Given a finite group \(G\), show that there are fields \(K\) and \(F\) such that \(K\) is Galois over \(F\), and \(G\) is isomorphic to the Galois group of \(K\) over \(F\).
Solution
Let \(m=|G|\). Take algebraically independent indeterminates \(x_g\), indexed by \(g\in G\), over \(\mathbb Q\), and put \[ K=\mathbb Q(x_g:g\in G). \] For \(h\in G\) let \(\sigma_h\) fix \(\mathbb Q\) and send \(x_g\) to \(x_{hg}\). Define the fixed field \[ F=\{a\in K:\sigma_h(a)=a\text{ for every }h\in G\}. \]
1. These maps give an injective homomorphism\(G\longrightarrow\operatorname{Aut}_F(K)\).
A permutation of algebraically independent indeterminates extends to an automorphism of the polynomial ring and its fraction field. On the generators, \(\sigma_h\sigma_k(x_g)=x_{hkg}\), so \(\sigma_h\sigma_k=\sigma_{hk}\). In particular \(\sigma_{h^{-1}}\) is the inverse of \(\sigma_h\). The action is faithful because \(\sigma_h(x_1)=x_h\), and different variables are different elements. The fixed elements are closed under addition, subtraction, multiplication, and inversion of nonzero elements; hence \(F\) is a field. Every \(\sigma_h\) fixes it by definition.
2. The extension degree satisfies \([K:F]\leq m\).
Take any \(a_1,\ldots,a_{m+1}\in K\). The homogeneous system \[ \sum_{j=1}^{m+1}\sigma_h(a_j)c_j=0\qquad(h\in G) \] has \(m\) equations in \(m+1\) unknowns over the field \(K\), so its kernel is nonzero. Choose a nonzero kernel vector \(c\) with the smallest possible number of nonzero coordinates, and scale it so that one such coordinate, say \(c_k\), is \(1\).
For \(t\in G\), the vector \(\sigma_t(c)\) also belongs to this kernel. Indeed, applying \(\sigma_t\) to the equation indexed by \(t^{-1}h\) gives the equation indexed by \(h\) for \(\sigma_t(c)\). Thus \(\sigma_t(c)-c\) is a kernel vector. Its support is contained in the support of \(c\), and its \(k\)th coordinate is zero. Minimality of the support forces \(\sigma_t(c)-c=0\). This holds for every \(t\), so each coefficient \(c_j\) lies in \(F\).
The equation for \(h=1\) is now the nontrivial \(F\)-linear relation \(\sum_j a_jc_j=0\). Every \(m+1\) elements of \(K\) are therefore linearly dependent over \(F\), proving the dimension bound and finiteness of the extension.
3. The extension is Galois and its full Galois group is thedisplayed copy of \(G\).
For any \(a\in K\), let \(O_a=\{\sigma_h(a):h\in G\}\), as a set without repetitions. The polynomial \[ p_a(T)=\prod_{b\in O_a}(T-b) \] has coefficients in \(F\): each \(\sigma_h\) permutes \(O_a\). It vanishes at \(a\) and splits into distinct linear factors in \(K\). The minimal polynomial of \(a\) over \(F\) divides \(p_a\), so it also splits into distinct linear factors in \(K\). Thus every element is separable, and every irreducible polynomial over \(F\) with a root in \(K\) splits there. These are separability and normality, so the finite extension \(K/F\) is Galois.
A finite extension of degree \(d\) has at most \(d\) automorphisms fixing its base field. To see the bound, choose a finite tower of simple adjunctions. At an adjunction of degree \(e\), an embedding of the preceding field has at most \(e\) extensions, because the new generator must map to a root of the corresponding minimal polynomial. Multiplying these bounds along the tower gives \(d\). Apply this with \(d=[K:F]\). Step 1 supplies \(m\) distinct automorphisms and step 2 gives \[ m\leq |\operatorname{Aut}_F(K)|\leq [K:F]\leq m. \] Equality holds throughout. Hence the injective homomorphism of step 1 is onto, proving \(\operatorname{Gal}(K/F)\cong G\). For the trivial group the construction has \(F=K\), so that case is included as well.