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By Franz Lemmermeyer

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Then χ(x) = x∈G #G if χ = 1l 0 if χ = 1l and χ(x) = χ∈X #G if x = 1 . 0 if x = 1 Proof. The first assertion is clear if χ = 1l. If χ = 1l, then there must be a y ∈ G such that χ(y) = 1. But now 44 3. Primes in Arithmetic Progression χ(y) χ(x) = x∈G χ(xy) = x∈G χ(x), x∈G proving our claim. The ‘dual’ assertion is reduced to the first case by identifying G and X(X(G)). Primitive Characters and Conductors Let χ be a Dirichlet character on (Z/nZ)× ; every integer m ∈ N such that a ≡ b mod m =⇒ χ(a) = χ(b) whenever a, b are prime to n is called a defining modulus for χ.

In particular, we have L(1, χ) = 0. 21. Let d > 0 be the discriminant of a real quadratic number field K. Then 1 πa L(1, χ) = − √ χ(a) log sin . 6) yields the following class number formula: h=− 1 2 log ε χ(a) log sin (a,d)=1 πa . d d Since ( ad ) = ( d−a ) and sin x = sin(π − x), this formula can be simplified slightly: πa 1 χ(a) log sin . h=− log ε d 1≤a

8 is odd. Proof. To be added soon. 10. Let p be an odd prime and χ = ( p· ). Then L(1, χ) > 0. 3 below we will show that ( ap )a = hp for all primes p ≡ √ 3 mod 4 with p > 3, where h is the class number of Q( −p ), and that the unit η is equal to η = ε2h , where ε is the fundamental unit and h the class √ number of Q( p ). 3 below. 2 Nonvanishing of Dirichlet’s L-functions In this section I will present an elementary proof that L(1, χ) = 0 for quadratic Dirichlet characters χ. The idea behind it is due to Gelfond [GL1965, pp.

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