Integral bases for an infinite family of cyclic quintic fields
For an integer parameter \(n\) consider Lehmer's family of cyclic quintic fileds \(K_n\) generated by a root \(\vartheta\) of the polynomial \[ \begin{multlined} f_n(x)=x^5+n^2x^4-(2n^3+6n^2+10n+10)x^3\\ +(n^4+5n^3+11n^2+15n+5)x^2+ (n^3+4n^2+10n+10)x+1.\end{multlined} \] Set \(m=n^4+5n^3+15n^2+25n+25\) for the conductor of \(K_n\) and let \(d=n^3+5n^2+10n+7\) and \(a=m^3-10m^2+5m\). Under the condition \(m\) is square-free apart from 5, the reviewer and \textit{M. Pohst} [Math. Comput. 66, 1689--1696 (1997; Zbl 0899.11064)] showed among others that an integer basis of \(K_n\) is given by \[ \{1,\vartheta,\vartheta^2,\vartheta^3,\omega_5\} \] where \[ \omega_5=\frac{1}{d} \biggl((n+2)+(2n^2+9n+9)\vartheta +(2n^2+4n-1)\vartheta^2+(-3n-4)\vartheta^3+\vartheta^4\biggr). \] Let \[ m=5^bPQ^2, \] where \(b=2\) if \(5| n\) and \(b=0\) otherwise, \(P,Q\) are coprime, square-free and not divisible by~5. The authors of this paper show that under the weaker assumption that \(m\) is cube-free, an integral basis is given by \[ \{1,\vartheta,\vartheta^2,v_4,v_5\}, \] where \[ v_4=\frac{1}{Q}\left(\vartheta-\frac{n^2}{5}(Q-1)\right) \] and \[ v_5=\frac{ad\omega_5+(1-a)Qv_4\vartheta}{dQ}. \] Under the original assumption (\(m\) is square-free apart from 5) this gives the integral basis of Gaál and Pohst.
scientific article; zbMATH DE number 5178021