Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 123 / 5

The decomposition group of the chosen prime is its stabilizer under the Galois group:The transitivity of the Galois action on primes identifies the primes of $M$ above $\mathfrak{p}$ with $G/D$ through $gD\mapsto g\mathfrak{P}$.

Put $H=\mathrm{Gal}(M/L)$. Contraction of prime ideals to ${\mathcal{O}}_L$ is constant on every $H$-orbit, because every element of $H$ fixes $L$ pointwise. It is surjective onto the primes of $L$ above $\mathfrak{p}$, since prime ideals extend in the integral extension ${\mathcal{O}}_M/{\mathcal{O}}_L$. Moreover, two primes of $M$ contracting to the same prime of $L$ are conjugate by $H$: the extension $M/L$ is Galois and the same transitivity theorem applies to it.

Thus the fibers of contraction are exactly the $H$-orbits on $G/D$. These orbits are the double cosets $H\backslash G/D$, proving the primes in an intermediate field as double cosets correspondenceExplicitly, replacing $g$ by $hgd$ does not change the contracted prime. Conversely, if $g\mathfrak{P}$ and $g^{\prime }\mathfrak{P}$ contract to the same prime, some $h\in H$ sends the former to the latter, so $g^{\prime -1}hg\in D$ and $H{g}^{\prime }D=HgD$. This also verifies the stated order of the two sides of the double coset.

A prime is completely split in an extension of degree $n$ exactly when there are $n$ primes above it. Indeed, the fundamental identity for prime decomposition is ${\sum }_{\mathfrak{q}\mid \mathfrak{p}}{e}_{\mathfrak{q}}{f}_{\mathfrak{q}}=n$, and every summand is a positive integer. Thus $n$ primes means that every ramification index and residue degree equals one, and conversely.

By part (a), the number of primes of $L$ above $\mathfrak{p}$ is the number of right $D$-orbits on $H\backslash G$. This set has $[G:H]=[L:K]$ elements. It has that many orbits exactly when every orbit is a singleton, equivalently whenThis condition says $gD{g}^{-1}\subseteq H$ for every $g$, orthe subgroup core.

Because $M$ is the Galois closure of $L/K$, it is generated by all the fields $g(L)$. The fixed field of $gH{g}^{-1}$ is $g(L)$, so the fixed field of the core is their compositum, namely $M$. The Galois correspondence therefore gives ${\mathrm{Core}}_G(H)=1$. HenceIn the Galois extension $M/K$, the number of primes above $\mathfrak{p}$ is $[G:D]$, so $D=1$ is equivalent to complete splitting in $M$. This proves that complete splitting is preserved by Galois closure:No prior unramifiedness assumption is needed: complete splitting already forces all ramification indices to be one, and the decomposition-group argument includes the ramified case.