By Guanghua Ji

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From the PREFACE. THE ebook of this tract has been not on time via numerous explanations, and i'm now pressured to factor it with out Dr Riesz's assist in the ultimate correction of the proofs. This has at any expense, one virtue, that it offers me the potential for asserting how unsleeping i'm that no matter what worth it possesses is due quite often to his contributions to it, and particularly to the actual fact, that it comprises the 1st systematic, account of his appealing idea of the summation of sequence by way of 'typical means'.

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Proof. We first note that (p, f1 (α))e1 · · · (p, fg (α))eg ⊂ pok . Thus it suffices to show that pi = (p, fi (α)) are prime ideals, with N (pi ) = pdi , di = deg fi . Now, since fi (x) is irreducible over Fp [X], then Fp [X]/(fi (x)) is a field. Also Z[X]/(p) ∼ = Fp [X] ⇒ Z[X]/(p, fi (x)) ∼ = Fp [X]/(fi (x)), and so Z[X]/(p, fi (x)) is a field. Consider the map ϕ : Z[X] → Z[α]/(p, fi (α)), Clearly (p, fi (x)) ⊂ ker(ϕ) = {n(x) : n(α) ∈ (p, fi (α))} If n(x) ∈ ker(ϕ), we can divide by fi (x) to get n(x) = q(x)fi (x) + ri (x), deg(ri ) < deg(fi ) We assume that ri is nonzero, for otherwise the result is trivial.

1 αn · · · σr1 αn σr1 +1 αn · · · is the absolute value of the determi- 23 σr1 +r2 α1 σr1 +r2 α2 σr1 +1 α1 · · · σr1 +1 α2 · · · σr1 +r2 αn σr1 +1 αn · · ·  σr1 +r2 α1 σr1 +r2 α2    σr1 +r2 αn By doing the column operators, we have V (σ(a)) = | det M | = (−2i)−r2 = 2−r2 dk/Q (α1 , α2 , · · · , αn ) |dk |N (a). Since det M = 0, show that σα1 , . . , σαn are R-linearly independent. It follows that σ(a) is a lattice in Rn . 2 The class number Let Pk denote the subgroup of Jk formed by the principal fractional ideals, that is, ideals of the form (α) = αok , for every α ∈ k × .

Since the λ(xi ) are linearly independent, so are the xi (provided that the notion of linear independence is translated to a multiplicative setting: x1 , . . , xs are multiplicatively independent if ms 1 xm = 1 implies that mi = 0 for all i, from which it follows that 1 · · · xs m1 ms x1 · · · xs = xn1 1 · · · xns s implies mi = ni for all i). The result follows. We now improve the estimate of s and show that s ≤ r1 + r2 − 1. so as above, λ(Uk ) is free Z-module of rank s ≤ r1 + r2 − 1. We call { 1 , .