A shortcut in Galois theory How could we prove Galois correspondence without using Dedekind’s Lemma on group characters, Artin’s lemma and the primitive element theorem ?
I just came across Meinolf Geck's article On the characterization of Galois extensions on arxiv, that you can find in Amer. Math. Monthly 121 (2014), no. 7, p. 637–639 which answers the question, and my answer - that I make a community wiki, will be inspired from this article.
Some remarks : Dedekind's lemma states that if $G$ is a monoid and $k$ a field then any familly of elements of $\textrm{Hom}_{\textrm{Mon.}}(G,k^{\times})$ is free over $k$. This implies that if $L/K$ is a finite extension then $\textrm{Aut}_K(L)$ is finite of order smaller than the degree $[L:K]$ of the extension (the degree inequality). This fact can be shown without Dedekind's lemma, this is what Geck does in the corollary 4 of the arxiv version of its paper. Artin's lemma states that if $L$ is a field a $G$ a finite group of field automorphisms of $L$ and $K = L^G$ then $L/K$ is a Galois extension of degree $|G|$. The Galois correspondance is the one from the corollary 1 in my answer, completed by the corollary 2 following it.
 A: Proposition 1. Let $K\to L$ be a field extension of finite degree. Then $L$ is not the union of finitely many subfields $M$ such that $K\to M\subsetneq L$.
Proof. 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 $\sum_{m=1}^{n-1} q^m =\frac{q-q^n}{1-q} < q^n$. Suppose now that $K$ 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 $y\in E_1$. This proves that $E\neq \bigcup_{i=1}^nE_i$. $\square$
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.
Proposition 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. 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. (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 $1\leq j\leq m$, 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 $\mathop{\rm Card}(\mathop{\rm Aut}_K(L))=[L:K]$. $\square$
Definition. If these equivalent conditions hold, we say that the finite extension $K\to L$ is Galois.
Corollary 1. 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, $H=\mathop{\rm Aut}_M(L)$. $\square$
Corollary 2. 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. $\square$
Proposition 3. Let $L$ be a field, let $G$ be a finite group of automorphisms of $L$ and let $K=L^G$. Then $K\to L$ is Galois, and $G=\mathop{\rm Aut}_K(L)$.
Proof. If $G$ is trivial it is trivial, if not, for each non trivial $g\in G$ pick an $a_g\in L$ such that $g(a_g)\not=a_g$ and note $X = \cup_{g\in G\backslash\{1\}} O(a_g)$. The set $X$ is clearly $G$ stable so that $M:=K(X)$ is also $G$-stable. Now each $a_g$ is algebraic over $K$, as it is root of $P(T) = \prod_{\lambda\in O(a_g)} (X-\lambda) \in L[X]$ which has in fact coefficients in $K$. You can express $P$'s coefficients as symmetric polynomials in the elements of $O(a_g)$ and this shows that they are fixed by $G$, and are therefore in $K$. This implies in fact that each element of $X$ is algebraic over $K$, and as $X$ is finite, $M$ is algebraic over $K$. By stability we have a restriction map $\varphi : G\to \textrm{Aut}_K (M)$ which is injective by definition of the $a_g$'s. Then you have $M^{\textrm{Aut}_K (M)} \subseteq M^G$ and the latter being included in $L^G = K$ you see that $M^{\textrm{Aut}_K (M)} = K$. As we saw that $M/K$ is finite, it is Galois, and then the correspondance shows that $G = \textrm{Aut}_K (M)$. Now take $a\in L$. The same argument shows that $M(O(a))/K$ is Galois with group $G$, the same group than $M/K$'s. The correspondance shows that $M(O(a)) = M$, and this is for any $a\in L$. Therefore $L = M$ is finite. Then as $G \subset \textrm{Aut}_K (L)$ we have $K\subseteq L^{\textrm{Aut}_K (L)} \subseteq L^G = K$ showing that $L^{\textrm{Aut}_K (L)} = K$ : the extension $L/K$ is Galois and $G = \textrm{Aut}_K (L)$. $\square$
