= Solution
The <minimal polynomial of an algebraic element>[minimal polynomial] of $\beta$ is
$$
g(T)=T^3-5T^2+11T-12,
$$
and part (b) showed that $\mathcal O_K=\mathbb Z[\beta]$. Modulo $3$,
$$
\overline g(T)
=T(T^2+T-1).
$$
The quadratic has no root in $\mathbb F_3$, so it is irreducible. The <Dedekind factorization theorem> therefore gives
$$
\boxed{
(3)=(3,\beta)(3,\beta^2+\beta-1)
}.
$$
The two factors are distinct <prime ideals>[prime ideal], with norms $3$ and $3^2=9$, respectively.
Modulo $2$,
$$
\overline g(T)
=T(T^2+T+1),
$$
and the quadratic is irreducible over $\mathbb F_2$. Hence
$$
\boxed{
(2)=(2,\beta)(2,\beta^2+\beta+1)
}.
$$
This is a product of two distinct proper prime ideals, so $(2)$ is not prime. These decompositions are collected in <prime ideals above two and three in the cubic field of discriminant minus 307>.
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