# Read e-book online Algebraic Number Fields PDF

By Ehud de Shalit

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Now compare the two experessions that we obtained for F (0). 2. Let f ∈ S. 12) f (n) = n∈Z f (n). n∈Z 1. 4. An important example. Let f (x) = e−πtx where t > 0 is a fixed parameter. This function belongs to S, and it’s Fourier transform is 2 1 f (y) = √ e−πy /t . 1. Prove this! You will have to use Cauchy’s theorem to shift the line of integration from iy/t + R back to R, and you will also have to use the fact 2 ∞ that −∞ e−πx dx = 1. Alternatively, if you only want to use real analysis, separate real and imaginary parts, and use integration by parts twice.

2. The decomposition field of P is the fixed field of GP . We denote it by ZP . 3. (i) ZP is an extension of degree g of K, and [L : ZP ] = ef. (ii) Gal(L/ZP ) = GP . (iii) Let PZ = ZP ∩ P be the prime of ZP below P. 2) e(PZ /p) = f (PZ /p) = 1. Proof. Parts (i) and (ii) are clear from Galois theory. For part (iii) note that for every σ ∈ GP , σ(P) = P by definition, and GP acts transitively on the primes above PZ . 3) e(P/PZ )f (P/PZ ) = [L : ZP ] = e(P/p)f (P/p). 4) e(P/PZ ) = e(P/p) and similarly for f.

Incidentally, the Euler product formula shows that ζ(s) = 0 for Re(s) > 1. 2. The Gamma function. 3) ∞ Γ(s) = 0 e−t ts dt . t Note that dt/t is invariant under the change of variables t → ct, c > 0. 4) Γ(s + 1) = sΓ(s). This wonderful formula has two immediate consequences. , because it is easy to see that Γ(1) = 1. Second, we may use the formula to meromorphically continue Γ(s) to the whole complex plane. It will have then simple poles at s = 0, −1, −2, ... and will be holomorphic elsewhere. Indeed, if it has already been defined on Re(s) > m, extend it to Re(s) > m − 1 by setting Γ(s) = Γ(s + 1)/s.