Let $R=\mathbb{Z}[\sqrt{-17}]=\lbrace m+n\sqrt{-17}|m,n \in \mathbb{Z} \rbrace$.
How to show that $R$ is not a principal ideal domain?
My way was:
Let $I \subsetneq R$ be an ideal of $R$, given by $I= \lbrace m+n\sqrt{-17}|m,n \in \mathbb{Z}, m \equiv n$ (mod$ \ 2) \rbrace$.
Now I showed that it's not a principal ideal:
Suppose it exists an $\alpha \in \mathbb{R}$ with $I=\langle \alpha \rangle$.
Since $2 \in I$ and $1+\sqrt{-17} \in I$, it exists $r_1,r_2 \in R$ with $2=r_1\alpha$ and $1+\sqrt{-17}=r_2\alpha$.
For the norm map $N: R \to \mathbb{N_0}$, given by $N(a+b\sqrt{-17})=a^2+17b^2$, it's $N(\alpha_1\alpha_2)=N(\alpha_1)N(\alpha_2)$
$\Rightarrow N(r_1)N(\alpha)=N(r_1\alpha)=N(2)=4$ and $N(r_2)N(\alpha)=N(1+\sqrt{-17})=18$
$\Rightarrow N(\alpha) \mid 4$ and $N(\alpha) \mid 18$
$\Rightarrow N(\alpha)=1$ or $N(\alpha)=2$
The equation $a^2+17b^2=2$ has no solutions with $a,b \in \mathbb{Z}$.
$a^2+17b^2=1$ has $(a,b)=(\pm 1, 0)$ as the only solution.
$\Rightarrow \alpha= \pm 1$ and $I= \langle \pm 1 \rangle =R$, which is a contradiction.
So $I$ is not a principal ideal and $\mathbb{Z}[\sqrt{-17}]$ is not a principal ideal domain.
Is this proof correct or is there something wrong or to improve?
I think it is correct (although it would be nice if you provide the proof that $I$ is indeed an ideal).
Another proof is that since $81=3^4=(8+\sqrt{-17})(8-\sqrt{-17})$, $\mathbb Z[\sqrt{-17}]$ is not a unique factorization domain (check with your norm). Hence it not a principal ideal domain.