## Algebraic Number Fields by Ehud de Shalit

By Ehud de Shalit

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In particular, if the discriminant vanishes for one basis, it always vanishes. 17) ∆(¯ ω1 , . . , ω ¯ n ) = ∆(ω1 , . . , ωn )modp. Observe that p ramifies in K if and only if R has nilpotent elements. Suppose first that p ramifies. Let ω ¯ 1, . . , ω ¯ n be a basis of R over Fp such that ω ¯ 1 ∈ R is nilpotent. Then ω ¯1ω ¯ j are nilpotent and T r(¯ ω1 ω ¯ j ) = 0. A whole row in the matrix whose determinant is ∆(¯ ω1 , . . , ω ¯ n ) then vanishes, so ∆(¯ ω1 , . . , ω ¯ n ) = 0. Now start with any basis {ωi } of OK over Z.

13) Let S = {P1 , . . , Ps } be a finite set of primes in K. An S-unit is u ∈ K × such that the fractional ideal (u) = P1e1 . . Pses for some ei ∈ Z. Show (i) the S-units × × form a multiplicative group OK,S . (ii) The ei define a homomorphism from OK,S to × × s Z whose kernel is OK . (iii) The group OK,S is a finitely generated abelian group whose rank is r1 (K) + r2 (K) + s − 1. 14) Let ζ = e2πi/7 and K = Q(ζ). 10) εa,b,c = (ζ − 1)a (ζ 2 − 1)b (ζ 4 − 1)c . (i) Show that these elements form a group C of units in K (it is called the group of circular units).

Even χ. This corresponds to Kχ real. 27) L(χ, 1) = − τ (χ) m bmodm χ(b) ¯ log |1 − ζ b |. Suppose that χ is non-trivial, quadratic (χ(b) = χ(b) ¯ = ±1) and even. 28) εχ = (1 − ζ b )χ(b) = bmodm sin bmodm πb m χ(b) . 29) L(χ, 1) = − τ (χ) log εχ . 30) = bmodm (1 − ζ a b )χ(b) (1 − ζ b )χ(ba −2 ) (1 − ζ b )χ(b) = εχ . It follows that εχ ∈ Kχ . In fact, it is a unit. 31) εχ = bmodm 1 − ζb 1−ζ χ(b) b and (1 − ζ )/(1 − ζ) is a unit. 3. Dedekind’s Zeta function Let K be a number field, [K : Q] = n, r1 and r2 as before, the number of real and pairs of complex embeddings, r = r1 +r2 −1 the unit rank.