Let $D=-43,-67,-163$.
Then $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ is not a Euclidean domain since it does not have a universal side divisor, but how do I prove that this is a PID?
Moreover, I have shown that $D=-1,-2,-3,-7,-11$ are Euclidean domains and $D=-19$ is a PID, and it's written in oeis that $D=-1,-2,-3,-7,-11,-19,-43,-67,-163$ are the only negative discriminants that the quadratic integer ring is a UFD. How do I prove this? And why is it important? What's an advantage of checking ring types of the quaratic integer rings?
If you know any book or web-article illustrating these concisely, please do me a favor, recommend me one. Thank you in advance.
For imaginary quadratic number fields there exists a very elementary algorithm to decide whether or not the ring of integers is a PID, namely to compute the class number. If it is $1$, then the ring is a PID. The algorithm is as follows:
Let $K$ be an imaginary quadratic number field with discrminant $d$. A complete residue system of the class group is given by the ideals $$ I=a\mathbb{Z}+\frac{b+\sqrt{d}}{2}\mathbb{Z}, $$ for $a,b\in \mathbb{Z}$ with $a\ge 1$, $4a\mid d-b^2$, $\mid b\mid \le a$, $4a^2\le b^2-d$, such that if equality holds with either $|b| = a$, or $4a^2=b^2-d$, then in addition $b\ge 0$.
Example: For $K=\mathbb{Q}(\sqrt{-67})$ this algorithm gives $3a^2\le -d=67$, hence $a\le 4$. Because of $|b|≤ a ≤ 4$ and $4\mid d-b^2$ we obtain $|b|=1$ or $|b|=3$. For $|b|=1$ we have $4a\mid d-1^2=-68$, hence $a\mid 17$. Because of $a\le 4$ this implies $a=1$. The other case is similar, and we get $a=b=1$, hence the ideal class group has exactly $1$ element, so the class number is equal to $1$, so the ring of integers in $K$ $$ \mathbb{Z}\left[\frac{1+\sqrt{-67}}{2} \right] $$ is a PID.
Similarly, one can do this for $K=\mathbb{Q}(\sqrt{-43})$ and $K=\mathbb{Q}(\sqrt{-163})$. Furthermore, rings of integers are PIDs if and only they are UFDs.
References: A book on algebraic number theory, e.g. by Milne.