= Ray class field of Q of square root minus six modulo square root minus six
{title2=$K_{(\sqrt{-6})}=\mathbb Q(\sqrt2,\sqrt{-3})$}
For $K=\mathbb Q(\sqrt{-6})$, reduction of $\mathbb Z[\sqrt{-6}]$ modulo $(\sqrt{-6})$ gives $\mathbb Z/6\mathbb Z$. Its <unit group> is covered by the units $\{1,-1\}$ of $K$. The <ray class exact sequence> therefore identifies the <ray class group> with the ordinary <ideal class group>, of order two. Consequently this <ray class field> is the <Hilbert class field of Q of square root minus six>. To verify its lack of <ramification>, the integral basis candidate $1,\sqrt2,b,\sqrt2b$, where $b=(1+\sqrt{-3})/2$, has <discriminant> $8^2(-3)^2=576$. The relative discriminant formula also forces the field discriminant to be divisible by $(-24)^2=576$. Equality proves that the relative <discriminant ideal> is trivial.
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