Extension of an absolute value

For a field extension $L/K$, an extension of an absolute value $|\cdot |$ on $K$ is an absolute value on $L$ whose restriction to $K$ is $|\cdot |$.

If $K$ is complete and $L/K$ is finite, an absolute value on $K$ extends uniquely to $L$. With the normalization that it restricts exactly to the given absolute value,Every $K$-automorphism of $L$ preserves this extended absolute value.

Let $K$ be complete for a Non-Archimedean absolute value and let $V$ be a finite-dimensional vector space with a norm satisfying $\Vert av\Vert =|a|\Vert v\Vert$ and the ultrametric inequality. All such norms are equivalent to the maximum of coordinates in any basis. The upper bound is immediate. For the lower bound, induct on dimension. The span $W$ of all but the last basis vector is complete by induction, hence closed for the given norm. Thus $\delta ={\inf }_{w\in W}\Vert e_n-w\Vert >0$. For $v=w+a{e}_n$, $\Vert v\Vert \ge |a|\delta$; also $\Vert w\Vert \le \max (\Vert v\Vert ,|a|\Vert e_n\Vert )$. The induction bound on coordinates of $w$ now bounds all coordinates by a constant times $\Vert v\Vert$. This proof needs no local compactness.

On a finite-dimensional vector space over ${\mathbb{Q}}_p$, the triangle inequality gives an upper bound by a coordinate maximum norm. The coordinate unit sphere is compact. The other norm is continuous and nonzero there, so has a positive minimum. Scaling by powers of $p$ gives the lower bound. The resulting equivalence of norms in finite dimensions proves completeness of finite extensions, and applying the comparison to powers of a field element proves equality of two extending multiplicative absolute values on a field.

Suppose a Non-Archimedean absolute value extends uniquely to every finite extension. If $f\in {\mathcal{O}}_K[X]$ is monic and irreducible, every root has absolute value at most one: otherwise its leading power dominates all lower terms, contradicting $f(\alpha )=0$ by the ultrametric inequality. In a splitting field, let $\phi$ be the minimal polynomial of an algebraic element of the residue of one root. Lift its coefficients to ${\mathcal{O}}_K$. The lift has absolute value less than one at that root and hence, by equal absolute values of algebraic conjugates, at every root. Consequently every root residue is a zero of $\phi$. Since reduction preserves the product of the linear factors with multiplicities, every irreducible factor of $\overline{f}$ is $\phi$, proving the displayed identity for some $m\ge 1$.

Suppose a Non-Archimedean absolute value on $K$ has a unique extension to every finite field extension. Any K-isomorphism $K({\alpha }_i)\to K({\alpha }_j)$ between conjugate roots pulls the absolute value back to an extension on the source field, which must be the given one. Thus $|h({\alpha }_i)|=|h({\alpha }_j)|$ for every $h\in K[X]$. This argument includes inseparable irreducible polynomials: no transitive Galois action or completeness assumption is needed.

Let $K$ be complete for a discrete valuation and let $L/K$ be finite. The valuation ring ${\mathcal{O}}_L$ is exactly the integral closure of ${\mathcal{O}}_K$ in $L$. One direction follows from a monic integral equation; for the other, the coefficients of the minimal polynomial are elementary symmetric polynomials in conjugates of absolute value at most one.

If $L/K$ is a finite normal extension, the Galois group acts transitively on the absolute values of $L$ extending a fixed absolute value of $K$. For number fields, this is equivalent to transitivity on the prime ideals above a fixed prime.

Give $\mathbb{Q}$ its $5$-adic valuation, take $L=\mathbb{Q}(i)$, and choose the extension associated with the prime $(2+i)$. Then $(2+i)/(2-i)$ lies in the chosen valuation ring but has negative valuation at the conjugate prime. It is therefore not in the integral closure of ${\mathbb{Z}}_{(5)}$ in $L$.

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 $A_3\cong C_3$.