Witt#5: Around the integrality criterion 9.93 [version 1.1 (21 April 2013), not completed, not proofread]

Transcription

1 Witt vectors. Part 1 Michiel Hazewinkel Sienotes by Darij Grinberg Witt#5: Aroun the integrality criterion 9.93 [version April 2013, not complete, not proofrea In [1, section 9.93, Hazewinkel states that The integrality aspects of the theory of Witt vectors can be evelope solely on the basis of this lemma The purpose of this note is to point out how this is one at least, by proving Theorem 9.73, even slightly generalize, to prove an exten Lemma 9.93 in [1 an to show some more of its applications. First, let us introuce some notation: 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. 1. 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 2. 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. 3 Note that if Ξ is the empty family, then Q [Ξ simply is the ring Q, an Z [Ξ simply is the ring Z. 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 1 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,... 2 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,... 3 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 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 4. Definition 6. If N is a set 5, 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. For any n N +, we efine a polynomial w n Z [X N n by w n n 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 n x k k N n n 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 n xk k N means wn x k k N n ; in other wors, w n xk k N n x n. 4 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. 5 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 The polynomials w 1, w 2, w 3,... are calle the big Witt polynomials or, simply, the Witt polynomials. 6 Definition 8. 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 9. 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 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. This Theorem 1 is part of the well-known Chinese Remainer Theorem for moules, which is proven in every book on commutative algebra; however, let us also give a quick proof of Theorem 1 here, in orer for this note to be self-containe. Proof of Theorem 1. We are going to prove Theorem 1 by inuction over N. First, the inuction base: The case of N 0 is obvious in this case, the assertion of Theorem 1 has to be interprete as M M, which is obviously true, an the case of N 1 is obvious as well in this case, the assertion of Theorem 1 simply states that I 1 M I 1 M, which is true. For the inuction step, let us fix some m N + such that m > 1, an let us assume that Theorem 1 is proven for N m 1. We want to prove that Theorem 1 hols for N m as well. In other wors, we want to prove that I 1 I 2...I m M I 1 M I 2 M... I m M for any m ieals I 1, I 2,..., I m of A which satisfy I i + I j A for any two elements i an j of {1, 2,..., m} satisfying i < j. 1 So let I 1, I 2,..., I m be m such ieals. For every i {1, 2,..., m 1}, we have I i +I m A ue to 1 applie to j m, since i < m; thus, there exist a i I i an b i I m such that a i + b i 1, an thus 1 a i + b i a i mo I m since b i I m. Therefore, 1 m 1 i1 1 a i mo I m m 1 i1 a i mo I m, 6 Caution: These polynomials are referre to as w 1, w 2, w 3,... most of the time in [1 beginning with Section 9. However, in Sections 5-8 of [1, Hazewinkel uses the notations w 1, w 2, w 3,... for some ifferent polynomials the so-calle p-aic Witt polynomials, efine by formula 5.1 in [1, which are not the same as our polynomials w 1, w 2, w 3,... though they are relate to them: namely, the polynomial enote by w k in Sections 5-8 of [1 is the polynomial that we are enoting by w p k here after a renaming of variables; on the other han, the polynomial that we call w k here is something completely ifferent. 3

4 so that 1 m 1 a i + I m. i1 But m 1 i1 a i I 1 I 2...I m 1 since a i I i for every i {1, 2,..., m 1}. Hence, 1 m 1 a i +I m yiels 1 I 1 I 2...I m 1 +I m. Thus, I 1 I 2...I m 1 + i1 I m A. But since Theorem 1 is proven for N m 1, we must have J 1 J 2...J m 1 M J 1 M J 2 M... J m 1 M for any m 1 ieals J 1, J 2,..., J m 1 of A which satisfy J i + J j A for any two elements i an j of {1, 2,..., m 1} satisfying i < j. 2 In particular, applying this to the ieals J 1 I 1, J 2 I 2,..., J m 1 I m 1 which satisfy 2 because of 1, we obtain I 1 I 2...I m 1 M I 1 M I 2 M... I m 1 M. Thus, I 1 M I 2 M... I m M I 1 M I 2 M... I m 1 M I m M I 1 I 2...I m 1 M I m M I 1 I 2...I m 1 M A I 1 I 2...I m 1 +I m I 1 I 2...I m 1 M I m M I 1 I 2...I m 1 + I m I 1 I 2...I m 1 M I m M I 1 I 2...I m 1 I m M since U + V UM V M UV M for any two ieals U an V of A, because U + V UM V M U UM V M +V UM V M UV M + V UM V M UM UV M + UV M UV M I 1 I 2...I m M. But clearly, I 1 I 2...I m M I 1 M I 2 M... I m M since I 1 I 2...I m M I i M for every i {1, 2,..., m}. Thus, I 1 I 2...I m M I 1 M I 2 M... I m M. This completes the inuction step, an thus Theorem 1 is verifie. A trivial corollary from Theorem 1 that we will use is: Corollary 2. Let A be an Abelian group written aitively. Let n N +. Then, na 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 vp i n i i1 p vpn n. p PF n Define an ieal I i of Z by I i p vp i 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 vp i n i an p vp j n j are coprime 7, an thus, by Bezout s theorem, there exist integers α an β such that 1 p vp i n i α+p vp j n j β in Z, an therefore 1 p vp i n i α + p vp j n j β I i +I j p vp i n i ZI i p vp j n j ZI j 7 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 4

5 in Z, an thus I i + I j Z. Hence, 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 I 1 I 2...I N A an N i1 I 1 A I 2 A... I N A I i p vp i n i Z N i1 A I i N i1 p vp i n i Z p vp i n i N A N Z A i1 p vp i n i since PF n {p 1, p 2,..., p N }, this becomes na proven. Another fact we will use: i1 p vp i n i } {{ } n Z A p PF n N i1 Z A nz A na p vp i n i A p PF n p v pn A. Corollary 2 is thus Lemma 3. Let A be a commutative ring with unity, an p N be a nonnegative integer 8. 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. Now we can start with the main theorem - an extension of Lemma 9.93 in [1: Theorem 4. Let N be a nest. 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. 3 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 vpn A. 4 Assertion D: There exists a family x n n N A N of elements of A such that bn w n xk k N for every n N. p v pn A This Theorem 4 is stronger than Lemma 9.93 in [1. In fact, if we set N N + in Theorem 4, an require the ring A to have characteristic zero, then we obtain Lemma 9.93 in [1 in a slightly ifferent formulation, however - for example, our Assertion C is the congruence 9.94 in [1 with n replace by n p. None of the requirements N N + an A has characteristic zero is necessary for Theorem 4 to hol; however, requiring 8 Though we call it p, we o not require it to be a prime in this lemma. 5

6 A to have characteristic zero woul make the family x n n N unique in Assertion D we will etail this later in Theorem 9. 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 n xk k N for every n N. 5 We want to prove that Assertion C hols, i. e., that every n N an every p PF n satisfies 4. 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 9. Thus, applying 5 to n p instea of n yiels b n p w n p xk k N. But wn p xk k N an w n xk k N x n. Now, 5 yiels n p x n p b n w n xk k N n n x n n; x n + n; x n. 6 n p n p But for any ivisor of n, the assertions n p an p vpn are equivalent 10. Thus, n; n p Thus, 6 becomes b n n; n p x n n; x n + n; On the other han, n p p vpn x n 0 mo p vpn A 0 mo p vpn A, since p vpn n; n p b n p w n p xk k N ϕ p b n p ϕ p x n p n p x n n; 0x n 0 mo p vpn A. p vpn x n +0 x n x n mo p vpn A. n; n p n p n p 9 because n N an because N is a nest 10 In fact, we have the following chain of equivalences: n p n n p / Z p x n p n p yiels / Z since n p n p p n here we use that n Z, since n 7 ϕ p x n p 8 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. 6

7 since ϕ p is a ring enomorphism. 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. Then, α + β v p n p + v p v p n p v p n v p p v p n 1. Besies, α v p n p yiels 1 p α n p, so that there exists some ν N such that n p p α ν. Finally, β v p yiels p β, so that there exists some κ N such that κp β. Applying Lemma 3 to the values k 1, l α, a ϕ p x an b x p which satisfy a b mo p k A because of 3, 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, 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 since p α ν n p x p n p x n mo p 1+α A. Multiplying this congruence with p β, we obtain In other wors, p β ϕ p x n p p β x n mo p 1+α+β A. p β ϕ p x n p p β x n mo p vpn A since 1 + α + β v p n. Now, multiplying this congruence with κ, we get v pn 1 which rewrites as κp β ϕ p x n p κp β x n mo p vpn A, ϕ p x n p x n mo p vpn A since κp β. Hence, 8 becomes ϕ p b n p ϕ p x n p n p x n mo p vpn A n p x n b n mo p vpn A by 7. This proves 4, 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 4. We will now recursively construct a family x n n N A N of elements of A which satisfies the equation x m 9 for every m N. b m m 7

8 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 9 hols for every m N {1, 2,..., n 1}. Now, we must construct an element x n A such that 9 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 11 an in {1, 2,..., n 1} 12, 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 13. Besies, n p {1, 2,..., n 1}. Hence, n p N {1, 2,..., n 1}. Since by our assumption the equation 9 hols for every m N {1, 2,..., n 1}, we can thus conclue that 9 hols for m n p. In other wors, b n p n p x n p. From this equation, we can conclue by the same reasoning as in the proof of the implication D C that ϕ p b n p x n mo p vpn A. n p Comparing this with 4, we obtain x n b n mo p vpn A. 10 n p Now, for any ivisor of n, the assertions n p an p vpn are equivalent 14. Thus, x n x n 0 mo p vpn A. n; Hence, n; n n; n p; n x n n; n p; n x n 0 mo p vpn A In other wors, n p p vpn ; n + n; n p; n 0 mo p vpn A, since p vpn x n n; x n n; 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 x n b n mo p vpn A by 10. b n n; n x n p vpn A. 11 because n N an because N is a nest 12 because is a ivisor of n satisfying n 13 because n N an because N is a nest 14 This has alreay been proven uring our proof of the implication D C. 8

9 This relation hols for every p PF n. Thus, b n x n p v pn A na by Corollary 2. n; p PF n n Hence, there exists an element x n of A that satisfies b n x n nx n. Fix such n; n an x n. We now claim that this element x n satisfies 9 for m n. In fact, n x n n; n x n + x n n; n nx n n n nx 1 n nxn n; n x n + nx n b n since b n x n nx n. Hence, 9 is satisfie for m n. This shows that we n; n can recursively construct a family x n n N A N of elements of A which satisfies the equation 9 for every m N. Therefore, this family satisfies b n x n by 9, applie to m n n w 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 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 15 : Theorem 5. Let N be a nest. 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 11 ϕ n ϕ m ϕ nm for every n N an every m N satisfying nm N. 12 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 vpn A. 13 Assertion E: There exists a family y n n N A N of elements of A such that b n n ϕ n y for every n N. 15 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. 9

10 Assertion F: Every n N satisfies µ ϕ b n na. Assertion G: Every n N satisfies φ ϕ b n na. n n Assertion H: Every n N satisfies n ϕ n gci,n bgci,n na. i1 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. 14 0, otherwise 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 15 0, otherwise φ n; 16 n µ [n 1 ; 17 n µ n φ n ; 18 n n µ φ µ n. 19 n Here, for{ any assertion κ, we enote by [κ the truth value of κ efine 1, if κ is true; by [κ 0, if κ is false. Proof of Theorem 6. First, let us prove the ientity 15. In fact, for every n N +, the assertions v p n 1 for every p PF n an n p are equivalent 16 ; hence, p PF n 15 follows irectly from 14. This proves In fact, if n p, then v p n 1 for every p PF n because n equals the prouct p PF n p PF n an every prime occurs only once in this prouct, an conversely, if v p n 1 for every p PF n, then n p because every p PF n satisfies v p n 1 an v p n 1 since p PF n yiels p PF n p n, so that v p n 1, an consequently, n p vpn p. p PF n p 1 p 10 p PF n p,

11 Next, let us show 16. Let n N +. Then, for every m {1, 2,..., n}, the number gc m, n is a ivisor of n. Hence, for every m {1, 2,..., n}, there exists one an only one ivisor of n such that gc m, n. Thus, {1, 2,..., n} n {m {1, 2,..., n} gc m, n }. Since the sets {m {1, 2,..., n} gc m, n } for varying are pairwise isjoint because gc m, n cannot equal two istinct numbers for one an the same m, this yiels that {1, 2,..., n} n {m {1, 2,..., n} gc m, n }. 20 For every ivisor of n, the map { { m 1, 2,..., n } m n } {m {1, 2,..., n} gc m, n }, x x is a bijection because this map is well-efine 17, injective 18 an surjective 19, so that { { m 1, 2,..., n } m n } {m {1, 2,..., n} gc m, n }. Since { { m 1, 2,..., n } m n } n φ by the efinition of φ, this becomes n φ {m {1, 2,..., n} gc m, n }. 21 { { 17 Proof. Let be a ivisor of n. For every x m 1, 2,..., n } m n { x 1, 2,..., n } an x n {, so that x {1, 2,..., n} since x 1, 2,..., n gc x, n gc x, n, an therefore x {m {1, 2,..., n} gc m, n }. }{{ } 1 since x n }, we have } an gc x, n 18 since 0 19 Proof. Let be a ivisor of n. Let y {m {1, 2,..., n} gc m, n }. Then, y {1, 2,..., n} y an gc y, n. Hence, Z since gc y, n y, so that y {1, 2,..., n } since y {1, 2,..., n} an y n y since gc, n gc y y, n gc y, n yiels gc, n y { { 1. Thus, m 1, 2,..., n } m n }. Of course, y y. Therefore, there exists some { { x m 1, 2,..., n } m n } such that y x namely, x y. In other wors, y lies in the image of our map. 11

12 Now, φ n φ φ here we substitute n for in the sum, since the map n N n N n N n N n, n is a bijection n φ {m {1, 2,..., n} gc m, n } by 21 n n {1, 2,..., n} by 20 n. Thus, 16 is proven. Let us now prove the remaining three ientities. Let us enote by P U the power set of any set U. We notice that for every finite set S of primes, we have µ p 1 S 22 p S 20. Recall also that every finite set U an every k N satisfy U {S P U S k}. 23 k This is a classical fact in elementary combinatorics, saying that the number of k- U element subsets of the finite set U is. Thus, it is easy to see that every finite k set U satisfies 1 S [ U 0 24 S PU 20 Proof. Let S be a finite set of primes. Set N p. Then, PF N PF p S N p p since S PF N. Now, 15 yiels p S p PF N 1 PF N, if N µ N 0, otherwise This rewrites as µ p PF N 1 S since PF N S. p p S p PF N 1 1 S since N p. p S p p S since N p PF N S. We have p 12

13 21. The map L : P PF n { N n µ 0 } efine by L S p S p for every S P PF n is well-efine 22, surjective since every element e of { N n µ 0 } satisfies e L S for some S P PF n, namely for S PF e 23 an injective 24. Hence, L is a bijection. Besies, every S P PF n satisfies µ L S 1 S since Proof of 24: Let U be a finite set. Then, S PU 1 S k N k N 0 U S PU; S k 1 k {S PU S k} 1 k U k 1 k U { 1, if U 0; 0, otherwise {S P U S k} 1 k k N U [ U 0. k by 23 by the binomial formula This proves Proof. Let S P PF n. Then, S is a subset of PF n. Hence, each element p of S is a prime ivisor of n. Therefore, the prouct p of these elements also ivies n. In other wors, p N n. p S p S Hence, the formula 22 yiels µ p 1 S 0. p S Thus, p { N n µ 0 } since p N n. p S p S Now, forget that we fixe S. We have thus shown that p { N n µ 0 } for each S P PF n. Hence, the map L is well-efine. 23 Proof. Let e { N n µ 0 }. We must prove that e L S for S PF e. In other wors, we must prove that e L PF e. From e { N n µ 0 }, we obtain that µ e 0. Hence, e p because otherwise, p PF e 1 PF e, if e p 15 woul yiel µ e p PF e 0, which woul contraict µ e 0. On the 0, otherwise other han, from e { N n µ 0 }, we obtain e N n, so that e n an thus PF e PF n. In other wors, PF e P PF n. Hence, L PF e is well-efine. The efinition of L PF e shows that L PF e p. p PF e Thus, e p L PF e. p PF e 24 since for every S P PF n, we have S PF p PF L S, an thus S can be uniquely reconstructe from L S p S p S 13

14 yiels µ L S µ µ n N n µ p p S N n ; µ 0 1 S, because S is a finite set of primes. Now, µ here, we have remove from the sum all aens with µ 0, S PPF n S PPF n [n 1 but these aens are all zero an thus on t change the sum { µ L S since L : P PF n N n µ 0 } is a bijection 1 S 1 S [ PF n 0 by 24, applie to U PF n 25. This proves 17. It remains to prove the remaining two ientities 18 an 19. First, let us show 18: For any p PF n, let us enote by U p the subset {m {1, 2,..., n} p m} of the set {1, 2,..., n}. We have {m {1, 2,..., n} m n} {1, 2,..., n} \ {m {1, 2,..., n} p m} p PF n since an element m {1, 2,..., n} satisfies m n if an only if there is no p PF n such that p m. In other wors, {m {1, 2,..., n} m n} {1, 2,..., n} \ U p 25 p PF n since {m {1, 2,..., n} p m} U p for every p PF n. But by the principle of inclusion an exclusion 26 applie to the family U p p PF n of subsets of the set {1, 2,..., n}, we have {1, 2,..., n} \ U p 1 S U p, p PF n S PF n 25 because for an integer n N +, the assertion PF n 0 is equivalent to n 1, since we have the following chain of equivalences: PF n 0 PF n n has no prime ivisors n 1 p S 26 The principle of inclusion an exclusion states that if X an U are finite sets, an U x x X P U X is a family of subsets of U, then U \ U x 1 S U x, where U x enotes x X the whole set U. We are applying this principle to the sets X PF n an U {1, 2,..., n} an the family U x x X U p p X P U X here. S X x S x 14

15 where U p enotes the whole set {1, 2,..., n}. Now, the efinition of φ yiels p φ n {m {1, 2,..., n} m n} {1, 2,..., n} \ p PF n U p by 25 1 S U p. 26 S PF n p S But for every S PF n, we have U p {m {1, 2,..., n} p m} m {1, 2,..., n} every p S satisfies p m p S p S this assertion is equivalent to { } m {1, 2,..., n} p m p S p m, since p is prime for every p S p S an thus { U p m {1, 2,..., n} p m} p S 27. Hence, 26 becomes p S n p S p φ n S PF n N n ; µ 0 1 S U p µ n N n µ n n µ n. p S S PF n S PPF n 1 S µls n p S p n L S, since LS p p S S PPF n µ L S n L S here, we have substitute for L S in the sum, since L : P PF n { N n µ 0 } is a bijection here, we have ae to the sum some aens with µ 0, but these aens are all zero an thus on t change the sum Thus, 18 is proven. 27 This is because p is a ivisor of n since each p S is a prime ivisor of n, an thus their p S prouct p is also a ivisor of n, an each ivisor of n satisfies {m {1, 2,..., n} m} n p S since there are exactly n elements of the set {1, 2,..., n} ivisible by, namely, 2, 3,..., n. 15

16 Now, we are going to prove the ientity 19 by strong inuction over n. So let m N be an integer, an assume that the ientity 19 hols for every n N + satisfying n < m. Then, we have to prove that 19 also hols for n m. In fact, we have e µ φ µ e 27 for every ivisor e of m satisfying e m 28. Now, e m e m; e e e µ φ e m m; e m e m; e Since every ivisor of m satisfies e φ e φ φ f e N m ; f N m e m; e e e µ φ µ e φ. m e m; e here, we substitute f for e in the sum, since the map { e N m e } N m, e e is a bijection because m φ f m by 16, with n an replace by m an f, f m this becomes e µ φ e m e µ e φ m µ m m e m; m m e m [m 1 m m m; m m/ { 1, if m 1; 0, if m 1 { m, if m 1; 0, if m 1 µ by 17 applie to m instea of n µ + m; m µ } {{ } µm µ + µ m. m; m µ m } {{ } [m1 by 17 applie to m instea of n { 1, if m 1; 0, if m 1 [m 1 28 Proof of 27: Let e be a ivisor of m satisfying e m. Thus, e < m. Also, clearly, e N +. But we have assume that the ientity 19 hols for every n N + satisfying n < m. Applying this to n e, we conclue that 19 hols for n e sinc e N + an e < m. In other wors, we have e µ φ µ e. This proves 27. e 16

17 Thus, µ + µ m e µ φ m; e m e m e µ φ + e µ φ e m; e e m; e e m em µe by 27 m µφ m µ e + e m; m e m here, we substitute for e in the first sum. Therefore, µ m m µ φ. m m µ φ µ + µ φ m; m m m In other wors, 19 hols for n m. This completes our inuction, an thus 19 is proven. Hence, the proof of Theorem 6 is now complete. 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 ϕ n y for every n N. 28 We want to prove that Assertion C hols, i. e., that every n N an every p PF n satisfies 13. 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 29. Thus, applying 28 to n p instea of n yiels b n p ϕ n p y. Now, 28 yiels n p b n n ϕ n y n; n p ϕ n y + n; n p ϕ n y. 29 But for any ivisor of n, the assertions n p an p vpn are equivalent 30. Thus, n; n p ϕ n y n; p vpn 0 mo p vpn A, since p vpn ϕ n y n; p vpn 29 because n N an because N is a nest 30 This has alreay been proven uring our proof of Theorem 4. 0ϕ n y 0 mo p vpn A. 17

18 Thus, 29 becomes b n n; n p n p ϕ n y + n; n p ϕ n y 0 mo p vpn A n; n p ϕ n y + 0 n; n p ϕ n y ϕ n y mo p vpn A. 30 On the other han, b n p ϕ p b n p ϕ p n p n p n p n p n p ϕ n p y ϕ p ϕn p y ϕ p ϕ n p y ϕp ϕ n p ϕ n p y yiels ϕ p n p ue to 12 ϕ p n p y ϕ n y n p since ϕ p is a group enomorphism ϕ n y b n mo p vpn A by 30. In other wors, 13 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 13. We will now recursively construct a family y n n N A N of elements of A which satisfies the equation b m ϕ m y 31 m 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 31 hols for every m N {1, 2,..., n 1}. Now, we must construct an element y n A such that 31 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 an element y A for every ivisor of n satisfying n in fact, every such ivisor of n must lie in N 31 an in {1, 2,..., n 1} 32, 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 33. Besies, n p {1, 2,..., n 1}. 31 because n N an because N is a nest 32 because is a ivisor of n satisfying n 33 because n N an because N is a nest 18

19 Hence, n p N {1, 2,..., n 1}. Since by our assumption the equation 31 hols for every m N {1, 2,..., n 1}, we can thus conclue that 31 hols for m n p. In other wors, b n p ϕ m p y. From this equation, we can conclue by m p the same reasoning as in the proof of the implication E C that ϕ p b n p ϕ n y. n p Comparing this with 13, we obtain ϕ n y b n mo p vpn A. 32 n p Now, for any ivisor of n, the assertions n p an p vpn are equivalent 34. Thus, ϕ n y ϕ n y 0 mo p vpn A. n; n p; n Hence, ϕ n y n; n n; n p; n ϕ n y 0 mo p vpn A n p + n; n p; n n; p vpn ; n ϕ n y 0 mo p vpn A, since p vpn n; n p; n ϕ n y n; n p ϕ n y 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 ϕ n y b n mo p vpn A by 32. In other wors, b n n; n ϕ n y p vpn A. This relation hols for every p PF n. Thus, b n ϕ n y p v pn A na by Corollary 2. n; p PF n n Hence, there exists an element y n of A that satisfies b n ϕ n y ny n. Fix n; n 34 This has alreay been proven uring our proof of Theorem 4. 19

20 such a y n. We now claim that this element y n satisfies 31 for m n. In fact, ϕ n y ϕ n y + ϕ n y ϕ n y + ny n b n n n; n; n; n n n nϕ n n y nnϕ 1 y nny n, ue to 11 since b n ϕ n y ny n. Hence, 31 is satisfie for m n. This shows that n; n we can recursively construct a family y n n N A N of elements of A which satisfies the equation 31 for every m N. Therefore, this family satisfies b n ϕ n y n for every n N by 31, applie to m n. So we have proven that there exists a family y n n N A N which satisfies b n n 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 28 hols. Then, every n N 20

21 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 since bn e ϕ n e y ϕ n e y n e e n n e by 28 applie to n e instea of n e n e n µ e n; n e n e n; n e n n n; n e n; n e n e ϕ eϕ n e y since ϕ e is a group enomorphism n e n; n e ϕ e ϕn e y ϕ e ϕ n e y e n µ e ϕ e ϕ n e y n ϕ e n e by 12 µ e ϕ n y n e n; e n e n µ e n; n e e n; n e µ e ϕ n y ϕ e ϕ n e y µ e 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 ϕ n y µ e ϕ n y e n n since 17 with n an replace by n an e yiels [n 0 since n ϕ n y + n; n [n ϕ n y } {{ } [nnnϕ n n y n since any ivisor of n satisfies either n or n e n 0ϕ n y + [n n nϕ n n y n [n n nϕ n n y n nϕ n n y n na. 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 µ e [n 1 [n 21

22 n N satisfies µ ϕ b n na. Thus, for every n N, there exists some y n A such that n ny n n µ ϕ b n. 33 Fix such a y n for every n N. Then, every n N satisfies ϕ n y eϕ n e y e here we substitute e for in the sum n e n ϕ n e ey e, since ϕ n e is a group enomorphism e n ϕ n e ey e e n ϕ n e µ ϕ b e e e µϕ n eϕ b e, since ϕ n e since ey e e is a group enomorphism µ ϕ b e by 33 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 12 µ ϕ n e ϕ b e µ ϕ n e b e n ϕ n e b e. 34 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 22

23 is a bijection. Thus, 34 becomes 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 17 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 hn [nnϕ 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 11 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 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 ϕ n y in the proof of the implication E F, n n e n only with µ replace by φ throughout the proof. Since every ivisor of n satisfies 23

24 e n φ e n by 16, with n an replace by n an e, this becomes φ ϕ b n n n φ e ϕ n y n ϕ n y n e n n n n n 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 na. Thus, for every n N, there exists some z n A such that n ϕ n y na. nz n n φ ϕ b n. 35 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 µ h ϕ h z n h. 24

25 Thus, ny n n h n h n h n h n h n h n µ h ϕ h z n h h n n µ h ϕ h z n h h n h hµ h n h ϕ h z n h hµ h ϕ h n h z n h h n ϕ hn hz n h since n h Z an since ϕ h is a group enomorphism hµ h ϕ h φ ϕ bn h n h n h φϕ hϕ b n h since ϕ h is a group enomorphism since the equation 35, applie to n h instea of n, yiels n h z n h φ ϕ bn h hµ h hµ h hµ h n h n h N n h N n h φ n h φ ϕ h ϕ bn h ϕ h ϕ b n h φ h ϕ h ϕ ϕ h by 12 h h ϕ h bn h. h b n h b n h Since every ivisor h of n satisfies h h N n h φ ϕ h bn h e N n ; h e 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 25

26 is a bijection, because h n, this becomes ny 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 n h n; h e h e e φ ϕ eb n e e N n ; h h e e h ϕ e b n e e n e hµ h φ h µe by 19, with an n replace by h an e h n; h e h e ϕ e b n e e n h e e hµ h φ ϕ e b n e h µ e ϕ e b n e e φ ϕ e b n e h µ ϕ b n here we substitute for e in the sum. In other wors, we have proven 33. 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 26

27 easy, since every n N satisfies φ ϕ b n φ ϕ b n φ n ϕ n b n n n N n N n b here we substitute n for in the sum, since the map N n N n, n is a bijection n φ ϕ n b n φ ϕ n b N n n n n {m {1, 2,..., n} gc m, n } ϕ n b m {1,2,...,n}; gcm,n m {1,2,...,n}; gcm,n n m {1,2,...,n}; gcm,n ϕ n b ϕ n b ϕ n gcm,nb gcm,n since gcm,n ϕ n gcm,n bgcm,n n m {1,2,...,n} m {1,2,...,n}; gcm,n because of 21 ϕ n gcm,n bgcm,n ϕ n gcm,n bgcm,n since every m {1, 2,..., n} satisfies gc m, n n n n ϕ n gcm,n bgcm,n ϕ n gci,n bgci,n m1 i1 here we substitute i for m in the sum. 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 10. 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: Theorem 7. 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 11 an 12 hol. 27

28 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 ϕ n y for every n N. 36 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 36. 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 ϕ n y for every n N an 37 b n n ϕ n y for every n N, 38 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 ϕ n y ϕ n y n; n; n n because y y hols for every ivisor of n satisfying n 35. But 37 yiels b n n ϕ n y n; n ϕ n y + n; n ϕ n y } {{ } nϕ n n y n nϕ 1 y nny n ue to 11 n; n ϕ n y + ny n an similarly 38 leas to b n n; n ϕ n y + ny n. 35 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. 28

29 Thus, n; n n; n ϕ n y + ny n b n ϕ n y + ny n. n; n ϕ n y from this equality, we obtain ny n ϕ n y n; n n y n y n ny n ny n Subtracting the equality ny n, so that ny n 0 an thus 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 36. 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 36. 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 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 3 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 n xk k N for every n N. 39 Again, having proven Theorem 4, the only thing we nee to o here is checking the following fact: Lemma 10. 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 39. 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 n xk k N for every n N an 40 bn w n x k k N for every n N, 41 29

30 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 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 36 b n w n xk k N an similarly 41 leas to n x n b n n; n n; n x n + n; n x n + nx n. x n nx n n n nx 1 nnx n x n n; n x n n; n. But 40 yiels x n + nx n Thus, x n +nx n n x n +nx n. Subtracting the equality x n n; n; n; n n n x n from this equality, we obtain nx n nx n, so that n x n x n nx n nx n n; n nx n 0 an thus x n x n 0 since 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 39. 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 39. In other wors, Assertion D hols. Hence, we have proven the implication D D. Together with D D, this yiels D D. 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 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 3, 11 an 12 are satisfie. 36 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. 30

31 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 37, an thus Theorem 11 is proven. An here comes the strengthening of Theorem 11 for torsionfree rings A: Theorem 12. 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 3, 11 an 12 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 38, 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: Theorem 13. 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 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 n xk k N for every n N. 37 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. 31

32 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 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 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 y Ξ n for every n N. Assertion F Ξ : Every n N satisfies µ b n Ξ nz [Ξ. n Assertion G Ξ : Every n N satisfies φ b n Ξ nz [Ξ. n Assertion H Ξ : Every n N satisfies n b gci,n Ξ n gci,n nz [Ξ. i1 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. 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 39. The 39 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 32

33 conition 3 is satisfie, since ϕ p a a Ξ p a p mo pz [Ξ by Lemma 14 hols for every a A. The conition 11 is satisfie as well since ϕ 1 P P Ξ 1 P Ξ P for every P Z [Ξ, an the conition 12 is also satisfie since ϕ n ϕ m ϕ nm for every n N an every m N satisfying nm N 40. Hence, the three conitions 3, 11 an 12 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 ; 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: 40 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. 33

34 Theorem 15. 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 vpn Z. Assertion D : There exists a family x n n N Z N of integers such that bn w 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 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 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 y for every n N. n Assertion F : Every n N satisfies µ b n nz. Assertion G : Every n N satisfies φ b n nz. n n Assertion H : Every n N satisfies n b gci,n nz. i1 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 34

35 itself. Every polynomial a Z satisfies a Ξ n a for every n N 41. 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 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 n x k k N+ for every n N +. b There exists one an only one family y n n N+ Z N + of integers such that q n n y for every n N +. c Every n N + satisfies Every n N + satisfies µ q n nz. n φ q n nz. n 41 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. 35