Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 20G (Number Fields)

Evaluation at $\alpha$ gives the ring isomorphismand reducing modulo $p$ givesFor each $i$,This quotient is a field because ${\overline{g}}_i$ is irreducible. Hence $P_i$ is a nonzero prime ideal; it is nonzero because it contains $p$. Its ideal norm is

The ideals are distinct. Indeed, if $P_i=P_j$ with $i\ne j$, then both ${\overline{g}}_i$ and ${\overline{g}}_j$ would vanish in the same quotient. Since distinct monic irreducibles are coprime, a Bezout identity between them would make $1$ vanish there, a contradiction.

Because the chosen lifts satisfyfor some $H\in \mathbb{Z}[X]$, evaluation at $\alpha$ givesEvery generator of the product ideal ${\prod }_i{P}_i^{e_i}$ either contains a factor $p$ or is the displayed product, soOn the other hand, multiplicativity of ideal norms and ${\sum }_i{e}_i\mathrm{deg}{\overline{g}}_i=\mathrm{deg}f=[L:\mathbb{Q}]$ giveTwo finite-index ideals with one contained in the other and equal index are equal. Thus the Dedekind factorization theorem is proved in this monogenic case:

Taking norms ingivesso $s=n$.

Suppose for contradiction that ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$, and let $f$ be the degree-$n$ minimal polynomial of $\alpha$. Part (a) identifies the prime ideals above $p$ with the distinct irreducible factors of $\overline{f}$. Since there are $n$ distinct prime ideals, all with norm $p$, the norm formula forces $\overline{f}$ to have $n$ distinct irreducible factors of degree one. But ${\mathbb{F}}_p[X]$ has only the $p$ distinct monic linear polynomialsThis is impossible when $n>p$. The completely split small-prime obstruction to monogenicity therefore gives