= 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 $(a-b)/2+b\omega$, proving
$$
\boxed{\mathcal O_K=\mathbb Z+\mathbb Z\omega,\qquad D_K=-23.}
$$
Both 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 gives
$$
(2)=\mathfrak p_2\overline{\mathfrak p}_2,\quad\mathfrak p_2=(2,\omega),\quad\overline{\mathfrak p}_2=(2,\omega-1),
$$
$$
(3)=\mathfrak p_3\overline{\mathfrak p}_3,\quad\mathfrak p_3=(3,\omega),\quad\overline{\mathfrak p}_3=(3,\omega-1).
$$
Their <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. Hence
$$
\boxed{\mathfrak p_2\mathfrak p_3=(\omega).}
$$
The 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 $b\ne0$ 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
$$
\boxed{h_K=3.}
$$
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