1 The implication from \(h^+\) to \(h^-\)
Let \(p\) be an odd prime. Suppose that
Then
If \(p=3\), Kummer’s range is empty, \(K^+ = \mathbb {Q}\), the group \(C^+\) is the full unit group \(\{ \pm 1\} \), and the index is \(1\). For \(p\ge 5\), theorem 3.7 identifies the Bernoulli nonvanishing hypothesis with the nonvanishing of the Kummer logarithm determinant. By theorem 3.12, the subgroup \(C^+\) is then \(p\)-saturated in the full unit group, and theorem 2.7 concludes that the index is prime to \(p\).
If \(p\nmid h^-(K)\), then \(p\nmid h^+(K)\).
By theorem 3.5, the hypothesis \(p\nmid h^-(K)\) implies that no Bernoulli numerator in Kummer’s range is divisible by \(p\), so theorem 1.1 gives
Suppose for contradiction that \(p\mid h^+(K)\). The prime-conductor index theorem (theorem 4.5) then gives \(p \mid [(\mathcal O_{K^+})^\times :C^+_{\mathrm{norm}}]\), and by theorem 1.6 this is equivalent to \(p\) dividing the ordinary \(C^+\) index — a contradiction.
If \(p\mid h^+(K)\), then \(p\mid h^-(K)\).
Contrapositive of theorem 1.2: if \(p\nmid h^-(K)\), that theorem gives \(p\nmid h^+(K)\).