set v p 0 = ; this way, our efinition of v p n extens to all n Z an not only to n Z \ {0}. Definition 8. Let n N +. We enote by PF n the set of all prime ivisors of n. By the unique factorization theorem, the set PF n is finite an satisfies n = p vpn. p PF n 5 In fact, there exists some n N since N is a nest an thus nonempty, an thus 1 N since 1 is a ivisor of n, an every ivisor of n must lie in N because N is a nest. 6 We will use this notation only for the case of N being a nest. However, it equally makes sense for any arbitrary set N. 2

3 Definition 9. A map F : P N N is sai to be pseuo-monotonous if it satisfies F p, 0 = 0 for every p P an 1 F p, a a F p, b b for every p P, a N an b N satisfying a b 2 If F : P N N is a pseuo-monotonous map, then we enote by F : N + N + the map efine by F n = p F p,vpn for every n N +. 7 We note that p PF n v p F n = F p, v p n for every n N + an every p PF n 3 since F n = p PF n p F p,vpn. Besies, F n n F for every n N + an every N n Note that F is always a multiplicative function, but not every multiplicative function from N + to N + can be written as F for some pseuo-monotonous map F : P N N. 8 In fact, we have p PF n p F p,vp = p PF p F p,vp = F by the efinition of F = F p PF n\pf p PF n\pf p F p,0 =1 since 1 yiels F p,0=0 an thus p F p,0 =p 0 =1 p F p,vp =p F p,0 since p PF n\pf yiels p / PF an thus v p=0 = F p PF n\pf since n yiels PF PF n 1 = F an p PF n p vpn = p PFn p vpn } {{ } =n p PF n\pfn since n n yiels PF n PF n = n 1 = n. p vpn =1 since p PF n\pfn yiels p / PFn an thus v pn =0, so that p vpn =p 0 =1 Now, for every p PF n, we have v p n = v p n = v p n +v p v p, an thus 2 0 3

4 Definition 10. Let F : P N N be a pseuo-monotonous map. For any n N +, we efine a polynomial w F,n Z [X N n by w F,n = n F X n. Hence, for every commutative ring A with unity, an for any family x k k N n A N n of elements of A, we have w F,n x k k N n = n F x n. As explaine in Definition 6, if N is a set containing N n, if A is a commutative ring with unity, an x k k N A N is a family of elements of A, then w F,n xk k N means wf,n x k k N n ; in other wors, w F,n xk k N = n F x n. The polynomials w F,1, w F,2, w F,3,... will be calle the big F -Witt polynomials or, simply, the F -Witt polynomials. First, here are two examples of pseuo-monotonous maps: Example 1: Define the map pr N : P N N by pr N p, k = k for every p P an k N. Then, pr N is a pseuo-monotonous map, an pr N = i since every n N + satisfies pr N n = p pr N } p,vpn = p {{} vpn = n. Hence, every n N + satisfies p PF n w prn,n = n pr N =i= =p vpn since pr N p,v pn=v pn X n = n X n applie to a = v p n an b = v p yiels p PF n F p, v p n v p n F p, v p v p, F p, v p n F p, v p +. Therefore, for every n N +, the polynomial so that v p n v p = F p, v p + v p n, =v pn +v p an consequently p F p,vpn p F p,vp+vpn. Hence, F n = p F p,vpn p F p,vp+vpn = p F p,vp p vpn p PF n p PF n p Fp,vp+vpn =p Fp,vp p PF n p PF n p vpn =n = F = F n = n F. 4

5 w prn,n is ientic with the polynomial w n efine in [4. Because of this, all the theorems that we will prove about the polynomials w F,1, w F,2, w F,3,... will generalize the corresponing theorems about the polynomials w 1, w 2, w 3,... in [4. Example 2: Define the map pra : P N N by pra p, k = { 0, if k = 0; 1, if k > 0 for every p P an k N. Then, pra is a pseuo-monotonous map 9, an the map pra is ientic with the map ra : N + N + efine by ra n = p for every n N + since every n N + p PF n 9 Proof. By the efinition of pseuo-monotonous, the map pra is pseuo-monotonous if an only if it satisfies pra p, 0 = 0 for every p P an 5 pra p, a a pra p, b b for every p P, a N an b N satisfying a b. 6 We will now prove that it inee satisfies these relations 5 an 6. First of all, every p P satisfies { 0, if 0 = 0; pra p, 0 = by the efinition of pra 1, if 0 > 0 = 0 since 0 = 0. Thus, 5 is proven. Next, let p P, a N an b N be given such that a b. We istinguish between three cases: Case 1: We have b = 0. Case 2: We have b > 0. Let us consier Case 1 first. In this case, b = 0. By the efinition of pra, we have { 0, if a = 0; pra p, a = 1, if a > 0 { a, if a = 0; = a, if a > 0 = a, { 0, if a = 0; a, if a > 0 since 0 = a in the case when a = 0 because 1 a in the case when a > 0 so that pra p, a a 0. Since b = 0, we have pra p, b b = pra p, 0 0 = 0. Thus, =0 pra p, a a 0 = pra p, b b. We have thus proven pra p, a a pra p, b b in Case 1. { Let us now consier Case 2. In this case, b > 0. Thus, the efinition of pra yiels pra p, b = 0, if b = 0; 1, if b > 0 pra p, a = = 1 since b > 0. On the other han, a b > 0. Hence, the efinition of pra yiels { 0, if a = 0; a pra p, b b. Thus, we 1, if a > 0 = 1 since a > 0. Now pra p, a =1=prap,b have proven pra p, a a pra p, b b in Case 2. Hence, we have proven pra p, a a pra p, b b in each of the cases 1 an 2. Since these two cases cover all possibilities, this yiels that pra p, a a pra p, b b always hols. Now, forget that we fixe p, a an b. We thus have proven that pra p, a a pra p, b b for every p P, a N an b N satisfying a b. In other wors, we have proven 6. Recall that the map pra is pseuo-monotonous if an only if it satisfies the relations 5 an 6. Since we have proven that it satisfies the relations 5 an 6, we thus conclue that the map pra is pseuo-monotonous, qe. b 5

6 satisfies pra n = p PF n p prap,vpn =p since p PF n yiels p n an thus v pn>0, so that prap,v pn=1 an thus p prap,vpn =p 1 =p = p PF n p = ra n. Hence, every n N + satisfies w pra,n = pra n =ra X n = n ra X n. Therefore, for every n N +, the polynomial w pra,n is ientic with the polynomial w n efine in [6. Because of this, all the theorems that we will prove about the polynomials w F,1, w F,2, w F,3,... will generalize the corresponing theorems about the polynomials w 1, w 2, w 3,... in [6. This is not to say that we will be able to generalize all results from [6 to our polynomials w F,1, w F,2, w F,3,... In fact, Theorem 4 in [6 oesn t follow from any of the theorems below. Now, we start by recalling some properties of primes an commutative rings: Theorem 1. Let A be a commutative ring with unity. Let M be an A-moule. Let N N. Let I 1, I 2,..., I N be N ieals of A such that I i + I j = A for any two elements i an j of {1, 2,..., N} satisfying i < j. Then, I 1 I 2...I N M = I 1 M I 2 M... I N M. We will not prove this Theorem 1 here, since it is ientic with Theorem 1 in [4 an was proven in [4. A trivial corollary from Theorem 1 that we will use is: Corollary Let A be an Abelian group written aitively. Let n N +. Let F : P N N be a pseuo-monotonous map. Then, F n A = p F p,v pn A. p PF n Proof of Corollary 2. Since PF n is a finite set, there exist N N an some pairwise N istinct primes p 1, p 2,..., p N such that PF n = {p 1, p 2,..., p N }. Thus, p Fp i,v pi n i = i=1 p F p,vpn = F n. p PF n Define an ieal I i of Z by I i = p Fp i,v pi n i Z for every i {1, 2,..., N}. Then, I i + I j = Z for any two elements i an j of {1, 2,..., N} satisfying i < j in fact, the integers p Fp i,v pi n i an p Fp j,v pj n j are coprime 11, an thus, by Bezout s theorem, there exist integers α an β such that 1 = p Fp i,v pi n i α + p Fp j,v pj n j β in Z, an therefore 1 = p Fp i,v pi n i α + p Fp j,v pj n j β I i + I j in Z, an thus I i + I j = Z. Hence, p Fp i,vp i n i Z=I i p F p j,vp j n j Z=I j 10 This is an analogue of Corollary 2 in [4 an can actually be easily erive from that Corollary 2 in [4, but here we will prove it ifferently. 11 since p i an p j are istinct primes because i < j an since the primes p 1, p 2,..., p N are pairwise istinct 6

7 Theorem 1 applie to Z an A instea of A an M, respectively yiels I 1 I 2...I N A = I 1 A I 2 A... I N A. Since an I 1 I 2...I N A = i=1 = N I i i=1 =p Fp i,vp i n i Z N i=1 p Fp i,v pi n i A = = F n N N I 1 A I 2 A... I N A = I i A = p Fp i,v pi n i Z A since PF n = {p 1, p 2,..., p N }, this becomes F n A = 2 is thus proven. Another fact we will use: i=1 N p Fp i,v pi n i Z i=1 A Z A = F n Z A = F n A = N p Fp i,v pi n i A i=1 p PF n = p PF n p F p,v pn A. Corollary Lemma 3. Let A be a commutative ring with unity, an p N be a nonnegative integer 12. Let k N an l N be such that k > 0. Let a A an b A. If a b mo p k A, then a pl b pl mo p k+l A. This lemma was proven in [3, Lemma 3. The following result generalizes Theorem 4 in [4: Theorem 4. Let N be a nest. Let F : P N N be a pseuo-monotonous map. Let A be a commutative ring with unity. For every p P N, let ϕ p : A A be an enomorphism of the ring A such that ϕ p a a p mo pa hols for every a A an p P N. 7 Let b n n N A N be a family of elements of A. Then, the following two assertions C an D are equivalent: Assertion C: Every n N an every p PF n satisfies ϕ p b n p b n mo p F p,vpn A. 8 Assertion D: There exists a family x n n N A N of elements of A such that bn = w F,n xk k N for every n N. 12 Though we call it p, we o not require it to be a prime in this lemma. p F p,v pn A 7

8 Proof of Theorem 4. Our goal is to show that Assertion C is equivalent to Assertion D. We will achieve this by proving the implications D = C an C = D. Proof of the implication D = C: Assume that Assertion D hols. That is, there exists a family x n n N A N of elements of A such that bn = w F,n xk k N for every n N. 9 We want to prove that Assertion C hols, i. e., that every n N an every p PF n satisfies 8. Let n N an p PF n. Then, p n, so that n p N +, an thus n p N since n p is a ivisor of n, an every ivisor of n lies in N 13. Thus, applying 9 to n p instea of n yiels b n p = w F,n p xk k N. But w F,n p xk k N = F x n p an w F,n xk k N = F x n. Now, 9 yiels n p b n = w F,n xk k N = n n; n p 0 mo p Fp,vpn A F x n = n; F x n + n; F x n. 10 n p n p But for any ivisor of n, the assertions n p an p vpn are equivalent 14. Hence, every ivisor of n which satisfies n p must satisfy F 0 mo p F p,vpn A 15. Thus, F 13 because n N an because N is a nest 14 In fact, we have the following chain of equivalences: n p n p / Z n x n n; 0x n = 0 mo p F p,vpn A. n p n p / Z since n p = n p p n here we use that n Z, since n v p n = 0 v p n 0 since v p n 0, because n Z v p n v p 0 since v p n = v p n v p v p n v p p vpn. 15 In fact, let be a ivisor of n satisfying n p. Then, p vpn since the assertions n p an p vpn are equivalent, so that v p v p n. Together with v p v p n which is because n yiels n n Z, thus v p 0 an now v p n = v p n n = v p + v p v p, }{{ } 0 this becomes v p = v p n. Hence, the equality v p F = F p, v p which follows from 3, applie to instea of n rewrites as v p F = F p, v p n, so that p F p,vpn F, an thus F 0 mo p F p,vpn A. 8

9 Thus, 10 becomes b n = n; n p = n p = n p On the other han, F x n + n; n p F x n 0 mo p Fp,vpn A n p F x n + 0 F x n mo p F p,vpn A. 11 b n p = w F,n p xk k N = ϕ p b n p = ϕ p n p since ϕ p is a ring enomorphism. n p F x n p = F x n p n p yiels F ϕ p x n p 12 Now, let be a ivisor of n p. Then, n p n, so that n Z an thus n v p 0. Let α = v p n p an β = v p F. Clearly, v p n = v p n = n v p + v p v p yiels F p, v p n v p n F p, v p v p by 2, }{{ } 0 applie to a = v p n an b = v p, an thus F p, v p F p, v p n v p n + v p. Since β = v p F = F p, v p by 3, applie to instea of n, this becomes β F p, v p n v p n + v p. Aing the equality α = v p n p to this inequality, we obtain α + β v p n p + F p, v p n v p n + v p = v p n p + v p +F p, v p n v p n =v pn p =v pn p =v pp n p =v pp+v pn p = F p, v p n v p p = F p, v p n 1, =1 so that 1 + α + β F p, v p n. Besies, α = v p n p yiels p α n p, so that there exists some ν N such that n p = p α ν. Finally, β = v p F yiels p β F, so that there exists some µ N such that F = µp β. Applying Lemma 3 to the values k = 1, l = α, a = ϕ p x an b = x p which satisfy a b mo pk A because of 7, applie to a = x yiels ϕ p x pα x p pα mo p 1+α A. Using the equation n p = p α ν, 9

10 we get ϕ p x n p = ϕ p x pαν = ϕ p x pα ν x p pα ν since ϕ p x pα x p pα mo p 1+α A = x p pαν = x p n p enn p α ν = n p = x p n p = x n mo p 1+α A. Multiplying this congruence with p β, we obtain p β ϕ p x n p p β x n mo p 1+α+β A. As a consequence of this, p β ϕ p x n p p β x n mo p F p,vpn A since 1 + α + β F p, v p n an hence p 1+α+β A p F p,vpn A. Now, multiplying this congruence with µ, we get which rewrites as µp β ϕ p x n p µp β x n mo p F p,vpn A, F ϕ p x n p F x n mo p F p,vpn A since µp β = F. Hence, 12 becomes ϕ p b n p = F ϕ p x n p n p F x n mo p Fp,vpn A n p F x n b n mo p F p,vpn A by 11. This proves 8, an thus Assertion C is proven. We have therefore shown the implication D = C. Proof of the implication C = D: Assume that Assertion C hols. That is, every n N an every p PF n satisfies 8. We will now recursively construct a family x n n N A N of elements of A which satisfies the equation b m = F x m 13 m for every m N. In fact, let n N, an assume that we have alreay constructe an element x m A for every m N {1, 2,..., n 1} in such a way that 13 hols for every m N {1, 2,..., n 1}. Now, we must construct an element x n A such that 13 is also satisfie for m = n. Our assumption says that we have alreay constructe an element x m A for every m N {1, 2,..., n 1}. In particular, this yiels that we have alreay constructe an element x A for every ivisor of n satisfying n in fact, every such ivisor of n must lie in N 16 an in {1, 2,..., n 1} 17, an thus it satisfies N {1, 2,..., n 1}. Let p PF n. Then, p n, so that n p N +, an thus n p N since n p is a ivisor of n, an every ivisor of n lies in N 18. Besies, n p {1, 2,..., n 1}. 16 because n N an because N is a nest 17 because is a ivisor of n satisfying n 18 because n N an because N is a nest 10

11 Hence, n p N {1, 2,..., n 1}. Since by our assumption the equation 13 hols for every m N {1, 2,..., n 1}, we can thus conclue that 13 hols for m = n p. In other wors, b n p = F x n p. From this equation, we can conclue by n p the same reasoning as in the proof of the implication D = C that ϕ p b n p F x n mo p F p,vpn A. n p Comparing this with 8, we obtain F x n b n mo p F p,vpn A. 14 n p Now, every ivisor of n which satisfies n p must satisfy F 0 mo p F p,vpn A 19. Thus, Hence, n; n F n; n p; 0 mo p Fp,vpn A n F x n = n; n p; In other wors, n F x n 0 mo p Fp,vpn A = n p x n n; 0x n = 0 mo p F p,vpn A. n p; n + n; n p; n F x n n; F x n = n; F x n n p; n p n since for any ivisor of n, the assertions n p an n an n p are equivalent, because if n p, then n since n n p F x n b n mo p F p,vpn A by 14. b n n; n F x n p F p,vpn A. This relation hols for every p PF n. Thus, b n F x n p F p,v pn A = F n A by Corollary 2. n; p PF n n Hence, there exists an element x n of A that satisfies b n F x n = F n x n. n; n Fix such an x n. We now claim that this element x n satisfies 13 for m = n. In fact, n F x n = n; n F x n + n; =n F x n = F nx n n n = F nx 1 n= F nx n = n; n F x n + F n x n = b n 19 This has alreay been proven uring our proof of the implication D = C. 11

12 since b n n; n F x n = F n x n. Hence, 13 is satisfie for m = n. This shows that we can recursively construct a family x n n N A N of elements of A which satisfies the equation 13 for every m N. Therefore, this family satisfies b n = F x n by 13, applie to m = n n = w F,n xk k N for every n N. So we have proven that there exists a family x n n N A N which satisfies b n = w F,n xk k N for every n N. In other wors, we have proven Assertion D. Thus, the implication C = D is proven. Now that both implications D = C an C = D are verifie, Theorem 4 is proven. Next, we will show a result similar to Theorem 4 20 : Theorem 5. Let N be a nest. Let F : P N N be a pseuo-monotonous map. Let A be an Abelian group written aitively. For every n N, let ϕ n : A A be an enomorphism of the group A such that ϕ 1 = i an 15 ϕ n ϕ m = ϕ nm for every n N an every m N satisfying nm N. 16 Let b n n N A N be a family of elements of A. Then, the following five assertions C, E, F, G an H are equivalent: Assertion C: Every n N an every p PF n satisfies ϕ p b n p b n mo p F p,vpn A. 17 Assertion E: There exists a family y n n N A N of elements of A such that b n = n F ϕ n y for every n N. Assertion F: Every n N satisfies µ ϕ b n F n A. n Assertion G: Every n N satisfies φ ϕ b n F n A. n Assertion H: Every n N satisfies n ϕ n gci,n bgci,n F n A. i=1 20 Later, we will unite it with Theorem 4 into one big theorem - whose conitions, however, will inclue the conitions of both Theorems 4 an 5, so it oes not replace Theorems 4 an 5. 12

13 Remark: Here, µ enotes the Möbius function µ : N + Z efine by { 1 PF n, if v µ n = p n 1 for every p PF n 0, otherwise. 18 Besies, φ enotes the Euler phi function φ : N + Z efine by φ n = {m {1, 2,..., n} m n}. We will nee some basic properties of the functions µ an φ: Theorem 6. Any n N + satisfies the five ientities { 1 PF n, if n = p µ n = p PF n 19 0, otherwise φ = n; 20 n µ = [n = 1 ; 21 n µ n = φ n ; 22 n n µ φ = µ n. 23 n Here, { for any assertion κ, we enote by [κ the truth value of κ efine by [κ = 1, if κ is true; 0, if κ is false. This Theorem 6 is exactly ientical to Theorem 6 in [4, an therefore we will not prove it here. Proof of Theorem 5. First, we are going to prove the equivalence of the assertions C an E. In orer to o this, we will prove the implications E = C an C = E. Proof of the implication E = C: Assume that Assertion E hols. That is, there exists a family y n n N A N of elements of A such that b n = n F ϕ n y for every n N. 24 We want to prove that Assertion C hols, i. e., that every n N an every p PF n satisfies 17. Let n N an p PF n. Then, p n, so that n p N +, an thus n p N since n p is a ivisor of n, an every ivisor of n lies in N 21. Thus, applying 24 to n p instea of n yiels b n p = F ϕ n p y. Now, 24 yiels n p b n = n F ϕ n y = n; n p F ϕ n y + n; n p F ϕ n y because n N an because N is a nest 13

14 But every ivisor of n which satisfies n p must satisfy F 0 mo p F p,vpn A 22. Thus, F ϕ n y 0ϕ n y = 0 mo p F p,vpn A. n; n p 0 mo p Fp,vpn A Thus, 25 becomes b n = n; n p = n p F ϕ n y + n; n p n; n p F ϕ n y 0 mo p Fp,vpn A n; n p F ϕ n y + 0 = n; n p F ϕ n y mo p F p,vpn A. 26 On the other han, b n p = ϕ p b n p = ϕ p = n p = n p = n p n p n p F ϕ n p y F ϕ p ϕn p y F =ϕ p ϕ n p y ϕp ϕ n p F ϕ n p y yiels =ϕ p n p ue to 16 F ϕ p n p y = =ϕ n y n p since ϕ p is a group enomorphism F ϕ n y b n mo p F p,vpn A by 26. In other wors, 17 is satisfie, an thus Assertion C is proven. We have therefore shown the implication E = C. Proof of the implication C = E: Assume that Assertion C hols. That is, every n N an every p PF n satisfies 17. We will now recursively construct a family y n n N A N of elements of A which satisfies the equation b m = F ϕ m y 27 m F ϕ n y for every m N. In fact, let n N, an assume that we have alreay constructe an element y m A for every m N {1, 2,..., n 1} in such a way that 27 hols for every m N {1, 2,..., n 1}. Now, we must construct an element y n A such that 27 is also satisfie for m = n. Our assumption says that we have alreay constructe an element y m A for every m N {1, 2,..., n 1}. In particular, this yiels that we have alreay constructe 22 This has alreay been proven uring our proof of Theorem 4. 14

15 an element y A for every ivisor of n satisfying n in fact, every such ivisor of n must lie in N 23 an in {1, 2,..., n 1} 24, an thus it satisfies N {1, 2,..., n 1}. Let p PF n. Then, p n, so that n p N +, an thus n p N since n p is a ivisor of n, an every ivisor of n lies in N 25. Besies, n p {1, 2,..., n 1}. Hence, n p N {1, 2,..., n 1}. Since by our assumption the equation 27 hols for every m N {1, 2,..., n 1}, we can thus conclue that 27 hols for m = n p. In other wors, b n p = F ϕ m p y. From this equation, we can conclue m p by the same reasoning as in the proof of the implication E = C that ϕ p b n p = F ϕ n y. n p Comparing this with 17, we obtain F ϕ n y b n mo p F p,vpn A. 28 n p Now, every ivisor of n which satisfies n p must satisfy F 0 mo p F p,vpn A 26. Thus, F ϕ n y 0ϕ n y = 0 mo p F p,vpn A. n; n p; 0 mo p Fp,vpn A n Hence, F ϕ n y = n; n = n; F ϕ n y n; n p; n F ϕ n y 0 mo p Fp,vpn A n; n p; n + n; n p; n F ϕ n y n; n p; n F ϕ n y n p = n p since for any ivisor of n, the assertions n p an n an n p are equivalent, because if n p, then n since n n p F ϕ n y b n mo p F p,vpn A by 28. In other wors, b n n; n F ϕ n y p F p,vpn A. 23 because n N an because N is a nest 24 because is a ivisor of n satisfying n 25 because n N an because N is a nest 26 This has alreay been proven uring our proof of Theorem 4. 15

16 This relation hols for every p PF n. Thus, b n F ϕ n y p F p,v pn A = F n A by Corollary 2. n; p PF n n Hence, there exists an element y n of A that satisfies b n F ϕ n y = F n y n. n; n Fix such a y n. We now claim that this element y n satisfies 27 for m = n. In fact, n F ϕ n y = n; n F ϕ n y + F ϕ n y n; =n = F nϕ n n y n= F nϕ 1 y n= F ny n, ue to 15 = n; n F ϕ n y + F n y n = b n since b n F ϕ n y = F n y n. Hence, 27 is satisfie for m = n. This n; n shows that we can recursively construct a family y n n N A N of elements of A which satisfies the equation 27 for every m N. Therefore, this family satisfies b n = F ϕ n y for every n N by 27, applie to m = n. So we have n proven that there exists a family y n n N A N which satisfies b n = n F ϕ n y for every n N. In other wors, we have proven Assertion E. Thus, the implication C = E is proven. Since both implications C = E an E = C are proven now, we can conclue that C E. Next we are going to show that E F. Proof of the implication E = F: Assume that Assertion E hols. That is, there exists a family y n n N A N of elements of A such that 24 hols. Then, every n N 16

17 satisfies µ ϕ b n = µ e ϕ e b n e here we substitute e for in the sum n e n = µ e ϕ e F ϕ n e y since b n e = F ϕ n e y n e e n n e by 24 applie to n e instea of n = e n = e n µ e n; n e = n e n; n e = n = n e n; n e = n e F ϕ eϕ n e y since ϕ e is a group enomorphism n e = n; n e F ϕ e ϕn e y =ϕ e ϕ n e y = e n µ e F ϕ e ϕ n e y = n =ϕ e n e by 16 µ e F ϕ n y = n e n; e n = e n µ e n; n e e n; n e µ e F ϕ n y F ϕ e ϕ n e y µ e F ϕ e n e y =ϕ n since for any n an any integer e, the assertion n e is equivalent to e n [n = F ϕ n y µ e F ϕ n y = e n n since 21 with n an replace by n an e yiels = [n = F ϕ n y + [n = F ϕ n y n; =0 since n n; n =n =[n=n F nϕ n n y n since any ivisor of n satisfies either n or = n e n µ e = [n = 1 = [n = = 0 F ϕ n y + [n = n F n ϕ n n y n = [n = n F n ϕ n n y n = F n ϕ n n y n F n A. n; =1 n =0 Thus, Assertion F is satisfie. Consequently, the implication E = F is proven. Proof of the implication F = E: Assume that Assertion F hols. That is, every 17

18 n N satisfies µ ϕ b n F n A. n Thus, for every n N, there exists some y n A such that F n y n = n µ ϕ b n. 29 Fix such a y n for every n N. Then, every n N satisfies F ϕ n y = F e ϕ n e y e here we substitute e for in the sum n e n = e n ϕ n e F e ye = e n =ϕ n e F ey e, since ϕ n e is a group enomorphism ϕ n e µ ϕ b e e = e µϕ n eϕ b e, since ϕ n e since F e y e = e is a group enomorphism µ ϕ b e by 29 applie to e instea of n = e n = n e = n; e µ e n; e µ ϕ n e ϕ b e = e n =ϕ n e ϕ b e e n; e = e N n ; e n; e = n e n; e =ϕ n e by 16 µ ϕ n e ϕ b e = µ ϕ n e b e n ϕ n e b e. 30 Now, for any ivisor of n, we have ϕ n e b e = =ϕ n e e N n ; e ϕ n e b e = h N n ϕ n h b h here we substitute h for e in the sum, since the map { e N n e } N n, e e e n; e 18

19 is a bijection. Thus, 30 becomes F ϕ n y = µ ϕ n e b e = µ n n e n; n = n µ h n; h n e ϕ n h b h = n = ϕ n h b h h N n h n; h n = h n n; h n µ ϕ n h b h = h n = µ ϕ n h b h h n n; n h = n h = µ ϕ n h b h = [n = h ϕ n h b h h n n h h n since 21 applie to n h instea of n yiels = [n = h ϕ n h b h + [n = h ϕ n h b h h n; =0 since h n h n; h n h=n =[n=nϕ n n b n h N n = = h n h n; h n n; h n ϕ n h b h µ ϕ n h b h since for any integer, the assertion h n is equivalent to n h n h since any ivisor h of n satisfies either h n or h = n = 0ϕ n h b h h n; h n } {{ } =0 + [n = n =1 ϕ n n =ϕ 1 =i by 15 b n = i b n = i b n = b n. µ = [n h = 1 = [n = h Therefore, Assertion E is satisfie. We have thus shown the implication F = E. Now we have proven both implications E = F an F = E. As a consequence, we now know that E F. Our next step will be to prove that E G. Proof of the implication E = G: Assume that Assertion E hols. Then, we can prove that every n N satisfies φ ϕ b n = φ e F ϕ n y n n e n this equation is proven in exactly the same way as we have shown the equation µ ϕ b n = µ e F ϕ n y in the proof of the implication E = n n e n F, only with µ replace by φ throughout the proof. Since every ivisor of n satisfies 19

20 e n φ e = n by 20, with n an replace by n an e, this becomes φ ϕ b n = n n n e n φ e F ϕ n y = n F ϕ n y n } {{ } =n F nz ue to 4 F n Zϕ n y = F n Zϕ n y F n A. n A Thus, Assertion G is satisfie. Consequently, the implication E = G is proven. Proof of the implication G = E: Assume that Assertion G hols. That is, every n N satisfies φ ϕ b n F n A. n Thus, for every n N, there exists some z n A such that F n z n = n φ ϕ b n. 31 Fix such a z n for every n N. For every n N, we efine an element y n A by y n = h n h F n h F n this is an integer, since F n h F n h in fact, 4, applie to =n h, yiels F n n n h F n h=h F n h µ h ϕ h z n h. 20

21 Then, F n y n = F n h n = h n = h n = h n = h n hµ h F n h ϕ h z n h h F n h µ h ϕ h z n h = F n h n =ϕ h F n hz n h, since n h Z an since ϕ h is a group enomorphism hµ h ϕ h n h = h n φ ϕ bn h = n h φϕ hϕ b n h, since ϕ h is a group enomorphism F n h F n h F n =h F n h hµ h ϕ h F n h zn h since the equation 31, applie to n h instea of n, yiels F n h z n h = φ ϕ bn h hµ h hµ h n h N n h φ n h φ ϕ h ϕ bn h = hµ h h n =ϕ h ϕ b n h h ϕ h bn h. h Since every ivisor h of n satisfies h h N n h φ ϕ h bn h = e N n ; h e n h = N n h here, we have substitute e for h in the sum, since the map e φ ϕ e b n e h N n h { e N n h e }, h µ h ϕ h z n h φ h = h ϕ h ϕ =ϕ h by 16 b n h =b n h 21

22 is a bijection, because h n, this becomes F n y n = h hµ h φ ϕ h bn h = hµ h h h n N n h h n e N n ; h e = e n; = h n hµ h φ e n; h e = e n = e n h n; h e µ e ϕ e b n e = n = e φ ϕ eb n e e N n ; h h e e h ϕ e b n e = e n h n; h e = h e h e e φ ϕ e b n e h e hµ h φ ϕ e b n e = h e n h e e hµ h φ h =µe by 23, with an n replace by h an e µ ϕ b n here we substitute for e in the sum. ϕ e b n e In other wors, we have proven 29. From this point, we can procee as in the proof of the implication F = E, an we arrive at Assertion E. Hence, we have shown the implication G = E. Now we have shown both implications E = G an G = E. Thus, the equivalence E G must hol. Finally, let us prove the equivalence between the assertions G an H. This is very easy, since every n N satisfies φ ϕ b n = n n ϕ n gci,n bgci,n i=1 27. Therefore, it is clear that G H. Altogether, we have now proven the equivalences C E, E F, E G, an G H. Thus, the five assertions C, E, F, G an H are equivalent. This proves Theorem 5. We can slightly exten Theorem 5 if we require our group A to be torsionfree. First, the efinition: Definition 11. An Abelian group A is calle torsionfree if an only if every element a A an every n N + such that na = 0 satisfy a = 0. A ring R is calle torsionfree if an only if the Abelian group R, + is torsionfree. Note that in [1, Hazewinkel calls torsionfree rings rings of characteristic zero - at least, if I unerstan him right, because he never efines what he means by ring of characteristic zero. Now, here comes the extension of Theorem 5: 27 A proof of this equality can be foun in [4 more precisely, in the proof of G H uring the proof of Theorem 5 in [4. 22

23 Theorem 7. Let F : P N N be a pseuo-monotonous map. Let N be a nest. Let A be a torsionfree Abelian group written aitively. For every n N, let ϕ n : A A be an enomorphism of the group A such that 15 an 16 hol. Let b n n N A N be a family of elements of A. Then, the six assertions C, E, E, F, G an H are equivalent, where the assertions C, E, F, G an H are the ones state in Theorem 5, an the assertion E is the following one: Assertion E : There exists one an only one family y n n N A N of elements of A such that b n = n F ϕ n y for every n N. 32 Obviously, most of Theorem 7 is alreay proven. The only thing we have to a is the following easy observation: Lemma 8. Uner the conitions of Theorem 7, there exists at most one family y n n N A N of elements of A satisfying 32. Proof of Lemma 8. In orer to prove Lemma 8, it is enough to show that if y n n N A N an y n n N A N are two families of elements of A satisfying b n = n F ϕ n y for every n N an 33 b n = n F ϕ n y for every n N, 34 then y n n N = y n n N. So let us show this. Actually, let us prove that y m = y m for every m N. We will prove this by strong inuction over m; so, we fix some n N, an try to prove that y n = y n, assuming that y m = y m is alreay proven for every m N such that m < n. But this is easy to o: We have F ϕ n y = n; n; n F ϕ n y because y = y hols for every ivisor of n satisfying n 28. But 33 yiels n b n = n F ϕ n y = n; n F ϕ n y + F ϕ n y n; =n = F nϕ n n y n = F nϕ 1 y n= F ny n ue to 15 = n; n F ϕ n y + F n y n 28 Proof. Let be a ivisor of n satisfying n. Then, < n. Moreover, every ivisor of n lies in N since n N an since N is a nest, so that N since is a ivisor of n. Now recall our assumption that y m = y m is alreay proven for every m N such that m < n. Applie to m =, this yiels y = y since N an < n. 23

24 an similarly 34 leas to Thus, n; n b n = n; n F ϕ n y + F n y n. F ϕ n y + F n y n = b n = n; n F ϕ n y + F n y n. Subtracting the equality F ϕ n y = F ϕ n y n; n; n n tain F n y n = F n y n, so that F n y n y n = F n y n from this equality, we ob- F n y n = 0 an thus = F ny n y n y n = 0 since the group A is torsionfree, so that y n = y n. This completes our inuction. Thus, we have proven that y m = y m for every m N. In other wors, y n n N = y n n N. This completes the proof of Lemma 8. Now the proof of Theorem 7 is trivial: Proof of Theorem 7. Theorem 5 yiels that the five assertions C, E, F, G an H are equivalent. In other wors, C E F G H. Besies, it is obvious that E = E. It remains to prove the implication E = E. Assume that Assertion E hols. In other wors, assume that there exists a family y n n N A N of elements of A satisfying 32. Accoring to Lemma 8, there exists at most one such family. Hence, there exists one an only one family y n n N A N of elements of A satisfying 32. In other wors, Assertion E hols. Hence, we have proven the implication E = E. Together with E = E, this yiels E E. Combining this with C E F G H, we see that all six assertions C, E, E, F, G an H are equivalent. This proves Theorem 7. Just as Theorem 7 strengthene Theorem 5 in the case of a torsionfree A, we can strengthen Theorem 4 in this case as well: Theorem 9. Let F : P N N be a pseuo-monotonous map. Let N be a nest. Let A be a torsionfree commutative ring with unity. For every p P N, let ϕ p : A A be an enomorphism of the ring A such that 7 hols. Let b n n N A N be a family of elements of A. Then, the three assertions C, D an D are equivalent, where the assertions C an D are the ones state in Theorem 4, an the assertion D is the following one: Assertion D : There exists one an only one family x n n N A N of elements of A such that bn = w F,n xk k N for every n N. 35 Again, having proven Theorem 4, the only thing we nee to o here is checking the following fact: Lemma 10. Let F : P N N be a pseuo-monotonous map. Let N be a nest. Let A be a torsionfree commutative ring with unity. Let b n n N A N be a family of elements of A. Then, there exists at most one family x n n N A N of elements of A satisfying

25 Proof of Lemma 10. In orer to prove Lemma 10, it is enough to show that if x n n N A N an x n n N A N are two families of elements of A satisfying bn = w F,n xk k N for every n N an 36 bn = w F,n x k k N for every n N, 37 then x n n N = x n n N. So let us show this. Actually, let us prove that x m = x m for every m N. We will prove this by strong inuction over m; so, we fix some n N, an try to prove that x n = x n, assuming that x m = x m is alreay proven for every m N F x n such that m < n. But this is easy to prove: We have n; n because x = x hols for every ivisor of n satisfying n 29 b n = w F,n xk k N = n an similarly 37 leas to F x n = n; n F x n + n; =n F x n = F nx n n n = F nx 1 n= F nx n = n; n F x n. But 36 yiels = n; n F x n + F n x n Thus, n; n F x n F x n b n = n; n + F n x n = b n = n; n equality = n; n; n n F n x n, so that F n x n x n = F n x n F x n + F n x n. F x n + F n x n. Subtracting the F x n from this equality, we obtain F n x n = F n x n = 0 an thus x n x n = 0 since = F nx n the ring A is torsionfree, so that x n = x n. This completes our inuction. Thus, we have proven that x m = x m for every m N. In other wors, x n n N = x n n N. This completes the proof of Lemma 10. Proving Theorem 9 now is immeiate: Proof of Theorem 9. Theorem 4 yiels that the two assertions C an D are equivalent. In other wors, C D. Besies, it is obvious that D = D. It remains to prove the implication D = D. Assume that Assertion D hols. In other wors, assume that there exists a family x n n N A N of elements of A satisfying 35. Accoring to Lemma 10, there exists at most one such family. Hence, there exists one an only one family x n n N A N of elements of A satisfying 35. In other wors, Assertion D hols. Hence, we have proven the implication D = D. Together with D = D, this yiels D D. 29 Proof. Let be a ivisor of n satisfying n. Then, < n. Moreover, every ivisor of n lies in N since n N an since N is a nest, so that N since is a ivisor of n. Now recall our assumption that x m = x m is alreay proven for every m N such that m < n. Applie to m =, this yiels x = x since N an < n. 25

26 Combining this with C D, we see that all three assertions C, D an D are equivalent. This proves Theorem 9. Let us recor, for the sake of application, the following result, which is a trivial consequence of Theorems 4 an 5: Theorem 11. Let F : P N N be a pseuo-monotonous map. Let N be a nest. Let A be a commutative ring with unity. For every n N, let ϕ n : A A be an enomorphism of the ring A such that the conitions 7, 15 an 16 are satisfie. Let b n n N A N be a family of elements of A. Then, the assertions C, D, E, F, G an H are equivalent, where the assertions C an D are the ones state in Theorem 4, an the assertions E, F, G an H are the ones state in Theorem 5. Proof of Theorem 11. Accoring to Theorem 4, the assertions C an D are equivalent. Accoring to Theorem 5, the assertions C, E, F, G an H are equivalent. Combining these two observations, we conclue that the assertions C, D, E, F, G an H are equivalent 30, an thus Theorem 11 is proven. An here comes the strengthening of Theorem 11 for torsionfree rings A: Theorem 12. Let F : P N N be a pseuo-monotonous map. Let N be a nest. Let A be a torsionfree commutative ring with unity. For every n N, let ϕ n : A A be an enomorphism of the ring A such that the conitions 7, 15 an 16 are satisfie. Let b n n N A N be a family of elements of A. Then, the assertions C, D, D, E, E, F, G an H are equivalent, where: the assertions C an D are the ones state in Theorem 4, the assertions E, F, G an H are the ones state in Theorem 5, the assertion D is the one state in Theorem 9, an the assertion E is the one state in Theorem 7. Proof of Theorem 12. Accoring to Theorem 9, the assertions C, D an D are equivalent. Accoring to Theorem 7, the assertions C, E, E, F, G an H are equivalent. Combining these two observations, we conclue that the assertions C, D, D, E, E, F, G an H are equivalent 31, an thus Theorem 12 is proven. We are now going to formulate the most important particular case of Theorem 12, namely the one where A is a ring of polynomials over Z: 30 Here, of course, we have use that the assertion C from Theorem 5 is ientic with the assertion C from Theorem Here, of course, we have use that the assertion C from Theorem 5 is ientic with the assertion C from Theorem 4. 26

27 Theorem 13. Let F : P N N be a pseuo-monotonous map. Let Ξ be a family of symbols. Let N be a nest, an let b n n N Z [Ξ N be a family of polynomials in the ineterminates Ξ. Then, the following assertions C Ξ, D Ξ, D Ξ, E Ξ, E Ξ, F Ξ, G Ξ an H Ξ are equivalent: Assertion C Ξ : Every n N an every p PF n satisfies b n p Ξ p b n mo p F p,vpn Z [Ξ. Assertion D Ξ : There exists a family x n n N Z [Ξ N of elements of Z [Ξ such that bn = w F,n xk k N for every n N. Assertion D Ξ : There exists one an only one family x n n N Z [Ξ N of elements of Z [Ξ such that bn = w F,n xk k N for every n N. Assertion E Ξ : There exists a family y n n N Z [Ξ N of elements of Z [Ξ such that b n = F y Ξ n for every n N. n Assertion E Ξ : There exists one an only one family y n n N Z [Ξ N of elements of Z [Ξ such that b n = n F y Ξ n for every n N. Assertion F Ξ : Every n N satisfies µ b n Ξ F n Z [Ξ. n Assertion G Ξ : Every n N satisfies φ b n Ξ F n Z [Ξ. n Assertion H Ξ : Every n N satisfies n b gci,n Ξ n gci,n F n Z [Ξ. i=1 Before we prove this result, we nee a lemma: Lemma 14. Let a Z [Ξ be a polynomial. Let p be a prime. Then, a Ξ p a p mo pz [Ξ. 27

28 This lemma is Lemma 4 a in [3 with ψ rename as a, so we on t nee to prove this lemma here. Proof of Theorem 13. Let A be the ring Z [Ξ this is the ring of all polynomials over Z in the ineterminates Ξ. Then, A is a torsionfree commutative ring with unity torsionfree because every element a Z [Ξ an every n N + such that na = 0 satisfy a = 0. For every n N, efine a map ϕ n : Z [Ξ Z [Ξ by ϕ n P = P Ξ n for every polynomial P Z [Ξ. It is clear that ϕ n is an enomorphism of the ring Z [Ξ 32. The conition 7 is satisfie, since ϕ p a = a Ξ p a p mo pz [Ξ by Lemma 14 hols for every a A. The conition 15 is satisfie as well since ϕ 1 P = P Ξ 1 = P Ξ = P for every P Z [Ξ, an the conition 16 is also satisfie since ϕ n ϕ m = ϕ nm for every n N an every m N satisfying nm N 33. Hence, the three conitions 7, 15 an 16 are satisfie. Therefore, Theorem 12 yiels that the assertions C, D, D, E, E, F, G an H are equivalent, where: the assertions C an D are the ones state in Theorem 4, the assertions E, F, G an H are the ones state in Theorem 5, the assertion D is the one state in Theorem 9, an the assertion E is the one state in Theorem 7. Now, comparing the assertions C, D, D, E, E, F, G an H with the respective assertions C Ξ, D Ξ, D Ξ, E Ξ, E Ξ, F Ξ, G Ξ an H Ξ, we notice that: we have C C Ξ since A = Z [Ξ an ϕ p b n p = b n p Ξ p ; we have D D Ξ since A = Z [Ξ; we have D D Ξ since A = Z [Ξ; we have E E Ξ since A = Z [Ξ an ϕ n y = y Ξ n ; 32 because ϕ n 0 = 0 Ξ n = 0, ϕ n 1 = 1 Ξ n = 1, an any two polynomials P Z [Ξ an Q Z [Ξ satisfy ϕ n P + Q = P + Q Ξ n = P Ξ n + Q Ξ n = ϕ n P + ϕ n Q ϕ n P Q = P Q Ξ n = P Ξ n Q Ξ n = ϕ n P ϕ n Q. an 33 Proof. Let n N an m N be such that nm N. Then, every P Z [Ξ satisfies ϕ n ϕ m P = ϕ n ϕ m P = ϕ n P Ξ m = P Ξ n m =P Ξ m =Ξ nm here, Ξ n m means the family of the m-th powers of all elements of the family Ξ n consiere as elements of Z [Ξ = P Ξ nm = ϕ nm P. Thus, ϕ n ϕ m = ϕ nm, qe. 28

29 we have E E Ξ since A = Z [Ξ an ϕ n y = y Ξ n ; we have F F Ξ since A = Z [Ξ an ϕ b n = b n Ξ ; we have G G Ξ since A = Z [Ξ an ϕ b n = b n Ξ ; we have H H Ξ since A = Z [Ξ an ϕ n gci,n bgci,n = bgci,n Ξ n gci,n. Hence, the equivalence of the assertions C, D, D, E, E, F, G an H yiels the equivalence of the assertions C Ξ, D Ξ, D Ξ, E Ξ, E Ξ, F Ξ, G Ξ an H Ξ. Thus, Theorem 13 is proven. Theorem 13 has a number of applications, incluing the existence of the Witt aition an multiplication polynomials. But first we notice the simplest particular case of Theorem 13: Theorem 15. Let F : P N N be a pseuo-monotonous map. Let N be a nest, an let b n n N Z N be a family of integers. Then, the following assertions C, D, D, E, E, F, G an H are equivalent: Assertion C : Every n N an every p PF n satisfies b n p b n mo p F p,vpn Z. Assertion D : There exists a family x n n N Z N of integers such that bn = w F,n xk k N for every n N. Assertion D : There exists one an only one family x n n N Z N of integers such that bn = w F,n xk k N for every n N. Assertion E : There exists a family y n n N Z N of integers such that b n = F y for every n N. n Assertion E : There exists one an only one family y n n N Z N of integers such that b n = n F y for every n N. Assertion F : Every n N satisfies µ b n F n Z. n 29

30 Assertion G : Every n N satisfies φ b n F n Z. n Assertion H : Every n N satisfies n b gci,n F n Z. i=1 Proof of Theorem 15. We let Ξ be the empty family. Then, Z [Ξ = Z because the ring of polynomials in an empty set of ineterminates over Z is simply the ring Z itself. Every polynomial a Z satisfies a Ξ n = a for every n N 34. Theorem 13 yiels that the assertions C Ξ, D Ξ, D Ξ, E Ξ, E Ξ, F Ξ, G Ξ an H Ξ are equivalent these assertions were state in Theorem 13. Now, comparing the assertions C Ξ, D Ξ, D Ξ, E Ξ, E Ξ, F Ξ, G Ξ an H Ξ with the respective assertions C, D, D, E, E, F, G an H, we notice that: we have C Ξ C since Z [Ξ = Z an b n p Ξ p = b n p ; we have D Ξ D since Z [Ξ = Z; we have D Ξ D since Z [Ξ = Z; we have E Ξ E since Z [Ξ = Z an y Ξ n = y ; we have E Ξ E since Z [Ξ = Z an y Ξ n = y ; we have F Ξ F since Z [Ξ = Z an b n Ξ = b n ; we have G Ξ G since Z [Ξ = Z an b n Ξ = b n ; we have H Ξ H since Z [Ξ = Z an b gci,n Ξ n gci,n = b gci,n. Hence, the equivalence of the assertions C Ξ, D Ξ, D Ξ, E Ξ, E Ξ, F Ξ, G Ξ an H Ξ yiels the equivalence of the assertions C, D, D, E, E, F, G an H. Thus, Theorem 15 is proven. We notice a simple corollary of Theorem 15: Theorem 16. Let F : P N N be a pseuo-monotonous map. Let q Z be an integer. Then: a There exists one an only one family x n n N+ Z N + of integers such that q n = w F,n x k k N+ for every n N In fact, a Ξ n is efine as the result of replacing every ineterminate by its n-th power in the polynomial a. But since there are no ineterminates, replacing them by their n-th powers oesn t change anything, an thus a Ξ n = a. 30

31 b There exists one an only one family y n n N+ Z N + of integers such that q n = F y for every n N +. n c Every n N + satisfies n µ q n F n Z. Every n N + satisfies n φ q n F n Z. e Every n N + satisfies n q gci,n F n Z. i=1 Note that this Theorem 16 is a generalization of Theorem 16 in [4, but the parts c, an e of our Theorem 16 are not stronger than the corresponing parts of Theorem 16 in [4, because F n n as quickly follows from 4, applie to = n. Still, we are going to prove the whole Theorem 16 here for the sake of completeness. Proof of Theorem 16. First we note that every n N + an every p PF n satisfies q n p q n mo p vpn Z 35. Since F p, v p n v p n because 2, applie to a = v p n an b = 0, yiels F p, v p n v p n F p, 0 0 = 0 yiels p F p,vpn p vpn =0 by 1 an thus p vpn Z p F p,vpn Z, this becomes q n p q n mo p F p,vpn Z. 38 Now let N be the nest N +. Define a family b n n N Z N by b n = q n for every n N. Accoring to Theorem 15, the assertions C, D, D, E, E, F, G an H are equivalent these assertions were state in Theorem 15. Since the assertion C is true for our family b n n N Z N because every n N an every p PF n satisfies b n p = q n p q n by 38 = b n mo p F p,vpn Z,, this yiels that the assertions D, D, E, E, F, G an H must also be true for our family b n n N Z N. But for the family b n n N Z N, 35 In fact, p vpn n, an thus there exists some u N + such that n = p vpn u. Since v p n 1 because p PF n, we have v p n 1 N, an thus can efine an element l N by l = v p n 1. Now, Fermat s little theorem yiels q u q u p = q up mo pz, an thus q u pl q up pl mo p 1+l Z by Lemma 3, applie to k = 1, a = q u, b = q up an A = Z. But n p = p vpn u p = p vpn 1 u = p l u = up l yiels q n p = q upl = q u pl, an n = n p p = up p l yiels q n = q up p l = q up pl. Finally, =up l 1 + l = 1 + v p n 1 = v p n. Hence, q u pl q up pl mo p 1+l Z becomes q n p q n mo p vpn Z since q n p = q u pl, q n = q up pl an 1 + l = v p n, qe. 31

32 assertion D is equivalent to Theorem 16 a since N = N + an b n = q n ; assertion E is equivalent to Theorem 16 b since N = N + an b n = q n ; assertion F is equivalent to Theorem 16 c since N = N + an b n = q n ; assertion G is equivalent to Theorem 16 since N = N + an b n = q n ; assertion H is equivalent to Theorem 16 e since N = N + an b gci,n = q gci,n. Hence, Theorem 16 a, Theorem 16 b, Theorem 16 c, Theorem 16 an Theorem 16 e must be true since the assertions D, E, F, G an H are true for the family b n n N Z N. This proves Theorem 16. Now here is a less-known analogue of Theorem 16: Theorem 17. In the following, for any u Z an any r Q, we efine u the binomial coefficient by r u = r 1 r 1 r! u k, if r N; k=0. 0, if r / N u In particular, if r Q \ Z, then is suppose to mean 0. r Let F : P N N be a pseuo-monotonous map. Let q Z an r Q. Then: a There exists one an only one family x n n N+ Z N + of integers such that qn = w F,n x k rn k N+ for every n N +. b There exists one an only one family y n n N+ Z N + of integers such that qn = F y for every n N +. rn n c Every n N + satisfies qn µ rn F n Z. n Every n N + satisfies qn φ rn F n Z. n e Every n N + satisfies n q gc i, n r gc i, n F n Z. i=1 32