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P. 67

VI. A “Natural” Proof of the Nonvanishing of L-Series
                        62
                         L 7 (z)          1            1             1            1    ,
                                      1 − p −z     1 + ip −z     1 − ip −z     1 + p −z
                                  p≡1          p≡3           p≡7           p≡9
                        and
                                          1            1            1             1
                          L 9 (z)                                                      .
                                       1 − p −z     1 + p −z     1 + p −z     1 − p −z
                                   p≡1          p≡3          p≡7          p≡9
                        (Here  z> 1 to insure convergence and the subscripting of the
                        characters is used to reflect the isomorphism of the dual group and
                        the original group.)
                           The generating function for the primes in the arithmetic pro-
                        gressions ((mod 10) in this case) are then linear combinations of
                        the logarithms of these L-series. And so indeed the crux is the
                        nonvanishing of these L-series.
                           What could be more natural or more in the spirit of Dirichlet, but
                        to prove these separate nonvanishings altogether? So we are led to
                        take the product of all the L-series! (Landau uses the same device to
                        prove nonvanishing of the L-series at point 1.)
                           The result is the Dirichlet series
                                                    1               1

                                   Z(z)
                                                      −z 4
                                                (1 − p )       (1 − p −4z )
                                           p≡1             p≡3
                                                      1               1
                                           ×                                 ,
                                                  (1 − p −4z )   (1 − p −2z 2
                                                                           )
                                              p≡7            p≡9
                        and the problem reduces to showing that Z(z) is zero-free on  z   1.
                                                                              1
                           Of course, this is equivalent to showing that       −z is zero-
                                                                        p≡1 1−p
                        free on  z   1, which seems, at first glance, to be a more attractive
                        form of the problem. This is misleading, however, and we are bet-
                        ter off with Z(z), which is the product of L-series and is an entire
                        function except possibly for a simple pole at z   1. (See the
                        appendix.)
                           Guided by the special cases let us turn to the general one. So let A
                        be a positive integer, and denote by G A the multiplicative group of
                        residue classes (mod A) which are prime to A. Set h   φ(Að , and
                        denote the group elements by 1   n 1 ,è 2 ,...,è h . Denote the dual
                                         ˆ                                       arranged
                        group of G A by G A and its elements by χ 1 ,χ n 2  ,...,χ n h
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