Prove that the ring $\mathcal O_K$ of algebraic integers of $K = \Bbb Q (\sqrt d)$ ($d$ is a square free integer) is a Dedekind domain.
I have taken an ideal $I \subseteq \mathcal O_K$. Consider the set $I' = \{a - b \sqrt {d} : a + b \sqrt d \in I \}$. Then it is easy to show that $I'$ is an ideal of $\mathcal O_K$. Our instructor has left as an exercise to prove that $II' = (n)$ for some $n \in \Bbb Z$ i.e. $II'$ is principal. So for any ideal $I$ there exists $(0) \neq I' \subseteq \mathcal O_K$ such that $II'$ is a principal ideal. This will prove that $\mathcal O_K$ is a Dedekind domain. But how can I show that $II'$ is principal?
Here's one way to see it. All you gotta show it's a Noetherian integrally closed domain of dimension $1$.
First of all it is an integrally closed domain by definition (integral closure of $\mathbb Z$ in $\mathbb Q(\sqrt d)$ ).
Next we observe $\mathcal O_K |_{\mathbb Z}$ is an integral extension and hence we have $\dim \mathcal O_K=\dim \mathbb Z =1$.
Thus all we have to show is it is a Noetherian ring. This is slightly tricky. Say $\alpha +\beta \sqrt d \in \mathcal O_K$ with $\alpha, \beta \in \mathbb Q$. Then the minimal polynomial is $X^2-2\alpha X +\alpha^2-d\beta^2$. Thus we have $2\alpha , \ \alpha^2-d\beta^2 \in \mathbb Z$
The upshot is $\alpha \in \frac{1}{2} \mathbb Z , \ \beta \in \frac{1}{2}\mathbb Z $. So we have $\mathcal O_k\subset \mathbb Z$-$\mathrm{span}(\frac{1}{2}, \frac{\sqrt d}{2})$ and hence $\mathcal O_k$ is a finitely generated $\mathbb Z$-module and hence a Noetherian ring.
Thus $\mathcal O_K$ is a Dedekind Domain.