 Solution

Since $\mathcal{O}=\mathbb{Z}[\theta ]$, reduction gives $\mathcal{O}/p\mathcal{O}\cong {\mathbb{F}}_p[X]/(\bar{g})$. The maximal ideals of this quotient are precisely those generated by the distinct irreducible factors ${\bar{g}}_i$. Their inverse images are thereforethe distinct prime ideals above $p$. In fact $\mathcal{O}/{\mathfrak{p}}_i\cong {\mathbb{F}}_p[X]/({\bar{g}}_i)$ is a field of size $p^{\mathrm{deg}{\bar{g}}_i}$. Every prime containing $p$ arises this way; independence of the choice of lift $g_i$ follows because another lift differs by a multiple of $p$.

To prove the ideal factorization, set $J={\prod }_i{\mathfrak{p}}_i^{e_i}$. A generator of the product either contains a factor $p$, and hence lies in $p\mathcal{O}$, or is the product ${\prod }_i{g}_i(\theta {)}^{e_i}$. The latter also lies in $p\mathcal{O}$, since its polynomial reduces to $\bar{g}$ and $g(\theta )=0$. Thus $J\subset p\mathcal{O}$. The ideal norm is multiplicative for nonzero ideals in the ring of integers; henceEqual finite indices together with the containment imply equality. ThereforeThis proves the Dedekind factorization theorem in the stated monogenic case by containment and norms, rather than assuming the desired factorization.