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By Guanghua Ji

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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 , .

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Algebraic Number Theory by Guanghua Ji


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