## Bernoulli Polynomials by Tony Forbes

By Tony Forbes

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Nn ) |d|1/2 N (J). Cancel N (J) to get the desired result. 6 Corollary The ideal class group is ﬁnite. Proof. 13), there are only ﬁnitely many integral ideals with a given norm. 5), we can associate with each ideal class an integral ideal whose norm is bounded above by a ﬁxed constant. If the ideal class group were inﬁnite, we would eventually use the same integral ideal in two diﬀerent ideal classes, which is impossible. 3. 7 7 Applications Suppose that a number ﬁeld L has a Minkowski bound on ideal norms that is less than 2.

1 This problem set will indicate how to ﬁnd the sign of the discriminant of the basis 1, α, . . , αn−1 of L = Q(α), where the minimal polynomial f of α has degree n. 1. Let c1 , . . , cr1 be the real conjugates of α, that is, the real roots of f , and let cr1 +1 , cr1 +1 , . . , cr1 +r2 , cr1 +r2 be the complex (=non-real) conjugates. Show that the sign of the discriminant is the sign of r2 (cr1 +i − cr1 +i )2 . i=1 2. Show that the sign of the discriminant is (−1)r2 , where 2r2 is the number of complex embeddings.

Yr1 , z1 , . . , zr2 ) ∈ Rr1 × Cr2 : |yi | ≤ ai , |zj | ≤ ar1 +j } where i ranges from 1 to r1 and j from 1 to r2 . We specify the ai as follows. Fix the positive real number b ≥ 2n−r1 (1/2π)r2 |d|1/2 . Given arbitrary positive real numbers a1 , . . , ar , where r = r1 + r2 − 1, we choose the positive real number ar+1 such that r1 +r2 r1 a2j = b. ai i=1 j=r1 +1 The set S is compact, convex, and symmetric about the origin, and its volume is r1 +r2 r1 πa2j = 2r1 π r2 b ≥ 2n−r2 |d|1/2 . 3)], to get S ∩ (H \ {0}) = ∅.