Let p be an odd prime number, and pn0{p^{n_0}} the highest power of p dividing 2 p−1 − 1. Let _boxclose_boxclose(_p^n+1){K_n={bf Q}(zeta_{p^{n+1}})} and Ln,j=Kn+(z2j+2){L_{n,j}=K_n^+(zeta_{2^{j+2}})} for j ≥ 0. Let hn*{h_n^*} be the relative class number of K n , and h n,j the class number of L n,j , respectively. Let n be an integer with n ≥ n 0. We prove that if the ratio hn*/hn-1*{h_n^*/h_{n-1}^*} is odd, then h n,j /h n−1,j is odd for any j ≥ 0.
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机译：令p为奇质数，p n 0 {p ^ {n_0}} p的最高幂除以2 p-1 − 1.让_boxclose_boxclose（_p ^ n + 1）{K_n = {bf Q}（zeta_ {p ^ {n + 1}}）}和L n，j = K n + （z 2 j + 2 ）{L_ {n，j} = K_n ^ +（zeta_ {2 ^ { j + 2}}）}的j≥0。令h n * {h_n ^ *}是K n 的相对类号。和h n，j 分别为L n，j 的类号。令n为n≥n 0 的整数。我们证明如果比率h n * / h n-1 * {h_n ^ * / h_ { n-1} ^ *}是奇数，那么对于任何j≥0的h n，j / h n-1，j 是奇数。
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