On the theorem of Gelfand-Mazur

In its classical statement, the theorem of Gelfand-Mazur is the following:

Theorem 1. — Let $K$ be a complete normed field which is an extension of the field $\mathbf C$ of complex numbers. Then $K=\mathbf C$.

As in the history of any theorem, the reality is a bit more complicated. The theorem proved by Stanislas Mazur in 1938 concerns all fields $F$, extensions of the field $\mathbf R$ of real numbers, which can be endowed with a norm:

Theorem 2 (S. Mazur, 1938). — Let $F$ be a field which is an extension of the field $\mathbf R$ of real numbers. If the field $F$ admits an algebra norm (that is, $\|ab\|\leq \|a\|, \|b\|$ for $a,b\in F$), then it is continuously isomorphic to $\mathbf R$, $\mathbf C$, or the field $\mathbf H$ of quaternions. If, moreover, the norm is multiplicative, then there is an isometric isomorphism.

This theorem is viewed by Stanislas Mazur as a generalization of Frobenius's theorem (1877) that treats the case where $F$ is finitely dimensional over $\mathbf R$: he replaces this assumption by the more general one that the field can be given an algebra norm. It is stated, with little arguments, in his 1938 note to Comptes rendus de l'Académie des sciences. In 2007, Pierre Mazet found why the proof had been skipped during the publication of this note and managed to recover it, see his paper in the Gazette de la Société mathématique de France.

In 1939, Israel Gelfand proved his version independently motivated by his theory of normed algebras. His statement, equivalent to the above one, is the following:

Theorem 3 (I. Gelfand, 1939). — The $A$ be normed algebra over the complex numbers. Then the residue field of every maximal ideal of $A$ is equal to $\mathbf C$.

There are well known classical proofs of these results in the literature, which most of the time rely on Liouville's theorem that bounded holomorphic entire functions are constant. The goal of this note is to record two proofs which I learned from browsing the online Bourbaki archives. I won't discuss the real or quaternionic aspects of that proof and concentrate on the first theorem.

1. A proof using covering theory

This proof is sketched in the tribu n°84 (p. 26 of the PDF file) of the Bourbaki group, where they record the decisions of their March 1972 meeting in Jerba (Tunisia), in relation with the preparation of their 11th chapter of Topologie générale. It goes as follows:

1) Démonstration de Gelfand-Mazur
Soit $K$ un corps normé complet sur $\mathbf C$, distinct de $\mathbf C$. Le groupe additif $K^+$ et le groupe multiplicatif $K^*$ sont simplement connexes et l'exponentielle fait de $K^+$ un revêtement de $K^*$, non trivial puisque $\exp(2\pi i)=1$. C'est idiot.

which, in English, gives the following: Both $K$ and $K^*$ are simply connected, and the exponential map makes $K$ a covering $K^*$; this covering is nontrivial since $\exp(2\pi i)=1$. And they add: This is dumb.

It may be worth giving a bit of detail to this very short argument.

They implicitly assume that $K$ is commutative when they consider the exponential map $$\exp : K \to K^*, \quad z \mapsto \sum_{n=0}^\infty\frac1{n!} z^n $$ and say that it is a covering. As over the complex numbers, one can prove that the series converges everywhere, and commutativity allows to prove the standard relation $\exp(a+b)=\exp(a)\exp(b),$ as well as show that the exponential is its own derivative. As a consequence, this derivative never vanishes and the exponential gives a local homeomorphism from $K$ to $K^*$. Moreover, this homeomorphism is locally trivial, because one can split it locally by way of the logarithm, which is given, for $\|z\|<1$ by the formula $$ \log(1+z)=\sum_{n=1}^\infty (-1)^{n-1} \frac1n z^n. $$ Again, the functional equation $$\exp(\log(1+z))=1+z$$ requires commutativity, using legitimate manipulation of power series and their evaluations.

The field $K$ itself is simply connected, because it is contractible.

The simple connectedness of $K^*$ will use the hypothesis that $K$ is not equal to $\mathbf C$. We need to prove that any loop in $K^*$ is homotopic to a constant loop. First, we observe that, by uniform continuity and local convexity, it is homotopic to a piecewise linear loop in $K^*$. Now, such a loop is homotopic to its projection to the unit sphere, and the image of the projected loop has empty interior — since $K$ is not equal to $\mathbf C$, that the unit sphere is not locally more 1-dimensional. In particular, our loop omits one point. Then, a stereographic projection turns the loop one that lives in the equatorial plane of the sphere, which is contractible, so that the loop is homotopic to a constant loop.

Finally, the theory of coverings shows that this covering has to be an isomorphism, which it isn't since $\exp(2\pi i)=1$. This contradiction implies that $K=\mathbf C$, Q.E.D.

In fact, a 1952 paper by Ernst Witt gives a closely related proof in a one-page paper. For the same reason that $K^*$ is simply connected, Witt writes that the differential relation $x^{-1} d x = d y$ gives a global isomorphism between the multiplicative group ($x \neq 0$) and the additive group ($y$). And he concludes that this isn't possible becasuse the multiplicative group $K^*$ admits the element~$-1$ of order~$2$, while $K$ has none since its characteristic is~$0$.

2. Tornheim's proof

There is an other, earlier, proof in these archives in a rédaction n° 283 written by Serge Lang where he sketches the proof given by Leonard Tornheim in 1951. (A variant of this proof, only valid for multiplicative norms, had been given by Alexander Ostrowski in 1917. It has been used recently by Dustin Clausen and Peter Scholze in theor development of complex analytic geometry in the condensed setting.) The classical proof by complex analysis argues by contradiction, considering an element $a\in K$ such that $a\notin \mathbf C$ and considers the function $z\mapsto 1/(z-a)$ from $\mathbf C$ to $K$. This function is holomorphic, bounded (it tends to $0$ at infinity), hence constant, which it clearly isn't. To avoid the theory of holomorphic functions with values in the field $K$, Tornheim argues as follows. The function $z\mapsto \| 1/(z-a)\|$ is continuous on $\mathbf C$, nonzero, and tends to $0$ at infinity, hence it attains it maximum $M$ on a nonempty closed subset $D$ of $\mathbf C$. It suffices to prove that $D$ is also open. By translation, it suffices to prove that if $0\in D$, then $D$ contains a small ball around $0$. Now, consider a strictly positive real number $r$, an integer $n\geq 1$ and a primitive $n$-root of unity $\omega\in\mathbf C.$ One sets $$ S_n(r) = \frac 1n \sum_{k=0}^{n-1} \frac 1{a-\omega^k r}.$$ Since the rational function $ \sum_{k=0}^{n-1} 1/(T-\omega^k r)$ is the logarithmic derivative of $\prod_{k=0}^{n-1}(T-\omega^k r)=T^n-r^n$, it is equal to $n T^{n-1}/(T^n-r^n).$ Consequently, $$ S_n (r)= \frac{a^{n-1}}{a^n - r^n} = \frac 1{a - r\cdot (r/a)^{n-1}} .$$ If $r<|a|$, we obtain $\lim_{n\to+\infty} S_n(r) = 1/a$. On the other hand, the assumption that $0\in D$ means that $M=1/\|a\|$, and the definition of $D$ implies that $ \| S_n(r)\| \leq M = 1/\| a\|$. Consequently, for $n$ large enough, all terms in the definition of $S_n(r)$ must be of norm $1/\| a\|.$ Taking $n$ arbitrarily large, the circle of radius $r$ centered at the origin contains a dense set of points of $D$. Since $D$ is closed, the whole circle is contained in $D$. This is valid for all $r<\|a\|$, and this proved that the ball of radius $\| a\|$ is contained in $D$. Q.E.D.