Recall that every finite group \(G\) is isomorphic to a subgroup of \(S_n\) for some \(n > 0\). Show that there are fields \(E \supseteq F\) such that the Galois group of \(E\) over \(F\) is isomorphic to \(G\).
Solution
Let \(m=|G|\), let \(\{x_g:g\in G\}\) be algebraically independent indeterminates over \(\mathbb Q\), and set \[ E=\mathbb Q(x_g:g\in G). \] For \(h\in G\), define \(\sigma_h\) on these generators by \(\sigma_h(x_g)=x_{hg}\), fixing \(\mathbb Q\). Set \[ F=\{a\in E:\sigma_h(a)=a\text{ for every }h\in G\}. \]
1. The map \(h\mapsto\sigma_h\) embeds \(G\) into the group of\(F\)-automorphisms of \(E\).
Permuting algebraically independent variables gives an automorphism of the polynomial ring and hence of its fraction field \(E\). The permutation associated with \(h^{-1}\) is its inverse. On each generator, \[ \sigma_h\sigma_k(x_g)=x_{hkg}=\sigma_{hk}(x_g), \] so this is a group homomorphism. It is injective, since \(\sigma_h(x_1)=x_h\), and distinct indeterminates are distinct elements of \(E\). Fixed elements are closed under the field operations, so \(F\) is a subfield of \(E\) containing \(\mathbb Q\). By its definition every \(\sigma_h\) fixes \(F\).
2. The extension \(E/F\) is finite and separable, and is a splittingfield over \(F\).
The polynomial \[ p(T)=\prod_{g\in G}(T-x_g) \] has coefficients in \(F\), because each \(\sigma_h\) permutes its factors. Thus every \(x_g\) is algebraic over \(F\). There are finitely many of them, and \(F(x_g:g\in G)=E\), since \(\mathbb Q\subseteq F\subseteq E\). Successively adjoining these algebraic elements therefore makes \(E/F\) finite. The displayed roots are distinct by algebraic independence. Their minimal polynomials over \(F\) divide the separable polynomial \(p\), so they are separable over \(F\). Hence \(E\) is precisely the splitting field of the separable polynomial \(p\) over \(F\).
3. The displayed \(m\) automorphisms are all the automorphisms of\(E/F\).
By the primitive element theorem for finite separable extensions [@DF04], there is \(u\in E\) with \(E=F(u)\). The polynomial \[ q(T)=\prod_{h\in G}(T-\sigma_h(u)) \] also has coefficients in \(F\): applying \(\sigma_k\) permutes its factors by \(h\mapsto kh\). It is monic of degree \(m\) and vanishes at \(u\). Therefore the minimal polynomial \(\mu_u\) of \(u\) over \(F\) divides \(q\), and \[ [E:F]=\deg\mu_u\leq m. \] Conversely, an \(F\)-automorphism of \(F(u)\) is uniquely determined by its value on \(u\), which must be a root of \(\mu_u\). A degree-\(\deg\mu_u\) polynomial has at most \(\deg\mu_u\) roots in a field. Together with the embedding in step 1 this gives \[ m\leq |\operatorname{Aut}_F(E)|\leq\deg\mu_u\leq m. \] All inequalities are equalities. Thus every \(F\)-automorphism is one of the \(\sigma_h\), and the faithful homomorphism in step 1 is an isomorphism \(G\cong\operatorname{Gal}(E/F)\). Step 2 also shows that the constructed extension is Galois in the finite, normal, separable sense.