Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 31 / 1 / i / Solution

Let $L/K$ be a finite Galois extension of number fields, with Galois group $G$, and let $\mathfrak{p}$ be a nonzero prime ideal of ${\mathcal{O}}_K$. Write its prime ideal factorization asThe Galois group acts transitively on the primes above $\mathfrak{p}$. One proof uses the Chinese remainder theorem. If $\mathfrak{Q}$ were outside the orbit of $\mathfrak{P}$, choose an integral element $a$ which is zero modulo $\mathfrak{Q}$ and one modulo every conjugate of $\mathfrak{P}$. Its field norm ${\prod }_{\sigma \in G}\sigma (a)$ belongs to $\mathfrak{Q}\cap {\mathcal{O}}_K=\mathfrak{p}$, hence to $\mathfrak{P}$. But each factor is one modulo $\mathfrak{P}$, a contradiction. Transitivity also makes all $e_j$ equal to a common ramification index $e$ and all $f_j$ equal to a common residue degree $f$.

Fix $\mathfrak{P}$ above $\mathfrak{p}$. Its decomposition group is the stabilizerThe orbit-stabilizer theorem gives $|D_{\mathfrak{P}}|=|G|/g$. Elements of $D_{\mathfrak{P}}$ act on the residue field $\kappa (\mathfrak{P})$ over $\kappa (\mathfrak{p})$; the group kernel is the inertia groupThus inertia measures the automorphisms invisible after reduction, whereas decomposition measures all automorphisms preserving the chosen prime.

The completion $L_{\mathfrak{P}}/K_{\mathfrak{p}}$ is a Galois extension with group $D_{\mathfrak{P}}$: preservation of $\mathfrak{P}$ lets an automorphism extend continuously to the completion, and the local embeddings of this normal extension arise in this way. The degree of a field extension formula is $[L_{\mathfrak{P}}:K_{\mathfrak{p}}]=ef$. Reduction gives the exact sequenceFor surjectivity, lift a primitive residue-field generator by the Hensel lemma to obtain the maximal unramified subextension of a local field extension of $L_{\mathfrak{P}}/K_{\mathfrak{p}}$. Its automorphisms are the residue-field automorphisms. They extend to $L_{\mathfrak{P}}$ because the latter is normal. Since a finite residue field extension has a cyclic Galois group of order $f$, the exact sequence yieldsThe fixed field of local inertia is the maximal unramified subextension of a local field extension, and the extension above it is totally ramified. Primes in the same global orbit have conjugate decomposition groups and inertia groups, so the orders do not depend on the chosen prime.

If $e=1$, inertia is trivial and $D_{\mathfrak{P}}$ is generated by the unique Frobenius automorphism acting on residues by $z\mapsto z^{N\mathfrak{p}}$. Its order is $f$. In a ramified extension this residue action determines only a coset modulo inertia. A prime splits completely exactly when $e=f=1$, equivalently its decomposition group is trivial. A totally ramified prime has $g=f=1$ and $I_{\mathfrak{P}}=D_{\mathfrak{P}}=G$; an unramified prime has $I_{\mathfrak{P}}=1$. In an abelian extension, conjugation does nothing, so the associated groups and unramified Artin symbol are independent of the prime above $\mathfrak{p}$.