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.