Computing the class group

1. Apr 27, 2012

Omukara

I have a pretty urgent question concerning the calculation of the class group, so any help will be very much appreciated:)

I'd like to illustrate my question with an example:

Calculate the ideal class group of Q(√-17), giving a representative ideal for each ideal class and a description of the group law.

My attempt at the question gets me up to the part:
(2) = (p_2)^2, where p_2 = (2, √-17 + 1) and N(p_2) = 2
(3) = (p_3)(q_3) where p_3 = (3, √-17 + 1), q_3 = (3, √-17 - 1) and N(p_3) = 3, N(q_3) = 3
where I've shown the other possibilities not to be viable as their norms are too large.

which leaves me with the ideals ϴ, p_2, p_3, q_3 which I've shown not to be principal.
The step proceeding this confuses me, it just says after this "we have N(1-√-17) = 18 = 2×3^2. Thus the possible decompositions of (1-√-17) are p_2×(p_3)^2, p_2×(q_3)^2 and p_2×(p_3)×(q_3)."

QUESTION i) I don't understand where the (1-√-17) comes from?

Furthermore, he goes on to give the relations of the group as (P_2)^2 = ϴ, (p_3)^2~(q_3)^(-2)~p_2 and finally (p_3)^4~ϴ, thus q_3~(q_3)^(-1)~(p_3)^3.

QUESTION ii) how did these relations come about?

Thanks again! I would be very grateful for any help given for either or both parts of my question:) if I missed something out or was unclear about something, please say:)

Last edited: Apr 27, 2012
2. Apr 27, 2012

Hurkyl

Staff Emeritus
Every number gives a relation on the class group: (x) = ϴ. (I assume ϴ is meant to be the zero element of the class group?)

(1-√-17) probably came from a brief search for numbers whose prime factorization included only the prime ideals you were already working with. This gives you a relation without having to go through the trouble of working with more prime ideals. It doesn't have any other special significance.

You've already written down some of the principal ideals that give you relations, haven't you? And you could always verify any relation I~J simply by computing the ideals and finding the x such that J=Ix. Or, equivalently, showing I/J principal.

(p.s. I think you've made at least two typos)

3. Apr 27, 2012

Omukara

Thank you very much for your reply! I did indeed use ϴ to denote the zero element (sorry!)

Just to clarify, I could've just as well considered (1+√-17)

And I really do apologise for my ignorance, but I don't see why for example we have the relation (p_3)^2~(q_3)^(-2)~p_2 and not just (p_3)~p_2 for example (without actually explicitly working the products of the ideals out)

4. Apr 27, 2012

Hurkyl

Staff Emeritus
Sure. And the corresponding relation would be the conjugate of the one you got from (1-√-17).

I hope it's clear from what you've already written that
• (p_2)^2 = ϴ
• (p_3)(q_3) = ϴ
• (p_2)(q_3)^2 = ϴ
Once you have this, everything else is just arithmetic.

5. Apr 27, 2012

Omukara

AH, thanks! I see it now!:D