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.

Joyal's proof of Cayley's formula for the enumeration of trees

That formula of Cayley asserts that there are $n^{n-2}$ trees with vertices labeled $1,\dots,n$. This is a chapter in graph theory, in which graphs are “simple”. More precisely, let's call a graph structure on a set $V$ (called “vertices”) is the datum of a subset $E$ of 2-element subsets of~$V$, called “edges”. So an edge is of the form $\{a,b\}$, where $a$ and $b$ are distinct elements of $V$, and we say that the edge links $a$ to $b$.

A path in this graph is a sequence $(a_0,a_1,\dots,a_n)$ where $a_0,\dots,a_n$ are vertices such that $\{a_{k-1},a_k\}$ is an edge, for each $k\in\{1,\dots,n\}$; this path has length $n$, and it connects $a_0$ to $a_n$. A graph is said to be connected if every two vertices are connected by some path.

A loop is a path whose first vertex $a_0$ equals the last one $a_n$. Examples of loops are paths of length $0,$ given by one vertex. Less trivial examples are obtained by taking an edge $\{a,b\}$ and considering the path $(a,b,a)$; a path that does not contain this pattern will be called a path without back and forth. (There's a translation issue between English and French, for the French wording for this expression is sans aller-retour, which seems more natural to me — you first go somewhere before returning back.) In any case, if every loop without back-and-forth in the graph has length $0$, then one says that the graph is a forest. Finally, a tree is a connected forest.

In a connected graph, any two vertices $a,b$ are connected by some path, and if the graph is a tree, then there is a unique such path which has no back-and-forth. Indeed, if there were two such paths $(a,a_1,\dots,a_{m-1},b)$ and $(a,b_1,\dots,b_{n-1},b)$ connecting $a$ to $b$, where, say, $m+n$ is minimal, then one can build the loop $(a,a_1,\dots,a_{m-1},b,b_{n-1},\dots,b_1,a)$. By definition of a tree, this loop must have some back-and-forth. It must be $(a_{m-1},b,b_{n-1})$, for otherwise the two given paths would have had some back-and-forth. In particular, $a_{m-1}=b_{n-1}$ and the two paths $(a,a_1,\dots,a_{m-1})$ and $(a,b_1,\dots,b_{m-1})$ have no back-and-forth, connect the same pair of points, and their lengths are smaller. This furnishes the desired contradiction.

Theorem (Cayley). — If $V$ is a finite set of cardinality $n\geq 2$, then there are $n^{n-2}$ structures of tree on $V$.

Here is the proof of this theorem that Joyal gives in his paper “Une théorie combinatoire des séries formelles” (Advances in Mathematics 42 (1): 1‑82). I thank François Lamarche for having urged me to read that paper.

Joyal's idea consists in considering the structure of an “animal” obtained by selecting two vertices of a tree. Given two vertices $a,b$, the vertices in the unique path linking $a$ to $b$ are called its vertebrae, and the set of vertices is called its spine. We now need to prove that there are $n^n$ structures of animal on a non-empty set with $n$ elements. (This combinatorial description does not work well for $n=0$, there are no animals with $0$ elements, but $0^0=1$.)

Now, each vertex $x$ in an animal is connected to some vertebra, and there is a preferred vertebra $v(x)$ for which there is a path $(x,\dots,v(x))$ that does not contain any other vertebra than $v(x)$. Note that all vertices $y$ in that path satisfy $v(y)=v(x)$. In particular, $v(v(x))=v(x)$, that is, the map $v$ is idempotent.

Given some vertebra $t$, let us consider the subgraph $T_t$ of the given tree with set of vertices $v^{-1}(t)$. It is again a graph, and is even a tree since each of its vertices is connected to $t$, and a subgraph of a forest is a forest.

As a conclusion, animals $(a,b,T)$ with vertex set $V$ are in bijection with ordered pairs consisting of the sequence $(a,\dots,b)$ of vertebrae forming the spine together with a family of disjoint trees indexed by the spine and whose union if $V$.

On the other hand, one can associate to any self-map $f\colon V\to V$ a similar structure of an idempotent map $v\colon V\to V$ together with a tree structure of $v^{-1}(t)$ for every $t\in V$. Indeed, let $S$ be the set of periodic points for $f$, that is, elements $x\in V$ for which there exists $n\geq 1$ with $f^n(x)=x$. If $x$ is periodic for $f$, then so is $f(x)$, so that $f$ induces a map $f_S$ from $S$ to $S$. That map is bijective. On the other hand, for every $x\in V$, the sequence $x, f(x),f^2(x),\dots$ will eventually repeat itself, because $V$ is finite, so that there exist integers $m$ and $n$ such that $0\leq m < n$ and $f^m(x)=f^n(x)$. In particular, $f^m(x)=f^{n-m}(f^m(x))$ is periodic for $f$. Let us then denote by $v(x)$ the first periodic point in the sequence $x,f(x),f^2(x),\dots$ of iterates of $x$. (Technically, $v(x)=f^m(x)$ where $m$ is the smallest integer such that $f^m(x)$ is periodic for $f$.

For each periodic point $s\in S$, let $V_s$ be the set $v^{-1}(s)$. We define a tree with set of vertices $V_s$ as follows: its edges are simply the pairs $\{x,f(x)\}$, for $x\in V_s\setminus\{s\}$. To convince oneself that this actually forms a tree, it maybe better to imagine its construction from the vertex set $s$. One has an edge from $s$ to all non-periodic points $x\in V$ such that $f(x)=s$. And then an edge from each of these points $x$ to the points $y\in V$ such that $f(y)=x$ (and such $y$ are distinct from $x$, and are non-periodic), etc. building the tree from its root $s$.

In conclusion, self-maps $f$ are in bijections with triples $(S,f_S, (V_s))$ where $S$ is a subset of $V$, $f_S$ is a bijection of $S$ and $(V_s)$ is a family of disjoint trees disjoint containing $s$ such that $V=\bigcup V_s$.

To conclude this proof of Cayley's theorem, it remains to observe that for each subset $S$ of $V$, there are $\operatorname{Card}(S) !$ ways to, either put the elements of $S$ in some linear order, or to define a bijection of $S$. In both cases, it remains to additional families of trees are equinumerous, so that the number of self-maps of $V$ is equal to the number of animals with vertex set $V$, as claimed.

Associated prime ideals and regular elements in polynomial rings

This post is here to record a fact that ought to be better known and for which I ought to have known a proper proof beforehand.

Let's start by defining two thirds of the terms of the title.

By polynomial ring, I simply mean a ring of polynomials $A[T_1,T_2,\dots]$ over a commutative ring $A$, in any number of indeterminates. The most important case here will however be that of a ring of polynomials in one single indeterminate, everything would follow from that one by “standard” arguments. For simplicity, I'll write $A[T]$ for such  a ring of polynomials.

Regular elements in a ring $R$ are those which are not zero divisors. Explicitly, $a\in R$ is regular if $ab=0$ implies $ b=0$ for any $b\in R$. Or $a$ is a zero divisor if there exists a nonzero $b\in R$ such that $ab=0$.

The classic result linking these two notions that I have in mind is the following result.

Proposition. — Let $A$ be a commutative ring. If $P\in A[T]$ is not regular, then there exists a nonzero $a\in A$ such that $aP=0$.

There is a classic, rather elementary, proof of that result which I recall first. Let us first treat the case where is only one indeterminate. By assumption, there exists a nonzero polynomial $Q\in Q[T]$ such that $PQ=0$, and we may choose $Q$ to be of minimal degree. Write $P=\sum_{k=0}^m a_k T^k$ and $Q=\sum_{k=0}^n b_k T^k$, where $m$ and $n$ are the degrees of $P$ and $Q$, so that $a_m\neq 0$ and $b_n\neq 0$.

Lemma. — One has $a_k Q=0$ for all $k$.

The result being obvious for $k>m$, we may argue by decreasing induction on $k$, assuming that $a_\ell Q=0$ for all $\ell>k$. Then, in the product $PQ$, the term of largest degree is $a_k b_n T^{k+n}$, so that $a_k b_n=0$. In particular, the polynomial $a_kQ$ has degree $<n$. Since $P(a_k Q)=a_k PQ=0$, and by the choice of $Q$, we have $a_kQ=0$, as claimed.

Given the lemma, we see that $a_k b_\ell=0$ for all integers $k$ and $\ell$. Choosing any nonzero coefficient $b_\ell$ of $Q$, we obtain $b_\ell P=0$. This concludes the proof of the proposition when there is only one indeterminate.

To treat the general case, it is probably possible to make an induction on the number of indeterminates that appear in an equality $PQ=0$, where $Q$ is a nonzero element of $A[T]$. Another possibility, closer to the above proof, would be to choose a graded monomial ordering, for example the `deglex' order (we first order monomials by total degree, and then by lexicographic order with respect to some linear ordering of the indeterminates). I leave the details to later thought.

The last concept of the title is that of associated prime ideal. There are in fact two different notions for this, and in absence of noetherian hypotheses, it is better to consider the one coined as “weakly associated” by N. Bourbaki (Algèbre commutative) or D. Lazard (“Autour de la platitude”, Bulletin de la S. M. F., tome 97 (1969), p. 81-128). This is also the choice I made in my book, (Mostly) commutative algebra, Universitext, 2021. And so let me skip the adverb “weakly”, defining a prime ideal $P$ of $A$ to be an associated prime to an $A$-module $M$ if there exists $m\in M$ such that $P$ is minimal among the set of prime ideals containing the annihilator $\operatorname{ann}_A(m)$ of $m$. (When $A$ is noetherian, more generally when $P$ is finitely generated, one can prove that there exists $m$ such that $P$ equals the annihilator of $m$, and this is the — sometimes — more classic definition of an associated prime ideal.) The set of associated prime ideals of $M$ is denoted by $\operatorname{Ass}_A(M)$ (or even $\operatorname{Ass}(A)$ when $M=A$), where the abbreviation stands for assassin — In mathematics, forcing something to vanish is often, too often, related to the vocabulary of murder…

This looks like it is a strange notion, but it has some algebraic content, as summaried by the following proposition for which proof I refer to Theorem 6.5.8 of my above mentioned book.

Proposition — Let $A$ be a commutative ring and let $M$ be an $A$-module.

1) An element $a\in A$ belongs to some associated prime ideal of $M$ if and only if there exists a nonzero $m\in M$ such that $am=0$.

2) An element $a$ belongs to every associated prime ideal of $M$ if and only if the localized module $M_a$ vanishes.

Associated primes also have some geometric meaning when $M=A/I$, for some ideal $I$ of $A$. Indeed, the minimal prime ideals of $A$ containing $I$ are associated, and they represent the irreducible components of the locus $\mathrm V(I)$ in $\operatorname{Spec}(A)$. When the ideal $I$ is equal to its radical, this is all there is, but in general there are more associated prime ideals, and the corresponding closed subsets of $\mathrm V(I)$ are called “embedded” components.

It is now a natural question to elucidate associate prime ideal in the context of polynomial rings. The answer is given by a theorem of J. W. Brewer and W. J. Heinzer in their paper “Associated primes of principal ideals”, Duke Math. Journal, 1974.

Theorem (Brewer & Heinzer). — Let $A$ be a commutative ring. The associated prime ideals of the polynomial ring $A[T]$ are the ideals of the form $p[T]$, where $p$ is an associated prime ideal of $A$.

This theorem implies the above proposition. Indeed, if $P\in A[T]$ is a zero divisor, there exists an associated prime ideal $q\in A[T]$ such that $P\in q$, and $q$ is of the form $p[T]$ for some associated prime ideal $p$ of $A$. By definition, there exists an element $a\in A$ such that $p$ is minimal among the prime ideals annihilating $a$. Then $ap=0$, which implies that $aP=0$. Since $p$ is prime, one has $a\neq 0$, hence the proposition.

It doesn't look so clear to me how to immediately deduce the theorem from the proposition, although that might well be possible, and even easy. Anyway, let us describe the proof given by Brewer and Heinzer (this is theorem 7 in their paper), adjusting the notation and terminology. In fact, I cannot really prove the theorem without using the proposition, but I might have overlooked some easy argument.

We consider an associated prime ideal $q$ of $A[T]$ and let $p=A \cap q$ be its trace on $A$. Note that $p$ is a prime ideal of $A$.

Lemma. — One has $q=p[T]$.

The inclusion $p[T]\subseteq q$ is obvious; let us show the other one. Let $f\in q$ and let us prove $f\in p[T]$. Localizing at $p$, we recall that $A_p$ is a local ring with maximal ideal $pA_p$: in that case, elements of $pA_p$ are zero divisors (because $pA_p$ is associated), and elements outside of $pA_p$ are units. In particular, elements of $A_p[T]$ outside of $pA_p[T]$ have at least one coefficient which is a unit, hence can't be zero-divisors in $A_p[T]$ (by the proposition). This implies that $qA_p=pA_p[T]$ after localization, hence there exists $a\notin p$ such that $af\in p[T]$. Using that $p$ is prime, we get $f\in p[T]$, as was to be shown.

Let us now show that $q$ is an associated prime ideal of $A$. We still assume that $A$ is local with maximal ideal $p$. Let $f\in A[T]$ be such that $q$ is a minimal prime ideal among those containing $\operatorname{ann}(f)$.

Lemma. — One has $q=\sqrt{\operatorname{ann}(f)}$.

The radical of an ideal is the intersection of the minimal prime ideals that contain it, and $q$ is among this set of prime ideals, so that it suffices to prove that $q$ is the only minimal prime ideal of $A[T]$ that contains $\operatorname{ann}(f)$. Let then $q'$ be such a prime ideal. By definition, it is an associated prime ideal of $A[T]$ so that, by what has already been proved of the theorem, there exists a prime ideal $p'$ of $A$ such that $q'=p'[T]$. Since $A$ is local with maximal ideal $p$, one has $p'\subseteq p$, hence $q'=p'[T]\subseteq p[T]=q$. Since $q$ was assumed to be a minimal prime ideal containing $\operatorname{ann}(f)$, we get $q'=q$, as claimed.

From this lemma, one deduces that $p=\sqrt{\operatorname{ann}(f)\cap A}$.

Let then $a$ be a nonzero coefficient of $f$. By definition, one has $\operatorname{ann}_{A[T]}(f)\cap A \subseteq \operatorname{ann}_A(a)$, so that $p = \sqrt{\operatorname{ann}_{A[T]}(f)\cap A} \subseteq \sqrt{\operatorname{ann}_A(a)}$. Since Since $A$ is local with maximal ideal $p$, this implies that $p=\sqrt{\operatorname{ann}_A(a)}$. Consequently, $p$ is the only prime ideal containing $\operatorname{ann}_A(a)$, which implies that is is an associated prime ideal of $A$, concluding the proof of the theorem of Brewer and Heinzer.

On the other hand, there is a slightly different approach to the theorem of Brewer and Heinzer, and I want to sketch it now. This will build on proposition II/3.2 of Lazard's paper mentioned above.

Proposition (Lazard). — Let $B$ be a flat $A$-algebra and let $M$ be a $B$-module. Then $\operatorname{Ass}_A(M)$ equals the image of $\operatorname{Ass}_B(M)$ under the natural continuous map from $\operatorname{Spec}(B)$ to $\operatorname{Spec}(A)$.

Now specialize that proposition to the case where $B=A[T]$ and $M=B$. This says that the image of $\operatorname{Ass}_B(B)$ (the associated prime ideals of $B$) in $\operatorname{Spec}(A)$ is exactly $\operatorname{Ass}_A(A[T])$. On the other hand, the freeness of $A[T]$ as an $A$-module implies that $\operatorname{Ass}_A(A[T])=\operatorname{Ass}_A(A)$. Indeed, one reduces to the case of finite free modules, and then apply the classic inclusion $$ \operatorname{Ass}_A(N) \subseteq \operatorname{Ass}_A(M) \subseteq \operatorname{Ass}_A(N) \cup \operatorname{Ass}_A(M/N) $$ for every submodule $N$ of $M$, which will show by induction that $\operatorname{Ass}_A(M)=\operatorname{Ass}_A(A)$ when $M$ is free. On the other hand, we have seen in the first (and easiest) part of the proof of the theorem of Brewer and Heinzer that every associated prime ideal of $A[T]$ is of the form $q=p[T]$ for some prime ideal $p$ of $A$. Since $p=q\cap A$, this shows that $p$ is an associated prime of $A$ and gives the alternative proof.

Lazard's proof is very similar to (but predates by 5 years) the arguments given by Brewer and Heinzer. I don't really know why they didn't use it in their paper.