If $f$ satisfies $(\mathbf{U}_A)$, then $f(1 + x) \leq A$ if $f(x) \leq 1$. Conversely, suppose that $f(x + 1) \leq A$ for the $x \in K$ such that $f(x) \leq 1$ (which implies that $A \geq f(1) = 1$); then, if $x = 0$ or $y = 0$, condition $(\mathbf{U}_A)$ is fulfilled; if on the other hand $x \neq 0$ and $y \neq 0$, we may assume for example that $f(y) \leq f(x)$, hence, by $(\mathbf{VA}_s)$, $f(yx^{-1}) \leq 1$ and therefore $f(1 + yx^{-1}) \leq A$, which gives, by virtue of $(\mathbf{VA}_{II})$, $f(x + y)f(x)^{-1} \leq A$; whence

If $f$ is an absolute value on $K$, then $f(n.1) \leq n$ by induction on the integer $n > 0$ starting from $(\mathbf{VA}_{III})$; conversely:

By induction on $r > 0$ we deduce from $(\mathbf{U}_A)$ the relation

for every family $(x_i)$ of $2^r$ elements of $K$. We set $n = 2^r - 1$; for all $x \in K$, we deduce from (1)

for $f\left( \binom{n}{i} \right) \leq C \binom{n}{i}$; therefore

Letting $r$ tend to $+\infty$, we obtain $f(1 + x) \leq 1 + f(x)$ for all $x \in K$; applying this inequality with $x$ replaced by $xy^{-1}$ (where $y \neq 0$) and taking account of $(\mathbf{VA}_{II})$, we obtain relation $(\mathbf{VA}_{III})$, which proves the proposition.

It is necessary, for $(\mathbf{VA}_{II})$ implies

Conversely, suppose $f$ satisfies $(\mathbf{VA}_I)$, $(\mathbf{VA}_s)$ and $(\mathbf{U}_2)$; for every integer $n > 0$, let $r$ be the least integer such that $2^r \geq n$; if in (1) $A$ is replaced by 2, the $x_i$ of index $i \leq n$ by 1 and the $x_i$ of index $i > n$ by 0, we obtain

then Proposition 2 may be applied with $C = 2$ and hence $f$ is an absolute value.

To say that $f$ satisfies $(\mathbf{U}_A)$ is equivalent to saying that $f^s$ satisfies $(\mathbf{U}_{A^s})$; as there exists $s > 0$ such that $A^s \leq 2$, Corollary 1 shows that for such a value of $s$, $f^s$ is an absolute value.

For every real number $c$ such that $0 < c < 1$, the mapping $t \mapsto c^t$ is an isomorphism of the ordered group $\mathbf{R}$ (with the opposite ordering to the usual ordering) on the ordered group $\mathbf{R}_+^*$; this shows the equivalence of (a) and (b). Clearly (a) implies (c); (c) implies (d), for we deduce from (c) that

for every integer $n > 0$ and Proposition 2 of no. 1 shows that $f''$ is an absolute value. Finally (d) implies (a): for iff" is an absolute value, it satisfies $(\mathbf{U}_2)$ and hence $f$ satisfies $(\mathbf{U}_{2^{1/s}})$ for all $s > 0$ and therefore also $(\mathbf{U}_1)$, letting $s$ tend to $+\infty$.

Every non-zero element $z = n.1$ ($n$ an integer $> 0$) belongs to the prime subfield $\mathbf{F}_p$ of $K$ and hence satisfies the relation $z^{p-1} = 1$, which implies $f(z) = 1$ and we may apply Proposition 3 (c).

Given a real number $c$ such that $0 < c < 1$, the formulae

therefore establish a one-to-one correspondence between ultrametric absolute values on $K$ and valuations on $K$ with real values. The improper valuation corresponds to the improper absolute value (General Topology, Chapter IX, § 3, no. 2). Let $v_1, v_2$ be two valuations on $K$ with real values and $f_1, f_2$ the corresponding absolute values; for $v_1$ and $v_2$ to be equivalent, it is necessary and sufficient that $f_1$ and $f_2$ be so: for to say that $v_1$ and $v_2$ are equivalent amounts to saying that the relations $v_1(x) \geq 0$ and $v_2(x) \geq 0$ are equivalent or again that the relations $f_1(x) \leq 1$ and $f_2(x) \leq 1$ are equivalent; it is therefore sufficient to apply Proposition 5 of General Topology, Chapter IX, § 3, no. 2. Moreover (loc. cit.) for the topologies defined on K by $f_1$ and $f_2$ to be identical, it is necessary and sufficient that $f_1$ and $f_2$ be equivalent.

(i) Either $f$ is the improper absolute value on $\mathbf{Q}$.

(ii) Or there exists a real number $a$ and a prime number $p$ such that $0 < a < 1$ and $f = a^{v_p}$, where $v_p$ is the $p$-adic valuation.

(iii) Or there exists $s > 0$ such that $f(x) = |x|^s$ for all $x \in \mathbf{Q}$.

In care (iii) for $f$ to be an absolute value on $\mathbf{Q}$, it is necessary and sufficient that $0 < s \leq 1$.

Suppose first that $f(n) \leq 1$ for every integer $n > 0$. By Proposition 3 of no. 2 there exist a real number $b$ and a valuation $v$ on $\mathbf{Q}$ such that $0 < b < 1$ and $f = b^v$. Now, we know (§ 3, no. 4, Example 4) that the only valuations on $\mathbf{Q}$ are (up to equivalence) the improper valuation and the $p$-adic valuations $v_p$; we therefore have either case (i) or case (ii).

Suppose from now on that there exists an integer $h > 0$ such that $f(h) > 1$; by no. 1, Corollary 2 to Proposition 2, there exists a number $\rho > 0$ such that $f^\rho$ is an absolute value; let us write

for every rational number $x \neq 0$. Let $a, b$ be two integers $\geq 2$; for every integer $n \geq 2$ let $q(n)$ denote the integral part of $n \log a/\log b$, in other words the least integer $m$ such that $a^n < b^{m+1}$; the expansion of $a^n$ to base $b$ is therefore

where $0 \leq c_i < b$ for $0 \leq i \leq q(n)$. As $f^\rho$ is an absolute value, $f^\rho(c_i) \leq c_i \leq b$ and we therefore deduce from (2) that

Taking logarithms on both sides of this inequality and dividing by $n \log a$, we obtain

Note now that as $n$ tends to $+\infty$, $q(n)/n$ tends to $\log a/\log b$; therefore $q(n)$ tends to $+\infty$ and

tends to 0 (Functions of a Real Variable, Chapter III, § 2, no. 1). Taking the limit in (3), we obtain

(4)

But $f(h) > 1$, whence $g(h) > 0$; if $a$ is replaced by $h$ in (4), we obtain $\sup(0, g(b)) > 0$ and hence

Then, for any integers $a, b$ at least equal to 2, $g(a) \leq g(b)$ and therefore $g(a) = g(b)$, exchanging the roles of $a$ and $b$. In other words, there exists a constant $\lambda$ such that $g(a) = A$ for every integer $a \geq 2$; if we write $s = \lambda / \rho$, then $f(a) = |a|^s$ for every integer $a \geq 2$. As $f(xy) = f(x)f(y)$ and $f(-x) = f(x)$, $f(x) = |x|^s$ for all $x \in Q$. Finally, if $0 < s \leq 1$, we know that $x \mapsto |x|^s$ is an absolute value (General Topology, Chapter IX, § 3, no. 2); conversely, if $s$ is such that $x \mapsto |x|^s$ is an absolute value on $Q$, then $(1 + 1)^s \leq 1^s + 1^s$, that is $2^s \leq 2$, whence $s \leq 1$.

Recall (loc. cit.) that it may always be assumed that $\|xy\| \leq \|x\| \cdot \|y\|$ for all $x, y$ in $K$. We shall give $K$ the topology (compatible with the algebra structure) defined by the norm.

(A) First case: $K$ is commutative and there exists $j \in K$ such that $j^2 = -1$

Then there exists an isomorphism $\sigma$ of the field $\mathbf{C}$ onto a subfield of $K$ such that $\sigma(\xi + i\eta) = \xi . 1 + \eta . j$ for $\xi, \eta$ in $\mathbf{R}$. We shall prove by reductio ad absurdum that $K = \sigma(\mathbf{C})$. Suppose then that there exists $x \in K - \sigma(\mathbf{C})$; for all $z \in \mathbf{C}, x - \sigma(z)$ is therefore invertible in $K$; let us write $F(z) = (x - \sigma(z))^{-1}$; as $\sigma$ is continuous and the inverse is continuous on $K$ (General Topology, Chapter IX, § 3, no. 7, Proposition 13 applied to the completion algebra of $K$), $F$ is a continuous mapping of $\mathbf{C}$ to $K$. Moreover, we may write for $z \neq 0$

But, as $(\sigma(z))^{-1} = \sigma(z^{-1})$ tends to 0 as $z$ tends to infinity in $\mathbf{C}$, it is seen that $F(z)$ tends to 0; in other words, $z \mapsto \|F(z)\|$ is a continuous real-valued function, $\geq 0$ on $\mathbf{C}$, tending to 0 at the point at infinity and which can therefore be considered as a continuous function on the compact space $\tilde{C}$ obtained by adjoining to $C$ a point at infinity. The least upper bound $a$ of $\| F \|$ on $C$ is therefore finite and $> 0$ and the set $P$ of complex numbers $z$ such that $\| F(z) \| = a$ is closed and non-empty (General Topology, Chapter IV, § 6, no. 1, Theorem 1).

Let $z \in P$; let us write $y = x - \sigma(z)$ and let $t$ be a complex number $\neq 0$ such that $\| \sigma(t) \| < \alpha^{-1}$, whence $\| \sigma(t) \cdot y^{-1} \| < 1$ by definition of $a$. The sequence of the $(\sigma(t) y^{-1})^n$ and that of the $n(\sigma(t) y^{-1})^n$ therefore tend to 0 in $K$ as $n$ tends to $+\infty$, for so do the corresponding sequences of norms in $R$. On the other hand note that for every polynomial $H(T) = \prod_{k=1}^p (T - \sigma(c_k))$, where the $c_k$ are distinct complex numbers, in the field $K(T)$ of rational functions

We apply this formula to the polynomial

where $\omega_n = \exp(2\pi i / n)$, and substitute for $T$ the element $y \in K$, which is distinct from all the $\sigma(\omega_n^k t)$. It follows (in the commutative field $K$) that

Taking account of the definitions of $F$ and $y$, we obtain

But by virtue of the choice of $t$ and the remarks made above, the last expression in (7) tends to 0 as $n$ tends to $+\infty$; hence

Now, $\| F(z) \| = \alpha$ and $\| F(z + \omega_n^k t) \| \leq \alpha$ by definition of $\alpha$, whence

Therefore by (8), letting n tend to $+\infty$, $\|F(z + t)\| \geq a$ and by definition of $a$ this implies

in other words $z + t \in P$. This proves that the set $P$ is open in $\mathbf{C}$; as it is also closed and non-empty and $\mathbf{C}$ is connected, $P = \mathbf{C}$ and $\|F\|$ is therefore constant on $\mathbf{C}$; as this function tends to 0 at the point at infinity, $\|F(z)\| = 0$ in $\mathbf{C}$ and in particular $\|F(0)\| = \|x^{-1}\| = 0$, which is absurd.

(B) Second case; K is commutative and $-1$ is not the square of an element of K Let L be the commutative field obtained by adjoining to K a root j of $T^2 + 1$; L is a vector space over K admitting $(1, j)$ as a basis and L is obviously an algebra over $\mathbf{R}$. Clearly the function $x + yj \to \|x\| + \|y\|$ is a norm on L compatible with its structure as a vector space over $\mathbf{R}$; on the other hand, for $z = x + yj, z' = x' + y'j$ in L,

The norm thus defined is consequently compatible with the R-algebra structure on L. By case (A) L is an R-algebra isomorphic to $\mathbf{C}$; now the only sub-R-algebra of $\mathbf{C}$ distinct from $\mathbf{C}$ is $\mathbf{R}$ and hence K is isomorphic to $\mathbf{R}$.

(C) Third case: K is not commutative Let Z be the centre of K and x an element of K not in Z; the subfield $Z(x)$ of K is commutative and the norm induced by that on K is compatible with the R-algebra structure on $Z(x)$; as $Z \neq Z(x)$ and Z and $Z(x)$ are R-algebras isomorphic to $\mathbf{R}$ or $\mathbf{C}$ by virtue of (A) and (B), Z is necessarily isomorphic to $\mathbf{R}$ and $Z(x)$ to $\mathbf{C}$. For all $x \in K, Z(x)$ is therefore of rank $\leq 2$ over Z. Now we have the following lemma:

We may obviously restrict our attention to the case where $D \neq L$. Then there exists in D a finite separable algebraic commutative extension E of L of degree $> 1$ (Algebra, Chapter VIII, § 10, no. 3, Lemma 1); as $E = L(x)$ for some suitable x in E (Algebra, Chapter V, § 7, no. 7, Proposition 12 and Chapter VII, § 5, no. 7), by hypothesis $[E : L] \leq m$. Suppose the separable extension E is taken such that $[E : L]$ is finite and as great as possible and consider the centralizer $E' \supset E$ of E in D, which is a field of centre E such that

(Algebra, Chapter VIII, § 10, no. 2, Theorem 2). If E $\#$ E', there would exist in $\mathbf{E}'$ a finite separable algebraic extension F of $\mathbf{E}$ of degree $> 1$ (Algebra, Chapter VIII, § 10, no. 3, Lemma 1); F would therefore be a finite separable algebraic extension of L (Algebra, Chapter V, § 7, no. 4, Proposition 7) of degree $> [\mathrm{E}: \mathrm{L}]$, contrary to the definition of E; therefore $\mathbf{E}' = \mathbf{E}$, whence $[\mathrm{D}: \mathrm{L}] = [\mathrm{D}: \mathrm{E}][\mathrm{E}: \mathrm{L}] \leq m^2$.

Applying this lemma to K with $m = 2$, it is seen that K is a non-commutative extension field of R of finite rank and hence isomorphic to the field of quaternions $\mathbf{H}$ (Algebra, Chapter VIII, § 11, no. 2, Theorem 2).

Remark (1) We shall give in the chapter devoted to normed algebras a shorter proof of the Gelfand-Mazur Theorem which is valid for every Hausdorff locally convex topological algebra K over R and whose principle is the following: it is reduced (as in cases (B) and (C)) to the case where K is a commutative algebra over $\mathbf{C}$; if $x \in \mathbf{K} - \mathbf{C}.1$, we consider as above the mapping $z \mapsto (x - z.1)^{-1}$ of $\mathbf{C}$ to K, which is continuous and differentiable on $\mathbf{C}$. For every element $x'$ of the dual $\mathbf{K}'$ of the locally convex space K, $z \mapsto \langle (x - z.1)^{-1}, x' \rangle$ is then a bounded integral function on $\mathbf{C}$ and therefore constant by Liouville's Theorem and we conclude as in part (A) of the proof of Theorem 1 that this necessarily implies $\langle (x - z.1)^{-1}, x' \rangle = 0$ for all $z \in \mathbf{C}$ and all $x' \in \mathbf{K}'$; the Hahn-Banach Theorem shows that this conclusion is absurd, since $(x - z.1)^{-1} \neq 0$. Note that the argument in part (A) of the proof of Theorem 1 differs from the above only in appearance, for this argument is only a special case of that which serves to prove the maximum principle for analytic functions, the summation over the roots of unity and and passing to the limit being equivalent to calculating the integral $\int_{\gamma} \frac{\mathbf{F}(z + t)}{t} dt$ along a circle of centre 0 and the use of Cauchy's formula being avoided here, thanks to the particular form of the function F.

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:

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.