I have an exercise in my class that I don't seem to find an answer to:
We are asked to tell whether the indicated subring of a number field is an order and whether it is a maximal order
a) $\mathbb{Z}(\cos(2\pi/7))\subset\mathbb{Q}(\cos(2\pi/7))$
b) $\mathbb{Z}(\alpha)\subset\mathbb{Q}(\alpha)$, where $\alpha$ is a root of $x^3+x^2-4x-1=0$
c) $\mathbb{Z}[\sqrt{-1},\sqrt{2}]\subset \mathbb{Q}(\sqrt{-1},\sqrt{2})$
Now I know that a subring of a number field is an order if it is finitely generated as a $\mathbb{Z}$-module and if it spans the number field as a $\mathbb{Q}$-vector space. Also the discriminant of any order is nonzero and the ring of integers is the unique maximal order in the number field (the latter lets me think that b) is a maximal order). However, I don't seem to be able to apply all of these.
I appreciate any help.
Thanks :)
edit: thanks to a hint I know, that a subring is an order iff all of its elements are alebraic integers. I can see that all $\cos(2\pi/7), \alpha, \sqrt{2}$ and $\sqrt{-1}$ are algebraic integers. Does this imply that $\mathbb{Z}[$ of this$]$ is also a ring of algebraic integers?
I got quite a bit ahead now.
a) is not an order since the minimal polynomial of $2\pi/7$ is $f(x)=x^3+1/2x^2-1/2x-1/8\notin \mathbb{Z}[x]$
b) I found a proposition claiming: If $\alpha$ is primitive and integral, then $\mathbb{Z}[\alpha]$ is an order of discriminant equal to the discriminant of the minimal polynomial of $\alpha$. And we have a polynomial with $\alpha$ as a root given. The discriminant is positive since it has no double root. and since for $f=(x-a_1)...(x-a_n)$ we have $disc(f)=\prod_{1\leq i<j\leq n} (a_i-a_j)$
Is it a maximal order though? I am still not sure how to figure that out...
c) It is not an order as I could find an algebraic integer in $\mathbb{Q}(\sqrt{-1}, \sqrt{2})$ which is not in $\mathbb{Z}[\sqrt{-1}, \sqrt{2}]$:
$\frac{1+i}{\sqrt{2}}=1/2 \sqrt{2} + 1/2 \sqrt{2}i$