Theorem 1 (Gelfand-Mazur)

Let $K$ be an algebra over the field $\mathbf{R}$ with the two following properties: (1) $K$ is a (not necessarily commutative) field. (2) There exists on $K$ a norm $x \mapsto \|x\|$ compatible with the algebra structure on $K$ (General Topology, Chapter IX, § 3, no. 7, Definition 9). Then the algebra $K$ is isomorphic to one of the algebras $\mathbf{R}, \mathbf{C}$ or $\mathbf{H}$.

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: