“…In particular, if K = K m is the simplest cyclic quartic field associated with the polynomial P m (x) = x 4 − mx 3 − 6x 2 + mx + 1, m 1, m = 3, of discriminant d m = 4∆ 3 m (where the odd part of ∆ m = m 2 + 16 is assumed to be square-free), then we have Reg Km 2,4,5,6,8,9,10,11,15, 24} (see [27], in which a more general problem is in fact solved). Notice that because of much weaker lower bounds for h Km , Lazarus could not solve in [10] and [11] this class number 1 problem for the simplest quartic fields.…”