Let \(K\) be a biquadratic field which contains \(\sqrt{-3}\). In this paper, the author gives a necessary and sufficient condition for the 3-class field tower of \(K\) to terminate at the first Hilbert 3-class field \(K^{(1)}\) of \(K\).
Let \(k_3= \mathbb{Q}(\sqrt{-3})\), and \(k_1\) denote the complex subfield of \(K\) different to \(k_3\), and \(k_2\) denote the real quadratic subfield of \(K\).
The author first notes that the problem is essentially reduced to the case where the 3-class groups \(\text{Cl}_{k_1}\) and \(\text{Cl}_{k_2}\) are cyclic, and that in this case there exist three intermediate cubic fields of \(k^{(1)}_i/\mathbb{Q}\) for \(i= 1, 2\).
Let \(F_i\) denote one of the three intermediate cubic fields of \(k^{(1)}_i/\mathbb{Q}\) for \(i= 1,2\). Let \(\{\varepsilon_0\}\) and \(\{\varepsilon_1, \varepsilon_2\}\) denote the fundamental units of \(F_1\) and \(F_2\), and put \(A_1= \{\varepsilon_0\}\) and \(A_2= \{\varepsilon_1, \varepsilon_2, \varepsilon_1\varepsilon_2, \varepsilon_1\varepsilon^2_2\}\). Under this assumptions the main result of the paper is the following. Assume that the class number \(h_K\neq 1\). Then the 3-class field tower of \(K\) terminates at \(K^{(1)}\) if and only if \(\text{Cl}_{k_1}\) is a cyclic group, and either (1) \(\text{Cl}_{k_2}\) is trivial, or (2) \(\text{Cl}_{k_2}\) is cyclic, and there are no \(\varepsilon\in A\) which satisfy \(\varepsilon^2\equiv 1\,(\text{mod\,}3\sqrt{-3}\cdot O_{L_j(\sqrt{-3})})\).
Furthermore, using the facts in the proof, an example with \(K^{(1)}= K^{(2)}\) and an example with \(K^{(1)}\neq K^{(2)}\) are given.