Absolute values - Local Fields - Cambridge III notes

1 Absolute values

Definition 1.1 (Absolute value). Let $K$ be a field. An absolute value on $K$ is a function $| ∙ | : K \to ℝ_{\geq 0}$ such that

(i) $| x | = 0$ if and only if $x = 0$ .
(ii) $| x y | = | x | | y |$ for all $x, y \in K$ .
(iii) $| x + y | \leq | x | + | y | \forall x, y \in K$ (triangly inequality).

We say $(K, | ∙ |)$ is a valued field.

Example.

$K = ℚ, ℝ, ℂ$ with usual absolute value $| a + i b | = \sqrt{a^{2} + b^{2}}$ . Write $| ∙ |_{\infty}$ for this absolute value.
$K$ any field. The trivial absolute value is
$| x | = {\begin{matrix} 0 & x = 0 \\ 1 & x \neq 0 \end{matrix}$

Although this is technically an absolute value, it is not useful or interesting, so should be ignored.

Definition 1.2 ( $p$ -adic absolute value). Let $K = ℚ$ , and $p$ be a prime. For $0 \neq x \in ℚ$ , write $x = p^{n} \frac{a}{b}$ , where $(a, p) = 1$ , $(b, p) = 1$ . The $p$ -adic absolute value is defined to be

| x |_{p} = {\begin{matrix} 0 & x = 0 \\ p^{- n} & x = p^{n} \frac{a}{b} \end{matrix}

Verification:

(i) Clear
(ii) Write $y = p^{m} \frac{c}{d}$ . Then
${| x y |}_{p} = {| p^{m + n} \frac{a c}{b d} |}_{p} = p^{- m - n} = {| x |}_{p} {| y |}_{p} .$
(iii) Without loss of generality, $m \geq n$ . Then
${| x + y |}_{p} = {| p^{n} \frac{a d + p^{m - n} b c}{b d} |}_{p} \leq p^{- n} = \max ({| x |}_{p}, {| y |}_{p}) .$

An absolute value $| ∙ |$ on $K$ induces a metric $d (x, y) = | x - y |$ on $K$ , hence a topology on $K$ .

Definition 1.3 (Place). Let $| ∙ |$ , ${| ∙ |}^{'}$ be absolute values on a field $K$ . We say $| ∙ |$ and ${| ∙ |}^{'}$ are equivalent if they induce the same topology. An equivalence class of absolute values is called a place.

Proposition 1.4. Assuming that:

$| ∙ |$ , ${| ∙ |}^{'}$ are (non-trivial) absolute values on $K$ .

Then the following are equivalent:

(i) $| ∙ |$ and ${| ∙ |}^{'}$ are equivalent.
(ii) $| x | < 1 ⟺ {| ∙ |}^{'} < 1$ for all $x \in K$ .
(iii) There exists $c \in ℝ_{> 0}$ such that ${| x |}^{c} = {| ∙ |}^{'}$ for all $x \in K$ .

Proof.

(i) $⟹$ (ii) $\begin{array}{l} | x | < 1 & ⟺ x^{n} \to 0 w.r.t | ∙ | \\ ⟺ x^{n} \to 0 w.r.t {| ∙ |}^{'} \\ ⟺ {| x |}^{'} < 1 \end{array}$
(ii) $⟹$ (iii) Note: ${| x |}^{c} = {| x |}^{'} ⟺ c \log | x | = \log {| x |}^{'}$ . Let $a \in K^{\times}$ such that $| a | > 1$ (exists since $| ∙ |$ is non-trivial). We need that $\forall x \in K^{\times}$ ,
$\frac{\log | x |}{\log | a |} = \frac{\log {| x |}^{'}}{\log {| a |}^{'}} .$

Assume that
$\frac{\log | x |}{\log | a |} < \frac{\log {| x |}^{'}}{\log {| a |}^{'}} .$

Choose $m, n \in ℤ$ (with $n > 0$ ) such that
$\frac{\log | x |}{\log | a |} < \frac{m}{n} < \frac{\log {| x |}^{'}}{\log {| a |}^{'}} .$

Then we have
$\begin{array}{l} n \log | x | & < m \log | a | \\ n \log {| x |}^{'} & > m \log {| a |}^{'} \end{array}$
Hence $| \frac{x^{n}}{a^{m}} | < 1$ and $| \frac{x^{n}}{a^{m}} | > 1$ , contradiction. Similarly for the case where
$\frac{\log | x |}{\log | a |} > \frac{\log {| x |}^{'}}{\log {| a |}^{'}} .$
(iii) $⟹$ (i) Clear.

□

Remark. $| ∙ |_{\infty}^{2}$ on $ℂ$ is not an absolute value by our definition. Some authors replace the triangle inequality by

{| x + y |}^{β} \leq {| x |}^{β} + {| y |}^{β}

for some fixed $β \in ℝ_{> 0}$ .

Definition 1.5 (Non-archimedean). An absolute value $| ∙ |$ on $K$ is said to be non-archimedean if it satisfies the ultrametric inequality:

| x + y | \leq \max (| x |, | y |) .

If $| ∙ |$ is not non-archimedean, then it is archimedean.

Example.

${| ∙ |}_{\infty}$ on $ℝ$ is archimedean.
${| ∙ |}_{p}$ is a non-archimedean absolute value.

Lemma 1.6. Assuming that:

$(K, | ∙ |)$ is non-archimedean
$x, y \in K$
$| x | < | y |$

Then

| x - y | = | y |

Proof.

| x - y | \leq \max (| x |, | y |) = | y |

and

| y | \leq \max (| x |, | x - y |) \leq | x - y | .

□

Proposition 1.7. Assuming that:

$(K, | ∙ |)$ is non-archimedean
${(x_{n})}_{n = 1}^{\infty}$ a sequence in $K$
$| x_{n} - x_{n + 1} | \to 0$

Then

{(x_{n})}_{n = 1}^{\infty}

is Cauchy. In particular, if

K

is in addition complete, then

{(x_{n})}_{n = 1}^{\infty}

converges.

Proof. For $𝜀 > 0$ , choose $N$ such that $| x_{n} - x_{n + 1} | < 𝜀$ for $n > N$ . Then $N < n < m$ ,

| x_{n} - x_{m} | = | (x_{n} - x_{n + 1}) + \dots + (x_{n - 1}) - x_{m}) | < 𝜀 .

The “In particular” is clear. □

Example. $p = 5$ , construct sequence ${(x_{n})}_{n = 1}^{\infty}$ in $ℤ$ such that

(i) $x_{n}^{2} + 1 \equiv 0 (m o d 5^{n})$
(ii) $x_{n} \equiv x_{n + 1} (m o d 5^{n})$

Take $x_{1} = 2$ . Suppose we have constructed $x_{n}$ . Let $x_{n}^{2} + 1 = a 5^{n}$ and set $x_{n + 1} = x_{n} + b 5^{n}$ . Then

\begin{array}{l} x_{n + 1}^{2} + 1 & = x_{n}^{2} + 2 b x_{n} 5^{n} + b^{2} 5^{2 n} + 1 \\ = a 5^{n} + 2 b x_{n} 5^{n} + b^{2} 5^{2 n} \end{array}

We choose $b$ such that $a + 2 b x_{n} \equiv 0 (m o d 5)$ . Then we have $x_{n + 1}^{2} + 1 \equiv 0 (m o d 5^{n + 1})$ . Now (ii) implies that ${(x_{n})}_{n = 1}^{\infty}$ is Cauchy. Suppose $x_{n} \to l \in ℚ$ . Then $x_{n}^{2} \to l^{2}$ . But (i) tells us that $x_{n}^{2} \to - 1$ , so $l^{2} = - 1$ , a contradiction. Thus $(ℚ, {| ∙ |}_{5})$ is not complete.

Definition 1.8. The $p$ -adic numbers $ℚ_{p}$ is the completion of $ℚ$ with respect to ${| ∙ |}_{p}$ .

Analogy with $ℝ$ :

Notation. As is usual when working with metric spaces, we will be using the notation:

\begin{array}{l} B (x, r) & = {y \in K | | x - y | < r} \\ \bar{B} (x, r) & = {y \in K | | x - y | \leq r} \end{array}

Lemma 1.9. Assuming that:

$(K, | ∙ |)$ is a non-archimedean valued field

Then

(i) If $z \in B (x, r)$ , then $B (z, r) = B (x, r)$ – so open balls don’t have a centre.
(ii) If $z \in \bar{B} (x, r)$ then $\bar{B} (x, r) = \bar{B} (z, r)$ .
(iii) $B (x, r)$ is closed.
(iv) $\bar{B} (x, r)$ is open.

Proof.

(i) Let $y \in B (x, r)$ . Then $| x - y | < r$ hence $\begin{array}{l} | z - y | & = | (z - x) + (x - y) | \\ \leq \max (| z - x |, | x - y |) \\ < r \end{array}$
Thus $B (x, r) \subseteq B (z, r)$ . $\supseteq$ follows by symmetry.
(ii) Same as (i).
(iii) Let $y \notin B (x, r)$ . If $z \in B (x, r) \cap B (y, r)$ then $B (x, r) = B (z, r) = B (y, r)$ Hence $y \in B (x, r)$ . Hence $B (x, r) \cap B (y, r) = \emptyset$ .
(iv) If $z \in \bar{B} (x, r)$ , then $B (z, r) \subseteq \bar{B} (z, r) = \bar{B} (x, r)$ . □