Exercise 12, § 3

Let $K$ be a separated topological field and non-discrete. Let $R$ be the set of $x \in K$ such that $\lim_{n \to \infty} x^n = 0$, and $N$ the complement of the set $R \cup R^{-1}$. Now prove that, in order that there exist on $K$ an absolute value defining the topology of $K$, it is necessary and sufficient that: 1° $R$ be open in $K$; 2° for every neighbourhood $V$ of 0 in $K$, there exist a neighbourhood $U$ of 0 in $K$ such that $R \cdot U \subset V$; 3° if $x \in R$ and $y \in R \cup N$, then $yx \in R$.

To establish that these conditions are sufficient, we shall show successively that:

(a) $N$ is a normal subgroup of the multiplicative group $K^*$ of the elements $\neq 0$ of $K$.

(b) In the quotient group $K^*/N$, put $\dot{x} \leq \dot{y}$ if there exist $x \in \dot{x}$ and $y \in \dot{y}$ such that $yx^{-1} \in R \cup N$; show that this relation is an ordered relation compatible with the group structure of $K^*/N$, and that the ordered group $K^*/N$ thus defined is isomorphic to a subgroup of the additive group $\mathbf{R}$ (use $V$, p. 16, exerc. 1 of § 3).

(c) Conclude using exerc. 11.

In particular, the topology of a locally compact non-discrete commutative field can be defined by an absolute value (cf. INT, VII, § 2, exerc. 4 and AC, VI, § 9).