According to SAGE the class number of Q(2^(1/n)) is 1 for n=2,...,18. For 19 it's still computing with a message: warning Zimmert's bound is large (1259129) certification will take a long time. Victor On Wed, Dec 3, 2014 at 2:47 PM, Warren D Smith <warren.wds@gmail.com> wrote:
Victor Miller victorsmiller at gmail.com According to SAGE the field Q(2^{1/12}) has class number 1, and so does have unique factorization. In addition Z[2^{1/12}] is the maximal order in this field.
--so: for which N does Q(2^{1/N}) have class number 1?
http://oeis.org/A005848 = {1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 24, 25, 27, 28, 32, 33, 35, 36, 40, 44, 45, 48, 60, 84} claims to answer the same question for Q(Nth root of unity) duplicating the Main Theorem of J. Myron Masley and Hugh L. Montgomery: Cyclotomic fields with unique factorization, J. Reine Angew. Math. 286/287 (1976) 248-256 which shows that all such N obey N<=84 and gives the full list. Amazingly enough, all the historically-used music-scale sizes {4,5,6,7,8,12} happen to be on this list (asterisk: M&M explain that if k is odd and on their list, then 2k automatically is also, so therefore they omit mentioning all 2k with k odd). Which actually is not so amazing since ALL numbers 1-12 are on the M&M list.
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