Azizi, Abdelmalek; Taous, Mohammed Determination of the fields \(K=\mathbb Q(\sqrt d,\sqrt{-1})\), given the 2-class groups are of type \((2,4)\) or \((2,2,2)\). (Déterminations des corps \(K=\mathbb Q(\sqrt d,\sqrt{-1})\) dont les 2-groupes de classes sont de type \((2,4)\) ou \((2,2,2)\).) (French. English summary) Zbl 1215.11107 Rend. Ist. Mat. Univ. Trieste 40, 93-116 (2008). Let \(d\) be a square-free positive integer, \(K=\mathbb Q(\sqrt{d},\sqrt{-1})\) and \(C_2\) the \(2\)-part of the class group of \(K\). The author gives the structure of all \(d\) such that \(C_2\cong \mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z\) or \(C_2\cong\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z\).This article is in on the continuity of articles [A. Azizi, Rend. Circ. Mat. Palermo, II. Ser. 48, No. 1, 71–92 (1999; Zbl 0920.11076); A. Azizi and I. Benhamza, Ann. Sci. Math. Qué. 29, No. 1, 1–20 (2005; Zbl 1217.11097); T. M. McCall, C. J. Parry and R. R. Ranalli, J. Number Theory 53, No. 1, 88–99 (1995; Zbl 0831.11059) and T. M. McCall, C. J. Parry and R. R. Ranalli, Can. J. Math. 49, No. 2, 283–300 (1997; Zbl 0884.11043)] on the same topic.In the sequel, \(p,p_i\) are primes \(\equiv 1\mod 4\), \(q,q_i\) are primes \(\equiv -1\mod 4\), \((\frac{\cdot}{\cdot})\) is for the quadratic symbol. The author gives the following theorem:Let \(d\) be a square-free natural integer, \(K=\mathbb Q(\sqrt{d}, \sqrt{-1})\). Then \(C_2\cong \mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z\) if and only if one of the following conditions is verified: (1) \(d=p_1p_2\) where \((\frac{p_1}{p_2})=-1,\;p_1\equiv p_2\equiv 1 \mod 8\) and \((\frac{2}{a+b})=-1\) with \(p_1p_2=a^2+b^2\) with \(a,b\in\mathbb Z\). (2) \(d=2p_1p_2\) where \(p_1\equiv p_2\equiv 1 \mod 4\) and at least two elements of \(\{(\frac{2}{p_1}), (\frac{2}{p_2}), (\frac{p_1}{p_2})\}\) take the value \(-1\). (3) \(d=2pq\) where \(p\equiv 1 \mod 8, q\equiv 3 \mod 8\) and \((\frac{p}{q})=-1\) (4) \(d=pq_1q_2\) where \((\frac{q_1}{q_2})=-(\frac{q_2}{q_1})=1\), \(p\equiv -q_1\equiv -q_2\equiv 1 \mod 4\) and \(d\) verifies one of the three conditions: (i) \((\frac{p}{q_1})\cdot(\frac{p}{q_2})=-1 ,(\frac{2}{p})=1 ,(\frac{2}{q_1})=(\frac{2}{q_2})=-1\). (ii) \((\frac{p}{q_1})=(\frac{p}{q_2})=-1 ,(\frac{2}{p})=1 ,(\frac{2}{q_1})\cdot(\frac{2}{q_2})=-1\). (iii) \((\frac{p}{q_1})\cdot(\frac{p}{q_2})=-1 ,(\frac{2}{p})=1 ,(\frac{2}{q_1})=1, (\frac{2}{q_2})=-1\). (5) \(d=p_1p_2q\) where \(p_1\) or \(p_2\equiv 5\mod 8\), \(q\equiv 3\mod 4\) and two or three of the symbols of \(\{(\frac{p_1}{p_2}), (\frac{p_1}{q}),\frac{p_2}{q})\}\) take the value \(-1\). The author gives also a theorem of same nature for \(C_2\cong\mathbb Z/2\mathbb Z\times\mathbb Z/4\mathbb Z\) more intricate and involving also some bi-quadratic symbols and the index of the group generated by the units of \(\mathbb Q(\sqrt{d}),\mathbb Q(\sqrt{-d}), \mathbb Q(\sqrt{-1})\) in the group of units of \(\mathbb Q(\sqrt{d},\sqrt{-1})\). Reviewer: Roland Quême (Brax) Cited in 1 ReviewCited in 13 Documents MSC: 11R29 Class numbers, class groups, discriminants 11R16 Cubic and quartic extensions Keywords:biquadratic extensions; \(2\)-class group; class number Citations:Zbl 0920.11076; Zbl 0831.11059; Zbl 0884.11043; Zbl 1217.11097 PDFBibTeX XMLCite \textit{A. Azizi} and \textit{M. Taous}, Rend. Ist. Mat. Univ. Trieste 40, 93--116 (2008; Zbl 1215.11107)