mersenneforum.org  

Go Back   mersenneforum.org > Math Stuff > Abstract Algebra & Algebraic Number Theory

Reply
 
Thread Tools
Old 2018-01-13, 05:09   #1
carpetpool
 
carpetpool's Avatar
 
"Sam"
Nov 2016

2×3×53 Posts
Post Ideal groupings in number fields

K is a number field,
h is its class number > 1,
P and Q are non-principal prime ideals in K, so are Pn and Qn,
[G, G2, G3,... Gn] (ideal groupings) are the groupings of all non-principal prime ideals such that the product of any two prime ideals P and Q in the same group Gn is principal.
d is the exponent on the class group generator of any prime ideal.

The number of groupings Gn is not necessarily the same as its class number.
However, the maximum number of groupings Gn is h-1.

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 non-principal 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 non-principal 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, G2]. 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 G2.

(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 h-1. 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, G2, and G3]

If d = 1 or 7, then P belongs in Group G.
If d = 2 or 6, then P belongs in Group G2.
If d = 3 or 5, then P belongs in Group G3.

This field is interesting because the number of ideal groupings (3) does not divide h-1 (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 K2-K19 (because h = 1), we have g = [G, G2, G3,... Gn] 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 Gn for each of the prime cyclotomic fields, as well as any other useful information.
carpetpool is offline   Reply With Quote
Old 2018-01-13, 10:24   #2
Nick
 
Nick's Avatar
 
Dec 2012
The Netherlands

2×751 Posts
Default

Quote:
Originally Posted by carpetpool View Post
K is a number field,
h is its class number > 1,
P and Q are non-principal prime ideals in K, so are Pn and Qn,
[G, G2, G3,... Gn] (ideal groupings) are the groupings of all non-principal prime ideals such that the product of any two prime ideals P and Q in the same group Gn is principal.
A field has only 2 ideals, both principal. Do you mean prime ideals of some subring of K such as its ring of integers?
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
Nick is online now   Reply With Quote
Old 2018-01-13, 18:07   #3
carpetpool
 
carpetpool's Avatar
 
"Sam"
Nov 2016

2·3·53 Posts
Post

Quote:
Originally Posted by Nick View Post
A field has only 2 ideals, both principal. Do you mean prime ideals of some subring of K such as its ring of integers?
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
Yes, Nick.

https://en.wikipedia.org/wiki/Ideal_class_group

I am not sure if this grasps the same concept I addressed:

Quote:
Originally Posted by Wikipedia;
In number theory, the ideal class group (or class group) of an algebraic number field K is the quotient group JK/PK where JK is the group of fractional ideals of the ring of integers of K, and PK is its subgroup of principal ideals. The class group is a measure of the extent to which unique factorization fails in the ring of integers of K. The order of the group, which is finite, is called the class number of K.
Quote:
Originally Posted by Wikipedia;
If R is an integral domain, define a relation ~ on nonzero fractional ideals of R by I ~ J whenever there exist nonzero elements a and b of R such that (a)I = (b)J. (Here the notation (a) means the principal ideal of R consisting of all the multiples of a.)
According to the second paragraph, for some number field K, and Ok is its ring of integers, we have a "sub-ring" of integers Ok/p such that an ideal of norm p divides every integer in the "sub-ring". We can ask which ideals Q multiplied by the ideal P (of norm p), will make PQ a principal ideal?

I.e. The Question I was asking is --- how many different, hence distinct "sub-rings" are contained in the field K?
carpetpool is offline   Reply With Quote
Old 2018-01-13, 18:13   #4
carpetpool
 
carpetpool's Avatar
 
"Sam"
Nov 2016

4768 Posts
Post

Quote:
Originally Posted by Nick View Post
A field has only 2 ideals, both principal. Do you mean prime ideals of some subring of K such as its ring of integers?
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
I didn't notice your question there, and yes this is true (I don't have a proof although I'm sure there is one someone already discovered).

Example:

K= Q(sqrt(-5)), the ideals P = <x-2,3>, Q = <x-3,7>, and R = <x-8,23> are non-principal.

It follows that PQ and PR are principal ideals, so QR is also a principal ideal.
carpetpool is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Unique Groupings found in the first 49 mersenne numbers ONeil ONeil 27 2018-12-03 01:52
Cyclic fields with class number h carpetpool Abstract Algebra & Algebraic Number Theory 0 2018-01-30 06:10
Basic Number Theory 19: Introducing rings and fields Nick Number Theory Discussion Group 0 2017-04-11 18:37
Questions about Number Fields Raman Miscellaneous Math 5 2013-06-12 13:54
Ideal climbing plant - mix michaf Hobbies 15 2008-03-28 02:48

All times are UTC. The time now is 16:24.

Sat Nov 28 16:24:12 UTC 2020 up 79 days, 13:35, 4 users, load averages: 1.84, 1.64, 1.51

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.