Absolute value
Let $K$ be a field.
Definition. A map $|\cdot|\to\mathbb{R}_{\geq 0}$ is called an absolute value when
- $|x|=0 \iff x=0$
- $|\cdot|$ restricted to $K^\ast$ is a homomorphism $K^\ast\to\mathbb{R}^\ast_{\geq 0}$.
- (Triangle inequality) $|x+y|\leq|x|+|y|$
Definition. An absolute value is called nonarchimedean if $$ |x+y|\leq\max(|x|,|y|). $$ Otherwise, it is called archimediean.
Equivalence
Definition. The absolute values $|\cdot|_1$ and $|\cdot|_2$ are called equivalent if either of the following holds:
- $|\cdot|_2=|\cdot|_1^t$ for some $t>0$
- They define the same topology on $K$.
Properties
- If $|\cdot|$ is nonarchimedean and $x\in K$ can be expressed as a finite sum of distinct $x_i$, then $|x|=\max_{i}|x_i|$.
- Consider the standard ring homomorphism $i:\mathbb{Z}\to K$. (i(\mathbb{Z}) is either $\mathbb{Z}$ or $\mathbb{F}_p$ depending on the characteristic of $K$.) Then $|\cdot|$ is nonarchimedian if and only if for all $x\in i(\mathbb{Z})$ we have $|x|\leq 1$.
- If $\text{char}(K)\neq 0$ then any absolute value on $K$ is nonarchimedean.
- Let $k=K(x)$, the rational polynomials in $K$. Suppose $|\cdot|$ is a nontrivial absolute value whose restriction to $K$ is trivial. Then it is nonarchimedean and either \begin{gathered}|x|=s^{-v_P(x)}\quad s\in\mathbb{R},s>1,P\in K[T]\text{ irreducible }, \text{deg}(P)\geq 1\text{, or} \ |x|=s^{\text{deg}(\text{numerator}(x))-\text{deg}(\text{denominator}(x))}\quad s\in\mathbb{R},s>1. \end{gathered}
- These two are in fact the same, differing only by an automorphism of $K$.
Proof of (2). For the forward direction, each $x\in i(\mathbb{Z})$ can be written as a sum of $1$s. Since $|1|=1$, the non-archimedean property ensures $|x|\leq 1$.
Proof of (2.1) Every $x$ is such that $x^p=1$, so $|x^p|=1$ and since 1 is the only root of unity in $\mathbb{R}_{\geq 0}$ it follows $|x|=1$.
Over a field
- Let $K$ be a field and $|\cdot|$ be an absolute value on it. Then
- Addition, subtraction, and multiplication, and absolute value are continuous functions.
- The absolute value may be extended to the completion of $K$, which is also a fieldd. It becomes an absolute value on the completion. If the original absolute value was nonarchimedean, so is the extended one.
- If the absolute value is nonarchimedean, then the set of absolute values of $K$ is the same as the set of absolute values of the completion of $K$.
- Let $|\cdot|$ be a nonarchimedean absolute value on $K$. Suppose $K$ is complete with respect to this absolute value. Then $$ \sum_{i=1}^\infty x_i,\quad x_i\in K $$ converges if and only if $\lim_{i\to\infty}x_i=0$.
- Let $K/k$ be a finite field extension, with absolute value $|\cdot|$ whose restriction to $k$ gives $k$ the structure of a complete metric space. Then the topology of the metric space $K$ coincides with the topology of $k^n$, i.e. the absolute value is entirely defined by its restriction to $k$.
Examples
- The trivial absolute value has $|0|=0$ and $|x|=1$ otherwise.