
Re: The irreducibility of 6sqrt(14) over Z[sqrt(14)]
Posted:
Dec 27, 2007 9:51 PM


On Dec 22, 12:19 pm, "I.M. Soloveichik" <ims...@sbcglobal.net> wrote: > As I recall it is true that Z[\sqrt{14}] is a UFD and even stronger it is Euclidean (but not wrt the norm)
It's not a UFD.

