Lemma 2

Let L be a (not necessarily commutative) field and K a subfield of L such that L is ajinite-dimensional left vector space over K. Let g be an absolute value on L andf its restriction to K. If K is complete and not discrete with respect to f, L is complete with respect to g; if further g' is another absolute value on L with the same restrictionf to K, then g' = g.

As the topology defined by g is Hausdorff and compatible with the left vector K-space structure on L, the first assertion follows from Topological Vector Spaces, Chapter I, § 2, no. 3, Theorem 2. Moreover the topologies on L defined by g and g' are identical (loc. cit.) ; there therefore exists a real number s > 0 such that g' = g^s (General Topology, Chapter IX, § 3, no. 2, Proposition 5). Let x be an element of K such that f(x) ≠ 1; the equation g'(x) = g(x) proves that s = 1.

Returning to the proof of Theorem 2, it is seen that, if j denotes the restriction of f to K, j is an isomorphism of K onto an everywhere dense subfield of R, C or H and g(x) = |j(x)| for x ∈ K, whence f(x) = |j(x)|^s.

Finally note that, iff is an absolute value on K, h is an absolute value on Q and s ≤ 1 by no. 3, Proposition 4; conversely, if s ≤ 1, f = g^s is an absolute value on K since g is (General Topology, Chapter IX, § 3, no. 2); this proves the last assertion of the statement.

Remarks

(2) If K is a field and a normed algebra over R, the norm is not necessarily an absolute value on K; for example, ξ + iη → |ξ| + |η| is a norm on C compatible with its R-algebra structure.

(3) For a proof of case (C) of Theorem 1 not using the general results of Algebra, Chapter VIII, see Exercise 2.