Let $d$ be a non-square integer and $\alpha$ be nonzero in $\mathbb{Z}[\sqrt d]$ with norm $N$, so $N = \alpha\overline{\alpha}$. Show that the principal ideal $(\alpha)$ in $\mathbb{Z}[\sqrt d]$ has index $|N|$. That is, show $\mathbb{Z}[\sqrt d]/(\alpha)$ has order $|N|$. (Hint: Consider the chain of ideals $\mathbb{Z}[\sqrt d] \supset (\alpha) \supset (N))$.
I really don't know where to start with this one; I've never really dealt with finding order of quotient rings, although I imagine it's similar to doing so in quotient groups. The hint makes me think of Artinian rings, but I can't see what this would have to do with this problem.
Here ideals are just subgroups. It is saying that the quotient groups $\Bbb{Z}[\sqrt{d}]/(\alpha),\Bbb{Z}[\sqrt{d}]/(\overline{\alpha})$ have the same number of elements $r$ and the latter is isomorphic to $\alpha\Bbb{Z}[\sqrt{d}]/\alpha(\overline{\alpha})=(\alpha)/(N)$,
since the index of subgroups satisfy $[A:C]=[A:B][B:C]$ then $\Bbb{Z}[\sqrt{d}]/(N)$ has $r^2$ elements
On the other hand it is easy to show that $\Bbb{Z}[\sqrt{d}]/(N)$ has $N^2$ elements