Witt vectors. Part 1 Michiel Hazewinkel Sidenotes by Darij Grinberg Witt#5c: The Chinese Remainder Theorem for Modules [not completed, not proofread] This is an auxiliary note; its goal is to prove a form of the Chinese Remainder Theorem that will be used in [2]. Definition 1. Let P denote the set of all primes. A prime means an integer n > 1 such that the only divisors of n are n and 1. The word divisor means positive divisor. Definition 2. We denote the set {0, 1, 2,...} by N, and we denote the set {1, 2, 3,...} by N +. Note that our notations conflict with the notations used by Hazewinkel in [1]; in fact, Hazewinkel uses the letter N for the set {1, 2, 3,...}, which we denote by N +. Now, here is the Chinese Remainder Theorem in one of its most general forms: Theorem 1. Let A be a commutative ring with unity. Let M be an A- module. Let N N. Let I 1, I 2,..., I N be N ideals of A such that I i +I j A for any two elements i and j of {1, 2,..., N} satisfying i < j. a Then, I 1 I 2...I N M I 1 M I 2 M... I N M. b Also, the map defined by Φ : M I 1 I 2...I N M N M I k M k1 Φ m + I 1 I 2...I N M m + I k M for every m M is a well-defined isomorphism of A-modules. c Let m k M N be a family of elements of M. Then, there exists an element m of M such that m k m mod I k M for every k {1, 2,..., N}. 1 Proof of Theorem 1. a Theorem 1 a occurred as Theorem 1 in [1], and we won t repeat the proof given there. b Let us first forget the definition of Φ made in Theorem 1 b until we have shown that it is indeed well-defined. For any integers i {1, 2,..., N} and j {1, 2,..., N} satisfying i < j, we can find an element a i,j of I i and an element a j,i of I j such that a i,j + a j,i 1 since 1
1 A I i + I j. Fix such elements a i,j and a j,i for all pairs of integers i {1, 2,..., N} and j {1, 2,..., N} satisfying i < j. Then, a i,j I i, a j,i I j and a i,j + a j,i 1 for any. 2 integers i {1, 2,..., N} and j {1, 2,..., N} satisfying i < j Consequently, we have a i,j I i, a j,i I j and a i,j + a j,i 1 for any integers i {1, 2,..., N} and j {1, 2,..., N} satisfying i j 3 1. Notice that I k for any l {1, 2,..., N} and k {1, 2,..., N} satisfying l k. 2 Also, 3 Now, define a map ϕ : M a i,k 1 + I k for every k {1, 2,..., N}. 5 N M I k M k1 ϕ m m + I k M for every m M. Clearly, ϕ is a homomorphism of A-modules. We have I 1 I 2...I N M Ker ϕ since 1 Proof of 3: Let i {1, 2,..., N} and j {1, 2,..., N} be two integers satisfying i j. Since i j, we must have either i < j and i > j. But in both of these cases, 3 can be derived from 2 in fact, if i < j, then 3 directly follows from 2, and if i > j, then 3 follows from 2 applied to j and i instead of i and j. Hence, 3 is proven. 2 Proof of 4: Let l {1, 2,..., N} and k {1, 2,..., N} satisfy l k. Then, 3 applied to i k and j l yields a k,l I k, a l,k I l and a k,l + a l,k 1. But the product contains the by factor a k,l because k l, and thus lies in I k since a k,l I k, and since I k is an ideal of A. This proves 4. 3 Proof of 5: Let k {1, 2,..., N}. Applying 3 to j k, we obtain the following: a i,k I i, a k,i I k and a i,k + a k,i 1 for any integer i {1, 2,..., N} satisfying i k. Thus, for any integer i {1, 2,..., N} satisying i k, we have 1 a i,k + a k,i 0 mod I k a i,k mod I k. Hence, since a k,i I k 1 a i,k mod I k. In other words, 1 1 mod I k, so that a i,k 1 + I k. This proves 5. a i,k 4 2
every m I 1 I 2...I N M satisfies ϕ m m } + {{ I km } I k M since m I 1 I 2...I N M I k M I km this is the zero of the A-module M I k M and thus m Ker ϕ. Hence, ϕ induces a homomorphism of A-modules satisfying Φ : M I 1 I 2...I N M N M I k M k1 0 0 Φ m + I 1 I 2...I N M m + I k M for every m M. This proves that the map Φ of Theorem 1 b is well-defined and a homomorphism of A-modules. We have yet to show that this Φ is an isomorphism. Define a map N Ψ : M I k M M I 1 I 2...I N M by k1 Ψ m k + I k M m l + I 1 I 2...I N M for every m k M N. This map Ψ is indeed well-defined, since the residue class N m l + I 1 I 2...I N M depends only on m k + I k M and not on m k because if m k M N and m k M N are two families satisfying m k + I k M m k + I km in N M I k M, then 4. m l + I 1 I 2...I N M k1 m l + I 1 I 2...I N M 4 Proof. In fact, m k + I k M m k + I km yields m k + I k M m k + I km 3
Every family m k M N satisfies Φ Ψ m k + I k M Φ Ψ m k + I k M Φ m l + I 1 I 2...I N M by the definition of Ψ m k + I k M m l + I k M by the definition of Φ. 6 for each k {1, 2,..., N}, and thus m k m k I km for each k {1, 2,..., N}. In other words, m l m l I lm for each l {1, 2,..., N}. Now, m l m l so that qed. I i m l m }{{ l } I l M I i I i i {1,2,...,N} I 1I 2...I N I 1 I 2...I N M since I 1 I 2...I N M is an A-module, m l + I 1 I 2...I N M I l M I 1 I 2...I N M m l + I 1 I 2...I N M, 4
Since every k {1, 2,..., N} satisfies m l and thus l {1,2,...,N} l {1,2,...,N} m l } {{ } I k by 4 I k m l + I k M + m k + I k M m l + a i,k lk m l } {{ } 1+I k by 5 I k m l + 1 + I k m k m k +I k m k the equation 6 becomes Φ Ψ m k + I k M m k a i,k m k I k m l I k M +m k + I k m k I k M I k M + I k M +m k m k + I k M I k M since I k M is an A-module m l + I k M m k + I k M, m l + I k M m k +I k M m k + I k M. 5
Since this holds for every m k M N, this yields Φ Ψ id because every N element of M I k M can be written in the form m k + I k M for some k1 m k M N. Now we are going to prove that the A-module homomorphism Φ is injective. In fact, let m M be such that Φ m + I 1 I 2...I N M 0. Then, 0 Φ m + I 1 I 2...I N M m + I k M, so that 0 m + I k M in M I k M for every k {1, 2,..., N}. This yields m I k M for every k {1, 2,..., N} because 0 m + I k M rewrites as m I k M, and thus m I 1 M I 2 M... I N M. Using Theorem 1 a, this rewrites as m I 1 I 2...I N M. Thus, we have proven that every m M such that Φ m + I 1 I 2...I N M 0 must satisfy m I 1 I 2...I N M. 7 5 Now, if α M I 1 I 2...I N M satisfies Φ α 0, then α 0. Thus, the homomorphism Φ is injective. Consequently, Φ is left cancellable, so that Φ Ψ Φ Φ}{{ Ψ} Φ Φ Φ id yields Ψ Φ id. id Since Φ Ψ id and Ψ Φ id, the map Ψ must be an inverse map of the map Φ. Hence, Φ is bijective. Since Φ is an A-module homomorphism, this yields that Φ is an A-module isomorphism, and thus Theorem 1 b is proven. c Let α Φ 1 m k + I k M where Φ 1 is a well-defined map, since Φ is an isomorphism. Then, α M I 1 I 2...I N M, and therefore α m + I 1 I 2...I N M for some m M. Consequently, Φ 1 m k + I k M α m + I 1 I 2...I N M, so that m k + I k M Φ m + I 1 I 2...I N M m + I k M by the definition of Φ. Hence, we have m k +I k M m+i k M for every k {1, 2,..., N}. This yields 1 since m k + I k M m + I k M is equivalent to m k m mod I k M. Thus, Theorem 1 c is proven. Here is a trivial corollary of Theorem 1 which is used in [2]: Corollary 2. Let M be an Abelian group written additively. Let P be a finite set of positive integers such that any two distinct elements of P are coprime. Let c p p P M P be a family of elements of M. Then, there exists an element m of M such that c p m mod pm for every p P. 5 In fact, we can find some m M such that α m + I 1 I 2...I N M by the definition of the factor module M I 1 I 2...I N M, and thus Φ α 0 becomes Φ m + I 1 I 2...I N M 0, so that 7 yields m I 1 I 2...I N M. In other words, m + I 1 I 2...I N M 0 in M I 1 I 2...I N M. Since α m + I 1 I 2...I N M, this rewrites as α 0, qed. 6
Proof of Corollary 2. Since P is a finite set of positive integers, it can be written in the form P {p 1, p 2,..., p N }, where p 1, p 2,..., p N are pairwise distinct positive integers and N P. Define a family m k M N of elements of M by m k c pk for every k {1, 2,..., N}. Now, let A be the ring Z. Then, M is a Z-module. For every k {1, 2,..., N}, define an ideal I k of Z by I k p k Z. Then, for any two elements i and j of {1, 2,..., N} satisfying i < j, we have I i + I j A 6. Hence, Theorem 1 c yields that there exists an element m of M such that m k m mod I k M for every k {1, 2,..., N}. 8 Hence, c p m mod pm for every p P 7. This proves Corollary 2. A yet more trivial consequence of Corollary 2: Corollary 3. Let M be an Abelian group written additively. Let P P be a finite set of primes. Let c p p P M P be a family of elements of M. Then, there exists an element m of M such that c p m mod pm for every p P. Proof of Corollary 3. Corollary 3 directly follows from Corollary 2, because any two distinct elements of P are coprime in fact, any two distinct elements of P are two distinct primes, and two distinct primes are always coprime. References [1] Darij Grinberg, Witt#5: Around the integrality criterion 9.93. http://www.cip.ifi.lmu.de/~grinberg/algebra/witt5.pdf [2] Darij Grinberg, Witt#5b: Some divisibilities for big Witt polynomials. http://www.cip.ifi.lmu.de/~grinberg/algebra/witt5b.pdf 6 In fact, let i and j be two elements of {1, 2,..., N} satisfying i < j. Then, p i and p j are distinct elements of P since i < j yields i j, and since p 1, p 2,..., p N are pairwise distinct. Hence, p i and p j are coprime because any two distinct elements of P are coprime. Thus, Bezout s Theorem yields that there exist u Z and v Z satisfying p i u + p j v 1. Hence, 1 p i u + p j v I i + I j and p izi i p jzi j thus I i + I j Z A. 7 Proof. Let p P. Then, there exists k {1, 2,..., N} such that p p k since P {p 1, p 2,..., p N }. Hence, 8 yields m k m mod I k M. Since m k c pk c p and I k M p k Z M p k M pm, this rewrites as c p m mod pm, qed. 7