Ideal class group of Q of square root 229 ($\mathrm{Cl}(\mathbb{Q}(\sqrt{229}))\cong C_3$)

Put $\omega =(1+\sqrt{229})/2$, $P=(3,\omega )$ and $Q=(5,\omega -4)$. The Minkowski bound for ideal classes is $\sqrt{229}/2<8$, so the ideal class group is generated by primes above three and five; two and seven are inert. The prime ideal factorizations $(6+\omega )=PQ$ and $(15+2\omega )=P^3$ reduce it to a cyclic group of order dividing three. It is not trivial: the unit $ϵ=(15+\sqrt{229})/2$ has field norm minus one. If an algebraic integer had norm of absolute value three, multiplying by a power of this unit would make both real absolute values at most $\sqrt{3ϵ}$. For $a=x+y\omega$, this bounds $|y|\sqrt{229}=|a-a^{\prime }|\le 2\sqrt{3ϵ}<\sqrt{229}$, forcing $y=0$, which cannot have norm of absolute value three.