By Ehud de Shalit

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Since K is a cyclic extension of degree p − 1, it has a unique quadratic subextension, the fixed field of the unique subgroup of index 2 in G. The only prime ramifying in it is p. It follows that the discriminant of F must be divisible only by p, and since a discriminant of a quadratic field is either 0 or 1 modulo 4, the sign is forced upon us as in the lemma. Alternatively, let a p be Legendre’s symbol, which is 1 if amodp is a square in × (Z/pZ) , -1 if amodp is a non-square in the same group, and 0 if p|a.

The decomposition group. Assume from now on that L/K is Galois with Gal(L/K) = G. Let σ ∈ G. Since σ induces an automorphism of OL , it carries any prime P to a prime σ(P), and σ(P)∩OK = σ(P∩OK ) = P∩OK , so σ induces a permutation of the primes dividing any given p of K. 1. G acts transitively on the primes dividing a given prime p. Proof. Let P be a prime dividing p and assume that Q is another prime dividing p which is not in the G-orbit of P. By the Chinese Remainder Theorem we may choose a ∈ OL such that a ≡ 1modσ(P) for every σ ∈ G, but a ≡ 0modQ.

9) F (s) = χ(a) ¯ log L(χ, s) χ where the sum is over the ϕ(m) Dirichlet characters modulo m, and s > 1. The sum is real, because together with χ there appears χ ¯ and the corresponding terms are complex conjugates. Also note that there are several branches of the logarithm that one can consider, but the Euler product for L(χ, s) allows one to make a natural choice as below. As s → 1 from the right, the trivial character gives log L(1, s) which, up to the logarithm of the Euler factors dividing m is just log ζ(s), hence tends to +∞.

### Algebraic Number Fields by Ehud de Shalit

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