Witt#5f: Ghost-Witt integrality for binomial rings

Witt vectors. Part 1 Michiel Hazewinkel

Sidenotes by Darij Grinberg

Witt#5f: Ghost-Witt integrality for binomial rings [version 1.1 (13 September 2014), not proofread]

§1. Definitions and basic results from [5]

The purpose of this note is applying results from [5] to the particular case of binomial rings, and extend them (in this particular case) by additional equivalent assertions.

We start by introducing notation that will be used. The following definitions 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 are copied from [5].

Definition 1. LetPdenote the set of all primes. (Aprime 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, ...} byN+. (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⁺.)

Definition 3. Let Ξ be a family of symbols. We consider the poly- nomial ring Q[Ξ] (this is the polynomial ring over Q in the indeter- minates Ξ; in other words, we use the symbols from Ξ as variables for the polynomials) and its subring Z[Ξ] (this is the polynomial ring over Zin the indeterminates Ξ). ¹. For anyn ∈N, let Ξⁿ mean the family of the n-th powers of all elements of our family Ξ (considered as elements of Z[Ξ]) ². (Therefore, whenever P ∈Q[Ξ] is a polyno- mial, then P(Ξⁿ) is the polynomial obtained from P after replacing every indeterminate by its n-th power.³)

Note that if Ξ is the empty family, then Q[Ξ] simply is the ring Q, and Z[Ξ] simply is the ring Z.

Definition 4. If m and n are two integers, then we write m ⊥ n if and only if m is coprime ton. If m is an integer and S is a set, then we write m⊥S if and only if (m⊥n for every n∈S).

1For instance, Ξ can be (X₀, X₁, X₂, ...), in which case Z[Ξ] means Z[X₀, X₁, X₂, ...].

Or, Ξ can be (X₀, X₁, X₂, ...;Y₀, Y₁, Y₂, ...;Z₀, Z₁, Z₂, ...), in which case Z[Ξ] means Z[X₀, X₁, X₂, ...;Y₀, Y₁, Y₂, ...;Z₀, Z₁, Z₂, ...].

2In other words, if Ξ = (ξ_i)_i∈I, then we define Ξⁿ as (ξ_iⁿ)_i∈I. For instance, if Ξ = (X0, X1, X2, ...), then Ξⁿ = (X₀ⁿ, X₁ⁿ, X₂ⁿ, ...).

If Ξ = (X0, X1, X2, ...;Y0, Y1, Y2, ...;Z0, Z1, Z2, ...), then Ξⁿ = (X₀ⁿ, X₁ⁿ, X₂ⁿ, ...;Y₀ⁿ, Y₁ⁿ, Y₂ⁿ, ...;Z₀ⁿ, Z₁ⁿ, Z₂ⁿ, ...).

3For instance, if Ξ = (X0, X1, X2, ...) and P(Ξ) = (X0+X1)²−2X3+ 1, then P(Ξⁿ) = (X₀ⁿ+X₁ⁿ)²−2X₃ⁿ+ 1.

Definition 5. A nest means a nonempty subset N of N+ such that for every element d ∈N, every divisor of d lies inN.

Here are some examples of nests: For instance, N⁺ itself is a nest.

For every prime p, the set {1, p, p², p³, ...} is a nest; we denote this nest by p^N. For any integer m, the set {n∈N+ |n⊥m} is a nest;

we denote this nest by N^⊥m. For any positive integer m, the set {n ∈N+|n ≤m} is a nest; we denote this nest by N^≤m. For any integer m, the set{n ∈N+|(n|m)}is a nest; we denote 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 set⁵, we shall denote by X_N the family (X_n)_n∈N of distinct symbols. Hence, Z[X_N] is the ring Z

(X_n)_n∈N (this is the polynomial ring over Z in |N| indeterminates, where the indeterminates are labelled X_n, where n runs through the ele- ments of the set N). For instance, Z

X_N+

is the polynomial ring Z[X₁, X₂, X₃, ...] (since N+={1,2,3, ...}), and Z

X{1,2,3,5,6,10}

is the polynomial ring Z[X₁, X₂, X₃, X₅, X₆, X₁₀].

IfAis a commutative ring with unity, ifN is a set, if (x_d)_d∈N ∈A^N is a family of elements ofAindexed by elements ofN, and ifP ∈Z[XN], then we denote by P (x_d)_d∈N

the element ofA that we obtain if we substitute x_d for X_d for every d ∈ N into the polynomial P. (For instance, if N ={1,2,5} and P =X₁² +X2X5−X5, and if x1 = 13, x₂ = 37 and x₅ = 666, thenP (x_d)_d∈N

= 13²+ 37·666−666.) We notice that whenever N and M are two sets satisfying N ⊆ M, then we canonically identify Z[X_N] with a subring of Z[X_M]. In particular, when P ∈Z[XN] is a polynomial, andA is a commutative ring with unity, and (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 x_m for m ∈ M \N are simply ignored when evaluating P (xm)_m∈M

.) In particular, ifN ⊆N+, and (x₁, x₂, x₃, ...)∈A^N+, thenP (x₁, x₂, x₃, ...) means P (x_m)_m∈N

.

Definition 7. For anyn∈N+, we define a polynomialw_n∈Z h

X_N|ni by

w_n=X

d|n

dX_d^nd.

Hence, for every commutative ring A with unity, and for any family (x_k)_k∈

N|n ∈A^N|n of elements of A, we have w_n

(x_k)_k∈

N|n

=X

d|n

dxⁿ_d^d.

4In fact, there exists some n∈N (since N is a nest and thus nonempty), and thus 1∈N (since 1 is a divisor ofn, and every divisor ofnmust lie inN becauseN is a nest).

5We will use this notation only for the case of N being a nest. However, it equally makes sense for any arbitrary setN.

As explained in Definition 6, ifN is a set containingN^|n, ifAis a com- mutative ring with unity, and (xk)_k∈N ∈ A^N is a family of elements of A, thenwn (xk)_k∈N

means wn

(xk)_k∈

N|n

; in other words, w_n (x_k)_k∈N

=X

d|n

dxⁿ_d^d.

The polynomialsw₁, w₂, w₃, ...are called thebig Witt polynomials or, simply, the Witt polynomials.⁶

Definition 8. Letn ∈ Z\ {0}. Let p∈ P. We denote by vp(n) the largest nonnegative integerm satisfyingp^m |n. Clearly, p^vp(n)|n and v_p(n)≥0. Besides, v_p(n) = 0 if and only if p-n.

We also set vp(0) = ∞; this way, our definition of vp(n) extends to all n∈Z (and not only to n ∈Z\ {0}).

Definition 9. Let n ∈ N+. We denote by PFn the set of all prime divisors of n. By the unique factorization theorem, the set PFn is finite and satisfies n = Q

p∈PFn

p^vp(n).

Definition 10. An Abelian group A is calledtorsionfree if and only if every element a ∈ A and every n ∈ N⁺ such that na = 0 satisfy a = 0.

A ring R is calledtorsionfree if and only if the Abelian group (R,+) is torsionfree.

Let us state a couple of theorems whose proofs we will mostly skip:

Theorem 1. Let N be a nest. Let A be a commutative ring with unity. For every p ∈ P∩N, let ϕp : A →A be an endomorphism of the ring A such that

(ϕ_p(a)≡a^pmodpA holds for every a ∈A and p∈P∩N). (1) Let (b_n)_n∈N ∈ A^N be a family of elements of A. Then, the following three assertions C, D and D^expl are equivalent:

Assertion C: Everyn ∈N and every p∈PFn satisfies

ϕ_p(b_np)≡b_nmodp^vp(n)A. (2) Assertion D: There exists a family (xn)_n∈N ∈ A^N of elements of A such that

b_n=w_n (x_k)_k∈N

for every n ∈N .

6Caution: These polynomials are referred to as w1, w2, w3, ... most of the time in [1]

(beginning with Section 9). However, in Sections 5-8 of [1], Hazewinkel uses the notations w1, w2, w3, ... for some different polynomials (the so-called p-adic Witt polynomials, defined by formula (5.1) in [1]), which are not the same as our polynomials w1, w2, w3, ... (though they are related to them: namely, the polynomial denoted by wk in Sections 5-8 of [1] is the polynomial that we are denoting byw_pk hereafter a renaming of variables; on the other hand,

Assertion D^expl: There exists a family (x_n)_n∈N ∈ A^N of elements of A such that



b_n=X

d|n

dxⁿ_d^d for every n∈N



.

Proof of Theorem 1. According to Theorem 4 of [5], the assertions C and D are equivalent.

On the other hand, if (x_n)_n∈N ∈ A^N is a family of elements of A, then every n ∈ N satisfies wn (xk)_k∈N

=P

d|n

dxⁿ_d^d. Therefore, the assertions D and D^expl are equivalent. Combining this with the fact that the assertions C and D are equivalent, we conclude that the three assertions C, D and D^expl are equivalent.

This proves Theorem 1.

Theorem 2. Let N be a nest. Let A be a torsionfree commuta- tive ring with unity. For every p ∈ P ∩N, let ϕ_p : A → A be an endomorphism of the ring A such that (1) holds.

Let (bn)_n∈N ∈A^N be a family of elements of A. Then, the five asser- tions C, D, D⁰, D^expl and D^expl0 are equivalent, where the assertions C, D and D^expl are the ones stated in Theorem 1, and the assertions D⁰ and D^expl0 are the following ones:

Assertion D⁰: There existsone and only one family (x_n)_n∈N ∈A^N of elements of A such that

b_n=w_n (x_k)_k∈N

for every n ∈N

. (3)

Assertion D^expl0: There existsone and only one family (x_n)_n∈N ∈A^N of elements of A such that



b_n=X

d|n

dx^nd_d for every n∈N



.

Proof of Theorem 2. Whenever (x_n)_n∈N ∈ A^N is a family of elements of A, every n ∈ N satisfies wn (xk)_k∈N

= P

d|n

dxⁿ_d^d. Hence, the assertions D⁰ and D^expl0 are equivalent.

But according to Theorem 9 of [5], the assertionsC,D and D⁰ are equivalent.

Combined with the fact that the assertions D⁰ and D^expl0 are equivalent, this yields that the four assertionsC,D,D⁰ andD^expl0 are equivalent. Combined with the fact that the assertionsC,Dand D^expl are equivalent (this is due to Theorem 1), this yields that the five assertions C, D, D⁰, D^expl and D^expl0 are equivalent.

This proves Theorem 2.

Theorem 3. Let N be a nest. Let A be a commutative ring with unity. For every n ∈N, let ϕn : A →A be an endomorphism of the ring A. Assume that

(ϕ₁ = id) and (4)

(ϕ_n◦ϕ_m =ϕ_nm for every n ∈N and everym∈N satisfying nm∈N). (5) Also, assume that (1) holds.

Let (b_n)_n∈N ∈A^N be a family of elements of A. Then, the assertions C, D, D^expl,E, F,G and H are equivalent, where the assertions C, D and D^expl are the ones stated in Theorem 1, and the assertions E, F, G and H are the following ones:

Assertion E: There exists a family (y_n)_n∈N ∈ A^N of elements of A such that



b_n=X

d|n

dϕ_nd(y_d) for every n ∈N



.

Assertion F: Everyn ∈N satisfies X

d|n

µ(d)ϕ_d(b_nd)∈nA.

Assertion G: Everyn ∈N satisfies X

d|n

φ(d)ϕ_d(b_nd)∈nA.

Assertion H: Every n∈N satisfies

n

X

i=1

ϕ_ngcd(i,n) b_gcd(i,n)

∈nA.

Proof of Theorem 3. According to Theorem 5 of [5], the five assertions C, E, F, G and H are equivalent. Combined with the fact that the three assertions C, D and D^expl are equivalent (this is due to Theorem 1), this yields that the assertions C, D, D^expl, E, F, G and H are equivalent. This proves Theorem 3.

Theorem 4. Let N be a nest. Let A be a torsionfree commutative ring with unity. For everyn∈N, letϕn:A→Abe an endomorphism of the ring A such that the conditions (1), (4) and (5) are satisfied.

Let (b_n)_n∈N ∈A^N be a family of elements of A. Then, the assertions C, D, D⁰, D^expl, D^expl0, E, E⁰, F, G and H are equivalent, where:

• the assertions D⁰ and D^expl0 are the ones stated in Theorem 2,

• the assertions E, F, G and H are the ones stated in Theorem 3, and

• the assertion E⁰ is the following one:

Assertion E⁰: There exists one and only one family (y_n)_n∈N ∈A^N of elements of A such that



b_n=X

d|n

dϕ_nd(y_d) for every n ∈N



. (6) Proof of Theorem 4. Theorem 7 of [5] yields that the six assertions C, E, E⁰, F,G andH are equivalent. Combined with the fact that the five assertionsC,D, D⁰,D^expl and D^expl0 are equivalent (this follows from Theorem 2), this yields that the assertionsC,D,D⁰,D^expl, D^expl0, E,E⁰, F, G andH are equivalent. Theorem 4 is thus proven.

§2. Binomial rings

So far we have done nothing but rewriting some results of [5]. We will now introduce the so-called binomial rings, and study the simplifications that occur in Theorem 4 when it is applied to such rings. The notion of binomial rings is a classical one (see [3] and [4], among other sources).

First, let us define binomial coefficients.

Definition 11. LetB be aQ-algebra with unity. For anyu∈B and any r∈Q, we define an element

u r

∈B by

u r

=





 1 r!

r−1

Q

k=0

(u−k), if r ∈N; 0, if r /∈N

.

In particular, if r ∈Q\Z, then u

r

is supposed to mean 0.

It is clear that Definition 11 generalizes the standard definition of binomial coefficients

u r

with u ∈ N and r ∈ N. As a consequence, we will refer to the elements

u r

defined in Definition 11 as ”binomial coefficients”. We will be mainly concerned with rings which are notQ-algebras but in which the binomial coefficients

u r

can still be defined.

Definition 12. Let A be a commutative ring with unity. We de- note by N^+A the subset {n·1A | n ∈N⁺} of A. This subset N^+A is multiplicatively closed, so a localization (N+A)⁻¹A of the ring A is defined. If A is torsionfree, then the canonical ring homomor- phism A → (N^+A)⁻¹A is injective (because if A is torsionfree, then each element of N+A is a non-zerodivisor in A). Hence, whenever A is torsionfree, we will regard A as a subring of its localization (N^+A)⁻¹A. It should be noticed that (N^+A)⁻¹A is aQ-algebra, since each element of N+A has been made invertible in (N+A)⁻¹A. Hence, whenever A is a torsionfree commutative ring with unity, an element u

r

∈(N+A)⁻¹A is well-defined for everyu∈A andr ∈Q(because every u ∈ A lies in (N^+A)⁻¹A). Of course, this element

u r

does not always lie in A (for example, if A = Z[X], r = 2 and u = X, then

u r

= X

2

= 1

2(X²−X) ∈ N⁺Z[X]

⁻¹

(Z[X]) does not lie in Z[X]).

Definition 13. LetAbe a commutative ring with unity. We say that Ais abinomial ring ifAis torsionfree and satisfies the following prop- erty: For any u ∈ A and any r ∈ N, the element

u r

∈ (N+A)⁻¹A lies in A.

The most important example of a binomial ring is:

Proposition 5. The ring Z is a binomial ring.

The proof of this hinges upon the following easy fact:

Proposition 6. LetA be aQ-algebra. Letu∈A. Let r ∈Z. Then, u

r

= (−1)^r

r−u−1 r

.

Proposition 6 is known as the upper negation formula.

Proof of Proposition 6. Ifr /∈N, then the equality u

r

= (−1)^r

r−u−1 r

is obvious by virtue of both binomial coefficients

u r

and

r−u−1 r

being zero. Hence, for the rest of this proof, we can WLOG assume thatr ∈N. Assume

this. Sincer ∈N, the definition of

r−u−1 r

yields r−u−1

r

= 1 r!

r−1

Y

k=0

((r−u−1)−k)

| {z }

=−(u−((r−1)−k))

=(−1)(u−((r−1)−k))

= 1 r!

r−1

Y

k=0

((−1) (u−((r−1)−k)))

| {z }

= _r−1

Q

k=0

(−1) _r−1

Q

k=0

(u−((r−1)−k))

= 1 r!

r−1

Y

k=0

(−1)

!

| {z }

=(−1)^r

r−1

Y

k=0

(u−((r−1)−k))

!

| {z }

=

r−1

Q

k=0

(u−k)

(here, we substitutedkfor (r−1)−k in the product)

= 1 r!(−1)^r

r−1

Y

k=0

(u−k) = (−1)^r 1 r!

r−1

Y

k=0

(u−k). Multiplying this identity with (−1)^r, we obtain

(−1)^r

r−u−1 r

= 1 r!

r−1

Y

k=0

(u−k).

On the other hand, since r∈N, the definition of u

r

yields u

r

= 1 r!

r−1

Y

k=0

(u−k).

Compared to (−1)^r

r−u−1 r

= 1 r!

r−1

Q

k=0

(u−k), this proves u

r

= (−1)^r

r−u−1 r

. Proposition 6 is proven.

Proof of Proposition 5. Clearly, the ring Z is torsionfree. Hence, in order to prove thatZis binomial, we only need to show that for anyu∈Zand anyr∈N, the element

u r

∈(N+Z)⁻¹Z lies in Z. So let r∈N. We need to prove that

u r

∈Z for every u∈Z. For every u ∈ N, the definition of

u r

yields u

r

= 1 r!

r−1

Q

k=0

(u−k) (since r ∈ N). Hence, for every u ∈ N, the number

u r

is the binomial coefficient ”u choose r” known from enumerative combinatorics. Thus, by a known fact from enumerative combinatorics, every u∈N satisfies

u r

= (the number of all r-element subsets of the set {1,2, ..., u})∈Z (7)

(because the cardinality of any finite set is ∈ Z). Thus, u

r

∈ Z is proven for every u∈N.

Now it remains to prove u

r

∈ Z for every u ∈ Z satisfying u /∈ N. So let u∈ Z satisfy u /∈ N. Since u /∈ N, we know that u is a negative integer, so that

−u is a positive integer. Thus, r−u−1∈N (since r ∈N). Hence, (7) (applied tor−u−1 instead ofu) yields

r−u−1 r

∈Z. But Proposition 6 (applied to A=Q) yields

u r

= (−1)^r

r−u−1 r

| {z }

∈Z

∈(−1)^rZ⊆Z.

We have thus proven that u

r

∈ Z for every u ∈ Z. As explained above, this concludes the proof that Z is binomial. Thus, Proposition 5 is proven.

For a less trivial example of a binomial ring, we can take the ring of all integer-valued polynomials:

Proposition 7. LetX be a symbol. The subring

{A ∈Q[X] | A(n)∈Z for every n∈Z}ofQ[X] is a binomial ring.

The proof of this proposition is easy and left to the reader. It is a known fact that the subring {A∈Q[X] | A(n)∈Z for every n∈Z} of Q[X] is the free Z-module with basis

X 0

,

X 1

,

X 2

, ...

; this, however, is not needed in the proof.

Of course, every commutative Q-algebra with unity itself is a binomial ring (because ifA is a commutative Q-algebra with unity, then (N^+A)⁻¹A=A).

A crucial property of binomial rings is that they satisfy a generalization of Fermat’s little theorem:

Theorem 8. LetA be a binomial ring. Letp∈P. Leta ∈A. Then, a^p ≡amodpA.

Theorem 8 is one of the fundamental properties of binomial rings. It appears in [4, Proposition 1.1], and also follows from the implication (1) =⇒(4) in Theorem 4.1 in Jesse Elliott’s paper [3]. We will reproduce the proof from [3] (in more details). The main ingredient of the proof of this theorem is the following fact about finite fields:

Proposition 9. Letp∈P.

(a) Consider the polynomial ring (Z(pZ)) [X] in one indeterminate X over Z(pZ). Then,

p−1

Q

k=0

(X−k) = X^p−X in (Z(pZ)) [X].

(b) Consider the polynomial ring Z[X] in one indeterminate X over Z. Then, there exists some Q∈Z[X] such that

p−1

Q

k=0

(X−k) =X^p−

Proof of Proposition 9. (a) Since p∈P, it is clear that Z(pZ) is a field.

Define a polynomial R∈(Z(pZ)) [X] by R =

p−1

Y

k=0

(X−k)−(X^p−X).

This polynomial R has degree degR ≤ p −1. (In fact, both polynomials

p−1

Q

k=0

(X−k) and X^p−X have degreepand leading termX^p; hence, their leading terms cancel upon subtraction, and their difference R is a polynomial of degree

≤p−1.)

Let π be the canonical projection Z → Z(pZ). Clearly, π is a ring homo- morphism, and we have Kerπ =pZ.

We recall the following known fact:

Fact Pf9.1: Let F be a field, and let P ∈ F [X] be a polynomial. If the polynomialP has more than degP roots in F, then P = 0.

Now, let λ ∈Z(pZ). Then, there exists some ` ∈ {0,1, ..., p−1} such that λ is the residue class of ` modulo p. Consider this `. Then, by the definition of π, we have π(`) = (the residue class of ` modulo p) = λ. Hence, λ−π(`) = 0.

But since`∈ {0,1, ..., p−1}, it is clear thatλ−π(`) is a factor in the product

p−1

Q

k=0

(λ−π(k)). Hence, at least one factor in the product

p−1

Q

k=0

(λ−π(k)) is 0 (since λ−π(`) = 0). This yields that the whole product

p−1

Q

k=0

(λ−π(k)) is 0 (because if one factor in a product is 0, then the whole product must be 0). We have thus shown that

p−1

Q

k=0

(λ−π(k)) = 0.

Also, `<sup>p</sup> ≡ `modp by Fermat’s Little Theorem. Thus, p | `<sup>p</sup> −`, so that

`<sup>p</sup>−`∈pZ= Kerπ, hence π(`<sup>p</sup>−`) = 0. Since

π(`<sup>p</sup>−`) =



π(`)

| {z }

=λ





p

−π(`)

| {z }

=λ

(since π is a ring homomorphism)

=λ^p−λ, this rewrites as λ^p−λ= 0.

Now, since R=

p−1

Y

k=0



X− k

|{z}

=π(k)



−(X^p−X) =

p−1

Y

k=0

(X−π(k))−(X^p−X), we have

R(λ) =

p−1

Y

k=0

(λ−π(k))

| {z }

=0

−(λ^p−λ)

| {z }

=0

= 0,

so that

λ∈ {x∈Z(pZ) | R(x) = 0}= (the set of roots of the polynomial R inZ(pZ)). Now forget that we fixed λ. We thus have shown that every λ ∈ Z(pZ) satisfies

λ∈(the set of roots of the polynomial R in Z(pZ)).

That is, everyλ∈Z(pZ) is a root of the polynomialRinZ(pZ). Hence, there exist at leastproots of the polynomialR inZ(pZ) (since there existpelements ofZ(pZ)). Sincep > p−1≥degR, this yields that the polynomialR has more than degR roots in Z(pZ). Therefore, applying Fact Pf9.1 to F = Z(pZ) and P =R, we obtain R = 0. Hence, 0 =R =

p−1

Q

k=0

(X−k)−(X^p−X), so that

p−1

Q

k=0

(X−k) = X^p−X in (Z(pZ)) [X]. This proves Proposition 9 (a).

(b) Letπ be the canonical projectionZ→Z(pZ). Clearly, Kerπ=pZ. Consider the polynomial ring Z[X] in one indeterminate X over Z, and the polynomial ring (Z(pZ)) [X] in one indeterminateX overZ(pZ). The canon- ical projection π :Z → Z(pZ) induces a ring homomorphism π[X] : Z[X] → (Z(pZ)) [X]. We have Ker (π[X]) =p·Z[X].

By the definition of π[X], we have (π[X]) (X) =X.

Sinceπ[X] is a ring homomorphism, the polynomial

p−1

Q

k=0

(X−k)−(X^p−X)∈ Z[X] satisfies

(π[X])

p−1

Y

k=0

(X−k)−(X^p−X)

!

=

p−1

Y

k=0



(π[X]) (X)

| {z }

=X

−k



−







(π[X]) (X)

| {z }

=X





p

−(π[X]) (X)

| {z }

=X





=

p−1

Y

k=0

(X−k)−(X^p−X) = 0

(since Proposition 9(a) yields

p−1

Q

k=0

(X−k) =X^p−X in (Z(pZ)) [X]). Hence, the polynomial

p−1

Q

k=0

(X−k)−(X^p−X)∈Z[X] satisfies

p−1

Y

k=0

(X−k)−(X^p−X)∈Ker (π[X]) = p·Z[X].

In other words, there exists someQ∈Z[X] such that

p−1

Q

k=0

(X−k)−(X^p−X) = pQ. In other words, there exists some Q ∈ Z[X] such that

p−1

Q

k=0

(X−k) = X^p−

Proof of Theorem 8. We know that A is a binomial ring. Hence, by the definition of a binomial ring, for any u∈ A and any r ∈ N, the element

u r

∈ (N+A)⁻¹A lies in A. Applied to u = a and r = p, this yields that the element a

p

∈(N+A)⁻¹A lies in A. By the definition of a

p

, we have a

p

=





 1 p!

p−1

Q

k=0

(a−k), if p∈N; 0, if p /∈N

= 1 p!

p−1

Y

k=0

(a−k) (sincep∈N), so that

1 p!

p−1

Y

k=0

(a−k) = a

p

∈A.

Multiplying this withp!, we obtain

p−1

Y

k=0

(a−k)∈ p!

|{z}

=p(p−1)!

A =p(p−1)!A

| {z }

⊆A

⊆pA.

Now, consider the polynomial ring Z[X] in one indeterminate X over Z. Due to Proposition 9 (b), there exists some Q ∈Z[X] such that

p−1

Q

k=0

(X−k) = X^p −X +pQ in Z[X]. Consider this Q. Evaluating the polynomial identity

p−1

Q

k=0

(X−k) = X^p−X+pQ atX =q, we obtain

p−1

Y

k=0

(a−k) =a^p−a+pQ(a), so that

a^p−a=

p−1

Y

k=0

(a−k)

| {z }

∈pA

−p Q(a)

| {z }

∈A

∈pA−pA⊆pA.

Hence, a^p ≡amodpA. This proves Theorem 8.

We will soon prove more properties of binomial rings. Let us first recall a known fact:

Proposition 10. LetAbe a commutative ring with unity. Letp∈P. Let a∈A and b ∈A. Then, (a+b)^p ≡a^p+b^pmodpA.

Proof of Proposition 10. It is known thatp| p

k

for everyk ∈ {1,2, ..., p−1}

(since p is prime). Thus, for every k ∈ {1,2, ..., p−1}, there exists some s ∈ Z such that

p k

=ps. Denote this s bys_k. Then, s_k ∈Z satisfies p

k

=ps_k for every k ∈ {1,2, ..., p−1}.

By the binomial formula, (a+b)^p =

p

X

k=0

p k

a^kb^p−k = p

0

| {z }

=1

a⁰

|{z}=1

b^p−0

|{z}

=b^p

+

p−1

X

k=1

p k

| {z }

=psk

(sincek∈{1,2,...,p−1})

a^kb^p−k+ p

p

| {z }

=1

a^p b^p−p

|{z}

=b⁰=1

=b^p+

p−1

X

k=1

ps_ka^kb^p−k

| {z }

≡0 modpA (sinceps_ka^kb^p−k∈pA)

+a^p

≡b^p+

p−1

X

k=1

0 +a^p =b^p +a^p =a^p +b^pmodpA.

This proves Proposition 10.

Lemma 11. Let A be a commutative ring with unity, and p∈Z be an integer⁷. Let k∈N and `∈N be such that k >0. Let a ∈A and b ∈A. If a≡bmodp<sup>k</sup>A, then a<sup>p`</sup> ≡b^{p`</sup>modp<sup>k+`}A.

Lemma 11 is exactly Lemma 3 in [9], and thus will not be proven here.

Now here is an important property of power series over binomial rings:

Theorem 12. Let Ξ be a family of symbols. Let A be a binomial ring. Let u ∈ A. Let A[[Ξ]] denote the ring of power series in the indeterminates Ξ overA(just asA[Ξ] denotes the ring of polynomials in the indeterminates Ξ over A). Let P ∈ A[[Ξ]] be a power series with constant term 1. Then, the canonical embeddingA→(N^+A)⁻¹A induces a canonical embedding A[[Ξ]]→ (N+A)⁻¹A

[[Ξ]], which we will regard as an inclusion. Clearly, P ∈ A[[Ξ]] ⊆ (N^+A)⁻¹A

[[Ξ]]

and u ∈ A ⊆ (N^+A)⁻¹A. Since (N^+A)⁻¹A is a Q-algebra, a power series P^u ∈ (N+A)⁻¹A

[[Ξ]] is thus defined. This power series P^u lies in A[[Ξ]].

Proof of Theorem 12. It is well-known that wheneverB is a commutative Q- algebra,v is an element ofB, andQ∈B[[Ξ]] is a power series with constant term 1, then a power seriesQ^v ∈B[[Ξ]] is defined. Applied to B = (N+A)⁻¹A

[[Ξ]], v = u and Q = P, this yields that a power series P^u ∈ (N^+A)⁻¹A

[[Ξ]] is defined. It remains to prove that this power series P^u lies in A[[Ξ]].

Let C be the power series P −1 ∈ A[[Ξ]]. Since the power series P has constant term 1, the power seriesP −1 has constant term 0. In other words, the power series C has constant term 0 (since C = P −1). Applying the binomial formula, we thus get

(1 +C)^u =X

r∈N

u r

C^r, (8)

7Though we call itp, we do not require it to be a prime!

where the sum on the right hand side converges because the power series C has constant term 0. But we know that

u r

∈A for every u ∈A and r ∈ N (since A is a binomial ring). Thus, everyr ∈N satisfies

u r

| {z }

∈A⊆A[[Ξ]]

C^r

|{z}

∈A[[Ξ]]

∈A[[Ξ]]·A[[Ξ]]⊆A[[Ξ]].

Hence, the equality (8) shows that (1 +C)^u is a convergent sum of power series in A[[Ξ]]. Hence, (1 +C)^uitself lies inA[[Ξ]]. Since 1+C=P (becauseC =P−1), we have thus shown that P^u lies in A[[Ξ]]. This proves Theorem 12.

Lemma 13. Let Ξ be a family of symbols. LetAbe a binomial ring.

Let n ∈Z. Let u ∈ A. Let A[[Ξ]] denote the ring of power series in the indeterminates Ξ over A.

Let P and Q be two power series in A[[Ξ]] with constant term 1.

Assume that P ≡QmodnA[[Ξ]]. Then, P^u ≡Q^umodnA[[Ξ]].

Proof of Lemma 13. The ideal nA[[Ξ]] is closed with respect to the (Ξ)-adic topology onA[[Ξ]]. Hence, every sequence of elements ofnA[[Ξ]] which converges inA[[Ξ]] has its limit lying innA[[Ξ]]. Thus, every convergent infinite sum whose addends lie in nA[[Ξ]] must itself lie in nA[[Ξ]].

Now, letC be the power series P −1∈A[[Ξ]]. Since the power series P has constant term 1, the power seriesP −1 has constant term 0. In other words, the power series C has constant term 0 (since C = P −1). Applying the binomial formula, we thus get

(1 +C)^u =X

r∈N

u r

C^r, (9)

where the sum on the right hand side converges because the power series C has constant term 0.

Also, let D be the power seriesQ−1∈A[[Ξ]]. Since the power series Q has constant term 1, the power seriesQ−1 has constant term 0. In other words, the power series D has constant term 0 (since D =Q−1). Applying the binomial formula, we thus get

(1 +D)^u =X

r∈N

u r

D^r, (10)

where the sum on the right hand side converges because the power seriesD has constant term 0.

Subtracting (10) from (9), we obtain (1 +C)^u−(1 +D)^u =X

r∈N

u r

C^r−X

r∈N

u r

D^r =X

r∈N

u r

(C^r−D^r). (11)

Thus, the infinite sum P

r∈N

u r

(C^r−D^r) converges.

SinceC = P

|{z}

≡QmodnA[[Ξ]]

−1≡Q−1 =DmodnA[[Ξ]], we haveC^r ≡D^rmodnA[[Ξ]]

for every r ∈ N. Thus, C^r−D^r ∈ nA[[Ξ]] for every r ∈ N. But we know that u

r

∈A for every u∈A and r∈N (since A is a binomial ring). Thus, u

r

| {z }

∈A

(C^r−D^r)

| {z }

∈nA[[Ξ]]

∈A·nA[[Ξ]] =n·A·A[[Ξ]]

| {z }

⊆A[[Ξ]]

⊆nA[[Ξ]]

for every r ∈ N. Hence, P

r∈N

u r

(C^r−D^r) ∈ nA[[Ξ]] (since every convergent infinite sum whose addends lie in nA[[Ξ]] must itself lie in nA[[Ξ]]). Due to (11), this rewrites as (1 +C)^u−(1 +D)^u ∈nA[[Ξ]]. Since 1 +C =P (because C=P−1) and 1+D=Q(becauseD=Q−1), this rewrites asP^u−Q^u ∈nA[[Ξ]].

In other words, P^u ≡Q^umodnA[[Ξ]]. Lemma 13 is thus proven.

Lemma 14. LetX be a symbol. LetAbe a binomial ring. Letp∈P. LetA[[X]] denote the ring of power series in the indeterminateXover A.

(a) The power series 1 +X and 1 +X^p have constant term 1. Thus, the power series (1 +X)^u and (1 +X^p)^u are well-defined and lie in A[[X]] for every u∈A.

(b) We have (1 +X^p)^qnp ≡ (1 +X)^qnmodp^vp(n)A[[X]] for every n ∈pN+ and q ∈A.

Proof of Lemma 14. It is clear that the power series 1 +X and 1 +X^p have constant term 1 (since p >0).

(a) Let u ∈ A. Applying Theorem 12 to P = 1 +X and Ξ = (X), we con- clude that the power series (1 +X)^u is well-defined and lies in A[[X]]. Applying Theorem 12 to P = 1 +X^p and Ξ = (X), we conclude that the power series (1 +X^p)^u is well-defined and lies inA[[X]]. This proves Lemma 14 (a).

(b) Let n ∈ pN⁺ and q ∈ A. We need to prove that (1 +X^p)^qnp ≡ (1 +X)^qnmodp^vp(n)A[[X]].

We definedv_p(n) as the largest nonnegative integermsatisfyingp^m |n. Thus, p^vp(n) | n. Hence, there exists a z ∈ Z such that n = zp^vp(n). Consider this z.

Sincezp^vp(n) =n∈pN+ ⊆N+, we have z ∈N+.

Since n ∈ pN+, we have np ∈ N+, so that v_p(np) ≥ 0. Thus, v_p(np) is a nonnegative integer. Denote this nonnegative integer v_p(np) by `. Then,

`=v_p(np)≥0.

Applying Proposition 10 to A[[X]], 1 and X instead of A, a and b, we ob- tain (1 +X)^p ≡ 1^p +X^pmodpA[[X]]. Since 1^p = 1 and p = p¹, this rewrites as (1 +X)^p ≡ 1 + X^pmodp¹A[[X]]. Hence, Lemma 11 (applied to A[[X]], 1, (1 +X)^p and 1 +X^p instead of A, k, a and b) yields that

((1 +X)^p)^{p`</sup> ≡(1 +X<sup>p</sup>)<sup>p`}modp^1+`A[[X]].

Since ((1 +X)^p)^{p`</sup> = (1 +X)<sup>pp`} = (1 +X)^{p1+`</sup> (because pp<sup>`} = p¹p^{`</sup> = p<sup>1+`}), this rewrites as

(1 +X)^{p1+`</sup> ≡(1 +X<sup>p</sup>)<sup>p`}modp^1+`A[[X]]. (12) Let Ξ be the one-element family (X) of indeterminates. Then, A[[Ξ]] = A[[X]]. Hence, (12) rewrites as

(1 +X)^{p1+`</sup> ≡(1 +X<sup>p</sup>)<sup>p`}modp^1+`A[[Ξ]].

Hence, Lemma 13 (applied to qz, p^{1+`</sup>, (1 +X)<sup>p1+`} and (1 +X^p)^p` instead of u, n, P and Q) yields

(1 +X)^p1+`qz

≡

(1 +X^p)^p`qz

modp^1+`A[[Ξ]]. Since

(1 +X)^p1+`qz

= (1 +X)^p1+`qz and

(1 +X^p)^p`qz

= (1 +X^p)^p`qz, this rewrites as

(1 +X)^{p1+`qz</sup> ≡(1 +X<sup>p</sup>)<sup>p`qz}modp^1+`A[[Ξ]]. (13) But

1

|{z}

=vp(p)

+ `

|{z}

=vp(np)

=v_p(p) +v_p(np) = v_p



p·(np)

| {z }

=n



=v_p(n), so that

p^1+`qz=p^vp(n)qz=q zp^vp(n)

| {z }

=n

=qn (14)

and thus

p^`

|{z}

=1 p^p

`+1

qz= 1

pp^`+1qz

| {z }

=qn

=qnp. (15)

Due to (14) and (15), the congruence (13) rewrites as

(1 +X)^qn ≡(1 +X^p)^qnpmodp^1+`A[[Ξ]].

Due to 1 + ` = vp(n) and A[[Ξ]] = A[[X]], this rewrites as (1 +X)^qn ≡ (1 +X^p)^qnpmodp^vp(n)A[[X]]. This proves Lemma 14(b).

Using Lemma 14, we can now show a congruence property of binomial coeffi- cients with ”numerator” in a binomial ring:

Lemma 15. LetAbe a binomial ring. Let n∈N⁺ and let p∈PFn.

Let q∈A and r ∈Q. Then, qnp

rnp

≡ qn

rn

modp^vp(n)A. (16)

This Lemma 15 is a generalization of Lemma 19 from [5]. In fact, since Z is a binomial ring, we can apply Lemma 15 toA=Z, and obtain precisely Lemma 19 from [5].

It should be said that Lemma 15 is nothing like a novel result. Indeed, it is well-known in the case when A =Z, and in the general case it follows from the known fact that, loosely speaking, any divisibility of a polynomial by an integer which holds everywhere in Z must hold everywhere in any binomial ring. This known fact is, e. g., a consequence of the implication (1) =⇒(2) of Theorem 4.1 in Elliott’s paper [3] (to which I also refer the reader for a precise statement).

Also, in most cases, the exponent v_p(n) in (16) can be replaced by larger numbers. Details can be found by searching the internet for ”Jacobsthal’s con- gruence”. Again, the caseA=Zis ”the worst case” in the sense that divisibilities that hold in this case must hold always in binomial rings. We will, however, never need these stronger results.

Proof of Lemma 15. Since p∈PFn, we know that p is a prime and satisfies p|n. Thus, p∈P (since p is a prime). Also, n ∈pN+ (since n∈ N+ and p|n), so thatnp∈N⁺.

Ifrn /∈N, then Lemma 15 is easily seen to be true.⁸ Therefore, we can WLOG assume thatrn∈N for the rest of the proof. Assume this.

Since rn∈N, we have rn≥0. Combined with n >0, this yields r ≥0.

Set m = qn. Lemma 14 yields (1 +X^p)^qnp ≡ (1 +X)^qnmodp^vp(n)A[[X]].

Sinceqn=m, this rewrites as (1 +X^p)^mp ≡(1 +X)^mmodp^vp(n)A[[X]]. Hence, for every λ∈N, we have

the coefficient of the power series (1 +X^p)^mp before X^λ

≡ the coefficient of the power series (1 +X)^m before X^λ

modp^vp(n)A. (17) But it is easy to see that

X

λ∈N

mp λp

X^λ = X

λ∈pN

mp λp

X^λ (18)

8Proof. Assume thatrn /∈N. Then,rnp /∈Nas well (sincep∈N+). Hence, both sides of (16) vanish. Thus, (16) holds, i. e., Lemma 15 is true, qed.

9. However, the binomial formula yields (1 +X^p)^mp

=X

µ∈N

mp µ

| {z }

=

mp pµp

(X^p)^µ

| {z }

=X^pµ

=X

µ∈N

mp pµp

X^pµ= X

λ∈pN

mp λp

X^λ

(here we substituted λ for pµ, since the map N→pN,µ7→pµ is a bijection)

=X

λ∈N

mp λp

X^λ (by (18)), and thus every λ∈N satisfies

the coefficient of the power series (1 +X^p)^mp beforeX^λ

=

mp λp

. (19) Besides, the binomial formula yields

(1 +X)^m =X

λ∈N

m λ

X^λ.

Hence, everyλ∈N satisfies

the coefficient of the power series (1 +X)^m before X^λ

= m

λ

. (20) Thus, everyλ∈N satisfies

mp λp

=

the coefficient of the power series (1 +X^p)^mp before X^λ

(by (19))

≡ the coefficient of the power series (1 +X)^m beforeX^λ

(by (17))

= m

λ

modp^vp(n)A (by (20)).

9Proof of (18): Every λ∈ N\(pN) satisfies λ /∈ pN. Hence, every λ ∈ N\(pN) satisfies λp /∈N. Thus, everyλ∈N\(pN) satisfies

mp λp

=





 1 (λp)!

λp−1

Q

k=0

(mp−k), ifλp∈N; 0, ifλp /∈N

= 0

(sinceλp /∈N). Thus, P

λ∈N\(pN)

mp λp

| {z }

=0 (sinceλ∈N\(pN))

X^λ= P

λ∈N\(pN)

0X^λ= 0.

Now, pN⊆N, so that the sum P

λ∈N

mp λp

X^λdecomposes as X

λ∈N

mp λp

X^λ= X

λ∈pN

mp λp

X^λ+ X

λ∈N\pN

mp λp

X^λ

| {z }

=0

= X

λ∈pN

mp λp

X^λ.

This proves (18).

Sincem =qn, this becomes qnp

λp

≡ qn

λ

modp^vp(n)A.

Applying this toλ =rn, we obtain (16), and thus Lemma 15 is proven.

Here comes a result similar to, but somewhat more interesting than, Lemma 15:

Lemma 16. LetAbe a binomial ring. Let n∈N+ and let p∈PFn.

Let q ∈ A and r∈ Q. Assume that there exist two integers α and β with vp(α)≥vp(β) and r= α

β. Then, qnp−1

rnp−1

≡

qn−1 rn−1

modp^vp(n)A. (21) This Lemma 16 is a generalization of Lemma 21 from [5]. In fact, since Z is a binomial ring, we can apply Lemma 16 toA=Z, and obtain precisely Lemma 21 from [5].

It seems impossible to prove Lemma 16 by generalizing the proof of Lemma 21 in [5]. However, the we can prove Lemma 16 in a different way. It requires two lemmas. The first one is a very basic one about binomial coefficients in binomial rings:

Lemma 17. LetA be a binomial ring. Let u∈A. Letr ∈Q. Then, u

r

=

u−1 r−1

+

u−1 r

.

When applied to A = Z, Lemma 17 yields the standard recursion of the binomial coefficients.

Proof of Lemma 17. If r /∈ N, then Lemma 17 is easily proven¹⁰. Hence, for the rest of this proof, we can WLOG assume thatr∈N. Assume this.

If r = 0, then Lemma 17 is also obvious¹¹. Hence, for the rest of this proof, we can WLOG assume that r6= 0. Assume this.

Since r∈N and r6= 0, we have r∈N+ and thusr−1∈N.

10Proof. Assume that r /∈N. Then, r−1 ∈/ Nas well. This causes the binomial coefficient u−1

r−1

to vanish, whiler /∈Nshows that the binomial coefficients u

r

and u−1

r

vanish as well. Hence, the equation that needs to be proven (

u r

= u−1

r−1

+ u−1

r

) reduces to 0 = 0 + 0, which is tautological. Thus, Lemma 17 is proven ifr /∈N.

11Proof. Assume thatr= 0. Then,r−1 =−1∈/N, so that the binomial coefficient u−1

r−1 vanishes. On the other hand,

u 0

and u−1

0

both equal 1, since we have x

0

= 1 for every

By the definition of u

r

, we have u

r

=





 1 r!

r−1

Q

k=0

(u−k), if r∈N; 0, if r /∈N

= 1 r!

r−1

Y

k=0

(u−k)

| {z }

=(u−0)

r−1

Q

k=1

(u−k)

(sincer ∈N)

= 1

r!(u−0)

| {z }

=u r−1

Y

k=1

(u−k) = 1 r!u

r−1

Y

k=1

(u−k) = 1 r!u

r−2

Y

k=0

(u−(k+ 1))

| {z }

=(u−1)−k

(here, we substituted k+ 1 for k in the product)

= 1 r!u

r−2

Y

k=0

((u−1)−k). (22)

By the definition of

u−1 r

, we have

u−1 r

=





 1 r!

r−1

Q

k=0

((u−1)−k), if r ∈N; 0, if r /∈N

= 1 r!

r−1

Y

k=0

((u−1)−k)

| {z }

=((u−1)−(r−1))

r−2

Q

k=0

((u−1)−k)

(sincer ∈N)

= 1

r!((u−1)−(r−1))

| {z }

=u−r

r−2

Y

k=0

((u−1)−k)

= 1

r!(u−r)

r−2

Y

k=0

((u−1)−k).

x∈A(this follows readily from the definition of x

0

). Now, sincer= 0, we have u

r

− u−1

r

= u

0

| {z }

=1

− u−1

0

| {z }

=1

= 1−1 = 0.

Compared with u−1

r−1

= 0, this yields u−1

r−1

= u

r

− u−r

r

. Thus, u

r

= u−1

r−1

+ u−1

r

. Hence, Lemma 17 is proven in the case whenr= 0.