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By Rainer Vogt

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L²-invariants: theory and applications to geometry and K-theory

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R=0 To prove that d0 > 0, it will suffice to prove that d1 < 1. Now any large prime p which has just one prime ideal factor of degree 1 in Q(η) will have some prime ideal factor of degree greater than 1 in Q(η), and so also in Q∗ (η). Since Q∗ (η) is normal, all prime ideal factors of p in Q∗ (η) will be of degree greater than 1, and the density of such p is exactly 1 − 1/n∗j , where n∗j denotes the degree of Q∗ (η) (Hasse [3], pp. 138–139). Hence 30 A. Diophantine equations and integral forms d1 1 − 1/n∗j , whence the result.

2] K. F. Roth, Rational approximations to algebraic numbers. , 168. [3] C. Runge, Über ganzzahlige Lösungen von Gleichungen zwischen zwei Veränderlichen. J. Reine Angew. Math. 100 (1887), 425–435. [4] C. L. Siegel, Über einige Anwendungen diophantischer Approximationen. Abh. Preuss. Akad. Wiss. -math. Kl. Nr. 1 (1929); Ges. Abhandlungen I, Springer, Berlin 1966, 209–266. [5] Th. Skolem, Über ganzzahlige Lösungen einer Klasse unbestimmter Gleichungen. Norsk. Mat. Forenings Skrifter (I) Nr. 10 (1922).

N − 1. Therefore q is a factor of θk with exponent en for k n − 1, and en−1 for k = n. The number q is a factor of (ϑk , θk ) with exponent ek for k n − 1 and en−1 for k = n. The number q is a factor of θn m/mn with the same exponent as that of m. d. (θk m/mk ) | m. ,n But m | ϑ − m and therefore the equation n (7) m+ ξk θk m/mk = ϑ k=1 has a solution. 24 A. ,n . We will show that for j = k we have mj ϑk | θj m k . (8) We distinguish two cases: 1◦ k 2◦ k = n. In this case q is a factor of θj with exponent en , of mj with exponent ej , therefore of the divisor with exponent ej + en , and of the dividend with exponent en + en−1 .

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