TITLE: Ideal class group of $\mathbb{Q}(\sqrt{-103})$
QUESTION [12 upvotes]: I want to calculate ideal class group of $\mathbb{Q}(\sqrt{-103})$. By Minkowsky bound every class has an ideal $I$ such that $N(I) \leq 6$. It is enough to consider prime ideals with the same property. Let me denote by $R$ the ring of integers. So we have
$2R= (2, \frac{1+\sqrt{-103}}{2})(2, \frac{1-\sqrt{-103}}{2})$, $3R$ and $5R$ are prime. Ideal class group is generated by $P=(2, \frac{1+\sqrt{-103}}{2})$. $N(P)=2$ and thus $P$ is not principal. Absolute norm $N(P^2)=4$ and if $P^2=(a)$ then $(a)=(2)$, but $P^2 \neq 2R$, because $2R=P \overline{P}$, $P \neq \overline{P}$. $N(P^3)$=8 and it is obvious that $P^3$ is not principal. How can I continue this argument for other powers of $P$ and when this stops?

REPLY [9 votes]: As Jyrki Lahtonen notes in his comment, the norm of the principal ideal $I=\left(\frac{5+\sqrt{-103}}{2}\right)$ is $2^5$. It will therefore be enough to show that $I=P^5$ or $I=\bar{P}^5$, since 5 is prime and therefore will have to be the exact order of $P$ in the class group.
Now, $P$ and $\bar{P}$ are the only two ideals of $R=\mathbb{Z}\left[\frac{1+\sqrt{-103}}{2}\right]$ above 2. So we immediately know that $I=P^n\bar{P}^{5-n}$ for some $n$ between 0 and 5. But if $n$ is neither 0 nor 5, then $P\bar{P}=2R$ divides (i.e. contains) $I$, which immediately leads to a contradiction, since the generator is not divisible by 2 in $R$. So $I\not\subseteq 2R$, thus $n=0$ or 5 and we are done.