Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 28 / 2 / c / Solution

The polynomial is an Eisenstein polynomial at $5$, so it is irreducible and $[K:\mathbb{Q}]=3$. Its polynomial discriminant isOne must remove the index at $2$ before using this as the field discriminant. Put $\beta =({\alpha }^2+\alpha )/2$. Direct reduction using ${\alpha }^3=5\alpha -20$ givesThus $B=\mathbb{Z}+\mathbb{Z}\alpha +\mathbb{Z}\beta$ is a ring finite over $\mathbb{Z}$, so its elements are integral. Its basis has discriminant of elements of a number field $-2575=-5^2\cdot 103$. The discriminant-index formula for an integral lattice says $\mathrm{disc}(B)=[{\mathcal{O}}_K:B{]}^2{\Delta }_K$, so only $5$ could divide the remaining index.

We use these standard different ideal facts: the norm of the different is the absolute field discriminant; its localization is the local different; an unramified prime has different exponent zero, and a tamely ramified prime of ramification index $e$ has different exponent $e-1$. The Eisenstein polynomial at $5$ gives total ramification of degree three, which is tame because $5\nmid 3$. Its contribution to $v_5({\Delta }_K)$ is therefore two. This excludes a further index factor $5$, proving ${\mathcal{O}}_K=B$ and $\boxed{{\Delta }_K=-2575}$.

Since the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]=2$ is prime to $5$ and $103$, Dedekind factorization theorem applies at both primes. Modulo $5$, $f=X^3$, giving a single prime $\mathfrak{p}=(5,\alpha )$ with residue degree one and ramification index three. Modulo $103$,Thus $\mathfrak{q}=(103,\alpha -6)$ has residue degree one and ramification index two; the other prime over $103$ is unramified. Both ramified primes are tame. No other prime ramifies because none divides ${\Delta }_K$. Consequently