**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:*

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

*The extension $K\to M$ is Galois;*- $\mathop{\rm Aut}_M(L)$ is a normal subgroup of $\mathop{\rm Aut}_K(L)$;
- 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 $\sigma(M)\subset L^G$. On the other hand, the hypothesis that $\mathop{\rm Aut}_M(L)$ is normal in $\mathop{\rm Aut}_K(L)$ implies that $G=\mathop{\rm Aut}_M(L)$, so that $L^G=M$. 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.

A typo: y \notin E_1 at the end of the proof of Prop. 1, instead of y in E_1...

ReplyDeleteThanks a lot! Corrected!

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