Let \(G\) be any finite group and \(F\) any field. Show that there exist fields \(L\) and \(E\) with \(F \subseteq L \subseteq E\), such that \(E\) is Galois over \(L\) with the Galois group of \(E/L\) being isomorphic to \(G\).
Solution
For each \(g\in G\), let \(x_g\) be an indeterminate, with all of the \(x_g\) algebraically independent over \(F\), and set \[ E:=F(x_g:g\in G). \]
1. The group \(G\) acts faithfully on \(E\) by \(F\)-automorphisms.
For \(h\in G\), define \[ \sigma_h(x_g):=x_{hg} \qquad(g\in G), \] and let \(\sigma_h\) fix \(F\) pointwise. Because the variables are algebraically independent, this permutation of the variables extends uniquely to an automorphism of the polynomial ring and hence of its fraction field \(E\). Moreover, \[ \sigma_h\sigma_k(x_g)=x_{hkg}=\sigma_{hk}(x_g), \] so \[ h\longmapsto\sigma_h \] is a homomorphism \[ G\longrightarrow\operatorname{Aut}_F(E). \] It is injective: if \(h\neq1\), then \[ \sigma_h(x_1)=x_h\neq x_1. \] Thus we identify \(G\) with a finite subgroup of \(\operatorname{Aut}_F(E)\).
2. Let a finite group \(H\) act faithfully by automorphisms on a field \(K\), and set \(K^H:=\{a\in K:\sigma(a)=a\text{ for all }\sigma\in H\}\).Then\[ [K:K^H]=|H| \]and\[ \operatorname{Gal}(K/K^H)=H. \]
Write \[ H=\{\sigma_1,\ldots,\sigma_m\}, \qquad m=|H|. \] We use the linear independence of distinct field homomorphisms: the maps \(\sigma_i:K\to K\) are linearly independent over \(K\) as functions. Indeed, if a nontrivial relation with the fewest nonzero coefficients were \[ \sum_{i=1}^r c_i\sigma_i=0, \qquad c_i\neq0, \] choose \(a\in K\) with \(\sigma_1(a)\neq\sigma_r(a)\). Evaluating at \(ab\) and subtracting \(\sigma_r(a)\) times the relation evaluated at \(b\) yields a shorter nontrivial relation, a contradiction.
This independence implies that there exist elements \[ a_1,\ldots,a_m\in K \] for which the matrix \[ A=(\sigma_i(a_j))_{i,j} \] is invertible. Otherwise every such evaluation matrix would have rank \(<m\), which would give a nonzero linear relation among the functions \(\sigma_i\).
Fix \(b\in K\). Because \(A\) is invertible, there are unique \(c_1,\ldots,c_m\in K\) such that \[ \sigma_i(b)=\sum_{j=1}^m c_j\sigma_i(a_j) \qquad(1\le i\le m). \] Let \(\tau\in H\). Applying \(\tau\) to all these equations replaces every \(\sigma_i\) by \(\tau\sigma_i\), which merely permutes the rows because \(H\) is a group. Therefore the coefficients \(\tau(c_j)\) solve the same linear system as the \(c_j\). By uniqueness, \[ \tau(c_j)=c_j \] for every \(j\) and every \(\tau\in H\). Hence \[ c_j\in K^H. \] Taking the row corresponding to the identity automorphism gives \[ b=\sum_{j=1}^m c_j a_j. \] Thus \(a_1,\ldots,a_m\) span \(K\) over \(K^H\), so \[ [K:K^H]\le m. \]
On the other hand, a finite field extension has at most its degree many embeddings into an algebraic closure over the base field. The \(m\) distinct automorphisms in \(H\) all fix \(K^H\), so \[ m\le [K:K^H]. \] Consequently \[ [K:K^H]=m=|H|. \] The extension has exactly its degree many \(K^H\)-automorphisms, hence is finite Galois, and all of its automorphisms are the elements of \(H\). Therefore \[ \operatorname{Gal}(K/K^H)=H. \]
3. Taking the fixed field of the action in 1 produces the required extension.
Set \[ L:=E^G. \] Because every element of \(G\) fixes \(F\) pointwise, \[ F\subseteq L\subseteq E. \] Applying 2 to the faithful action of \(G\) on \(E\) gives \[ [E:L]=|G| \] and shows that \(E/L\) is Galois with \[ \operatorname{Gal}(E/L)=G. \] Thus every finite group occurs as a Galois group over some extension of the prescribed field \(F\).