20180113, 05:09  #1 
"Sam"
Nov 2016
2^{2}·83 Posts 
Ideal groupings in number fields
K is a number field,
h is its class number > 1, P and Q are nonprincipal prime ideals in K, so are P_{n} and Q_{n}, [G, G_{2}, G_{3},... G_{n}] (ideal groupings) are the groupings of all nonprincipal prime ideals such that the product of any two prime ideals P and Q in the same group G_{n} is principal. d is the exponent on the class group generator of any prime ideal. The number of groupings G_{n} is not necessarily the same as its class number. However, the maximum number of groupings G_{n} is h1. Let's take a look at some examples: Lemma I: If the only ideal grouping in K are G, then the product of any two nonprincipal ideals is principal. The exponent on the class group generator of P is 1. For K=Q(sqrt(5)), h = 2, and the groupings in K are [G]. P and Q must be in this group and d = 1. Since there is only ideal grouping G, this implies that the product of any two nonprincipal ideals are principal (a restate of Lemma I). In fact, this is true for all fields K with class number 2, and some other fields with class number h > 2. K=Q(sqrt(23)) has class number h = 3, and there is also one ideal grouping G, hence Lemma I is true here. K=Q(sqrt(47)) has class number h = 5, however Lemma I is not true. There are two ideal groupings [G, G_{2}]. One can determine which group P belongs in. If d = 1 or 4, then P belongs in Group G. If d = 2 or 3, then P belongs in Group G_{2}. (P is principal otherwise) One common conclusion to come to is if K is a number field with class number h, then the number of ideal groupings in K divides h1. The short and easy answer to this is no, this is not always true. (It is sometimes.) The field K = Q(sqrt(95)) has class number h = 8. The groupings are [G, G_{2}, and G_{3}] If d = 1 or 7, then P belongs in Group G. If d = 2 or 6, then P belongs in Group G_{2}. If d = 3 or 5, then P belongs in Group G_{3}. This field is interesting because the number of ideal groupings (3) does not divide h1 (7), yet the distribution of prime ideals in these groups are equal. It is obvious that no prime ideal will have exponent d = 4 on its class group generator. For quadratic fields, it seems pretty easy to work out. What about for the nth cyclotomic fields Kn, for prime n? Excluding K2K19 (because h = 1), we have g = [G, G_{2}, G_{3},... G_{n}] the number of ideal groupings (as I have first defined) in Kn, and (n,g): (23,1) (29,1) (31,2) (37,1) (41,3) (43,5) (47,15) I have also computed the series of exponents d in each of the following ideal groupings for Kn. (Private Message me if you want any of these references.) I spend most of my PC power currently trying to classify the ideal groupings for larger cyclotomic fields. It would be nice to know if there is already a list of the number of ideal groupings G_{n} for each of the prime cyclotomic fields, as well as any other useful information. 
20180113, 10:24  #2  
Dec 2012
The Netherlands
1,759 Posts 
Quote:
You appear to be assuming you have an equivalence relation here. Is that really so? For example, if PQ and QR are principal, does it follow that PR is? What about PP? More generally, if you are interested in class groups, it would help to learn a little group theory. You could start here: http://www.mersenneforum.org/showthread.php?t=21877 

20180113, 18:07  #3  
"Sam"
Nov 2016
2^{2}·83 Posts 
Quote:
https://en.wikipedia.org/wiki/Ideal_class_group I am not sure if this grasps the same concept I addressed: Quote:
Quote:
I.e. The Question I was asking is  how many different, hence distinct "subrings" are contained in the field K? 

20180113, 18:13  #4  
"Sam"
Nov 2016
14C_{16} Posts 
Quote:
Example: K= Q(sqrt(5)), the ideals P = <x2,3>, Q = <x3,7>, and R = <x8,23> are nonprincipal. It follows that PQ and PR are principal ideals, so QR is also a principal ideal. 

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