Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 32 / 2 / i / Solution

Write $G=\mathrm{Gal}(F/K)$ and fix a prime ideal $\mathfrak{q}$ above $\mathfrak{p}$. Suppose there is another prime $\mathfrak{r}$ above $\mathfrak{p}$ outside the $G$-orbit of $\mathfrak{q}$. Distinct nonzero prime ideals in the ring of integers are distinct maximal ideals, hence pairwise comaximal. The Chinese remainder theorem supplies $a\in {\mathcal{O}}_F$ satisfying $a\equiv 0(\mathrm{mod}\mathfrak{r})$ and $a\equiv 1(\mathrm{mod}\sigma \mathfrak{q})$ for every $\sigma \in G$. Its field norm isThe factor $a$ puts $b$ in $\mathfrak{r}$, hence in $\mathfrak{r}\cap {\mathcal{O}}_K=\mathfrak{p}$. But every $\sigma (a)$ is congruent to $1$ modulo $\mathfrak{q}$, since $a\equiv 1(\mathrm{mod}{\sigma }^{-1}\mathfrak{q})$. Thus $b\equiv 1(\mathrm{mod}\mathfrak{q})$, contradicting $b\in \mathfrak{p}\subset \mathfrak{q}$. This proves transitivity of the Galois action on primes without any unramified hypothesis.

Let $D={\mathrm{Stab}}_G(\mathfrak{q})$ be the decomposition group and $H=\mathrm{Gal}(F/L)$. Transitivity identifies the primes of $F$ above $\mathfrak{p}$ with the coset space $G/D$. Two such primes contract to the same prime in $L$ exactly when they are in the same $H$-orbit: apply the transitivity just proved to the Galois extension $F/L$. Therefore primes in an intermediate field as double cosets givesThe second equality follows by inversion of representatives. It allows one to count $D$-orbits on the embeddings of $L$ into $F$, represented by $G/H$. If $H$ is normal, this simplifies to $[G:HD]$; the double-coset formulation also handles nonnormal intermediate fields.