$\mathbb{Z}[\sqrt{5}]$ is not Dedekind? I am considering the following example, but confused to convince that it is not Dedekind.

The ring $\mathbb{Z}[\sqrt{5}]=\{ a+b\sqrt{5}:a,b\in\mathbb{Z}\}$ is not Dedekind, because it is not integrally closed in its field of Fractions.

Justification. The field of fractions of $\mathbb{Z}[\sqrt{5}]$ is $\mathbb{Q}[\sqrt{5}]$, and the element $(1+\sqrt{5})/2$ is integral over $\mathbb{Z}[\sqrt{5}]$ but it is not in this domain, hence this domain is not integrally closed.
Q.1 Is above fact correct with this justification?
If yes, then I have next question. It is known that the quotient of a Dedekind domain by a proper non-zero ideal is [Edited] Principal ideal ring (not necessarily domain-pointed out in answer.)
Q.2 What is the non-zero proper ideal $I$ of $\mathbb{Z}[\sqrt{5}]$ such that quotient is not principal ideal ring?

Edit: The following is an exercise from Isaacs' Algebra. This is reference for Q.2.





Q1: you are correct.
Q2: the quotient of a Dedekind domain by a proper nonzero ideal is in general not a domain, let alone a PID. Are you confusing quotients with
localisations?
[Added after OP's edit]
Try $I=(4)$.
