Relative integral basis for algebraic number fields
For an arbitrary number field, \(F\), let \(\mathfrak O_F\) denote the ring of integers of \(F\). If \(K/k\) is an extension of number fields of relative degree \(n\), then it is shown in \textit{W. Narkiewicz} [Elementary and analytic theory of algebraic numbers (1974; Zbl 0276.12002), p. 24 and p. 356] that \(\mathfrak O_K \approx \mathfrak O_k^{n-1}\oplus I\) where \(I\) is an ideal of \(\mathfrak O_k\). Hence a relative integral basis exists for the extension \(K/k\) if and only if \(I\) is a principal ideal. Moreover, if \(C\) is the ideal class of \(k\) containing \(I\) then \(C^2\) contains the relative discriminant of \(K/k\). In this article, this criterion is first used to show that \(K=\mathbb Q(\sqrt{2}, \sqrt{-7})\) and \(k=\mathbb Q(\sqrt{-14})\) gives an extension with no integral basis. The latter parts of the article consider the extensions \(K/k\) with \(K=\mathbb Q(\sqrt [3]{n}, \sqrt{-3})\) and either \(k=\mathbb Q(\sqrt{-3})\) or \(k=\mathbb Q(\sqrt[3]{n})\). In the former case, \(\mathfrak O_k\) is a PID, so a relative integral basis always exists. An explicit relative integral basis is given, but it sometimes involves an integer \(\alpha\) of \(\mathbb Q(\sqrt[3]{n})\) which has norm \(3\). When \(k=\mathbb Q(\sqrt [3]{n})\), a relative integral basis is given only when \(3\) does not divide the class number of \(k\). Here the basis also involves an integer \(\alpha\) as mentioned above.
scientific article; zbMATH DE number 3952893