Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 26 / 2

Write $G=\mathrm{Gal}(F/K)$ andEvery member of this decomposition group preserves ${\mathcal{O}}_F$ and $\mathfrak{q}$, so reduction defines a homomorphismIts kernel is the inertia group. We prove surjectivity of decomposition group reduction without presupposing a lift of Frobenius.

If the two residue fields coincide, surjectivity is immediate. Otherwise choose a nonzero $\overline{a}\in k_{\mathfrak{q}}$ generating it over $k_{\mathfrak{p}}$; a generator exists by the primitive element theorem for finite fields. Use the Chinese remainder theorem in ${\mathcal{O}}_F$ to choose $a$ whose residue is $\overline{a}$ at $\mathfrak{q}$ and zero at every other prime above $\mathfrak{p}$. ConsiderIts coefficients are integral and fixed by $G$, hence lie in ${\mathcal{O}}_K$: an element of $K$ integral over ${\mathcal{O}}_K$ belongs to this integrally closed ring of integers.

Reduce at $\mathfrak{q}$. When $\sigma \notin D_{\mathfrak{q}}$, the chosen residue of $a$ at ${\sigma }^{-1}\mathfrak{q}$ is zero, so $\sigma a$ reduces to zero at $\mathfrak{q}$. When $\sigma \in D_{\mathfrak{q}}$, its residue is $\overline{\sigma }(\overline{a})$, which is nonzero. ThereforeAlthough this product is written in $k_{\mathfrak{q}}[T]$, its coefficients belong to $k_{\mathfrak{p}}$. Put $q=|k_{\mathfrak{p}}|$. The finite-field Frobenius automorphism $b\mapsto b^q$ therefore permutes its nonzero roots. In particular, ${\overline{a}}^q$ is $\overline{\sigma }(\overline{a})$ for some $\sigma \in D_{\mathfrak{q}}$. Since $\overline{a}$ generates $k_{\mathfrak{q}}$, that $\overline{\sigma }$ agrees with Frobenius on the whole field. The Galois group of a finite field extension is cyclic and generated by Frobenius, so its generator lies in the image. Thus

Put $a=\sqrt[3]{2}$ and $\zeta ={\zeta }_3$. The polynomial $X^3-2$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $2$, so $[\mathbb{Q}(a):\mathbb{Q}]=3$. This field is real and cannot contain the nonreal $\zeta$. Adjoining $\zeta$ gives a quadratic extension and hence $[F:\mathbb{Q}]=6$. Since $F$ is the splitting field of $X^3-2$, its Galois group acts faithfully on $a,\zeta a,{\zeta }^{2}a$ and has order six. Consequently

For a prime $\mathfrak{q}$ over a rational prime $p$, its absolute residue degree is the order of the cyclic quotient group $D_{\mathfrak{q}}/I_{\mathfrak{q}}$ from part (i). If that degree were six, then $D_{\mathfrak{q}}$, a subgroup of $S_3$, would have order at least six, so $D_{\mathfrak{q}}=S_3$ and $I_{\mathfrak{q}}=1$. This would make $S_3$ cyclic, a contradiction. The argument applies also at ramified primes. Thus

For the second request printed with the same subpart label, work over $E=\mathbb{Q}(\zeta )$. The Galois group of $F/E$ is cyclic of order three, with generator $\sigma (a)=\zeta a$ and $\sigma (\zeta )=\zeta$. The ideal norm of $(\zeta +3)$ is seven, sinceIn its residue field we have $\zeta \equiv -3\equiv 4(\mathrm{mod}{\mathfrak{p}}_1)$. The polynomial $X^3-2$ has no root in ${\mathbb{F}}_7$, because its cube residues are $0,1,6$. It is therefore irreducible there. Its discriminant $-108$ is a unit at ${\mathfrak{p}}_1$. To check the local index condition, the trace pairing matrix of $1,a,a^2$ has this unit determinant. For any integral element $b$ of $F$, its traces against these three basis elements are integral; solving with this invertible matrix shows that its basis coefficients are integral locally. Thus the relative ring of integers equals ${\mathcal{O}}_E[a]$ at ${\mathfrak{p}}_1$. The relative Kummer-Dedekind theorem gives one unramified prime above ${\mathfrak{p}}_1$, of residue degree three.

Use arithmetic Frobenius, characterized by $b\mapsto b^7$ on this residue field. As $a^3=2$,The three possible images $a,\zeta a,{\zeta }^{2}a$ remain distinct modulo $\mathfrak{q}$, since $a$ is nonzero and $1,4,2$ are distinct modulo seven. This congruence therefore identifies the actual automorphism uniquely:The geometric Frobenius convention would give its inverse. Since $F/E$ is abelian, the arithmetic element is independent of the choice of prime above ${\mathfrak{p}}_1$.