Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 123 / 1

The prime ideal factorization of $pB$ iswhich defines the ramification index of a prime ideal $e_{q/p}=v_q(pB)$. The residue-field degree isLocalizing at $p$ makes $B_p$ a free $R_p$-module of rank $[L:K]$, so $B/pB$ has dimension $[L:K]$ over the residue field $R/p$. The Chinese remainder theorem and the filtration by powers of each $q$ decompose this vector space into $e_{q/p}$ successive quotients isomorphic to $B/q$, each of dimension $f_{q/p}$. Therefore the fundamental identity for prime decomposition gives

The completion of a number field at a prime ideal does not change either invariant. More precisely, ${\hat{B}}_q$ is the valuation ring of $L_q$, its maximal ideal is $q{\hat{B}}_q$, andThus the completed extension has the same ramification index and the same residue-field extension, hence the same residue-field degree $f_{q/p}$. This is completion preserves ramification and residue data.

Write $G=\mathrm{Gal}(L/K)$. The decomposition group and inertia group at $q$ areThe transitivity of the Galois action on primes gives $G/D_{q/p}$ as the set of primes above $p$. Completion induces the canonical isomorphismand reduction gives the exact sequenceConsequently $|I_{q/p}|=e_{q/p}$, $|D_{q/p}/I_{q/p}|=f_{q/p}$ and, if $g$ is the number of primes above $p$,

LetOver ${\mathbb{Q}}_2$, the Newton polygon has three length-one segments of slopes $-2,-1,0$. The resulting roots can also be obtained directly from Hensel lemma. There is one unit root because $F(1)$ is even and $F^{\prime }(1)$ is odd. For the two remaining roots, putwhose parenthesized polynomial has a simple root $Y\equiv 1(\mathrm{mod}2)$, andwhose parenthesized polynomial has a simple root $Z\equiv 1(\mathrm{mod}2)$. Thus $F$ splits completely over ${\mathbb{Q}}_2$, with roots of valuations $0,1,2$. There are three primes $q$ above $2$, and for each one

Over ${\mathbb{Q}}_5$, $F$ is an Eisenstein polynomial. Hence there is one prime $q$ above $5$, and if $\alpha$ is the chosen root thenwith $e=3$ and $f=1$.

The polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $5$. Its polynomial discriminant iswhich is not a square number, so the Galois group of an irreducible cubic isAt $2$, all three roots already lie in ${\mathbb{Q}}_2$, so the local splitting field is trivial andAt $5$, the cubic is totally and tamely ramified. The square class of its discriminant is represented by $-13807\equiv 3(\mathrm{mod}5)$, a nonsquare unit, so the quadratic resolvent field of a cubic is the unramified quadratic extension of ${\mathbb{Q}}_5$. The local splitting field therefore has degree six, withThese calculations are summarized by local factorization of X3 plus 25X2 minus 50X plus 40.