Local factorization and extended absolute values

If $L=K(\alpha )$ is a finite separable extension and $v$ is an absolute value on $K$, the extensions of $v$ to $L$ correspond to the irreducible factors of the minimal polynomial of $\alpha$ over the completion of $K$ at $v$.

This polynomial is Eisenstein at $2$. Over ${\mathbb{Q}}_3$, its Newton polygon has a single slope $-2/5$, so it is irreducible and a root generates a totally ramified extension of degree five. Over ${\mathbb{Q}}_7$, its factor degrees are $1,1,3$: the simple residue roots $1,-1$ lift, and the remaining roots have valuation $1/3$. Thus the degree-five number field generated by a root has one prime above $3$ and three above $7$.

For a finite extension of number fields $L/K$ and a nonzero prime ideal $\mathfrak{p}$ of $K$, the primes of $L$ above $\mathfrak{p}$ correspond to the orbits ofAn embedding determines a prime by pulling back the local absolute value on a field. Embeddings with the same prime extend to embeddings of the same completion and are conjugate. If $L=K(\alpha )$, these orbits are the irreducible factors of its minimal polynomial over $K_{\mathfrak{p}}$.

For a number field $K$ and a nonzero prime ideal $\mathfrak{p}\subset {\mathcal{O}}_K$, the completion $K_{\mathfrak{p}}$ is the completion for the $\mathfrak{p}$-adic absolute value. It is a finite extension of ${\mathbb{Q}}_p$, where $p\mathbb{Z}=\mathfrak{p}\cap \mathbb{Z}$.

For a number field $K$ and a rational prime $p$, the natural diagonal map gives an isomorphism of ${\mathbb{Q}}_p$-algebras

If $L/K$ is a finite Galois extension of number fields, its Galois group acts transitively on the prime ideals of ${\mathcal{O}}_L$ lying above a fixed prime ideal of ${\mathcal{O}}_K$.

For a prime $\mathcal{P}$ of a finite Galois extension $L/K$ above $\mathfrak{p}$, the decomposition group is the stabilizer

For a finite Galois extension of number fields, choose a nonzero residue field generator at $\mathfrak{q}$ and lift it to an integral element $a$ whose residues at the other primes above $\mathfrak{p}$ are zero, using the Chinese remainder theorem. The orbit polynomial ${\prod }_{\sigma \in G}(T-\sigma a)$ has coefficients in the base ring of integers. Modulo $\mathfrak{q}$, its nonzero roots are precisely the orbit of $\overline{a}$ under the decomposition group. Its coefficients belong to $k_{\mathfrak{p}}$, so finite-field Frobenius automorphism permutes these roots. Consequently some decomposition-group element acts as Frobenius on the generator, and therefore on the entire residue field. Frobenius generates the finite-field Galois group, proving surjectivity.

Let $M/K$ be a finite Galois extension, $L=M^H$, and $D$ the decomposition group of a prime $\mathfrak{P}$ above $\mathfrak{p}$. The bijectionfollows from transitivity on primes for both $M/K$ and $M/L$.

If $M/K$ is the Galois closure of a finite extension $L/K$, a prime splits completely in $L$ exactly when it splits completely in $M$. Complete splitting means every right action of $D$ on $H\backslash G$ fixes each coset, so $D$ lies in the subgroup core of $H$. That core is trivial because the conjugates of $L$ generate $M$.

Restriction to completions gives a canonical isomorphismConsequently the number of primes above $\mathfrak{p}$ is the index of the decomposition group.

For a prime $\mathfrak{q}$ of a finite separable extension $L/K$ above $\mathfrak{p}$, completion preserves the ramification index and residue-field extension:

The polynomialsplits into three linear factors over ${\mathbb{Q}}_2$. Over ${\mathbb{Q}}_5$ it is an Eisenstein polynomial, so adjoining a root gives a totally ramified cubic extension. Its splitting field over ${\mathbb{Q}}_5$ has decomposition group $S_3$ and inertia group $A_3\cong C_3$.