Dedekind domains and extensions - Local Fields

10 Dedekind domains and extensions

Let $L ∕ K$ be a finite extension. For $x \in L$ , we write ${Tr}_{L ∕ K} (x) \in K$ for the trace of the $K$ -linear map $L \to L$ , $y \mapsto x y$ .

If $L ∕ K$ is separable of degree $n$ and $σ_{1}, \dots, σ_{n} : L \to \bar{K}$ denotes the set of embeddings of $L$ into an algebraic closure $\bar{K}$ , then ${Tr}_{L ∕ K} (x) = \sum_{i = 1}^{n} σ_{i} (x) \in K$ .

Lemma 10.1. Assuming that:

$L ∕ K$ a finite separable extension of fields

Then the symmetric bilinear pairing

\begin{array}{l} (∙, ∙) & \to K \\ (x, y) & \mapsto {Tr}_{L ∕ K} (x y) \end{array}

is non-degenerate.

Proof. $L ∕ K$ separable tells us that $L = K (α)$ for some $α \in L$ . Consider the matrix $A$ for $(∙, ∙)$ in the $K$ -basis for $L$ given by $1, α, \dots α^{n - 1}$ .

Then $A_{i j} = {Tr}_{L ∕ K} (α^{i + j}) = {[B B^{⊤}]}_{i j}$ where

B = (\begin{matrix} 1 & 1 & \dots & 1 \\ σ_{1} (α) & σ_{2} (α) & \dots σ_{n} (α) \\ ⋮ & ⋮ & ⋱ & ⋮ \\ σ_{1} (α^{n - 1}) & σ_{2} (α^{n - 1}) & \dots & σ_{n} (α^{n - 1}) \end{matrix})

\det A = \det {(B)}^{2} = {[\prod_{1 \leq i < j \leq n} (σ_{i} (α) - σ_{j} (α))]}^{2}

(Vandermonde determinant), which is non-zero since $σ_{i} (α) \neq σ_{j} (α)$ for $i \neq j$ by separability. □

Exercise: On Example Sheet 3 we will show that a finite extension $L ∕ K$ is separable if and only if the trace form is non-degenerate.

Theorem 10.2. Assuming that:

$O_{K}$ a Dedekind domain
$L$ a finite separable extension of $K = Frac (O_{K})$

Then the integral closure

O_{L}

O_{K}

L

is a Dedekind domain.

Proof. $O_{L}$ a subring of $L$ , hence $O_{L}$ is an integral domain.

Need to show:

(i) $O_{L}$ is Noetherian.
(ii) $O_{L}$ is integrally closed in $L$ .
(iii) Every $\neq 0$ prime ideal $P$ in $O_{L}$ is maximal.

Proofs:

(i) Let $e_{1}, \dots, e_{n} \in L$ be a $K$ -basis for $L$ . Upon scaling by $K$ , we may assume $e_{i} \in O_{L}$ for all $i$ .
Let $f_{i} \in L$ be the dual basis with respect to the trace form $(∙, ∙)$ . Let $x \in O_{L}$ , and write $x = \sum_{i = 1}^{n} λ_{i} f_{i}, λ_{i} \in K$ . Then $λ_{i} = {Tr}_{L ∕ K} (x e_{i}) \in O_{K}$ .

(For any $z \in O_{L}$ , ${Tr}_{L ∕ K} (z)$ is a sum of elements in $\bar{K}$ which are integral over $O_{K}$ . Hence ${Tr}_{L ∕ K} (z) \in K$ is integral over $O_{K}$ , hence ${Tr}_{L ∕ K} (z) \in O_{K}$ .)

Thus $O_{L} \subseteq O_{K} f_{1} + \dots + O_{K} f_{n} \subseteq L$ . Since $O_{K}$ is Noetherian, $O_{L}$ is finitely generated as an $O_{K}$ -module, hence $O_{L}$ is Noetherian.
(ii) Example Sheet 2.
(iii) Let $P$ be a non-zero prime ideal of $O_{L}$ , and $p : = P \cap O_{K}$ be a prime ideal of $O_{K}$ . Let $0 \neq x \in P$ . Then $x$ satisfies an equation
$x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0} = 0, a_{i} \in O_{K},$

with $a_{0} \neq 0$ . Then $a_{0} \in P \cap O_{K}$ is a non-zero element of $p$ , hence $p$ is non-zero, hence $p$ is maximal.

We have $O_{K} ∕ p ↪ O_{L} ∕ P$ , and $O_{L} ∕ P$ is a finite dimensional vector space over $O_{K} ∕ p$ . Since $O_{L} ∕ P$ is an integral domain and finite, it is a field. □

Remark. Theorem 10.2 holds without the assumption that $L ∕ K$ is separable.

Corollary 10.3. The ring of integers of a number field is a Dedekind domain.

Convention: $O_{K}$ is the ring of integers of a number field – $p \leq O_{K}$ a non-zero prime ideal. We normalise $| ∙ |_{p}$ (absolute value associated to $v_{p}$ , as defined in Definition 9.6) by $| x |_{p} = {(N p)}^{- v_{p} (x)}$ , where $N_{p} = | O_{K} ∕ p |$ .

In the following theorems and lemmas we will have:

$O_{K}$ a Dedekind domain
$K = Frac (O_{K})$
$L ∕ K$ finite separable
$O_{L}$ the integral closure of $O_{K}$ in $L$ (which is a Dedekind domain by Theorem 10.2).

Lemma 10.4. Assuming that:

$0 \neq x \in O) K$

Then

(x) = \prod_{\begin{array}{c} p \neq 0 \\ prime ideal \end{array}} p^{v_{p} (x)} .

Proof. $x O_{K,_{(p)}} = {(p O_{K, (p)})}^{v_{p} (x)}$ by definition of $v_{p} (x)$ .

Lemma follows from property of localisation

I = J ⟺ I O_{K, (p)} = J O_{K, (p)}

for all prime ideals $p$ . □

Notation. $P \leq O_{L}$ , $p \leq O_{K}$ non-zero prime ideals. We write $P | p$ if $p O_{L} = P_{1}^{e_{1}} \dots P_{r}^{e_{r}}$ and $P \in {P_{1}, \dots, P_{r}}$ ( $e_{i} > 0$ , $P$ distinct).

Theorem 10.5. Assuming that:

$O_{K}$ , $O_{L}$ , $K$ , $L$ as usual
for $p$ a non-zero prime ideal of $O_{K}$ , we write $p O_{L} P_{1}^{e_{1}} \dots P_{r}^{e_{r}}$

Then the absolute values on

L

extending

| ∙ |_{p}

(up to equivalence) are precisely

| ∙ |_{P_{1}}, \dots, | ∙ |_{P_{L}}

Proof. By Lemma 10.4 for any $0 \neq x \in O_{K}$ and $i = 1, \dots, r$ we have $v_{P_{i}} (x) = e_{i} v_{p} (x)$ . Hence, up to equivalence, $| ∙ |_{P_{i}}$ extends $| ∙ |_{p}$ .

Now suppose $| ∙ |$ is an absolute value on $L$ extending $| ∙ |_{p}$ . Then $| ∙ |$ is bounded on $ℤ$ , hence is non-archimedean. Let $R = {x \in L | | x | \leq 1} \leq L$ be the valuation ring for $L$ with respect to $| ∙ |$ . Then $O_{K} \subseteq R$ , and since $R$ is integrally closed in $L$ (Lemma 6.8), we have $O_{L} \subseteq R$ . Set

\begin{array}{l} P & : = {x \in O_{L} | | x | < 1} \\ = m_{R} \cap O_{L} \end{array}

(where $m_{R}$ is the maximal ideal of $R$ ).

Hence $P$ a prime ideal in $O_{L}$ . It is non-zero since $p \subseteq P$ . Then $O_{L, (p)} \subseteq R$ , since $s \in O_{L} ∖ P ⟹ | s | = 1$ .

But $O_{L, (p)}$ is a discrete valuation ring, hence a maximal subring of $L$ , so $O_{L, (p)} = R$ . Hence $| ∙ |$ is equivalent to $| ∙ |_{p}$ . Since $| ∙ |$ extends $| ∙ |_{p}$ , $P \cap O_{K} = p$ so $P_{1}^{e_{1}} \dots P_{r}^{e_{r}} \subseteq P$ , so $P = P_{i}$ for some $i$ . □

Let $K$ be a number field. If $σ : K \to ℝ, ℂ$ is a real or complex embedding, then $x \mapsto | σ (x) |_{\infty}$ defines an absolute value on $K$ (Example Sheet 2) denoted $| ∙ |_{σ}$ .

Corollary 10.6. Let $K$ be a number field with ring of integers $O_{K}$ . Then any absolute value on $K$ is equivalent to either

(i) $| ∙ |_{p}$ for some non-zero prime ideal of $O_{K}$ .
(ii) $| ∙ |_{σ}$ for some $σ : K \to ℝ, ℂ$ .

Proof. Case 1: $| ∙ |$ non-archimedean. Then $| ∙ | |_{ℚ}$ is equivalent to $| ∙ |_{p}$ for some prime $p$ by Ostrowski’s Theorem. Theorem 10.5 gives that $| ∙ |$ is equivalent to $| ∙ |_{p}$ for some $𝔭 \subseteq O_{K}$ a prime ideal with $𝔭 | p$ .

Case 2: $| ∙ |$ archimedean. See Example Sheet 2. □

10.1 Completions

$O_{K}$ a Dedekind domain, $L ∕ K$ a finite separable extension.

Let $𝔭 \subseteq O_{K}$ , $P \subseteq O_{L}$ be non-zero prime ideals with $P | 𝔭$ .

We write $K_{𝔭}$ and $L_{P}$ for the completions of $K$ and $L$ with respect to the absolute values $| ∙ |_{𝔭}$ and $| ∙ |_{P}$ respectively.

Lemma 10.7.

(i) The natural $π_{P} : L \otimes_{K} K_{𝔭} \to L_{P}$ is surjective.
(ii) $[L_{P} : K_{P}] \leq [L : K]$ .

Proof. Let $M = L K_{𝔭} = Im (π_{P}) \subseteq L_{P}$ .

Write $L = K (α)$ then $M = K_{𝔭} (α)$ . Hence $M$ is a finite extension of $K_{𝔭}$ and $[M : K_{𝔭}] \leq [L : K]$ . Moreover $M$ is complete (Theorem 6.1) and since $L \subseteq M \subseteq L_{P}$ , we have $M = L_{P}$ . □

Lemma 10.8 (Chinese remainder theorem). Assuming that:

$R$ a ring
$I_{1}, \dots, I_{n} \subseteq R$ ideals
$I_{i} + I_{j} = R$ for all $i \neq j$

Then

(i) $⋂_{i = 1}^{n} = \prod_{i = 1}^{n} I_{i}$ ( $= I$ say).
(ii) $R ∕ I ≅ \prod_{I = 1}^{n} R ∕ I_{i}$ .

Proof. Example Sheet 2. □

Theorem 10.9. The natural map

L \otimes_{K} K_{𝔭} \to \prod_{P | 𝔭} L_{P}

is an isomorphism.

Proof. Write $L = K (α)$ and let $f (X) \in K [X]$ be the minimal polynomial of $α$ . Then we have

f (X) = f_{1} (X) \dots f_{r} (X) \in K_{𝔭} [X]

where $f_{i} (X) \in K_{𝔭} [X]$ are distinct irreducible (separable). Since $L ≅ K [X] ∕ f (X)$ ,

L \otimes_{K} K_{𝔭} ≅ K_{𝔭} [X] ∕ f_{i} (X) ≅ \prod_{i = 1}^{r} K_{𝔭} [X] ∕ f_{i} (X) .

Set $L_{i} = K_{𝔭} [X] ∕ f_{i} (X)$ a finite extension of $K_{𝔭}$ . Then $L_{i}$ contains both $K_{𝔭}$ and $L$ (use $K [X] ∕ f (x) \to K_{𝔭} [X] ∕ f_{i} (X)$ injective since morphism of fields). Moreover $L$ is dense inside $L_{i}$ (approximate coefficients of $K_{𝔭} [X] ∕ f_{i} (X)$ with an element of $K [X] ∕ f_{i} (X)$ ).

The theorem follows from the following three claims:

(1) $L_{i} ≅ L_{P}$ for some prime $P$ of $O_{L}$ dividing $𝔭$ .
(2) Each $P$ appears at most once.
(3) Each $P$ appears at least once.

Proof of claims:

(1) Since $[L_{i} : K_{𝔭}] < \infty$ , there is a unique absolute value on $L_{i}$ extending $| ∙ |_{𝔭}$ . Theorem 10.5 gives us that $| ∙ | |_{L}$ is equivalent to $| ∙ |_{P}$ for some $P | 𝔭$ . Since $L$ is dense in $L$ and $L_{i}$ is complete, we have $L_{i} ≅ L_{P}$ .
(2) Suppose $φ : L_{i} \to L_{j}$ is an isomorphism preserving $L$ and $K_{𝔭}$ ; then
$φ : K_{𝔭} [X] ∕ f_{i} (X) \to K_{𝔭} [X] ∕ f_{i} (X)$

takes $x$ to $x$ and hence $f_{i} = f_{i}$ .
(3) By Lemma 10.7, the natural map $π_{P} : L \otimes_{K} K_{𝔭} \to L_{P}$ is surjective for any prime $P | 𝔭$ .
Since $L_{P}$ is a field, $π_{P}$ factors through $L_{i}$ for some $i$ , and hence $L_{i} ≅ L_{P}$ by surjectivity of $π_{P}$ . □

Example. $K = ℚ$ , $L = ℚ (i)$ , $f (X) = X^{2} + 1$ . Hensel’s Lemma version 1 gives us that $\sqrt{- 1} \in ℚ_{5}$ . Hence $(5)$ splies in $ℚ (i)$ , i.e. $5 O_{L} = 𝔭_{1} 𝔭_{2}$ .

Corollary 10.10. Let $0 \neq 𝔭 \subseteq O_{K}$ a prime ideal. For $x \in L$ we have

N_{L ∕ K} (x) = \prod_{P | 𝔭} N_{L_{P} ∕ L_{𝔭}} (x) .

Proof. Let $B_{1}, \dots, B_{r}$ be bases for $L_{P_{1}}, \dots, L_{P_{r}}$ as $K_{𝔭}$ -vector spaces. Then $B = ⋃_{i} B_{i}$ is a basis for $L \otimes_{K} K_{𝔭}$ over $K_{𝔭}$ . Let ${[mult (x)]}_{B}$ (respectively ${[mult (x)]}_{B_{i}}$ ) denote the matrix for $mult (x) : L \otimes_{K} K_{𝔭} \to L \otimes_{K} K_{𝔭}$ (respectively $L_{P_{i}} \to L_{P_{i}}$ ) with respect to the basis $B$ (respectively $B_{i}$ ). Then

{[mult (x)]}_{B} = (\begin{matrix} {[mult (x)]}_{B_{1}} \\ ⋱ \\ {[mult (x)]}_{B_{r}} \end{matrix})

hence

\begin{array}{l} N_{L ∕ K} (x) & = \det ({[mult (x)]}_{B}) \\ = \prod_{i = 1}^{r} \det {[mult (x)]}_{B_{i}} \\ = \prod_{i = 1}^{r} N_{L_{P_{i}} ∕ K_{𝔭}} (x) □ \end{array}