1. ## Algebraic Integers trouble...

I'm having trouble showing that a particular number is an algebraic integer in the cubic number field $\mathbb{Q}(\alpha)$ where $\alpha^3=m=hk^2\in\mathbb{Z}$ and $m$ is cube-free and $h$ and $k$ are both square-free. There is also the extra condition that $m\equiv1\mod9$.

I'm trying to show that $\frac{\alpha^2+\alpha k^2+k^2}{3k}$ is an algebraic integer, given that $\frac{\alpha^2}{k}$ and $\frac{(\alpha-1)^2}{3}$ are both algebraic integers...

I've tried adding, subtracting and multiplying and fiddling around with the given alg. integers, and even got maple to do some of the work for me, but to no avail... Any ideas anyone?

2. Originally Posted by hg305
I'm having trouble showing that a particular number is an algebraic integer in the cubic number field $\mathbb{Q}(\alpha)$ where $\alpha^3=m=hk^2\in\mathbb{Z}$ and $m$ is cube-free and $h$ and $k$ are both square-free. There is also the extra condition that $m\equiv1\mod9$.

I'm trying to show that $\frac{\alpha^2+\alpha k^2+k^2}{3k}$ is an algebraic integer, given that $\frac{\alpha^2}{k}$ and $\frac{(\alpha-1)^2}{3}$ are both algebraic integers...

I've tried adding, subtracting and multiplying and fiddling around with the given alg. integers, and even got maple to do some of the work for me, but to no avail... Any ideas anyone?
this is one little tricky problem! we have: $\frac{\alpha^2+\alpha k^2+k^2}{3k}=\frac{\alpha^2(\alpha - 1)^2}{3k} + \frac{k \alpha (1-h)}{3} + \frac{k(2h +1)}{3}.$ now the result follows because $h \equiv 1 \mod 3.$ why?

3. Thanks that's brilliant! How did you get the answer so fast?!?!? I've been trying for days!

Is it true that h must be congruent to 1 mod 3 because modular arithemtic is multiplicative? I mean: $[h][k^2]=[h][k]^2=[1]$ and since $[k]^2=[1]$ (unless $[k]=[0]$ but this gives a contradiction anyway...) we must have $[h]=[1]$...

4. Originally Posted by hg305

Thanks that's brilliant! How did you get the answer so fast?!?!? I've been trying for days!
i guess i just got lucky! isn't this problem a part of a bigger problem? i'm asking this because there are some unnecessary assumptions in the problem that are not used in the solution.

Is it true that h must be congruent to 1 mod 3 because modular arithemtic is multiplicative? I mean: $[h][k^2]=[h][k]^2=[1]$ and since $[k]^2=[1]$ (unless $[k]=[0]$ but this gives a contradiction anyway...) we must have $[h]=[1]$...
correct.

5. Do you know Marcus' book "Number Fields"? It's brilliant. It's an exercise from there. We find the ring of alg. integers by determining an integral basis. It's exercise 41 at the end of chapter 2. It's a pretty hefty exercise, with 12 parts, but it's good to see an example of a number field with degree >2 and be able to know exactly what its ring of integers is!

Btw what didn't we use? I suppose if I really think about it we didn't really need that h and k are square-free...

Thanks!

6. Originally Posted by hg305

Do you know Marcus' book "Number Fields"? It's brilliant. It's an exercise from there. We find the ring of alg. integers by determining an integral basis. It's exercise 41 at the end of chapter 2. It's a pretty hefty exercise, with 12 parts, but it's good to see an example of a number field with degree >2 and be able to know exactly what its ring of integers is!

Btw what didn't we use? I suppose if I really think about it we didn't really need that h and k are square-free...

Thanks!
also $m \equiv 1 \mod 9$ is not necessary. we only need to have $m \equiv 1 \mod 3.$