Valuation Rings - Local Fields - Cambridge III notes

Definition 2.1 (Valuation). Let $K$ be a field. A valuation on $K$ is a function $v : K^{\times} \to ℝ$ such that

(i) $v (x y) = v (x) + v (y)$
(ii) $v (x + y) \geq \min (v (x), v (y))$

Remark.

Ignore the trivial valuation $v (x) = 0$ .
Say $v_{1}, v_{2}$ are equivalent if there exists $c \in ℝ > 0$ such that $v_{1} (x) = c v_{2} (x)$ for all $x \in K^{\times}$ .

Example.

$K = ℚ$ , $v_{p} (x) = - \log_{p} {| x |}_{p}$ is known as the $p$ -adic valuation.
If $k$ is a field, consider $K = k (t) = Frac (k [t])$ the rational function field. Then define
$v (t^{n} \frac{f (t)}{g (t)}) = n$

for $f, g \in k [t]$ with $f (0), g (0) \neq 0$ . We call this the $t$ -adic valuation.
$K = k ((t)) = Frac (k [[t]]) = {\sum_{i = n}^{\infty} a_{i} t^{i} | a_{i} \in k, n \in ℤ}$ , known as the field of formal Laurent series over $k$ . Then we can define
$v (\sum_{I} a_{i} t^{i}) = \min {i | a_{i} \neq 0}$

is the $t$ -adic valuation on $K$ .

Definition 2.2. Let $(K, | ∙ |)$ be a non-archimedean valued field. The valuation ring of $K$ is defined to be

\begin{array}{l} O_{K} & = {x \in K | | x | \leq 1} (= \bar{B} (0, 1)) \\ ( & = {x \in K^{\times} | v (x) \geq 0} \cup {0}) \end{array}

Proposition 2.3.

(i) $O_{K}$ is an open subring of $K$
(ii) The subsets ${x \in K | | x | \leq r}$ and ${x \in K | | x | < r}$ for $r \leq 1$ are open ideals in $O_{K}$ .
(iii) $O_{K}^{\times} = {x \in K | | x | = 1}$ .

Proof.

(i) $| 0 | = 0$ , $| 1 | = 1$ so $0, 1 \in O_{K}$ . If $x \in O_{K}$ , then $| - x | = | x |$ hence $- x \in O_{K}$ . If $x, y \in O_{K}$ , then
$| x + y | \leq \max (| x |, | y |) \leq 1 .$

Hence $x + y \in O_{K}$ . If $x, y \in O_{K}$ , then $| x y | = | x | | y | \leq 1$ , hence $x y \in O_{K}$ . Thus $O_{K}$ is a ring. Since $O_{K} = \bar{B} (0, 1)$ , it is open.
(ii) Similar to (i).
(iii) Note that $| x | | x^{- 1} | = | x x^{- 1} | = 1$ . Thus $\begin{array}{l} | x | = 1 & ⟺ | x^{- 1} | = 1 \\ ⟺ x, x^{- 1} \in O_{K} \\ ⟺ x \in O_{K}^{\times} □ \end{array}$

Notation.

$m : = {x \in O_{K} | | x | < 1}$ is a max ideal of $O_{K}$ .
$k : = O_{K} ∕ m$ is the residue field.

Proof. Let $m^{'}$ be a maximal ideal. Suppose $m^{'} \neq m$ . Then there exists $x \in m^{'} ∖ m$ . Using part (iii) of Proposition 2.3, we get that $x$ is a unit, hence $m^{'} = O_{K}$ , a contradiction. □

Example. $K = ℚ$ with ${| ∙ |}_{p}$ . Then

O_{K} = ℤ_{(p)} = {\frac{a}{b} \in ℚ | p ∤ b},

and $m = p ℤ_{(p)}$ , $k = 𝔽_{p}$ .

Definition 2.5. Let $v : K^{\times} \to ℝ$ be a valuation. If $v (K^{\times}) ≅ ℤ$ , we say $v$ is a discrete valuation. $K$ is said to be a discretely valued field. An element $π \in O_{K}$ is uniformiser if $v (π) > 0$ and $v (π)$ generates $v (K^{\times})$ .

Example.

$K = ℚ$ with $p$ -adic valuation is a discrete valuation ring.
$K = k (t)$ with $t$ -adic valuation is a discrete valuation ring.
$K = k (t) (t^{1 ∕ 2}, t^{1 ∕ 4}, t 1 ∕ 8, \dots)$ . Here, the $t$ -adic valuation is not discrete.

Remark. If $v$ is a discrete valuation, can replace with equivalent one such that $v (K^{\times}) = ℤ$ > Call such a $v$ normalised valuations (then $v (π) = 1$ if and only if $π$ is a unit).

Lemma 2.6. Assuming that:

$v$ is a valuation on $K$

Then the following are equivalent:

(i) $v$ is discrete
(ii) $O_{K}$ is a PID
(iii) $O_{K}$ is Noetherian
(iv) $m$ is principal

Proof.

(i) $⟹$ (ii) $O_{K}$ is an integral domain since it is a subset of $K$ , which is an integral domain.
Let $I \subseteq O_{K}$ be a non-zero ideal. Let $x \in I$ such that $v (x) = \min {v (a) | a \in I}$ , which exists since $v$ is discrete. Then we claim
$x O_{K} = {a \in O_{K} | v (a) \geq v (x)}$

is equal to $I$ .
- $\subseteq$ ( $I$ is an ideal)
- $\supseteq$ Let $y \in I$ . Then $v (x^{- 1} y) \geq 0$ . Hence $y = x (x^{- 1} y) \in x O_{K}$ .
(ii) $⟹$ (iii) Clear.
(iii) $⟹$ (iv) Write $m = x_{1} O_{K} + \dots + x_{n} O_{K}$ . Without loss of generality,
$v (x_{1}) \leq v (x_{2}) \leq \dots \leq v (x_{n}) .$

Then $x_{2}, \dots, x_{n} \in x_{1} O_{K}$ . Hence $m = x_{1} O_{K}$ .
(iv) $⟹$ (i) Let $m = π O_{K}$ for some $π \in O_{K}$ and let $c = v (π)$ . Then if $v (x) > 0$ , $x \in m$ hence $v (x) \geq c$ . Thus $v (K^{\times}) \cap (0, c) = \emptyset$ . Since $v (K^{\times})$ is a subgroup of $(ℝ, +)$ , we deduce $v (K^{\times}) = ℤ$ . □

Suppose

v

is a discrete valuation on

K

π \in O_{K}

a uniformiser. For

x \in K^{\times}

, let

n \in ℤ

such that

v (x) = n v (π)

. Then

u = π^{- n} x \in O_{K}^{\times}

and

x = u π^{n}

. In particular,

K = O_{K} [\frac{1}{π}]

and hence

K = Frac (O_{K})

Definition 2.7 (Discrete valuation ring). A ring $R$ is called a discrete valuation ring (DVR) if it is a PID with exactly one non-zero prime ideal (necessarily maximal).

Lemma 2.8.

(i) Let $v$ be a discrete valuation on $K$ . Then $O_{K}$ is a discrete valuation ring.
(ii) Let $R$ be a discrete valuation ring. Then there exists a valuation on $K : = Frac (R)$ such that $R = O_{K}$ .

Proof.

(i) $O_{K}$ is a PID by Lemma 2.6. Hence any non-zero prime ideal is maximal and hence $O_{K}$ is a discrete valuation ring since it is a local ring.
(ii) Let $R$ be a discrete valuation ring, with maximal ideal $m$ . Then $m = (π)$ for some $π \in R$ . Since PIDs are UFDs, we may write any $x \in R ∖ {0}$ uniquely as $π^{n} u$ with $n \geq 0$ , $u \in R^{\times}$ . Then any $y \in K^{\times}$ can be written uniquely as $π^{m} u$ with $u \in R^{\times}$ , $m \in ℤ$ . Define $v (π^{m} u) = m$ ; check $v$ is a valuation and $O_{K} = R$ . □

2 Valuation Rings