On coefficient valuations of Eisenstein polynomials

M. Künzer, E. Wirsing

January 10, 2006

Abstract

Let p ⩾ 3 be a prime, let n > m ⩾ 1. Let π_{n} be the norm of ζ_{p^{n}} − 1 under C_{p−1}, so that Z_{(p)}[π_{n}]|Z_{(p)} is a purely ramified extension of discrete valuation rings of degree p^{n−1}. The minimal polynomial of π_{n} over Q(π_{m}) is an Eisenstein polynomial; we give lower bounds for its coefficient valuations at π_{m}. The function field analogue, as introduced by Carlitz and Hayes, is studied as well.

Contents

0 Introduction  2
0.1 Problem and methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
0.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
0.2.1 The number field case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
0.2.2 The function field case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
0.3 Notations and conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1 A polynomial lemma  5
2 Consecutive purely ramified extensions  6
2.1 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.2 Characteristic 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.3 As an illustration: cyclotomic polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.4 Characteristic p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3 Towers of purely ramified extensions  9
4 Galois descent of a divisibility  11
5 Cyclotomic number fields  11
5.1 Coefficient valuation bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
5.2 A different proof of (5.3. i, i′) and some exact valuations . . . . . . . . . . . . . . . . . . . . 12
5.3 Some traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
5.4 An upper bound for N(p) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
6 Cyclotomic function fields, after Carlitz and Hayes  17
6.1 Notation and basic facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
6.2 Coefficient valuation bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
6.3 Some exact valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
6.4 A simple case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

AMS subject classification: 11R18, 11R60.

0 Introduction

0.1 Problem and methods

Consider a primitive p^{n}th root of unity ζ_{p^{n}} over Q, where p is a prime and n ⩾ 2. One has Gal(Q(ζ_{p^{n}})|Q) ≃ C_{p^{n−1}} × C_{p−1}. To isolate the p-part of this extension, let π_{n} be the norm of ζ_{p^{n}} − 1 under C_{p−1}; that is, the product of the Galois conjugates (ζ_{p^{n}} − 1)^{σ}, where σ runs over the subgroup C_{p−1}. Then

[figure: Q(π_{n}); p^{n−1}; Q; p−1; p−1; Q(ζ_{p^{n}}); p^{n−1}; Q(ζ_{p})]

We ask for the minimal polynomial μ_{π_{n},Q}(X) = ∑_{j∈[0,p^{n−1}]} a_{j}X^{j} ∈ Z[X] of π_{n} over Q. By construction, it is an Eisenstein polynomial; that is, v_{p}(a_{j}) ⩾ 1 for j ∈ [0, p^{n−1} − 1], and v_{p}(a_{0}) = 1, where v_{p} denotes the valuation at p.

More is true, though. Our basic objective is to give lower bounds bigger than 1 for these p-values v_{p}(a_{j}), except, of course, for v_{p}(a_{0}). As a byproduct of our method of proof, we shall also obtain congruences between certain coefficients for varying n.

A consideration of the trace Tr_{Q(π_{n})|Q}(π_{n}) yields additional information on the second coefficient of μ_{π_{n},Q}(X). By the congruences just mentioned, this also gives additional information for certain coefficients of the minimal polynomials μ_{π_{l},Q}(X) with l > n; these coefficients no longer appear as traces.

Finally, a comparison with the different ideal 𝔇_{Z_{(p)}[π_{n}]|Z_{(p)}} = Z_{(p)}[π_{n}] μ′_{π_{n},Q}(π_{n}) then yields some exact coefficient valuations, not just lower bounds.

Actually, we consider the analogous question for the coefficients of the slightly more general relative minimal polynomial μ_{π_{n},Q(π_{m})}(X), where n > m ⩾ 1, which can be treated using essentially the same arguments. Note that π_{1} = p.

Except for the trace considerations, the whole investigation carries over mutatis mutandis to the case of cyclotomic function field extensions, as introduced by Carlitz [1] and Hayes [5].

As an application, we mention the Wedderburn embedding of the twisted group ring (with trivial 2-cocycle)

Z_{(p)}[π_{n}] ≀ C_{p^{n−1}} [figure: ω] End_{Z_{(p)}}Z_{(p)}[π_{n}] ≃ Z_{(p)}^{p^{n−1}×p^{n−1}} ,

to which we may reduce the problem of calculating Z_{(p)}[ζ_{p^{n}}] ≀ (C_{p^{n−1}} × C_{p−1}) by means of Nebe decomposition. The image ω(π_{n}) is the companion matrix of μ_{π_{n},Q}(X). To describe the image ω(Z_{(p)}[π_{n}]≀C_{p^{n−1}}) of the whole ring, we may replace this matrix modulo a certain ideal. To do so, we need to know the valuations of its entries, i.e. of the coefficients of μ_{π_{n},Q}(X), or at least a lower bound for them. So far, this could be carried through only for n = 2 [10].

In this article, however, we restrict our attention to the minimal polynomial itself.

0.2 Results

0.2.1 The number field case

Let p ⩾ 3 be a prime, and let ζ_{p^{n}} denote a primitive p^{n}th root of unity over Q in such a way that ζ_{p^{n+1}}^{p} = ζ_{p^{n}} for all n ⩾ 1. Put

\begin{aligned}F_{n} & = & Q(ζ_{p^{n}}) \\ E_{n} & = & Fix_{C_{p−1}}F_{n} , \end{aligned}

so [E_{n} : Q] = p^{n}. Letting

π_{n} = N_{F_{n}|E_{n}}(ζ_{p_{n}} − 1) = ∏_{j∈[1,p−1]}(ζ_{p^{n}}^{j^{p^{n−1}}} − 1) ,

we have E_{n} = Q(π_{n}). In particular, E_{m+i} = E_{m}(π_{m+i}) for m, i ⩾ 1. We fix m and write

μ_{π_{m+i}, E_{m}}(X) = ∑_{j∈[0,p^{i}]} a_{i,j}X^{j} = X^{p^{i}} + ( ∑_{j∈[1,p^{i}−1]} a_{i,j}X^{j}) − π_{m} ∈ Z_{(p)}[π_{m}][X] .

Theorem (5.3, 5.5, 5.8).

(i) We have p^{i} | ja_{i,j} for j ∈ [0, p^{i}].

(i′) If j < p^{i}(p − 2)/(p − 1), then p^{i}π_{m} | ja_{i,j}.

(ii) We have a_{i,j} ≡_{p^{i+1}} a_{i+β,p^{β}j} for j ∈ [0, p^{i}] and β ⩾ 1.

(ii′) If j < p^{i}(p − 2)/(p − 1), then a_{i,j} ≡_{p^{i+1}π_{m}} a_{i+β,p^{β}j} for β ⩾ 1.

(iii) The element p^{i−β} exactly divides a_{i,p^{i}−(p^{i}−p^{β})/(p−1)} for β ∈ [0, i − 1].

(iv) We have μ_{π_{n},Q}(X) ≡_{p^{2}} X^{p^{n−1}} + pX^{(p−1)p^{n−2}} − p for n ⩾ 2.

Assertion (iv) requires the computation of a trace, which can be reformulated in terms of sums of (p − 1)th roots of unity in Q_{p} (5.6). Essentially, one has to count the number of subsets of μ_{p−1} ⊆ Q_{p} of a given cardinality whose sum is of a given valuation at p. We have not been able to go much beyond this reformulation, and this seems to be a problem in its own right — see e.g. (5.9).

To prove (i, i′, ii, ii′), we proceed by induction. Assertions (i, i′) also result from the different 𝔇_{Z_{(p)}[π_{m+i}]|Z_{(p)}[π_{m}]} = (μ′_{π_{m+i},E_{m}}(π_{m+i})) = (p^{i}π_{m+i}^{p^{i}−1−(p^{i}−1)/(p−1)}). Moreover, (ii) yields (iii) by an argument using the different. (In the function field case below, we will no longer be able to use the different for the assertion analogous to (i, i′), and we will have to resort to induction).

Suppose m = 1. Let us call an index j ∈ [1, p^{i} − 1] exact, if either j < p^{i}(p − 2)/(p − 1) and p^{i}π_{m} exactly divides ja_{i,j}, or j ⩾ p^{i}(p − 2)/(p − 1) and p^{i} exactly divides ja_{i,j}. If i = 1 and e.g. p ∈ {3, 19, 29, 41}, then all indices j ∈ [1, p − 1] are exact. If i ⩾ 2, we propose to ask whether the number of non-exact indices j asymptotically equals p^{i−1} as p → ∞.

0.2.2 The function field case

Let p ⩾ 3 be a prime, ρ ⩾ 1 and r = p^{ρ}. We write Z = F_{r}[Y] and Q = F_{r}(Y). We want to study a function field analogue over Q of the number field extension Q(ζ_{p^{n}})|Q. Since 1 is the only p^{n}th root of unity in an algebraic closure Q̄, we have to proceed differently, following Carlitz [1] and Hayes [5]. First of all, the power operation of p^{n} on Q̄ becomes replaced by a module operation of f^{n} on Q̄, where f ∈ Z is an irreducible polynomial. The group of p^{n}th roots of unity

μ_{p^{n}} = {ξ ∈ Q̄ : ξ^{p^{n}} = 1}

becomes replaced by the annihilator submodule

λ_{f^{n}} = {ξ ∈ Q̄ : ξ^{f^{n}} = 0} .

Instead of choosing a primitive p^{n}th root of unity ζ_{p^{n}}, i.e. a Z-linear generator of that abelian group, we choose a Z-linear generator θ_{n} of this Z-submodule. A bit more precisely speaking, the element θ_{n} ∈ Q̄ plays the role of ϑ_{n} := ζ_{p^{n}} − 1 ∈ Q̄. Now Q(θ_{n})|Q is the function field analogue of Q(ϑ_{n})|Q. See also [3, sec. 2].

To state the result, let f(Y) ∈ Z be a monic irreducible polynomial and write q = r^{deg f}. Let ξ^{Y} := Y ξ + ξ^{r} define the Z-linear Carlitz module structure on an algebraic closure Q̄, and choose a Z-linear generator θ_{n} of ann_{f^{n}} Q̄ in such a way that θ_{n+1}^{f} = θ_{n} for all n ⩾ 1. We write F_{n} = Q(θ_{n}), so that Gal(F_{n}|Q) ≃ (Z/f^{n})^{∗}. Letting E_{n} = Fix_{C_{q−1}}F_{n}, we get [E_{n} : Q] = q^{n}. Denoting ϖ_{n} = N_{F_{n}|E_{n}}(θ_{n}) = ∏_{e∈(Z/f)^{∗}} θ_{n}^{e^{q^{n−1}}} , we obtain E_{n} = Q(ϖ_{n}). In particular, E_{m+i} = E_{m}(ϖ_{m+i}) for m, i ⩾ 1. We fix m and write

μ_{ϖ_{m+i}, E_{m}}(X) = ∑_{j∈[0,q^{i}]} a_{i,j}X^{j} = X^{q^{i}} + ( ∑_{j∈[1,q^{i}−1]} a_{i,j}X^{j}) − ϖ_{m} ∈ Z_{(f)}[ϖ_{m}][X] .

Let v_{q}(j) := max{α ∈ Z_{⩾0} : q^{α} | j }.

Theorem (6.6, 6.7, 6.9).

(i) We have f^{i−v_{q}(j)} | a_{i,j} for j ∈ [0, q^{i}]. (i′) If j < q^{i}(q − 2)/(q − 1), then f^{i−v_{q}(j)}ϖ_{m} | a_{i,j}. (ii) We have a_{i,j} ≡_{f^{i+1}} a_{i+β,q^{β}j} for j ∈ [0, q^{i}] and β ⩾ 1. (ii′) If j < q^{i}(q − 2)/(q − 1), then a_{i,j} ≡_{f^{i+1}ϖ_{m}} a_{i+β,q^{β}j} for β ⩾ 1. (iii) The element f^{i−β} exactly divides a_{i,q^{i}−(q^{i}−q^{β})/(q−1)} for β ∈ [0, i − 1]. (iv) If f = Y, then μ_{ϖ_{m+i}, E_{m}}(X) ≡_{Y^{2}} X^{q^{i}} + Y X^{(q−1)q^{i−1}} − ϖ_{m} . A comparison of the assertions (iv) in the number field case and in the function field case indicates possible generalizations — we do not know what happens for μ_{π_{m+i},E_{m}}(X) for m ⩾ 2 in the number field case; moreover, we do not know what happens for f ≠ Y in the function field case.

0.3 Notations and conventions

(o) Within a chapter, the lemmata, propositions etc. are numbered consecutively.

(i) For a, b ∈ Z, we denote by [a, b] := {c ∈ Z : a ⩽ c ⩽ b} the interval in Z.

(ii) For m ∈ Z ∖ {0} and a prime p, we denote by m[p] := p^{v_{p}(m)} the p-part of m, where v_{p} denotes the

valuation of an integer at p. (iii) If R is a discrete valuation ring with maximal ideal generated by r, we write v_{r}(x) for the valuation

of x ∈ R ∖ {0} at r, i.e. x/r^{v_{r}(x)} is a unit in R. In addition, v_{r}(0) := +∞. (iv) Given an element x algebraic over a field K, we denote by μ_{x,K}(X) ∈ K[X] the minimal polynomial

of x over K.

(v) Given a commutative ring A and an element a ∈ A, we sometimes denote the quotient by A/a := A/aA — mainly if A plays the role of a base ring. For b, c ∈ A, we write b ≡_{a} c if b − c ∈ aA.

(vi) For an assertion X, which might be true or not, we let {X} equal 1 if X is true, and equal 0 if X is

false.

Throughout, let p ⩾ 3 be a prime.

1 A polynomial lemma

We consider the polynomial ring Z[X, Y]. Lemma 1.1 We have (X + pY)^{k} ≡_{k[p]·p^{2}Y^{2}} X^{k} + kX^{k−1}pY for k ⩾ 1. Since (\begin{matrix}k \\ j \end{matrix}) = k/j · (\begin{matrix}k−1 \\ j−1 \end{matrix}), we obtain for j ⩾ 2 that

\begin{aligned}v_{p}(p^{j} (\begin{matrix}k \\ j \end{matrix})) & ⩾ & j + v_{p}(k) − v_{p}(j) \\ & ⩾ & v_{p}(k) + 2 , \end{aligned}

where the second inequality follows from j ⩾ 2 if v_{p}(j) = 0, and from j ⩾ p^{v_{p}(j)} ⩾ 3^{v_{p}(j)} ⩾ v_{p}(j) + 2 if v_{p}(j) ⩾ 1.

Corollary 1.2 We have (X + pY)^{k} ≡_{k[p]·pY} X^{k} for k ⩾ 1.

Corollary 1.3 For l ⩾ 1 and x, y ∈ Z such that x ≡_{p^{l}} y, we have x^{k} ≡_{k[p]·p^{l}} y^{k} for k ⩾ 1.

Corollary 1.4 We have (X + Y)^{p^{β+α}} ≡_{p^{α+1}} (X^{p^{β}} + Y^{p^{β}})^{p^{α}} for all α, β ⩾ 0.

The assertion follows by (1.2) since f(X, Y) ≡_{p} g(X, Y) implies that f(X, Y)^{p^{α}} ≡_{p^{α+1}} g(X, Y)^{p^{α}}, where f(X, Y), g(X, Y) ∈ Z[X, Y].

2 Consecutive purely ramified extensions

2.1 Setup

Let T |S and S|R be finite and purely ramified extensions of discrete valuation rings, of residue characteristic char R/rR = p. The maximal ideals of R, S and T are generated by r ∈ R, s ∈ S and t ∈ T, and the fields of fractions are denoted by K = frac R, L = frac S and M = frac T, respectively. Denote m = [M : L] and l = [L : K]. We may and will assume s = (−1)^{m+1}N_{M|L}(t) and r = (−1)^{l+1}N_{L|K}(s).

We have S = R[s] with

μ_{s,K}(X) = X^{l} + ( ∑_{j∈[1,l−1]} a_{j}X^{j}) − r ∈ R[X] ,

and T = R[t] with

μ_{t,K}(X) = X^{lm} + ( ∑_{j∈[1,lm−1]} b_{j}X^{j}) − r ∈ R[X] .

Cf. [9, I.§7, prop. 18]. The situation can be summarized in the diagram

[figure: rR ⊆ R; sS ⊆ S = R[s]; tT ⊆ T = S[t] = R[t]; K; l; L; m; M]

Note that r | p, and that for z ∈ M, we have v_{t}(z) = m · v_{s}(z) = ml · v_{r}(z).

2.2 Characteristic 0

In this section, we assume char K = 0. In particular, Z_{(p)} ⊆ R. Assumption 2.1 Suppose given x, y ∈ T and k ∈ [1, l − 1] such that

(i) p | y and t^{m} ≡_{y} s, (ii) x | ja_{j} for all j ∈ [1, l − 1], and (iii) xr | ja_{j} for all j ∈ [1, k − 1]. Put c := gcd(xys^{k−1}, yls^{l−1}) ∈ T. Lemma 2.2 Given (2.1), we have c | μ_{s,K}(t^{m}) . We may decompose

\begin{aligned}μ_{s,K}(t^{m}) & = & μ_{s,K}(t^{m}) − μ_{s,K}(s) \\ & = & (t^{ml} − s^{l}) + ( ∑_{j∈[1,k−1]} a_{j}(t^{mj} − s^{j})) + ( ∑_{j∈[k,l−1]} a_{j}(t^{mj} − s^{j})) . \end{aligned}

Now since t^{m} = s + zy for some z ∈ T by (2.1.i), we have

t^{mj} ≡^{(1.1)}_{jy^{2}} s^{j} + js^{j−1}zy ≡_{js^{j−1}y} s^{j}

for any j ⩾ 1, so that s^{j−1} | r | p | y gives t^{mj} ≡_{js^{j−1}y} s^{j}. In particular, yls^{l−1} | t^{ml} − s^{l}. Moreover, xys^{l} | ∑_{j∈[1,k−1]} a_{j}(t^{mj} − s^{j}) by (2.1.iii). Finally, xys^{k−1} | ∑_{j∈[k,l−1]} a_{j}(t^{mj} − s^{j}) by (2.1.ii). The following proposition will serve as inductive step in (3.2). Proposition 2.3 Given (2.1), we have t^{−j}c | b_{j} if j ≢_{m} 0 and t^{−j}c | (b_{j} − a_{j/m}) if j ≡_{m} 0, where j ∈ [1, lm − 1]. From (2.2) we take

∑_{j∈[1,lm−1]} (b_{j} − {j ≡_{m} 0} a_{j/m}) t^{j} = −μ_{s,K}(t^{m}) ≡_{c} 0 .

Since the summands have pairwise different valuations at t, we obtain

(b_{j} − {j ≡_{m} 0} a_{j/m}) t^{j} ≡_{c} 0

for all j ∈ [1, lm − 1].

2.3 As an illustration: cyclotomic polynomials

For n ⩾ 1, we choose primitive roots of unity ζ_{p^{n}} over Q in such a manner that ζ_{p^{n+1}}^{p} = ζ_{p^{n}}. We abbreviate ϑ_{n} = ζ_{p^{n}} − 1.

We shall show by induction on n that writing

μ_{ϑ_{n},Q}(X) = Φ_{p^{n}}(X + 1) = ∑_{j∈[0,p^{n−1}(p−1)]} d_{n,j}X^{j}

with d_{n,j} ∈ Z, we have p^{n−1} | jd_{n,j} for j ∈ [0, p^{n−1}(p − 1)], and even p^{n} | jd_{n,j} for j ∈ [0, p^{n−1}(p − 2)].

This being true for n = 1 since Φ_{p}(X + 1) = ((X + 1)^{p} − 1)/X, we assume it to be true for n − 1 and shall show it for n, where n ⩾ 2. We apply the result of the previous section to R = Z_{(p)}, r = −p, S = Z_{(p)}[ϑ_{n−1}], s = ϑ_{n−1} and T = Z_{(p)}[ϑ_{n}], t = ϑ_{n}. In particular, we have l = p^{n−2}(p−1) and μ_{s,K}(X) = Φ_{p^{n−1}}(X+1); we have m = p and μ_{t,L}(X) = (X+1)^{p}−1−ϑ_{n−1}; finally, we have μ_{t,K}(X) = Φ_{p^{n}}(X + 1).

We may choose y = pϑ_{n}, x = p^{n−2} and k = p^{n−2}(p−2)+1 in (2.1). Hence c = p^{n−1}ϑ_{n}^{p^{n}−2p^{n−1}+1}. By (2.3), we obtain that p^{n−1}ϑ_{n}^{p^{n}−2p^{n−1}+1−j} divides d_{n,j} − d_{n−1,j/p} if j ≡_{p} 0 and that it divides d_{n,j} if j ≢_{p} 0. Since the coefficients in question are in R, we may draw the following conclusion.

(I) \{ \begin{matrix}If j ≡_{p} 0, then p^{n} | d_{n,j} − d_{n−1,j/p} if j ⩽ p^{n−1}(p − 2), \\ and p^{n−1} | d_{n,j} − d_{n−1,j/p} if j > p^{n−1}(p − 2); \\ if j ≢_{p} 0, then p^{n} | d_{n,j} if j ⩽ p^{n−1}(p − 2), \\ and p^{n−1} | d_{n,j} if j > p^{n−1}(p − 2). \end{matrix}

By induction, this establishes the claim.

Using (1.4), assertion (I) also follows from the more precise relation

(II) Φ_{p^{n}}(X + 1) − Φ_{p^{n−1}}(X^{p} + 1) ≡_{p^{n}} X^{p^{n−1}(p−2)} ((X^{p} + 1)^{p^{n−2}} − (X + 1)^{p^{n−1}})

for n ⩾ 2, which we shall show now. In fact, by (1.4) we have (X + 1)^{p^{n}} ≡_{p^{n}} (X^{p} + 1)^{p^{n−1}} as well as (X^{p} + 1)^{p^{n−2}} − (X + 1)^{p^{n−1}} ≡_{p^{n−1}} 0, and so

((X + 1)^{p^{n}} − 1) ((X^{p} + 1)^{p^{n−2}} − 1) − ((X^{p} + 1)^{p^{n−1}} − 1) ((X + 1)^{p^{n−1}} − 1)
≡_{p^{n}} ((X^{p} + 1)^{p^{n−1}} − 1) ((X^{p} + 1)^{p^{n−2}} − 1) − ((X^{p} + 1)^{p^{n−1}} − 1) ((X + 1)^{p^{n−1}} − 1)
= ((X^{p} + 1)^{p^{n−1}} − 1) ((X^{p} + 1)^{p^{n−2}} − (X + 1)^{p^{n−1}})
≡_{p^{n}} X^{p^{n}} ((X^{p} + 1)^{p^{n−2}} − (X + 1)^{p^{n−1}})
= X^{p^{n−1}(p−2)} ((X^{p} + 1)^{p^{n−2}} − (X + 1)^{p^{n−1}}) · X^{p^{n−1}} · X^{p^{n−1}}
≡_{p^{n}} X^{p^{n−1}(p−2)} ((X^{p} + 1)^{p^{n−2}} − (X + 1)^{p^{n−1}}) ((X + 1)^{p^{n−1}} − 1) ((X^{p} + 1)^{p^{n−2}} − 1) ,

and the result follows by division by the monic polynomial

((X + 1)^{p^{n−1}} − 1) ((X^{p} + 1)^{p^{n−2}} − 1) .

Finally, we remark that writing F_{n}(X) := Φ_{p^{n}}(X + 1) + X^{p^{n}−2p^{n−1}}(X + 1)^{p^{n−1}}, we can equivalently reformulate (II) to

(II′)  F_{n}(X) ≡_{p^{n}} F_{n−1}(X^{p}) .

2.4 Characteristic p

In this section, we assume char K = p.

Assumption 2.4 Suppose given x, y ∈ T and k ∈ [1, l − 1] such that

(i) t^{m} ≡_{ys} s,

(ii) x | a_{j}y^{j[p]} for all j ∈ [1, l − 1], and

(iii) xr | a_{j}y^{j[p]} for all j ∈ [1, k − 1].

Let c := gcd(xs^{k}, y^{l[p]}s^{l}) ∈ T.

Lemma 2.5 Given (2.4), we have c | μ_{s,K}(t^{m}) .

We may decompose

\begin{aligned}μ_{s,K}(t^{m}) & = & μ_{s,K}(t^{m}) − μ_{s,K}(s) \\ & = & (t^{ml} − s^{l}) + ( ∑_{j∈[1,k−1]} a_{j}(t^{mj} − s^{j})) + ( ∑_{j∈[k,l−1]} a_{j}(t^{mj} − s^{j})) . \end{aligned}

Now since t^{m} ≡_{ys} s, we have t^{mj} ≡_{y^{j[p]}s^{j}} s^{j} for any j ⩾ 1.

In particular, y^{l[p]}s^{l} | t^{ml} − s^{l}.

Moreover, xs^{l} | ∑_{j∈[1,k−1]} a_{j}(t^{mj} − s^{j}) by (2.4.iii).

Finally, xs^{k} | ∑_{j∈[k,l−1]} a_{j}(t^{mj} − s^{j}) by (2.4.ii).

Proposition 2.6 Given (2.4), we have t^{−j}c | b_{j} if j ≢_{m} 0 and t^{−j}c | (b_{j} − a_{j/m}) if j ≡_{m} 0 for j ∈ [1, lm − 1].

This follows using (2.5), cf. (2.3).

3 Towers of purely ramified extensions

Suppose given a chain

R_{0} ⊆ R_{1} ⊆ R_{2} ⊆ ⋯

of finite purely ramified extensions R_{i+1}|R_{i} of discrete valuations rings, with maximal ideal generated by r_{i} ∈ R_{i}, of residue characteristic char R_{i}/r_{i}R_{i} = p, with field of fractions K_{i} = frac R_{i}, and of degree [K_{i+1} : K_{i}] = p^{κ} = q for i ⩾ 0, where κ ⩾ 1 is an integer stipulated to be independent of i. We may and will suppose that N_{K_{i+1}|K_{i}}(r_{i+1}) = r_{i} for i ⩾ 0. We write

μ_{r_{i},K_{0}}(X) = X^{q^{i}} + ( ∑_{j∈[1,q^{i}−1]} a_{i,j}X^{j}) − r_{0} ∈ R_{0}[X] .

For j ⩾ 1, we denote v_{q}(j) := max{α ∈ Z_{⩾0} : j ≡_{q^{α}} 0}. That is, v_{q}(j) is the largest integer below v_{p}(j)/κ. We abbreviate g := (q − 2)/(q − 1).

Assumption 3.1 Suppose given f ∈ R_{0} such that r_{i}^{q−1}f | r_{i}^{q} − r_{i−1} for all i ⩾ 0. If char K_{0} = 0, then suppose p | f | q. If char K_{0} = p, then suppose r_{0} | f.

Proposition 3.2 Assume (3.1).

(i) We have f^{i−v_{q}(j)} | a_{i,j} for i ⩾ 1 and j ∈ [1, q^{i} − 1].

(i′) If j < q^{i}g, then f^{i−v_{q}(j)}r_{0} | a_{i,j}.

(ii) We have a_{i,j} ≡_{f^{i+1}} a_{i+β,q^{β}j} for i ⩾ 1, j ∈ [1, q^{i} − 1] and β ⩾ 1.

(ii′) If j < q^{i}g, then a_{i,j} ≡_{f^{i+1}r_{0}} a_{i+β,q^{β}j} for β ⩾ 1.

Consider the case char K_{0} = 0. To prove (i, i′), we perform an induction on i, the assertion being true for i = 1 by (3.1). So suppose given i ⩾ 2 and the assertion to be true for i − 1. To apply (2.3), we let R = R_{0}, r = r_{0}, S = R_{i−1}, s = r_{i−1}, T = R_{i} and t = r_{i}. Furthermore, we let y = r_{i}^{q−1}f, x = f^{i−1} and k = q^{i−1} − (q^{i−1} − 1)/(q − 1), so that (2.1) is satisfied by (3.1) and by the inductive assumption. We have c = f^{i}r_{i}^{qk−1}.

Consider j ∈ [1, q^{i} − 1]. If j ≢_{q} 0, then (2.3) gives

v_{r_{i}}(a_{i,j}/f^{i}) ⩾ qk − 1 − j ,

whence f^{i} divides a_{i,j}; f^{i} strictly divides a_{i,j} if j < q^{i}g, since 0 < (qk−1)−q^{i}g = 1/(q−1) < 1.

If j ≡_{q} 0, then (2.3) gives

v_{r_{i}}((a_{i,j} − a_{i−1,j/q})/f^{i}) ⩾ qk − 1 − j ,

whence f^{i} divides a_{i,j} −a_{i−1,j/q}; strictly, if j < q^{i}g. By induction, f^{i−1−v_{q}(j/q)} divides a_{i−1,j/q}; strictly, if j/q < q^{i−1}g. But a_{i−1,j/q} ≡_{f^{i}} a_{i,j}, and therefore f^{i−v_{q}(j)} divides also a_{i,j}; strictly, if j < q^{i}g. This proves (i, i′).

The case β = 1 of (ii, ii′) has been established in the course of the proof of (i, i′). The general case follows by induction.

Consider the case char K_{0} = p. To prove (i, i′), we perform an induction on i, the assertion being true for i = 1 by (3.1). So suppose given i ⩾ 2 and the assertion to be true for i − 1. To apply (2.6), we let R = R_{0}, r = r_{0}, S = R_{i−1}, s = r_{i−1}, T = R_{i} and t = r_{i}. Furthermore, we let y = r_{i}^{−1}f, x = r_{i}^{−1}f^{i} and k = q^{i−1} − (q^{i−1} − 1)/(q − 1), so that (2.4) is satisfied by (3.1) and by the inductive assumption. In fact, xy^{−j[p]} = r_{i}^{j[p]−1}f^{i−j[p]} divides f^{i−1−v_{q}(j)} both if j ≢_{p} 0 and if j ≡_{p} 0; in the latter case we make use of the inequality p^{α−1}(p − 1) ⩾ α + 1 for α ⩾ 1, which needs p ⩾ 3. We obtain c = f^{i}r_{i}^{qk−1}.

Using (2.6) instead of (2.3), we may continue as in the former case to prove (i, i′), and, in the course of this proof, also (ii, ii′).

4 Galois descent of a divisibility

Let

[figure: S; G; S̃; m; m; T; G; T̃]

be a commutative diagram of finite, purely ramified extensions of discrete valuation rings. Let s ∈ S, t ∈ T, s̃ ∈ S̃ and t̃∈ T̃ generate the respective maximal ideals. Let L = frac S, M = frac T, L̃ = frac S̃ and M̃ = frac T̃ denote the respective fields of fractions. We assume the extensions M|L and L̃|L to be linearly disjoint and M̃ to be the composite of M and L̃. Thus m := [M : L] = [M̃ : L̃] and [L̃ : L] = [M̃ : M]. We assume L̃|L to be galois and identify G := Gal(L̃|L) = Gal(M̃|M) via restriction. We may and will assume that s = N_{L̃|L}(s̃), and that t = N_{M̃|M}(t̃).

Lemma 4.1 In T̃, the element 1 − t̃^{m}/s̃ divides 1 − t^{m}/s .

Let d̃= 1 − t̃^{m}/s̃, so that t̃^{m} = s̃(1 − d̃). We conclude

t^{m} = N_{M̃|M}(t̃^{m})
= N_{L̃|L}(s̃) · ∏_{σ∈G}(1 − d̃^{σ})
≡_{sd̃} s .

5 Cyclotomic number fields

5.1 Coefficient valuation bounds

For n ⩾ 1, we let ζ_{p^{n}} be a primitive p^{n}th root of unity over Q. We make choices in such a manner that ζ_{p^{n}}^{p} = ζ_{p^{n−1}} for n ⩾ 2. We denote ϑ_{n} = ζ_{p^{n}} − 1 and F_{n} = Q(ζ_{p^{n}}). Let E_{n} = Fix_{C_{p−1}}F_{n}, so [E_{n} : Q] = p^{n−1}. Let

π_{n} = N_{F_{n}|E_{n}}(ϑ_{n}) = ∏_{j∈[1, p−1]}(ζ_{p^{n}}^{j^{p^{n−1}}} − 1) .

The minimal polynomial μ_{ϑ_{n},F_{n−1}}(X) = (X +1)^{p}−ϑ_{n−1}−1 shows that N_{F_{n}|F_{n−1}}(ϑ_{n}) = ϑ_{n−1}, hence also N_{E_{n}|E_{n−1}}(π_{n}) = π_{n−1}. Note that π_{1} = p and E_{1} = Q.

Let O be the integral closure of Z_{(p)} in E_{n}. Since N_{E_{n}|Q}(π_{n}) = π_{1} = p, we have Z_{(p)}/pZ_{(p)} ^{\overline{}^{∼}} O/π_{n}O. In particular, the ideal π_{n}O in O is prime. Now π_{n}^{p^{n−1}}O = pO, since π_{n}^{p^{n−1}}/p = π_{n}^{p^{n−1}}/N_{E_{n}|Q}(π_{n}) ∈ Z_{(p)}[ϑ_{n}]^{∗} ∩ E_{n} = O^{∗}. Thus O is a discrete valuation ring, purely ramified of degree p^{n−1} over Z_{(p)}, and so O = Z_{(p)}[π_{n}] [9, I.§7, prop. 18]. In particular, E_{n} = Q(π_{n}).

Remark 5.1 The subring Z[π_{n}] of Q(π_{n}), however, is not integrally closed in general. For example, if p = 5 and n = 2, then μ_{π_{2},Q}(X) = X^{5} − 20X^{4} + 100X^{3} − 125X^{2} + 50X − 5 has discriminant 5^{8} · 7^{6}, which does not divide the discriminant of Φ_{5^{2}}(X), which is 5^{35}.

Lemma 5.2 We have π_{n}^{p} ≡_{π_{n}^{p−1}p} π_{n−1} for n ⩾ 2.

First of all, ϑ_{n}^{p} ≡_{ϑ_{n}p} ϑ_{n−1} since (X −1)^{p} −(X^{p} −1) is divisible by p(X −1) in Z[X]. Letting T̃ = Z_{(p)}[ϑ_{n}] and (t̃, s̃, t, s) = (ϑ_{n}, ϑ_{n−1}, π_{n}, π_{n−1}), (4.1) shows that 1 − ϑ_{n}^{p}/ϑ_{n−1} divides 1 − π_{n}^{p}/π_{n−1}. Therefore, ϑ_{n}p ϑ_{n−1}^{−1}π_{n−1} | π_{n−1} − π_{n}^{p}.

Now suppose given m ⩾ 1. To apply (3.2), we let f = q = p, R_{i} = Z_{(p)}[π_{m+i}] and r_{i} = π_{m+i} for i ⩾ 0. We keep the notation

μ_{π_{m+i}, E_{m}}(X) = μ_{r_{i}, K_{0}}(X) = X^{p^{i}} + ( ∑_{j∈[1,p^{i}−1]} a_{i,j}X^{j}) − π_{m} ∈ R_{0}[X] = Z_{(p)}[π_{m}][X] .

Theorem 5.3

(i) We have p^{i} | ja_{i,j} for i ⩾ 1 and j ∈ [1, p^{i} − 1].

(i′) If j < p^{i}(p − 2)/(p − 1), then p^{i}π_{m} | ja_{i,j}.

(ii) We have a_{i,j} ≡_{p^{i+1}} a_{i+β,p^{β}j} for i ⩾ 1, j ∈ [1, p^{i} − 1] and β ⩾ 1.

(ii′) If j < p^{i}(p − 2)/(p − 1), then a_{i,j} ≡_{p^{i+1}π_{m}} a_{i+β,p^{β}j}.

Assumption (3.1) is fulfilled by virtue of (5.2), whence the assertions follow by (3.2).

Example 5.4 For p = 5, m = 1 and i = 2, we have

μ_{π_{3},Q}(X) = X^{25} − 4 · 5^{2}X^{24} + 182 · 5^{2}X^{23} − 8 · 5^{6}X^{22} + 92823 · 5^{2}X^{21}
− 6175454 · 5X^{20} + 12194014 · 5^{2}X^{19} − 18252879 · 5^{3}X^{18}
+ 4197451 · 5^{5}X^{17} − 466901494 · 5^{3}X^{16} + 8064511079 · 5^{2}X^{15}
− 4323587013 · 5^{3}X^{14} + 1791452496 · 5^{4}X^{13} − 113846228 · 5^{6}X^{12}
+ 685227294 · 5^{5}X^{11} − 15357724251 · 5^{3}X^{10} + 2002848591 · 5^{4}X^{9}
− 4603857997 · 5^{3}X^{8} + 287207871 · 5^{4}X^{7} − 291561379 · 5^{3}X^{6}
+ 185467152 · 5^{2}X^{5} − 2832523 · 5^{3}X^{4} + 121494 · 5^{3}X^{3} − 514 · 5^{4}X^{2}
+ 4 · 5^{4}X − 5 .

Now v_{5}(a_{3,22}) = 6 ≠ 5 = v_{5}(a_{4,5·22}), so the valuations of the coefficients considered in (5.3.ii) differ in general. This, however, does not contradict the assertion a_{3,22} ≡_{5^{4}} a_{4, 5·22} from loc. cit.

5.2 A different proof of (5.3. i, i′) and some exact valuations

Let m ⩾ 1 and i ⩾ 0. We denote R_{i} = Z_{(p)}[π_{m+i}], r_{i} = π_{m+i}, K_{i} = frac R_{i}, R̃_{i} = Z_{(p)}[ϑ_{m+i}] and r̃_{i} = ϑ_{m+i}. Denoting by 𝔇 the respective different [9, III.§3], we have 𝔇_{R̃_{i}|R̃_{0}} = (p^{i}) and 𝔇_{R̃_{i}|R_{i}} = (r̃_{i}^{p−2}) [9, III.§3, prop. 13], whence

(∗)  𝔇_{R_{i}|R_{0}} = (μ′_{r_{i},K_{0}}(r_{i})) = 𝔇_{R̃_{i}|R̃_{0}}𝔇_{R̃_{0}|R_{0}}𝔇_{R̃_{i}|R_{i}}^{−1} = (p^{i}r_{i}^{p^{i}−1−(p^{i}−1)/(p−1)}) ,

cf. [9, III.§3, cor. 2]. Therefore, p^{i}r_{i}^{p^{i}−1−(p^{i}−1)/(p−1)} divides ja_{i,j}r_{i}^{j−1} for j ∈ [1, p^{i} − 1], and (5.3. i, i′) follow.

Moreover, since only for j = p^{i} − (p^{i} − 1)/(p − 1) the valuations at r_{i} of p^{i}r_{i}^{p^{i}−1−(p^{i}−1)/(p−1)} and ja_{i,j}r_{i}^{j−1} are congruent modulo p^{i}, we conclude by (∗) that they are equal, i.e. that p^{i} exactly divides a_{i,p^{i}−(p^{i}−1)/(p−1)}.

Corollary 5.5 The element p^{i−β} exactly divides a_{i,p^{i}−(p^{i}−p^{β})/(p−1)} for β ∈ [0, i − 1].

This follows by (5.3.ii) from what we have just said.

E.g. in (5.4), 5^{1} exactly divides a_{2,25−5} = a_{2,20}, and 5^{2} exactly divides a_{2,25−5−1} = a_{2,19}.

5.3 Some traces

Let μ_{p−1} denote the group of (p − 1)st roots of unity in Q_{p}. We choose a primitive (p − 1)st root of unity ζ_{p−1} ∈ μ_{p−1} and may thus view Q(ζ_{p−1}) ⊆ Q_{p} as a subfield. Note that [Q(ζ_{p−1}) : Q] = φ(p−1), where φ denotes Euler’s function. The restriction of the valuation v_{p} at p on Q_{p} to Q(ζ_{p}), is a prolongation of the valuation v_{p} on Q to Q(ζ_{p−1}) (there are φ(p − 1) such prolongations).

Proposition 5.6 For n ⩾ 1, we have

Tr_{E_{n}|Q}(π_{n}) = p^{n}s_{n} − p^{n−1}s_{n−1} ,

where

s_{n} := \frac{1}{p − 1} ∑_{H ⊆ μ_{p−1}}(−1)^{#H}\{v_{p}( ∑_{ξ∈H} ξ) ⩾ n\} for n ⩾ 0 .

We have s_{0} = 0, and s_{n} ∈ Z for n ⩾ 0. The sequence (s_{n})_{n} becomes stationary at some minimally chosen N_{0}(p). We have

N_{0}(p) ⩽ N(p) := max_{H ⊆ μ_{p−1}} {v_{p}(∑_{ξ∈H} ξ): ∑_{ξ∈H} ξ ≠ 0} + 1 .

An upper estimate for N(p), hence for N_{0}(p), is given in (5.13).

Proof of (5.6). For j ∈ [1, p − 1] the p -adic limits

ξ(j) := lim_{n→∞} j^{p^{n}}

exist since j^{p^{n−1}} ≡_{p^{n}} j^{p^{n}} by (1.3). They are distinct since ξ(j) ≡_{p} j, and, thus, form the group μ_{p−1} = {ξ(j) | j ∈ [1, p − 1]}. Using the formula

Tr_{F_{n}|Q}(ζ_{p^{n}}^{m}) = p^{n}{v_{p}(m) ⩾ n} − p^{n−1}{v_{p}(m) ⩾ n − 1}

and the fact that j^{p^{n−1}} ≡_{p^{n}} ξ(j), we obtain

Tr_{F_{n}|Q}(π_{n}) = Tr_{F_{n}|Q}( ∏_{j∈[1, p−1]} (1 − ζ_{p^{n}}^{j^{p^{n−1}}}))
= ∑_{J ⊆ [1, p−1]}(−1)^{#J} Tr_{F_{n}|Q}(ζ_{p^{n}}^{∑_{j∈J} j^{p^{n−1}}})
= ∑_{J ⊆ [1, p−1]}(−1)^{#J}(p^{n} {v_{p}( ∑_{j∈J} ξ(j)) ⩾ n}
− p^{n−1}{v_{p}( ∑_{j∈J} ξ(j)) ⩾ n − 1})
= (p − 1)(p^{n}s_{n} − p^{n−1}s_{n−1}) ,

whence

Tr_{E_{n}|Q}(π_{n}) = p^{n}s_{n} − p^{n−1}s_{n−1} .

Now s_{0} = 0 ∈ Z by the binomial formula. Therefore, by induction, we conclude from p^{n}s_{n} − p^{n−1}s_{n−1} ∈ Z that p^{n}s_{n} ∈ Z. Since (p − 1)s_{n} ∈ Z, too, we obtain s_{n} ∈ Z.

As soon as n ⩾ N(p), the conditions v_{p}(∑_{ξ∈H} ξ) ⩾ n and v_{p}(∑_{ξ∈H} ξ) = +∞ on H ⊆ μ_{p−1} become equivalent, and we obtain

s_{n} = \frac{1}{p − 1} ∑_{H⊆μ_{p−1}}(−1)^{#H}\{∑_{ξ∈H} ξ = 0\} ,

which is independent of n. Thus N_{0}(p) ⩽ N(p).

Lemma 5.7 We have s_{1} = 1. In particular, Tr_{E_{2}|Q}(π_{2}) ≡_{p^{2}} −p .

Since Tr_{E_{1}|Q}(π_{1}) = Tr_{Q|Q}(p) = p, and since s_{0} = 0, we have s_{1} = 1 by (5.6). The congruence for Tr_{E_{2}|Q}(π_{2}) follows again by (5.6).

Corollary 5.8 We have

μ_{π_{n},Q}(X) ≡_{p^{2}} X^{p^{n−1}} + pX^{(p−1)p^{n−2}} − p

for n ⩾ 2.

By dint of (5.7), this ensues from (5.3. i′, ii).

Example 5.9 The last n for which we list s_{n} equals N(p), except if there is a question mark in the next column. The table was calculated using Pascal (p ⩽ 53) and Magma (p ⩾ 59). In the last column, we list the upper bound for N(p) calculated below (5.13).

s_{n} n = 0 1  2  3  4  5  6  7 upper bound ; for N(p)
p = 3  0 1
5  0 1  1
7  0 1  1
11  0 1  3  3
13  0 1  3  3
17  0 1  8  16  5
19  0 1  10  12  4
23  0 1  33  89  93  7
29  0 1  377  571  567  8
31  0 1  315  271  259  6
37  0 1  107  940  1296  9
41  0 1  6621  51693  18286  20186  20250  12
43  0 1  1707  4767  6921  6665  9
47  0 1  2250 272242  173355  181481  182361  16
53  0 1  71201 363798 1520045 1350049 1292229 1289925  18
59  0 1  1276  ?  21
61  0 1  2516  ?  12
67  0 1  407186  ?  15
71  0 1  5816605  ?  18
73  0 1  8370710  ?  18
79  0 1  169135  ?  18
83  0 1  632598  ?  30
89  0 1  26445104  ?  30
97  0 1  282789  ?  24
101  0 1  25062002  ?  31
103  0 1  56744199  ?  25
107  0 1  1181268305  ?  40
109  0 1  91281629  ?  28
113  0 1  117774911422  ?  37
127  0 1  6905447  ?  28
131  0 1  2988330952791  ?  37
137  0 1  1409600547  ?  50
139  0 1  3519937121  ?  34
149  0 1  25026940499  ?  56
151  0 1  164670499159  ?  31
157  0 1  51594129045351  ?  38
163  0 1  288966887341  ?  42
167  0 1  1205890070471  ?  64
173  0 1  17802886165762  ?  66
179  0 1  1311887715966  ?  69
181  0 1  128390222739  ?  38
191  0 1 233425263577158  ?  57
193  0 1 306518196952028  ?  51
197  0 1 347929949728221  ?  66
199  0 1  9314622093145  ?  48
211  0 1  12532938009082  ?  39

So for example if p = 31, then Tr_{Q(π_{3})|Q}(π_{3}) = 271 · 31^{3} − 315 · 31^{2}, whereas Tr_{Q(π_{7})|Q}(π_{7}) = 259 · 31^{7} − 259 · 31^{6}. Moreover, N_{0}(31) = N(31) = 4 ⩽ 6.

Remark 5.10 Vanishing (resp. vanishing modulo a prime) of sums of roots of unity has been studied extensively. See e.g. [2], [6], where also further references may be found.

Remark 5.11 Neither do we know whether s_{n} ⩾ 0 nor whether Tr_{E_{n}|Q}(π_{n}) ⩾ 0 always hold. Moreover, we do not know a prime p for which N_{0}(p) < N(p).

Remark 5.12 We calculated some further traces appearing in (5.3), using Maple and Magma. For p = 3, n ∈ [2, 10], we have Tr_{E_{n}|E_{n−1}}(π_{n}) = 3 · 2. For p = 5, n ∈ [2, 6], we have Tr_{E_{n}|E_{n−1}}(π_{n}) = 5 · 4. For p = 7, n ∈ [2, 5], we have Tr_{E_{n}|E_{n−1}}(π_{n}) = 7 · 6. For p = 11, we have Tr_{E_{2}|E_{1}}(π_{2}) = 11 · 32, whereas

Tr_{E_{3}|E_{2}}(π_{3})
= 22 · (15 + ζ^{2} + 2ζ^{3} − ζ^{5} + ζ^{6} − 2ζ^{8} − ζ^{9} + 2ζ^{14} − ζ^{16} + ζ^{18} − ζ^{20} − 2ζ^{24}
+2ζ^{25} − 2ζ^{26} − ζ^{27} − ζ^{31} + 2ζ^{36} − ζ^{38} + ζ^{41} − ζ^{42} − 2ζ^{43} + 2ζ^{47} − 3ζ^{49}
−ζ^{53} + ζ^{54} + 2ζ^{58} − ζ^{60} − ζ^{64} + ζ^{67} + 2ζ^{69} − ζ^{71} − 2ζ^{72} − ζ^{75} − 2ζ^{78}
+3ζ^{80} − ζ^{82} − ζ^{86} + 2ζ^{91} − ζ^{93} − 2ζ^{95} − 3ζ^{97} + 2ζ^{102} + ζ^{103} − ζ^{104} − ζ^{108})
= 22 · 2014455354550939310427^{−1} · (34333871352527722810654
+1360272405267541318242502π − 31857841148164445311437042π^{2}
+135733708409855976059658636π^{3} − 83763613130017142371566453π^{4}
+20444806599344408104299252π^{5} − 2296364631211442632168932π^{6}
+117743741083866218812293π^{7} − 2797258465425206085093π^{8}
+27868038642441136108π^{9} − 79170513243924842π^{10}) ,

where ζ := ζ_{11^{2}} and π := π_{2}.

5.4 An upper bound for N(p)

We view Q(ζ_{p−1}) as a subfield of Q_{p}, and now, in addition, as a subfield of C. Since complex conjugation commutes with the operation of Gal(Q(ζ_{p−1})|Q), we have |N_{Q(ζ_{p−1})|Q}(x)| = |x|^{φ(p−1)} for x ∈ Q(ζ_{p−1}). We abbreviate Σ(H) := ∑_{ξ∈H} ξ for H ⊆ μ_{p−1}. Since |Σ(H)| ⩽ p − 1, we have |N_{Q(ζ_{p−1})|Q}(Σ(H))| ⩽ (p − 1)^{φ(p−1)}. Hence, if Σ(H) ≠ 0, then

v_{p}(Σ(H)) ⩽ v_{p}(N_{Q(ζ_{p−1})|Q}(Σ(H))) < φ(p − 1) ,

and therefore N(p) ⩽ φ(p − 1). We shall ameliorate this bound by a logarithmic term.

Proposition 5.13 We have

N(p) ⩽ φ(p − 1) (1 − \frac{log π}{log p} ) + 1

for p ⩾ 5.

It suffices to show that |Σ(H)| ⩽ p/π for H ⊆ μ_{p−1}. We will actually show that

max_{H⊆μ_{p−1}} |Σ(H)| = \frac{1}{sin \frac{π}{p−1}} ,

from which this inequality follows using sin x ⩾ x − x^{3}/6 and p ⩾ 5.

Choose H ⊆ μ_{p−1} such that |Σ(H)| is maximal. Since p − 1 is even, the (p − 1)st roots of unity fall into pairs (η, −η). The summands of Σ(H) contain exactly one element of each such pair, since |Σ(H) + η|^{2} + |Σ(H) − η|^{2} = 2|Σ(H)|^{2} + 2 shows that at least one of the inequalities |Σ(H) + η| ⩽ |Σ(H)| and |Σ(H) − η| ⩽ |Σ(H)| fails.

By maximality, replacing a summand η by −η in Σ(H) does not increase the value of |Σ(H)|, whence

|Σ(H)|^{2} ⩾ |Σ(H) − 2η|^{2} = |Σ(H)|^{2} − 4 Re(η · \overline{Σ(H)}) + 4 ,

and thus

Re(η · \overline{Σ(H)}) ⩾ 1 > 0 .

Therefore, the (p − 1)/2 summands of Σ(H) lie in one half-plane, whence the value of |Σ(H)|.

6 Cyclotomic function fields, after Carlitz and Hayes

6.1 Notation and basic facts

We shall give a brief review while fixing notation.

Let ρ ⩾ 1 and r := p^{ρ}. Write Z := F_{r}[Y] and Q := F_{r}(Y), where Y is an independent variable. We fix an algebraic closure Q̄ of Q. The Carlitz module structure on Q̄ is defined by the F_{r}-algebra homomorphism given on the generator Y as

\begin{matrix}Z & \overline{} & End_{Q}Q̄ \\ Y & \overline{} & (ξ \overline{}ξ^{Y} := Y ξ + ξ^{r} ) . \end{matrix}

We write the module product of ξ ∈ Q̄ with e ∈ Z as ξ^{e}. For each e ∈ Z, there exists a unique polynomial P_{e}(X) ∈ Z[X] that satisfies P_{e}(ξ) = ξ^{e} for all ξ ∈ Q̄. In fact, P_{1}(X) = X, P_{Y} (X) = Y X + X^{r}, and P_{Y^{i+1}} = Y P_{Y^{i}}(X) + P_{Y^{i}}(X^{r}) for i ⩾ 1. For a general e ∈ Z, the polynomial P_{e}(Y) is given by the according linear combination of these.

Note that P_{e}(0) = 0, and that P′_{e}(X) = e, whence P_{e}(X) is separable, i.e. it decomposes as a product of distinct linear factors in Q̄[X]. Let

λ_{e} = ann_{e}Q̄ = {ξ ∈ Q̄ : ξ^{e} = 0} ⊆ Q̄

be the annihilator submodule. Separability of P_{e}(X) shows that #λ_{e} = deg P_{e}(X) = r^{deg e}. Given a Q-linear automorphism σ of Q̄, we have (ξ^{e})^{σ} = P_{e}(ξ)^{σ} = P_{e}(ξ^{σ}) = (ξ^{σ})^{e}. In particular, λ_{e} is stable under σ. Therefore, Q(λ_{e}) is a Galois extension of Q.

Since #annẽλ_{e} = #λẽ = r^{deg ẽ} for ẽ | e, we have λ_{e} ≃ Z/e as Z-modules. It is not possible, however, to distinguish a particular isomorphism.

We shall restrict ourselves to prime powers now. We fix a monic irreducible polynomial f = f(Y) ∈ Z and write q := r^{deg f}. For n ⩾ 1, we let θ_{n} be a Z-linear generator of λ_{f^{n}}. We make our choices in such a manner that θ_{n+1}^{f} = θ_{n} for n ⩾ 1. Note that Z[λ_{f^{n}}] = Z[θ_{n}] since the elements of λ_{f^{n}} are polynomial expressions in θ_{n}.

Suppose given two roots ξ, ξ̃ ∈ Q̄ of

Ψ_{f^{n}}(X) := P_{f^{n}}(X)/P_{f^{n−1}}(X) ∈ Z[X] ,

i.e. ξ, ξ̃ ∈ λ_{f^{n}} ∖ λ_{f^{n−1}}. Since ξ is a Z-linear generator of λ_{f^{n}}, there is an e ∈ Z such that ξ̃ = ξ^{e}. Since ξ^{e}/ξ = P_{e}(X)/X|_{X=ξ} ∈ Z[θ_{n}], ξ̃ is a multiple of ξ in Z[θ_{n}]. Reversing the argument, we see that ξ̃ is in fact a unit multiple of ξ in Z[θ_{n}].

Lemma 6.1 The polynomial Ψ_{f^{n}}(X) is irreducible.

We have Ψ_{f^{n}}(0) = \frac{P_{f^{n}}(X)/X}{P_{f^{n−1}}(X)/X} |_{X=0} = f. We decompose Ψ_{f^{n}}(X) = ∏_{i∈[1,k]} F_{i}(X) in its

distinct monic irreducible factors F_{i}(X) ∈ Z[X]. One of the constant terms, say F_{j}(0), is thus a unit multiple of f in Z, while the other constant terms are units. Thus, being conjugate under the Galois action, all roots of F_{j}(X) in Q[θ_{n}] are non-units in Z[θ_{n}], and the remaining roots of Ψ_{f^{n}}(X) are units. But all roots of Ψ_{f^{n}}(X) are unit multiples of each other. We conclude that Ψ_{f^{n}}(X) = F_{j}(X) is irreducible.

By (6.1), Ψ_{f^{n}}(X) is the minimal polynomial of θ_{n} over Q. In particular, [Q(θ_{n}) : Q] = q^{n−1}(q − 1), and so

Z[θ_{n}]θ_{n}^{(q−1)q^{n−1}} = Z[θ_{n}]N_{Q(θ_{n})|Q}(θ_{n}) = Z[θ_{n}]f .

In particular, Z_{(f)}[θ_{n}] is a discrete valuation ring with maximal ideal generated by θ_{n}, purely ramified of index q^{n−1}(q −1) over Z_{(f)}, cf. [9, I.§7, prop. 18]. There is a group isomorphism

\begin{matrix}(Z/f^{n})^{∗} & ^{\overline{}^{∼}} & Gal(Q(θ_{n})|Q) \\ e & \overline{} & (θ_{n} \overline{}θ_{n}^{e}) , \end{matrix}

well defined since θ_{n}^{e} is a root of Ψ_{f^{n}}(X), too; injective since θ_{n} generates λ_{f^{n}} over Z; and surjective by cardinality.

Note that the Galois operation on Q(θ_{n}) corresponding to e ∈ (Z/f^{n})^{∗} coincides with the module operation of e on the element θ_{n}, but not everywhere. For instance, if f ≠ Y, then the Galois operation corresponding to Y sends 1 to 1, whereas the module operation of Y sends 1 to Y + 1.

The discriminant of Z[θ_{n}] over Z is given by Δ_{Z[θ_{n}]|Z} = N_{Q(θ_{n})|Q}(Ψ′_{f^{n}}(θ_{n})) = N_{Q(θ_{n})|Q} (P′_{f^{n}}(θ_{n})/P_{f^{n−1}}(θ_{n})) = N_{Q(θ_{n})|Q} (f^{n}/θ_{1}) = f^{q^{n−1}(nq−n−1)} .

Lemma 6.2 The ring Z[θ_{n}] is the integral closure of Z in Q(θ_{n}).

Let e ∈ Z be a monic irreducible polynomial different from f. Write O_{0} := Z_{(e)}[θ_{n}] and let O be the integral closure of O_{0} in Q(θ_{n}). Let

\begin{aligned}O_{0}^{+} & := & \{ξ ∈ Q(θ_{n}) : Tr_{Q(θ_{n})|Q}(ξO_{0}) ⊆ Z_{(e)}\} \\ O^{+} & := & \{ξ ∈ Q(θ_{n}) : Tr_{Q(θ_{n})|Q}(ξO) ⊆ Z_{(e)}\} . \end{aligned}

Then O_{0} ⊆ O ⊆ O^{+} ⊆ O_{0}^{+}. But O_{0} = O_{0}^{+}, since the Z_{(e)}-linear determinant of this embedding is given by the discriminant Δ_{Z[θ_{n}]|Z}, which is a unit in O_{0}.

We resume.

Proposition 6.3 ([1],[5], cf. [3, p. 115]) The extension Q(θ_{n})|Q is galois of degree [Q(θ_{n}) : Q] = (q − 1)q^{n−1}, with Galois group isomorphic to (Z/f^{n})^{∗}. The integral closure of Z in Q(θ_{n}) is given by Z[θ_{n}]. We have Z[θ_{n}]θ_{n}^{[Q(θ_{n}):Q]} = Z[θ_{n}]f. In particular, θ_{n} is a prime element of Z[θ_{n}], and the extension Z_{(f)}[θ_{n}]|Z_{(f)} of discrete valuation rings is purely ramified.

6.2 Coefficient valuation bounds

Denote F_{n} = Q(θ_{n}). Let E_{n} = Fix_{C_{q−1}}F_{n}, so [E_{n} : Q] = q^{n−1}. Let

ϖ_{n} = N_{F_{n}|E_{n}}(θ_{n}) = ∏_{e∈(Z/f)^{∗}} θ_{n}^{e^{q^{n−1}}} .

The minimal polynomial μ_{θ_{n}, F_{n−1}}(X) = P_{f}(X) − θ_{n−1} together with X | P_{f}(X) shows that N_{F_{n}|F_{n−1}}(θ_{n}) = θ_{n−1}, whence N_{E_{n}|E_{n−1}}(ϖ_{n}) = ϖ_{n−1}. Note that ϖ_{1} = ∏_{e∈(Z/f)^{∗}} θ_{1}^{e} = Ψ_{f}(0) = f.

The extension Z_{(f)}[ϖ_{n}] is a discrete valuation ring with maximal ideal generated by ϖ_{n}, purely ramified of index q^{n−1} over Z_{(f)}. In particular, E_{n} = Q(ϖ_{n}).

Example 6.4 Let r = 3 and f(Y) = Y^{2} + 1, so q = 9. A Magma calculation shows that

ϖ_{2} = θ_{2}^{60} − Y θ_{2}^{58} + Y^{2}θ_{2}^{56} + (−Y^{9}−Y^{3}−Y)θ_{2}^{42} + (Y^{10}+Y^{4}+Y^{2}+1)θ_{2}^{40} + (−Y^{11}−Y^{5}−Y^{3}+Y)θ_{2}^{38} + (−Y^{6}−Y^{4}−Y^{2})θ_{2}^{36} + (Y^{7}+Y^{5}+Y^{3}+Y)θ_{2}^{34} + (−Y^{8}−Y^{6}+Y^{4}−Y^{2}−1)θ_{2}^{32} + (−Y^{5}+Y^{3}−Y)θ_{2}^{30} + (Y^{18}−Y^{12}−Y^{10}+Y^{6}−Y^{4}+Y^{2})θ_{2}^{24} + (−Y^{19}+Y^{13}+Y^{11}+Y^{9}−Y^{7}+Y^{5}+Y)θ_{2}^{22} + (Y^{20}−Y^{14}−Y^{12}+Y^{10}+Y^{8}−Y^{6}−Y^{4}+Y^{2}+1)θ_{2}^{20} + (−Y^{15}−Y^{13}−Y^{11}−Y^{9}+Y^{7}+Y^{5}−Y^{3})θ_{2}^{18} + (Y^{16}+Y^{14}+Y^{12}−Y^{10}−Y^{8}−Y^{2})θ_{2}^{16} + (−Y^{17}−Y^{15}+Y^{13}+Y^{11}+Y^{7}+Y^{5}−Y^{3}+Y)θ_{2}^{14} + (−Y^{14}−Y^{12}+Y^{10}−Y^{8}−Y^{6}−Y^{4}+Y^{2}+1)θ_{2}^{12} + (−Y^{13}+Y^{11}−Y^{7}+Y^{3})θ_{2}^{10} + (Y^{14}−Y^{12}−Y^{10}+Y^{6}+Y^{4})θ_{2}^{8} + (−Y^{11}−Y^{7}+Y^{5}+Y^{3}+Y)θ_{2}^{6} + (Y^{8}+Y^{6}+Y^{2}+1)θ_{2}^{4} .

With regard to §6.4, we remark that ϖ_{2} ≠ ± θ_{2}^{q−1}.

Lemma 6.5 We have ϖ_{n}^{q} ≡_{ϖ_{n}^{q−1}f} ϖ_{n−1} for n ⩾ 2.

We claim that θ_{n}^{q} ≡_{θ_{n}f} θ_{n−1}. In fact, the non-leading coefficients of the Eisenstein polynomial Ψ_{f}(X) are divisible by f, so that the congruence follows by θ_{n−1} − θ_{n}^{q} = P_{f}(θ_{n}) − θ_{n}^{q} = θ_{n}(Ψ_{f}(θ_{n}) − θ_{n}^{q−1}). Letting T̃ = Z_{(f)}[θ_{n}] and (t̃, s̃, t, s) = (θ_{n}, θ_{n−1}, ϖ_{n}, ϖ_{n−1}), (4.1) shows that 1 − θ_{n}^{q}/θ_{n−1} divides 1 − ϖ_{n}^{q}/ϖ_{n−1}. Therefore, θ_{n}fθ_{n−1}^{−1}ϖ_{n−1} | ϖ_{n−1} − ϖ_{n}^{q}.

Now suppose given m ⩾ 1. To apply (3.2), we let R_{i} = Z_{(f)}[ϖ_{m+i}] and r_{i} = ϖ_{m+i} for i ⩾ 0. We continue to denote

(#)  \begin{matrix}μ_{ϖ_{m+i}, E_{m}}(X) = μ_{r_{i}, K_{0}}(X) = X^{q^{i}} + ( ∑_{j∈[1,q^{i}−1]} a_{i,j}X^{j}) − ϖ_{m} \\ ∈ R_{0}[X] = Z_{(f)}[ϖ_{m}][X] , \end{matrix}

and v_{q}(j) = max{α ∈ Z_{⩾0} : j ≡_{q^{α}} 0 }.

Theorem 6.6

(i) We have f^{i−v_{q}(j)} | a_{i,j} for i ⩾ 1 and j ∈ [1, q^{i} − 1].

(i′) If j < q^{i}(q − 2)/(q − 1), then f^{i−v_{q}(j)}ϖ_{m} | a_{i,j}.

(ii) We have a_{i,j} ≡_{f^{i+1}} a_{i+β,q^{β}j} for i ⩾ 1, j ∈ [1, q^{i} − 1] and β ⩾ 1.

(ii′) If j < q^{i}(q − 2)/(q − 1), then a_{i,j} ≡_{f^{i+1}ϖ_{m}} a_{i+β,q^{β}j} for β ⩾ 1.

Assumption (3.1) is fulfilled by virtue of (6.5), whence the assertions follow by (3.2).

6.3 Some exact valuations

Let m ⩾ 1 and i ⩾ 0. We denote R_{i} = Z_{(f)}[ϖ_{m+i}], r_{i} = ϖ_{m+i}, K_{i} = frac R_{i}, R̃_{i} = Z_{(f)}[θ_{m+i}] and r̃_{i} = θ_{m+i}. We obtain 𝔇_{R̃_{i}|R̃_{0}} = (f^{i}) and 𝔇_{R̃_{i}|R_{i}} = (r̃_{i}^{q−2}) [9, III.§3, prop. 13], whence

(∗∗)  𝔇_{R_{i}|R_{0}} = (μ′_{r_{i},K_{0}}(r_{i})) = (f^{i}r_{i}^{q^{i}−1−(q^{i}−1)/(q−1)}) .

Therefore, f^{i}r_{i}^{q^{i}−1−(q^{i}−1)/(q−1)} divides ja_{i,j}r_{i}^{j−1} for j ∈ [1, q^{i} −1], which is an empty assertion if j ≡_{p} 0. Thus (6.6. i, i′) do not follow entirely.

However, since only for j = q^{i} − (q^{i} − 1)/(q − 1) the valuations at r_{i} of f^{i}r_{i}^{q^{i}−1−(q^{i}−1)/(q−1)} and ja_{i,j}r_{i}^{j−1} are congruent modulo q^{i}, we conclude by (∗∗) that they are equal, i.e. that f^{i} exactly divides a_{i,q^{i}−(q^{i}−1)/(q−1)}.

Corollary 6.7 The element f^{i−β} exactly divides a_{i,q^{i}−(q^{i}−q^{β})/(q−1)}.

This follows by (6.6.ii) from what we have just said.

6.4 A simple case

Suppose that f(Y) = Y and m ⩾ 1. Note that

ϖ_{m+1} = ∏_{e∈F_{q}^{∗}} θ_{m+1}^{e} = ∏_{e∈F_{q}^{∗}} eθ_{m+1} = −θ_{m+1}^{q−1} .

Lemma 6.8 We have

μ_{ϖ_{m+1}, E_{m}}(X) = −ϖ_{m} + ∑_{j∈[1,q]} Y^{q−j}X^{j} .

Using the minimal polynomial μ_{θ_{m+1}, F_{m}}(X) = P_{Y} (X) − θ_{m} = X^{q} + Y X − θ_{m}, we get

−ϖ_{m} + ∑_{j∈[1,q]} Y^{q−j}ϖ_{m+1}^{j}
= θ_{m}^{q−1} + (Y^{q+1} − θ_{m+1}^{q^{2}−1})/(Y + θ_{m+1}^{q−1}) − Y^{q}
= (Y θ_{m}^{q−1}θ_{m+1} + θ_{m}^{q−1}θ_{m+1}^{q} − θ_{m+1}^{q^{2}} − Y^{q}θ_{m+1}^{q})/(θ_{m+1}(Y + θ_{m+1}^{q−1}))
= 0 .

Corollary 6.9 Let m, i ⩾ 1. We have

μ_{ϖ_{m+i}, E_{m}}(X) ≡_{Y^{2}} X^{q^{i}} + Y X^{(q−1)q^{i−1}} − ϖ_{m} .

This follows from (6.8) using (6.6.ii).

Remark 6.10 The assertion of (6.8) also holds if p = 2.

Conjecture 6.11 Let m, i ⩾ 1. We use the notation of (#) above, now in the case f(Y) = Y. For j ∈ [1, q^{i}], we write q^{i} − j = ∑_{k∈[0,i−1]} d_{k}q^{k} with d_{k} ∈ [0, q − 1]. Consider the following conditions.

(i) There exists k ∈ [0, i − 2] such that d_{k+1} < d_{k}.

(ii) There exists k ∈ [0, i − 2] such that v_{p}(d_{k+1}) > v_{p}(d_{k}).

If (i) or (ii) holds, then a_{i,j} = 0. If neither (i) nor (ii) holds, then

v_{ϖ_{m}}(a_{i,j}) = q^{m−1} · ∑_{k∈[0,i−1]} d_{k} .

Remark 6.12 We shall compare (6.7) with (6.11). If j = q^{i} − (q^{i} − q^{β})/(q − 1) for some β ∈ [0, i − 1], then q^{i} − j = q^{i−1} + ⋯ + q^{β}. Hence ∑_{k∈[0,i−1]} d_{k} = i − β, and so according to (6.11), v_{ϖ_{m}}(a_{i,j}) should equal q^{m−1}(i − β), which is in fact confirmed by (6.7).

References

[1] Carlitz, L., A class of polynomials, Trans. Am. Math. Soc. 43 (2), p. 167–182, 1938.

[2] Dvornicich, R.; Zannier, U., Sums of roots of unity vanishing modulo a prime, Arch. Math. 79, p. 104–108, 2002.

[3] Goss, D., The arithmetic of function fields. II. The ”cyclotomic” theory, J. Alg. 81 (1), p. 107–149, 1983.

[4] Goss, D., Basic structures of function field arithmetic, Springer, 1996.

[5] Hayes, D. R., Explicit class field theory for rational function fields, Trans. Am. Math. Soc. 189 (2), p. 77–91, 1974.

[6] Lam, T. Y.; Leung, K. H., On Vanishing Sums of Roots of Unity, J. Alg. 224, p. 91–109, 2000.

[7] Neukirch, J., Algebraische Zahlentheorie, Springer, 1992.

[8] Rosen, M., Number Theory in Function Fields, Springer GTM 210, 2000.

[9] Serre, J.P., Corps Locaux, Hermann, 1968.

[10] Weber, H., Künzer, M., Some additive galois cohomology rings, math.NT/0102048, to appear in Comm. Alg., 2004.

Matthias Künzer Universität Ulm

Abt. Reine Mathematik

D-89069 Ulm

kuenzer@mathematik.uni-ulm.de

Eduard Wirsing Universität Ulm Fak. f. Mathematik

D-89069 Ulm

ewirsing@mathematik.uni-ulm.de

(M. K., 24.02.2010.) Recently, A. Lbekkouri [L, On the construction of normal wildly ramified extensions over Q_{p} (p ≠ 2), Arch. Math. 93, p. 331–344, 2009], using work of M. Krasner [K, Sur la primitivité des corps 𝔭-adiques, Math., Cluj 13, p. 72–191, 1937], has derived congruences for the coefficients of the Eisenstein polynomial under consideration from the normality of the extension, extending a result of Ø. Ore [O, Abriß einer arithmetischen Theorie der Galoisschen Körper, Math. Ann. 100 p. 650– 673, 1928]; cf. [O; Satz 6], [L; Th. 4.1, Th. 4.6]. In [L; Th. 4.6], nonlinear congruences appear ! Furthermore, cf. [L; Prop. 3.2].