Bielefeld, June 1997

Chinese Leftovers.

A finite family F of ideals in a ring R will be called strongly distributive if

(I + J) ∩ G = I ∩ G + J ∩ G ,  (∗)

whenever I, J ∈ F and G is the intersection over a subfamily G ⊂ F which excludes I and J. Clearly this property is inherited by subfamilies. An F with only two members is always strongly distributive.

Lemma 1: Let F be the intersection of all members of a strongly distributive family F of ideals in a ring R. Then R/F is the inverse limit of the system

R/I ⟶ R/(I + J) ⟵ R/J  (†)

as I, J runs over all pairs in F. Proof. We have a natural injection from R/F into that limit, and must prove its surjectivity. Given a system {b_{I} ∈ R | I ∈ F} such that b_{I} ≡ b_{J} modulo I + J for every pair I, J, we must obtain an a ∈ R such that a ≡ b_{I} modulo I for all I.

The case |F| = 2 being well-known and easy, we may suppose that F = {I, J} ∪ G with G non-empty. By induction hypothesis, the lemma holds for the subfamilies {I}∪G and {J} ∪ G. We therefore have an a_{I} ∈ R which agrees with b_{I} modulo I and with all the b_{K} for K ∈ G modulo their respective K. Analogously we have an a_{J} ∈ R which agrees with b_{J} modulo J and all the b_{K} as above.

Taking G to be the intersection over G, we have F = R/I ∩ J ∩ G and the pull-back diagram

R/F ⟶  R/J ∩ G
↓  ↓
R/I ∩ G ⟶ R/(I ∩ G) + (J ∩ G)

associated with the family consisting of I ∩ G and J ∩ G only. The aforementioned elements a_{I} and a_{J} sit in its lower left and upper right, respectively, and have a common pre-image a in the upper left, provided they agree modulo (I ∩ G) + (J ∩ G).

Since F is strongly distributive, it suffices to prove that a_{I} and a_{J} are congruent modulo the (normally larger) ideal (I + J) ∩ G. Well, they do agree modulo G by construction, and modulo (I + J) because b_{I} and b_{J} do so by hypothesis.

Cyclotomic Polynomials.

Let (k, p) = 1. The following identity is obtained by comparing degrees and substituting primitive k-th as well as kpth roots of unity.

Φ_{k}(t)Φ_{kp}(t) = Φ_{k}(t^{p}) .  (1)

The next identity likewise with kp^{s}-th roots for s > 1:

Φ_{kp^{s}}(t) = Φ_{kp}(t^{p^{s−1}}) ,  (2)

whence by induction Φ_{k}(t^{p^{s}}) = Φ_{kp^{s}}(t) ⋯ Φ_{kp}(t)Φ_{k}(t). In particular, t^{kp^{s−1}} − 1 = Φ_{k}(t^{p^{s−1}})Ψ(t) where Ψ(t) is the product of all Φ_{d}(t) with d | k and d ≠ k.

Sums. The reason for saying all this is the following formula about ideal sums.

(Φ_{kp^{s}}(t) , t^{kp^{s−1}} − 1) = (p , Φ_{k}(t^{p^{s−1}})) .  (3)

To prove it, let us abbreviate kp^{s} = n, kp^{s−1} = m, and note that the ideal on the left contains p, since it includes (t^{n} − 1)/(t^{m} − 1) = 1 + t^{m} + ⋯ t^{m(p−1)} as well as t^{m} − 1. Hence it suffices to show:

Φ_{k}(t^{p^{s−1}}) is the greatest common divisor of Φ_{n}(t) and t^{m} − 1 over F_{p}. Let us agree to write τ for t modulo p. Since F_{p}[τ] is a principal ideal domain, we must prove that Φ_{n}(τ) and τ^{m} − 1 are both p^{s−1} times divisible by Φ_{k}(τ) and have no other common factors. By (2) the polynomial Φ_{n}(τ) is p^{s−1} times divisible by the factor Φ_{kp}(τ), and by (1) the latter is the (p − 1)-st power of Φ_{k}(τ). On the other hand, τ^{m} − 1 has p^{s−1} factors Φ_{k}(τ) and a bunch of Φ_{d}(τ) with d ≠ k, as we have already seen. The proof is finished by remarking that Φ_{k}(τ) and Φ_{d}(τ) are relatively prime. Indeed, over a suitable extension of F_{p} both split into distinct linear factors τ − a and τ − b, where the a’s are of order k and the b’s of order d. (Note: for this argument it is essential, that k and d be prime to the characteritic of the ground field.)

A second ideal sum of interest to us is much easier: for (a, b) = d we have

(t^{a} − 1 , t^{b} − 1) = (t^{d} − 1) .  (4)

Indeed, ar = d + bs (in positive integers) yields t^{d} − 1 = (t^{ar} − 1) − t^{d}(x^{bs} − 1).

Cyclic Group Rings.

To thin out the notation, let us abbreviate Φ_{m}(t) = φ_{m} and t^{m} − 1 = ψ_{m} for all natural numbers m. Fixing a natural number n > 1, we consider the family F_{n} of principal ideals in R = Z[t] which consists of (φ_{n}) as well as all (ψ_{n/p}) as p runs over the prime divisors of n.

Lemma 2: F_{n} is strongly distributive. Proof. If n has a single prime divisor, F_{n} has just two members, and the proof is trivial. Otherwise, the left hand side of (∗) takes one of the forms

(ψ_{n/l}, ψ_{n/p}) ∩ (γ) or (φ_{n}, ψ_{n/p}) ∩ (γ) ,  (5)

where γ is the least common multiple of whatever makes up the subfamily in question (remember that R has unique factorization). For the sake of clarity, let us discuss the proof of (α, β) ∩ (γ) ⊆ (α) ∩ (γ) + (β) ∩ (γ) in abstracto.

If (α, β) = (δ), let γ′ equal γ divided by the greatest common divisor of δ and γ. Then (δ) ∩ (γ) = (δ) · (γ′) = (α, β) · (γ′) = (αγ′, βγ′), and all is well because both α and β — being multiples of δ — contain the missing factors of γ. By (4), this is the case of the first alternative in (5).

By (3), the second alternative shapes up as (α, β) ∩ (γ) = (p, φ_{k}^{q}) ∩ (γ), where q = p^{s−1} and n = kp^{s} as in the preceding paragraph. We shall eventually prove that (p, φ_{k}^{q}) ∩ (γ) = (p, φ_{k}^{q}) · (γ) = (α, β) · (γ) and thereby solve our problem. To that avail, we first work over F_{p} and show

(φ̄_{k}^{q}) ∩ (γ̄) = (φ̄_{k}^{q} · γ̄) ,  (6)

where the bar indicates reduction modulo p. The rest then follows easily: given ε ∈ (p, φ_{k}^{q}) ∩ (γ), we can find u ∈ R such that ε̄ = ū · φ̄_{k}^{q} · γ̄. Thus ε − uφ_{k}^{q}γ is a multiple of p and of γ, hence of the form vpγ.

It remains to establish (6) — which means that φ̄_{k} and γ̄ have no common divisor. By definition of strong distributivity, the subfamily G ⊂ F_{n} whose intersection produces (γ) consists entirely of ψ_{n/l} with l ≠ p. Hence all prime divisors of γ are of the form φ_{m} with ord_{l}(m) < ord_{l}(k) = ord_{l}(n) for at least one prime l ≠ p. For every such m = dp^{r}, equations (1) and (2) show that φ̄_{m} is a power of φ̄_{d}, and the argument following equation (3) proves that φ̄_{d} is relatively prime to φ̄_{k}.

Conclusion. Let n = p_{1}^{s_{1}} ⋯ p_{r}^{s_{r}}. For each i between 1 and r, define the divisors m_{i} = n/p_{i}, k_{i} = n/p_{i}^{s_{i}}, and q_{i} = p_{i}^{s_{i}−1}; for each pair i, j of indices, put m_{ij} = n/p_{i}p_{j}. Then we have the pull-back

R/(ψ_{n})  ⟶  R/(φ_{n})
↓  ↓
∏_{i}^{⋆} R/(ψ_{m_{i}}) ⟶ ∏_{i} R/(p_{i}, φ_{k_{i}}^{q_{i}}) ,

where ∏^{⋆} means that components in slots i and j must always agree modulo ψ_{m_{ij}}.

Et pax domini sit semper vobiscum.