Let K = ℚ(pd24)$\begin{array}{} \displaystyle (\sqrt[4]{pd^{2}}) \end{array}$ be a real pure quartic number field and k = ℚ(p$\begin{array}{} \displaystyle \sqrt{p} \end{array}$) its real quadratic subfield, where p ≡ 5 (mod 8) is a prime integer and d an odd square-free integer coprime to p . In this work, we calculate r 2 ( K ), the 2-rank of the class group of K , in terms of the number of prime divisors of d that decompose or remain inert in ℚ(p$\begin{array}{} \displaystyle \sqrt{p} \end{array}$), then we will deduce forms of d satisfying r 2 ( K ) = 2. In the last case, the 4-rank of the class group of K is given too.

中文翻译：

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在一些实纯四次数域的类组的2级和4级上

令K =ℚ（pd24）$ \ begin {array} {} \ displaystyle（\ sqrt [4] {pd ^ {2}}）\ end {array} $是一个真正的纯四次数字段，而k =ℚ（p $ \ begin {array} {} \ displaystyle \ sqrt {p} \ end {array} $）其实数二次子字段，其中p≡5（mod 8）是素数整数，d是无奇数平方无整数。在这项工作中，我们根据ℚ（p $ \ begin {array} {}中可分解或保持惰性的d的素数除数来计算k 2类组的2秩r 2（K）。 \ displaystyle \ sqrt {p} \ end {array} $），那么我们将推导出满足r 2（K）= 2的d的形式。在最后一种情况下，也给出了K类组的4级。