= Ideal class group of Q of square root thirty-five
{c}
{title2=$\operatorname{Cl}(\mathbb Q(\sqrt{35}))\simeq\mathbb Z/2\mathbb Z$}
The <ring of integers> is $\mathbb Z[\sqrt{35}]$ and the <field discriminant> is $140$. The <Minkowski bound for ideal classes> is $\sqrt{35}<6$. At three the defining <polynomial> is irreducible, so the only <prime ideals> needed to generate the <ideal class group> are $P_2=(2,\sqrt{35}-1)$ and $P_5=(5,\sqrt{35})$, of <ideal norms> two and five. They satisfy $P_2^2=(2)$, $P_5^2=(5)$ and $P_2P_5=(5+\sqrt{35})$, so one class of <order of a group element> at most two generates. The <field norm> equation $a^2-35b^2=\pm2$ is impossible modulo five, proving $P_2$ is not a <principal ideal>. Hence the <ideal class group> has exactly two elements.
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