Theorem 2 (Ostrowski)

Let K be a (not necessarily commutative) field and f an element $\mathscr{V}(K)$ which is not an ultrametric absolute value. Then there exist a unique real number $s > 0$ and an isomorphism j of K onto an everywhere dense subfield of one of the fields $\mathbf{R}, \mathbf{C}$ or $\mathbf{H}$ such that $f(x) = |j(x)|^s$ for all $x \in K$ (*). For f to be an absolute value on K, it is necessary and sufficient that $s \leq 1$.

By no. 2, Corollary to Proposition 3, K is of characteristic 0 and hence an algebra over $\mathbf{Q}$; for all $x \in \mathbf{Q}$ we write $h(x) = f(x.1)$; clearly $h \in \mathscr{V}(\mathbf{Q})$ and therefore Proposition 4 of no. 3 may be applied; neither of cases (i) and (ii) of the statement of this proposition can hold, for this would imply $f(n.1) \leq 1$ for every integer $n > 0$ and f would be an ultrametric absolute value by virtue

(*) On $\mathbf{H}$ we write $|z|^2 = z.\bar{z} = \bar{z}.z$, $\bar{z}$ being the conjugate quaternion of $z$.

of no. 2, Proposition 3. Then there exists a real number s > 0 such that h(x) = |x|^s for all x ∈ Q, that is f(x.1) = |x|^s; we write g = f^{1/s}. Then g ∈ V(K) and g(n.1) = n for every integer n; Proposition 2 of no. 1 therefore shows that g is an absolute value on K.

For x ∈ Q and y ∈ K, g(xy) = |x|g(y) and hence g is a norm on K compatible with its Q-algebra structure (with the usual absolute value on Q). The completion K of K is therefore a normed algebra over Q = R (General Topology, Chapter IX, § 3, no. 7); let ĝ be the norm on K the continuous extension of g. As g is an absolute value on K, K is a field and ĝ an absolute value on K (General Topology, Chapter IX, § 3, no. 3, Proposition 6). By Theorem 1 there exists an R-algebra isomorphismf of K̂ onto one of the fields R, C or H and g'(x) = |j(x)| is therefore an absolute value on K̂; as K̂ is finite-dimensional over R and g' and ĝ coincide on the subfield R . 1 of K̂, g' = ĝ by the following lemma: