LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Number field completions as local polynomial factors

Statement

Let L/K be a finite separable extension of number fields, $L = K (α)$ with monic minimal polynomial F, and p a finite prime of K. Factor F over $K_{p}$ into distinct monic irreducibles $F_{i}$ . Then $L \otimes_{K} K_{p} ≅ \prod_{i} K_{p} [T] / (F_{i}) ≅ \prod_{P ∣ p} L_{P}, \sum_{P ∣ p} [L_{P} : K_{p}] = [L : K] .$ Use extending absolute values on each factor; their positive powers give the normalized number-field completions. Under this product, local multiplication matrices give $N_{L / K} (x) = \prod_{P ∣ p} N_{L_{P} / K_{p}} (x)$ and ${Tr}_{L / K} (x) = \sum_{P ∣ p} {Tr}_{L_{P} / K_{p}} (x)$ , with values embedded in $K_{p}$ .

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Completion of a number field at a prime: For a nonzero prime P of $O_{K}$ , let $K_{P}$ be the completion at $∣ x ∣_{P} = (N P)^{- {ord}_{P} x}$ . Its valuation ring has residue field $O_{K} / P$ , since the original valuation ring is $(O_{K})_{P}$ and completion preserves residues. In L/K with $P ∣ p$ , the normalized value restricts as $∣ \cdot ∣_{P} ∣_{K} = ∣ \cdot ∣_{p}^{e f}$ , because $N P = (N p)^{f}$ and ${ord}_{P} ∣_{K} = e {ord}_{p}$ . When a literal extension of $∣ \cdot ∣_{p}$ is needed use $∣ \cdot ∣_{P}^{1 / (e f)}$ . Positive powers define the same topology and completion.

[F2]

Unique extension of a nonarchimedean absolute value: For every finite field extension L/K with K complete nonarchimedean, the unique extending absolute value is $∣ x ∣_{L} = ∣ N_{L / K} (x) ∣_{K}^{1 / [L : K]} .$ It is nonarchimedean and makes L complete. Separability and discreteness are not assumed; the trivial valuation is included.

[F3]

Finite dimensional norm equivalence over a complete valued field: Let F be complete for a multiplicative absolute value and V a finite-dimensional normed F-vector space. For any basis $v_{1}, \dots, v_{n}$ , its coordinate sup norm $∥ \sum a_{i} v_{i} ∥_{\infty} = \max_{i} ∣ a_{i} ∣$ is bounded above and below by positive multiples of the given norm. For $n = 0$ both norms are zero. Consequently V is complete and every linear subspace is closed.

[F4]

A finite extension generated by elements all but possibly one of which are separable is simple: Let $E = F (α_{1}, \dots, α_{r})$ be a finite extension. If all but possibly one of the generators are separable over $F$ , then $E / F$ is simple. In particular, every finite separable extension is simple.

[F5]

Chinese remainder theorem for pairwise comaximal ideals: Let $R$ be a commutative ring and let $I_{1}, \dots, I_{r}$ be pairwise comaximal ideals, where $r \geq 1$ . Then the canonical map $R ⟶ \prod_{i = 1}^{r} R / I_{i}, x ⟼ (x + I_{1}, \dots, x + I_{r})$ is surjective, its kernel is $⋂_{i = 1}^{r} I_{i}$ , and $⋂_{i = 1}^{r} I_{i} = \prod_{i = 1}^{r} I_{i} .$ Equivalently, $R / \prod_{i = 1}^{r} I_{i} ≅ \prod_{i = 1}^{r} R / I_{i} .$

[F6]

Number field places classification: The nonarchimedean places of a number field are in bijection with the nonzero primes of its ring of integers, with representative $∣ x ∣_{P} = (N P)^{- {ord}_{P} x}$ .

Proof

1.1

The primitive-element theorem supplies alpha if needed. The presentation $L = K [T] / (F)$ remains $K_{p} [T] / (F)$ after scalar extension, as is seen on the power basis. Separability gives a Bezout identity for F,F' over K, hence over $K_{p}$ , so the irreducible factors remain distinct. Polynomial CRT gives the product of factor fields.

F4F5

2.1

Each factor E has the unique extending absolute value and is complete. Its element $α_{i} = T m o d F_{i}$ generates E over $K_{p}$ . Approximating each coefficient of a finite polynomial in $α_{i}$ by elements of K shows that the image of L in E is dense. Thus E is the completion of the induced nonarchimedean place on L. Its restriction to K is the p-adic place, so [F6] classifies it by a unique prime P of $O_{L}$ above p.

F1F2F3F6step 1.1

3.1

Conversely the inclusion K into $L_{P}$ , using the extending power normalization, extends to $K_{p}$ . The natural algebra map $L \otimes_{K} K_{p} \to L_{P}$ has finite-dimensional image over $K_{p}$ . With its inherited norm this image is complete and therefore closed; it also contains the dense L. Hence the map is surjective onto the field $L_{P}$ , and so factors through exactly one of the displayed factor fields. Two factors cannot induce the same place: equivalent extending values agree on K and hence have exponent one, so the completion isometry fixes K and alpha and, by density, $K_{p}$ ; the minimal polynomial of alpha over $K_{p}$ would then be the same factor. This establishes the bijection.

F1F3step 1.1step 2.1

4.1

Dimensions in the finite product add to the degree of F. For x in L, scalar extension of its multiplication matrix preserves its determinant and trace; in the product it becomes block diagonal with the local multiplication matrices. The determinant of a block diagonal matrix is the product of its block determinants and its trace their sum. This proves the norm and trace formulas, including x=0 and degree one.

step 1.1step 3.1algebra∎

Depends on

Used by

Decomposition group and completion Theorem

Dependency tree · two levels

36 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Chapter 8, Propositions 8.1–8.2, pp.135–136; Conrad Lemma 7.2, pp.14–15 (standard reference, not scraped)