Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 20G / b

The polynomial of $\alpha$ satisfies the Eisenstein criterion at $p$. Let $\mathfrak{p}$ be any prime ideal of ${\mathcal{O}}_K$ above $p$, normalize its ideal valuation by $v_{\mathfrak{p}}(\mathfrak{p})=1$, and write $e=v_{\mathfrak{p}}(p)$ for its ramification index of a prime ideal. Put $t=v_{\mathfrak{p}}(\alpha )$. Inthe constant term has valuation $e$, while every other $a_j{\alpha }^j$ has valuation at least $e+jt$. The non-Archimedean valuation property then forcesIndeed, either strict inequality would leave a unique term of least valuation in the equation. Since $t$ is a positive integer and $e\le n$, it follows that $t=1$ and $e=n$.

The fundamental inequality ${\sum }_{\mathfrak{p}\mid p}{e}_{\mathfrak{p}}{f}_{\mathfrak{p}}\le n$ now shows that $\mathfrak{p}$ is the unique prime above $p$ and has residue degree one. At this prime,while $(p,\alpha )$ has no factor away from $p$. Unique prime ideal factorization therefore givesFinally $v_P(\alpha )=1$, soThis is total ramification from an Eisenstein polynomial.

Suppose that $p$ divided the finite index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$. Then the quotient would contain nonzero $p$-torsion, so there would be $\beta \in {\mathcal{O}}_K\setminus \mathbb{Z}[\alpha ]$ withwhere not all $b_j$ are divisible by $p$. Let $k$ be the least index for which $p\nmid b_k$. Since $v_P(\alpha )=1$ and $v_P(p)=n$, the term $b_k{\alpha }^k$ is the unique term on the right of least $P$-valuation. ThusBut $\beta$ is an algebraic integer, so $v_P(\beta )\ge 0$ anda contradiction. Therefore

The polynomial $T^3+3T+3$ is Eisenstein at $3$, so part (ii) givesFor a cubic $T^3+aT+b$, the polynomial discriminant is $-4a^3-27b^2$. HereThe discriminant-index formula for an integral lattice saysAny prime dividing the index must therefore occur to at least the second power in $351$; the only possibility is $3$, which has already been excluded. The index is one, and hence