By Ehud de Shalit

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12) T rL/K f (X)αj (X − α)f (α) = Xj (0 ≤ j ≤ n − 1). However, both sides of the equation are polynomials of degree ≤ n − 1 having n distinct roots (all the conjugates of α), so they must coincide. It follows that the inverse of DL/K,p (the dual of OL,p ) is spanned over OK,p i . Solving successively for the bi we get that the OK,p -span of the bi is the by f b(α) same as the OK,p -span of the αi , and is therefore OL,p . The theorem follows. 3. GALOIS EXTENSIONS 35 3. 1. The decomposition group.

These are the smooth (complex valued) functions f such that |x|n f (m) (x) → 0 as |x| → ∞ for any m, n ≥ 0. 8) f (y) = ∞ f (x)e2πixy dx. −∞ We recall its well-known properties. For the moment denote f also by Ff • If f ∈ S then Ff ∈ S too and FFf (x) = f (−x). • Let Mf (x) = 2πixf (x) and Df (x) = f (x). 9) FMf = DFf, FDf = −MFf. • If g(x) = f (x + a) then g(y) = e−2πiay f(y). Let f ∈ S. The function F (x) = m∈Z f (x + m) is smooth an 1-periodic. 10) an e2πinx . 11) ∞ F (x)e−2πinx dx = an = 0 f (x)e−2πinx dx = f(−n).

If ω1 , . . , ωn is a basis of OK and ω1 , . . t. B, then it spans OK . 22) ωi = mij ωj then B(ωi , ωj ) = mij and therefore dK = det(mij ) = [OK : OK ] = [OK : DK ] = N DK . We have proven the following lemma. 7. The different is an ideal of K whose norm is the discriminant. We can now refine the theorem saying that p ramifies if and only if it divides the discriminant to deal with primes of K. We shall state it without a proof. 2. 8. A prime p is ramified in K if and only if it divides the different DK .