Let \(K\) be a field of characteristic zero and \(f\in K[x]\) an irreducible polynomial of degree \(n\). Let \(L\) be a splitting field for \(f\). Let \(G\) be the group of automorphisms of \(L\) which act trivially on \(K\).
Show that \(G\) embeds in the symmetric group \(S_n\).
For each \(n\), give an example of a field \(K\) and polynomial \(f\) such that \(G=S_n\).
What are the possible groups \(G\) when \(n=3\). Justify your answer.
Solution
1. The natural action of \(G\) on the roots of \(f\) defines an injective homomorphism\[ G\hookrightarrow S_n. \]
Because \(\operatorname{char}K=0\), the irreducible polynomial \(f\) is separable, so it has \(n\) distinct roots in its splitting field \(L\). Every \(K\)-automorphism \(\sigma\) of \(L\) sends a root \(\alpha\) of \(f\) to another root, since \[ f(\sigma(\alpha))=\sigma(f(\alpha))=0. \] Thus \(G\) acts on the set of \(n\) roots and yields a homomorphism \(G\to S_n\).
If an automorphism lies in the kernel, it fixes every root of \(f\). Since \(L\) is generated over \(K\) by those roots, it fixes all of \(L\). Hence the kernel is trivial, so the homomorphism is injective.
2. For every positive integer \(n\), there is an example with Galois group \(S_n\).
Let \[ L=\mathbb Q(x_1,\dots,x_n) \] be the rational function field in \(n\) algebraically independent variables. Let \(S_n\) act on \(L\) by permuting the variables, and define \[ K=L^{S_n}. \] By Artin’s fixed-field theorem for a finite group of field automorphisms, \[ L/K \] is a finite Galois extension and \[ \operatorname{Gal}(L/K)=S_n. \]
Now set \[ f(T)=\prod_{i=1}^n(T-x_i). \] Its coefficients are symmetric rational functions in the \(x_i\), hence are fixed by \(S_n\) and therefore lie in \(K\). Thus \(f\in K[T]\). It splits in \(L\), and \[ L=K(x_1,\dots,x_n), \] so \(L\) is its splitting field.
It remains to show that \(f\) is irreducible over \(K\). Let \(m(T)\) be the minimal polynomial of \(x_1\) over \(K\). For every \(i\), choose \(\sigma\in S_n\) with \(\sigma(1)=i\). Since \(\sigma\) fixes \(K\), \[ m(x_i)=m(\sigma(x_1))=\sigma(m(x_1))=0. \] Thus the \(n\) distinct elements \(x_1,\dots,x_n\) are all roots of \(m\), so \[ \deg m\ge n. \] But \(m\) divides the degree-\(n\) polynomial \(f\), hence \(\deg m=n\) and \(m=f\). Therefore \(f\) is irreducible and its Galois group is \(S_n\).
3. If \(n=3\), the only possibilities are\[ G\cong A_3\cong C_3 \qquad\text{or}\qquad G\cong S_3. \]
By 1, \(G\) is a subgroup of \(S_3\). Because \(f\) is irreducible and separable, \(G\) acts transitively on its three roots: any \(K\)-embedding from the field generated by one root to the field generated by another extends to a \(K\)-automorphism of the normal splitting field \(L\).
For a transitive action on three points, orbit-stabilizer implies that \(3\) divides \(|G|\). Since \(G\le S_3\), Lagrange’s theorem gives \(|G|\mid6\). Hence \[ |G|=3\quad\text{or}\quad |G|=6. \] The subgroup of order \(3\) is \(A_3\), and a subgroup of order \(6\) is all of \(S_3\).
4. Both cubic possibilities occur over \(\mathbb Q\).
For \(S_3\), take \[ f(x)=x^3-2. \] It is irreducible over \(\mathbb Q\) by Eisenstein at \(2\). If \(\alpha=\sqrt[3]{2}\) and \(\zeta_3\) is a primitive cube root of unity, its splitting field is \[ \mathbb Q(\alpha,\zeta_3). \] The field \(\mathbb Q(\alpha)\) is real of degree \(3\), while \(\zeta_3\notin\mathbb R\), so adjoining \(\zeta_3\) has degree \(2\). Thus the splitting field has degree \(6\), and its Galois group, being a subgroup of \(S_3\) of order \(6\), is \(S_3\).
For \(A_3\), take \[ g(x)=x^3-3x-1. \] It is irreducible over \(\mathbb Q\) by the rational-root test. Its discriminant is \[ \Delta=-4(-3)^3-27(-1)^2=81=9^2. \] If \(r_1,r_2,r_3\) are its roots, then \[ \delta=(r_1-r_2)(r_1-r_3)(r_2-r_3) \] satisfies \(\delta^2=\Delta\). Since \(\Delta=81\), we have \(\delta=\pm9\in\mathbb Q\). An automorphism permuting the roots multiplies \(\delta\) by the sign of the induced permutation. Because every Galois automorphism fixes \(\delta\), every induced permutation is even. Hence the Galois group lies in \(A_3\). By 3 it is transitive, so its order is divisible by \(3\); therefore it is exactly \(A_3\).