When Baire meets Krasner

Here is a well-but-ought-to-be-better known theorem.

Theorem. — Let $\ell$ be a prime number and let $G$ be a compact subgroup of $\mathop{\rm GL}_d(\overline{\mathbf Q_\ell})$. Then there exists a finite extension $E$ of $\mathbf Q_\ell$ such that $G$ is contained in $\mathop{\rm GL}_d(E)$.

Before explaining its proof, let us recall why such a theorem can be of any interest at all. The keyword here is Galois representations.

It is now a well-established fact that linear representations are an extremly useful tool to study groups. This is standard for finite groups, for which complex linear representations appear at one point or another of graduate studies, and its topological version is even more classical for the abelian groups $\mathbf R/\mathbf Z$ (Fourier series) and $\mathbf R$ (Fourier integrals). On the other hand, some groups are extremly difficult to grasp while their representations are ubiquitous, namely the absolute Galois groups $G_K=\operatorname{Gal}(\overline K/K)$ of fields $K$.

With the notable exception of real closed fields, these groups are  infinite and have a natural (profinite) topology with open subgroups the groups $\operatorname{Gal}(\overline K/L)$, where $L$ is a finite extension of $K$ lying in $\overline K$. It is therefore important to study their continuous linear representations. Complex representations are important but since $G_K$ is totally discontinuous, their image is always finite. Therefore, $\ell$-adic representations, namely continuous morphisms from $G_K$ to $\mathop{\rm GL}_d(\mathbf Q_\ell)$, are more important. Here $\mathbf Q_\ell$ is the field of $\ell$-adic numbers.

Their use goes back to Weil's proof of the Riemann hypothesis for curves over finite fields, via the action on $\ell^\infty$-division points of its Jacobian variety. Here $\ell$ is a prime different from the characteristic of the ground field. More generally, every Abelian variety $A$ over a field $K$ of characteristic $\neq\ell$ gives rise to a Tate module $T_\ell(A)$ which is a free $\mathbf Z_\ell$-module of rank $d=2\dim(A)$, endowed with a continuous action $\rho_{A,\ell}$ of  $G_K$. Taking a basis of $T_\ell(A)$, one thus has a continuous morphism $G_K\to \mathop{\rm GL}_d(\mathbf Z_\ell)$, and, embedding $\mathbf Z_\ell$ in the field of $\ell$-adic numbers,  a continuous morphism $G_K\to\mathop{\rm GL}_d(\mathbf Q_\ell)$. Even more generally, one can consider the $\ell$-adic étale cohomology of algebraic varieties over $K$.

For various reasons, such as the need to diagonalize additional group actions, one can be led to consider similar representations where $\mathbf Q_\ell$ is replaced by a finite extension of $\mathbf Q_\ell$, or even by the algebraic closure $\overline{\mathbf Q_\ell}$. Since $G_K$ is a compact topological groups, its image by a continuous representation $\rho\colon G_K\to\mathop{\rm GL}_d(\overline{\mathbf Q_\ell}$ is a compact subgroup of $\mathop{\rm GL}_d(\overline{\mathbf Q_\ell}$ to which the above theorem applies.

This being said for the motivation, one proof (attributed to Warren Sinnott)  is given by Keith Conrad in his short note, Compact subgroups of ${\rm GL}_n(\overline{\mathbf Q}_p)$. In fact, while browsing at his large set of excellent expository notes,  I fell on that one and felt urged to write this blog post.

The following proof had been explained to me by Jean-Benoît Bost almost exactly 20 years ago. I believe that it ought to be much more widely known.

It relies on the Baire category theorem and on Krasner's lemma.

Lemma 1 (essentially Baire). — Let $G$ be a compact topological group and let $(G_n)$ be an increasing sequence of closed subgroups of $G$ such that $\bigcup G_n=G$. There exists an integer $n$ such that $G_n=G$.

Proof. Since $G$ is compact Hausdorff, it satisfies the Baire category theorem and there exists an integer $m$ such that $G_m$ contains a non-empty open subset $V$. For every $g\in V$, then $V\cdot g^{-1}$ is an open neighborhood of identity contained in $G_m$. This shows that $G_n$ is open in $G$. Since $G$ is compact, it has finitely many cosets $g_iG_m$ modulo $G_m$; there exists an integer $n\geq m$ such that $g_i\in G_n$ for every $i$, hence $G=G_n$. QED.

Lemma 2 (essentially Krasner). — For every integer $d$, the set of all extensions of $\mathbf Q_\ell$ of degree $d$, contained in $\overline{\mathbf Q_\ell}$, is finite.

Proof. Every finite extension of $\mathbf Q_\ell$ has a primitive element whose minimal polynomial can be taken monic and with coefficients in $\mathbf Z_\ell$; its degree is the degree of the polynomial. On the other hand, Krasner's lemma asserts that for every such irreducible polynomial $P$, there exist a real number $c_P$ for every monic polynomial $Q$ such that the coefficients of $Q-P$ have absolute values $<c_P$, then $Q$ has a root in the field $E_P=\mathbf Q_\ell[T]/(P)$. By compactness of $\mathbf Z_\ell$, the set of all finite subextensions of given degree of $\overline{\mathbf Q_\ell}$ is finite. QED.

Let us now give the proof of the theorem. Let $(E_n)$ be a increasing sequence of finite subextensions of $\overline{\mathbf Q_\ell}$ such that $\overline{\mathbf Q_\ell}=\bigcup_n E_n$ (lemma 2; take for $E_n$ the subfield generated by $E_{n-1}$ and all the subextensions of degree $n$ of $\overline{\mathbf Q_\ell}$). Then $G_n=G\cap \mathop{\rm GL}_d(E_n)$ is a closed subgroup of $G$, and $G$ is the increasing union of all $G_n$. By lemma 1, there exists an integer $n$ such that $G_n=G$. QED.
 

When Lagrange meets Galois

Jean-Benoît Bost told me a beautiful proof of the main ingredient in the proof of Galois correspondence, which had been published by Lagrange in his 1772 “Réflexions sur la résolution des résolutions algébriques”, almost 60 years before Galois. (See Section 4 of that paper, I think; it is often difficult to recognize our modern mathematics in the language of these old masters.)

In modernized notations, Lagrange considers the following situation. He is given a polynomial equation $ T^n + a_{n-1} T^{n-1}+\cdots + a_0 = 0$, with roots $x_1,\dots,x_n$, and two “rational functions” of its roots  $f(x_1,\dots,x_n)$ and $\phi(x_1,\dots,x_n)$. (This means that $f$ and $\phi$ are the evaluation at the $n$-tuple $(x_1,\dots,x_n)$ of two rational functions in $n$ variables.) Lagrange says that $f$ and $\phi$ are similar (“semblables”) if every permutation of the roots which leaves $f(x_1,\dots,x_n)$ unchanged leaves $\phi(x_1,\dots,x_n)$ unchanged as well (and conversely). He then proves that $\phi(x_1,\dots,x_n)$ is a rational function of $a_0,\dots,a_{n-1}$ and $f(x_1,\dots,x_n)$.

Let us restate this in a more modern language. Let $K\to L$ be a finite Galois extension of fields, in the sense that $K= L^{G}$, where $G=\mathop{\rm Aut}_K(L)$. Let $a, b\in L$ and let us assume that every element $g\in G$ which fixes $a$ fixes $b$ as well; then Lagrange proves that $b\in K(a)$.

Translated in our language, his proof could be as follows. In formula, the assumption is that $g\cdot a=a$ implies $g\cdot b=b$; consequently, there exists a unique *function* $\phi\colon G\cdot a\to G\cdot b$ which is $G$-equivariant and maps $a$ to $b$. Let $d=\mathop{\rm Card}(G\cdot a)$ and let us consider Lagrange's interpolation polynomial —the unique polynomial $P\in L[T]$ of degree $d$ such that $P(x)=\phi(x)$ for every $x\in G\cdot a$. If $h\in G$, the polynomial $P^h$ obtained by applying $h$ to the coefficients of $P$ has degree $d$ and coincides with $\phi$; consequently, $P^h=P$. By the initial assumption, $P$ belongs to $K[T]$ and $b=P(a)$, hence $b\in K(a)$, as claimed.

Combined with the primitive element theorem, this allows to give another short, and fairly elementary, presentation of the Galois correspondence.

Galois Theory, Geck's style

This note aims at popularizing a short note of Meinolf Geck, “On the characterization of Galois extensions”, Amer. Math. Monthly 121 (2014), no. 7, 637–639 (Article, Math Reviews, arXiv), that proposes a radical shortcut to the treatment of Galois theory at an elementary level. The proof of the pudding is in the eating, so let's see how it works. The novelty lies in theorem 2, but I give the full story so as to be sure that I do not hide something under the rug.

Proposition 1. Let $K\to L$ be a field extension. Then $L$ is not the union of finitely many subfields $M$ such that $K\to M\subsetneq L$.

Proof. It splits into two parts, according whether $K$ is finite or infinite.

Assume that $K$ is finite and let $q=\mathop{\rm Card}( K)$. Then $L$ is finite as well, and let $n=[L:K]$ so that $\mathop{\rm Card}(L)=q^n$. If $M$ is a subextension of $L$, then $\mathop{\rm Card}( L)=q^m$, for some integer $m$ dividing $n$; moreover, $x^{q^m}=x$ for every $x\in L$. Then the union of all strict sub-extensions of $L$ has cardinality at most $\sum_{m=1}^{n-1} q^m =\frac{q^n-q}{q-1}<q^n$.

It remains to treat the case where $K$ is infinite; then the proposition follows from the fact that a finite union of strict subspace of a $K$-vector space $E$ is not equal to $E$. Let indeed $(E_i)_{1\leq i\leq n}$ be a family of strict subspaces of $E$ and let us prove by induction on $n$ that $E\neq \bigcup_{i=1}^n E_i$. The cases $n\leq1$ are obvious. By induction we know that for every $j\in\{1,\dots,n\}$, the union $\bigcup_{i\neq j}E_i$ is distinct from $E$, hence select an element $x\in E$ such that $x\not\in E_2\cup \dots\cup E_n$. The desired result follows if, by chance, $x\not\in E_1$. Otherwise, choose $y\in E\setminus E_1$. For $s\neq t\in K$, and $i\in\{2,\dots,n\}$, observe that $y+sx$ and $y+tx$ cannot both belong to $E_i$, for this would imply that $(s-t)x\in E_i$, hence $x\in E_i$ since $s\neq t$. Consequently, there are at most $n-1$ elements $s\in K$ such that $y+sx\in \bigcup_{i=2}^nE_i$. Since $K$ is infinite, there exists $s\in K$ such that $y+sx\not\in\bigcup_{i=2}^n E_i$. Then $y+sx\not\in E_1$, neither, since $x\in E_1$ and $y\not\in E_1$. This proves that $E\neq \bigcup_{i=1}^nE_i$.

Let $K\to L$ be a field extension and let $P\in K[T]$. We say that $P$ is split in $L$ if it is a product of linear factors in $L[T]$. We say that $P$ is separable if all of its roots (in some extension where it is split) have multiplicity $1$. We say that $K\to L$ is a splitting extension of $P$ if $P$ is split in $L$ and if $L$ is the subextension of $K$ generated by the roots of $P$ in $L$. Finally, we let $\mathop{\rm Aut}_K(L)$ be the set of $K$-linear automorphisms of $L$; it is a group under composition.

Theorem 2. Let $K\to L$ be a finite extension of fields and let $G=\mathop{\rm Aut}_K(L)$. Then $\mathop{\rm Card}( G)\leq [L:K]$. Moreover, the following conditions are equivalent:

1. One has $\mathop{\rm Card}( G)=[L:K]$;
2. There exists an irreducible separable polynomial $P\in K[T]$ such that $\deg(P)=[L:K]$ and which is split in $L$;
3. The extension $K\to L$ is a splitting extension of a separable polynomial in $K[T]$;
4. One has $K=L^G$.

Remark 3. In the conditions of (2), let us fix a root $z\in L$ of $P$. One has $L=K(z)$. Moreover, the map $f\mapsto f(z)$ is a bijection from $\mathop{\rm Aut}_K(L)$ to the set of roots of $P$ in $L$.

Proof of Theorem 2.
(a) Let us prove that $\mathop{\rm Card} (G)\leq [L:K]$. Let $m\in\mathbf N$ be such that $m\leq \mathop{\rm Card}( G)$ and let $\sigma_1,\dots,\sigma_m$ be distinct elements of $G$. For $1\leq i<j\leq m$, let $M_{i,j}$ be the subfield of $L$ consisting of all $x\in L$ such that $\sigma_i(x)=\sigma_j(x)$. It is a strict subextension of $L$ because $\sigma_i\neq\sigma_j$. Consequently, $L$ is not the union of the subfields $M_{i,j}$ and there exists an element $z\in L$ such that $\sigma_i(z)\neq \sigma_j(z)$ for all $i\neq j$. Let $P$ be the minimal polynomial of $z$. Then the set $\{\sigma_1(z),\dots,\sigma_m(z)\}$ consists of distinct roots of $P$, hence $\deg(P)\geq m$. In particular, $m\leq [L:K]$. Since this holds for every $m\leq \mathop{\rm Card}( G)$, this shows that $\mathop{\rm Card}( G)\leq [L:K]$.

(b) If one has $\mathop{\rm Card}( G)=[L:K]$, then taking $m=\mathop{\rm Card}( G)$, we get an irreducible polynomial $P\in K[T]$ of degree $m$, with $m$ distinct roots in $L$. Necessarily, $P$ is separable and split in $L$. This gives (1)$\Rightarrow$(2).

The implication (2)$\Rightarrow$(3) is obvious.

(1)$\Rightarrow$(4). Let $M=L^G$. One has $\mathop{\rm Aut}_K(L)=\mathop{\rm Aut}_M(L)=G$. Consequently, $\mathop{\rm Card}(G)\leq [L:M]$. Since $\mathop{\rm Card}( G)=[L:K]=[L:M][M:K]$, this forces $M=K$.

(4)$\Rightarrow$(3). There exists a $G$-invariant subset $A$ of $L$ such that $L=K(A)$. Then $P=\prod_{a\in A}(T-a)$ is split in $L$, and is $G$-invariant. Consequently, $P\in K[T]$. By construction, $P$ is separable and $L$ is a splitting extension of $P$.

(3)$\Rightarrow$(1). Let $M$ be a subextension of $L$ and let $f\colon M\to L$ be a $K$-morphism. Let $a\in A$ and let $Q_a$ be the minimal polynomial of $a$ over $M$. The association $g\mapsto g(a)$ defines a bijection between the set of extensions of $f$ to $M(a)$ and the set of roots of $Q_a$ in $L$. Since $P(a)=0$, the polynomial $Q_a$ divides $P$, hence it is separable and split in $L$. Consequently, $f$ has exactly $\deg(Q_a)=[M(a):M]$ extensions to $M(a)$.

By a straightforward induction on $\mathop{\rm Card}(B)$, for every subset $B$ of $A$, the set of $K$-morphisms from $K(B)$ to $L$ has cardinality $[K(B):K]$. When $B=A$, every such morphism is surjective, hence $\mathop{\rm Card}(\mathop{\rm Aut}_K(L))=[L:K]$.

If these equivalent conditions hold, we say that the finite extension $K\to L$ is Galois.

Corollary 4. Let $K\to L$ be a finite Galois extension. The maps $H\to L^H$ and $M\to \mathop{\rm Aut}_M(L)$ are bijections, inverse one of the other, between subgroups of $\mathop{\rm Aut}_K(L)$ and subextensions $K\to M\subset L$.

Proof. a) For every subextension $K\to M\subset L$, the extension $M\subset L$ is Galois. In particular, $M=L^{\mathop{\rm Aut}_M(L)}$ and $\mathop{\rm Aut}_M(L)=[L:M]$.

b) Let $H\subset\mathop{\rm Aut}_K(L)$ and let $M=L^H$. Then $M\to L$ is a Galois extension and $[L:M]=\mathop{\rm Aut}_M(L)$; moreover, one has $H\subset\mathop{\rm Aut}_M(L)$ by construction. Let us prove that $H=\mathop{\rm Aut}_M(L)$. Let $z\in L$ be any element whose minimal polynomial $P_z$ over $M$ is split and separable in $L$. One has $\mathop{\rm Card}(\mathop{\rm Aut}_M(L))=\deg(P_z)$. On the other hand, the polynomial $Q_z=\prod_{\sigma\in H}(T-\sigma(z))\in L[T]$ divides $P_z$ and is $H$-invariant, hence it belongs to $L^H[T]=M[T]$. This implies that $P_z=Q_z$, hence $\mathop{\rm Card}(H)=\deg(P_z)=\mathop{\rm Card}(\mathop{\rm Aut}_M(L))$. Consequently, $H=\mathop{\rm Aut}_M(L)$.

Corollary 5. Let $K\to L$ be a Galois extension and let $K\to M\to L$ be an intermediate extension. The extension $M\to L$ is Galois too. Moreover, the following assertions are equivalent:

1. The extension $K\to M$ is Galois;
2. $\mathop{\rm Aut}_M(L)$ is a normal subgroup of $\mathop{\rm Aut}_K(L)$;
3. For every $\sigma\in\mathop{\rm Aut}_K(L)$, one has $\sigma(M)\subset M$.
   

Proof. (a) Let $P\in K[T]$ be a separable polynomial of which $K\to L$ is a splitting field. Then $M\to L$ is a splitting extension of $P$, hence $M\to L$ is Galois.

(b) (1)$\Rightarrow$(2): Let $\sigma\in \mathop{\rm Aut}_K(L)$. Let $z$ be any element of $M$ and let $P\in K[T]$ be its minimal polynomial. One has $P(\sigma(z))=\sigma(P(z))=0$, hence $\sigma(z)$ is a root of $P$; in particular, $\sigma(z)\in M$. Consequently, the restriction of $\sigma$ to $M$ is a $K$-morphism from $M$ to itself; it is necessarily a $K$-automorphism. We thus have defined a map from $\mathop{\rm Aut}_K(L)$ to $\mathop{\rm Aut}_K(M)$; this map is a morphism of groups. Its kernel is $\mathop{\rm Aut}_M(L)$, so that this group is normal in $\mathop{\rm Aut}_K(L)$.

(2)$\Rightarrow$(3): Let $\sigma\in\mathop{\rm Aut}_K(L)$ and let $H=\sigma\mathop{\rm Aut}_M(L)\sigma^{-1}$. By construction, one has $\sigma(M)\subset L^H$. On the other hand, the hypothesis that $\mathop{\rm Aut}_M(L)$ is normal in $\mathop{\rm Aut}_K(L)$ implies that $H=\mathop{\rm Aut}_M(L)$, so that $L^H=M$. We thus have proved that $\sigma(M)\subset M$.

(3)$\Rightarrow$(1): Let $A$ be a finite subset of $M$ such that $M=K(A)$ and let $B$ be its orbit under $\mathop{\rm Aut}_K(L)$. The polynomial $\prod_{b\in B}(T-b)$ is separable and invariant under $\mathop{\rm Aut}_K(L)$, hence belongs to $K[T]$. By assumption, one has $B\subset M$. This implies that $K\to M$ is Galois.

Remark 6. Let $L$ be a field, let $G$ be a finite group of automorphisms of $L$ and let $K=L^G$. Every element $a$ of $L$ is algebraic and separable over $K$; inded, $a$ is a root of the separable polynomial $\prod_{b\in G\cdot a}(T-b)=0$, which is $G$-invariant hence belongs to $K[T]$. There exists a finite extension $M$ of $K$, contained in $L$, such that $G\cdot M=M$ and such that the map $\mathop{\rm Aut}_K(L)\to \mathop{\rm Aut}_K(M)$ is injective. Then $K\to M$ is Galois, and $G=\mathop{\rm Aut}_K(M)$. Indeed, one has $G\subset\mathop{\rm Aut}_K(M)$, hence $K\subset M^{\mathop{\rm Aut}_K(M)}\subset M^G\subset L^G=K$. This implies that $K\to M$ is Galois and the Galois correspondence then implies $G=\mathop{\rm Aut}_K(M)$. The argument applies to every finite extension of $K$ which contains $M$. Consequently, they all have degree $\mathop{\rm Card}(G)$; necessarily, $L=M$.

Remark 7 (editions). Matt Baker points out that the actual novelty of the treatment lies in theorem 2, the rest is standard. Also, remark 6 has been edited following an observation of Christian Naumovic that it is not a priori obvious that the extension $K\to L$ is finite.