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

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.

Write $\alpha ={\lambda }^{1/5}$, $M=\mathbb{Q}(\alpha )$ and $E=\mathbb{Q}({\zeta }_5)$. The Eisenstein criterion at $\lambda$ gives $[M:\mathbb{Q}]=5$, whereas the prime cyclotomic field degree gives $[E:\mathbb{Q}]=4$. The splitting field $F=E(\alpha )$ is a Galois extension of $E$, and an automorphism fixing $E$ sends $\alpha$ to ${\zeta }_5^b\alpha$. Thus its Galois group embeds in $C_5$. It is nontrivial, since $M$ cannot be contained in the degree-four field $E$. Therefore $[F:E]=5$ and $[F:\mathbb{Q}]=20$.

Index the five roots by $j\in {\mathbb{F}}_5$, with root ${\zeta }_5^j\alpha$. Every automorphism has the form ${\zeta }_5\mapsto {\zeta }_5^u$, $\alpha \mapsto {\zeta }_5^b\alpha$ and acts on indices as $j\mapsto uj+b$, with $u\in {\mathbb{F}}_5^{\times }$. This is a faithful inclusion of $G=\mathrm{Gal}(F/\mathbb{Q})$ into the affine group over a finite field, and both groups have order $20$, so it is an isomorphism. This is the affine Galois group of a prime-radical splitting field.

Let $D$ be the specified decomposition group. Its nonidentity element is an involution. If its linear coefficient were $1$, it would be a nontrivial translation, which has order $5$. Hence it acts as $j\mapsto -j+b$. There is exactly one fixed root, at $j=b/2$, and the other four roots form two pairs. By involution orbits in an affine group of odd prime degree, $D$ has three orbits on the five roots. Those roots identify with $G/\mathrm{Gal}(F/M)$, so the double-coset formula gives three primes in $M$.

For $E$, the embeddings identify with $G/\mathrm{Gal}(F/E)\simeq {\mathbb{F}}_5^{\times }$. The image of $D$ is $\{1,-1\}$ and acts by multiplication, giving the two orbits $\{1,-1\}$ and $\{2,-2\}$. ThereforeThe orbit argument counts primes whether or not $p$ is ramified; cyclicity of the given order-two decomposition group does not require that it be generated by an unramified Frobenius element.