= Ideal class group of Q of square root 229
{title2=$\operatorname{Cl}(\mathbb Q(\sqrt{229}))\cong C_3$}
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 $\epsilon=(15+\sqrt{229})/2$ 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 $\sqrt{3\epsilon}$. For $a=x+y\omega$, this bounds $|y|\sqrt{229}=|a-a'|\leq2\sqrt{3\epsilon}<\sqrt{229}$, forcing $y=0$, which cannot have norm of absolute value three.
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