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Soient $\mathbf{K}=\mathbf{Q}\left(\sqrt{-pq(2+\sqrt{2})}\right)$ où $p$ et $q$ deux nombres premiers différents tels que $p\equiv q\equiv \pm 5mod8$, ${\mathbf{K}}_2^{\left(1\right)}$ le $2$-corps de classes de Hilbert de $\mathbf{K}$, ${\mathbf{K}}_2^{\left(2\right)}$ le $2$-corps de classes de Hilbert de ${\mathbf{K}}_2^{\left(1\right)}$ et $G$ le groupe de Galois de ${\mathbf{K}}_2^{\left(2\right)}/K$. D’après [], la $2$-partie $C_{2,\mathbf{K}}$ du groupe de classes de $\mathbf{K}$ est de type $(2,2)$, par suite ${\mathbf{K}}_2^{\left(1\right)}$ contient trois extensions ${\mathbf{F}}_i/\mathbf{K}$ ; $i=1,2,3$. Dans ce papier, on s’interesse au problème de capitulation des $2$-classes d’idéaux de $\mathbf{K}$ dans ${\mathbf{F}}_i$ $(i=1,2,3)$ et à déterminer la structure de $G$.

Let $K=k\left(\sqrt{-p\epsilon \sqrt{l}}\right)$ with $k=\mathbb{Q}\left(\sqrt{l}\right)$ where $l$ is a prime number such that $l=2$ or $l\equiv 5mod8$, $\epsilon$ the fundamental unit of $k$, $p$ a prime number such that $p\equiv 1mod4$ and ${\left(\frac{p}{l}\right)}_4=-1$, $K_2^{\left(1\right)}$ the Hilbert $2$-class field of $K$, $K_2^{\left(2\right)}$ the Hilbert $2$-class field of $K_2^{\left(1\right)}$ and $G=Gal(K_2^{\left(2\right)}/K)$ the Galois group of $K_2^{\left(2\right)}/K$. According to E. Brown and C. J. Parry [7] and [8], $C_{2,K}$, the Sylow $2$-subgroup of the ideal class group of $K$, is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, consequently $K_2^{\left(1\right)}/K$ contains three extensions $F_i/K$ $(i=1,2,3)$ and the tower of the Hilbert $2$-class field of $K$ terminates at either $K_2^{\left(1\right)}$ or $K_2^{\left(2\right)}$. In this work, we are...

Let $K=\mathbb{Q}\left(\theta \right)$ be a pure cubic field, with ${\theta }^3=D$, where $D$ is a cube-free integer. We will determine the reduced ideals of the order $𝒪=\mathbb{Z}\left[\theta \right]$ of $K$ which coincides with the maximal order of $K$ in the case where $D$ is square-free and $\neg \equiv \pm 1(mod9)$.