A problem in Algebraic Number Theory

[lost_dm]

Hello,
I have a question I'm trying to solve for several days now, and I'm really stuck.

let a,b>0 positive integers so that ab != 1 and is free from squares.
let \alpha = cube roobe of (a*b^2) and let \beta = cube root of (a^2*b).
Mark K=Q(\alpha)=Q(\beta) - the rationals with alpha\beta.
Let O_k be the algebraic integers in K.
Mark the following subgroups of [tex]O_k:
I = Z+ Z\alpha + Z(\alpha)^2,
J = Z+ Z\beta + Z(\beta)^2,
L= Z+Z\alpha +Z\beta.

Prove that:disc_k = -3^n*a^2*b^2, where n is 1 or 3.

I showed that:
disc(1,alpha,alpha^2)=3^3*a^2*b^4
disc(1,beta,beta^2)=-3^3*a^4*b^2
disc(1,alpha,beta)=-3^3*a^2*b^2

and that: if A is a base conversion matrix, that is: the matrix representing the identity transformation from the base (1,alpha,alpha^2) to an integral basis B=(x,y,z) then:

disc_k = disc(B) = 1/(det(A)^2)*-3^3*a^2*b^4.
I also know that b divides det(A), and that if I can show that: a,b^2 don't divide det(A) I can finish the proof.

I know it's somewhat complicated, but I would really appreciate help.

Also, sorry for not putting it in LaTex but the tex button failed me. I could attach an image of a PDF if it's really unclear.

Thanks in advance!

[Sisyphus]

The PDF might help.

[jshort]

The Tex button seems to be broken. The above red X is what appeared after I surrounded a number with tex.

[jshort]

btw, I don't know the answer to your question. I have a textbook on the subject which I could read through, but I'm lazy, and I suspect you may already have one yourself.

[lost_dm]

Thanks, I found the answer yesterday. It was a bit tricky.
In case someone is interested, I needed to prove two small theorems in the process:

1. Every sum module N of the Integer Domain O_k which is of degree n (the degree of the expansion) has a well defined Discriminant, that is - for every two bases of N the discriminant is equal.
2. The discriminant of such N is equal to the discriminant of O_k times the index of N in O_k as an additive subgroup (O_k ; N).

Thanks for willing to help, I think I'll try to stick around and maybe help answer some questions ^_^