 Solution

The element $\omega$ satisfies ${\omega }^2-\omega +6=0$, so is an algebraic integer. Conversely, write an algebraic integer as $(a+b\sqrt{-23})/2$ with $a,b$ rational. Its trace gives $a\in \mathbb{Z}$, and its field norm implies $23b^2\in \mathbb{Z}$. If $b=p/q$ in lowest terms, $q^2$ divides 23, forcing $q=1$. The integral field norm condition modulo four then makes $a,b$ have the same parity. Therefore it equals $(a-b)/2+b\omega$, provingBoth reductions of $x^2-x+6$ modulo two and three split into the distinct roots zero and one. The index is one, so ideal factorization givesTheir norms are two and three respectively. The product ${\mathfrak{p}}_2{\mathfrak{p}}_3$ is generated by $6,2\omega ,3\omega ,{\omega }^2$. All are multiples of $\omega$, using $6=\omega \overline{\omega }$; and $\omega =3\omega -2\omega$ belongs to the product. HenceThe Minkowski bound is $(2/\pi )\sqrt{23}<4$, in fact about $3.05$. Every class therefore has a representative of ideal norm one, two or three. The displayed primes and their conjugates exhaust those possibilities. Their classes are all $1,c,c^{-1}$, with $c=[{\mathfrak{p}}_2]$, because $[{\mathfrak{p}}_3]=c^{-1}$ and conjugation inverts a quadratic ideal class. There are at most three classes.

There is no element of field norm two: $N(a+b\omega )=a^2+ab+6b^2=(a+b/2{)}^2+23b^2/4$, which exceeds two for $b\ne 0$ and is a square for $b=0$. Thus ${\mathfrak{p}}_2$ is not principal. Moreover $N(\omega -2)=8$, and $\omega -2$ belongs to ${\mathfrak{p}}_2$ but not its conjugate, so $(\omega -2)={\mathfrak{p}}_2^3$. Consequently $c$ has order three, and