Let \(F\) be a splitting field over \(K\) of a polynomial \(f\in K[x]\), and write in \(F\) \[ f=\prod_{i=1}^k (x-u_i)^{n_i}, \] where the \(u_i\) are distinct and each \(n_i\geq1\). Let \[ g(x)=\prod_{i=1}^k(x-u_i)=v_0+v_1x+\cdots+v_kx^k \] and let \[ E=K(v_0,\ldots,v_k). \] Then show that the following hold:
\(F\) is a splitting field of \(g\) over \(E\).
\(F\) is Galois over \(E\).
\(\mathrm{Aut}_E(F) = \mathrm{Aut}_K(F)\).
Solution
Because \(F\) is the splitting field of \(f\) over \(K\), it is generated over \(K\) by the distinct roots: \[ F=K(u_1,\ldots,u_k). \] Each coefficient \(v_j\) of \(g\) is a symmetric polynomial in the \(u_i\), so \(v_j\in F\) and therefore \[ K\subseteq E\subseteq F. \]
1. \(F\) is the splitting field of \(g\) over \(E\).
The polynomial \(g\) belongs to \(E[x]\) by definition of \(E\), and in \(F[x]\) it splits as \[ g(x)=\prod_{i=1}^k(x-u_i). \] Moreover, \[ E(u_1,\ldots,u_k) \supseteq K(u_1,\ldots,u_k) =F. \] The reverse inclusion is immediate because \(E\subseteq F\) and every \(u_i\) is in \(F\). Hence \[ E(u_1,\ldots,u_k)=F. \] Thus \(F\) is generated over \(E\) by the roots of \(g\) and \(g\) splits in \(F\); therefore \(F\) is the splitting field of \(g\) over \(E\).
2. The extension \(F/E\) is Galois.
The roots \(u_1,\ldots,u_k\) of \(g\) are pairwise distinct. Hence \(g\) is separable over \(E\): equivalently, every irreducible factor of \(g\) has only simple roots in its splitting field.
By step 1, \(F\) is the splitting field over \(E\) of the separable polynomial \(g\). Therefore \(F/E\) is finite, normal, and separable, hence Galois.
3. Every \(E\)-automorphism of \(F\) is a \(K\)-automorphism.
Since \(K\subseteq E\), any automorphism fixing \(E\) pointwise also fixes \(K\) pointwise. Thus \[ \operatorname{Aut}_E(F)\subseteq\operatorname{Aut}_K(F). \]
4. Every \(K\)-automorphism of \(F\) fixes \(E\) pointwise.
Let \[ \sigma\in\operatorname{Aut}_K(F). \] For each root \(u_i\) of \(f\), \[ f(\sigma(u_i)) =\sigma(f(u_i)) =0, \] because the coefficients of \(f\) lie in \(K\) and are fixed by \(\sigma\). Hence \(\sigma\) permutes the finite set of distinct roots \[ \{u_1,\ldots,u_k\}. \] Consequently \[ \begin{aligned} \sigma(g) &=\prod_{i=1}^k(x-\sigma(u_i))\\ &=\prod_{i=1}^k(x-u_i)\\ &=g. \end{aligned} \] Thus \(\sigma\) fixes every coefficient \(v_0,\ldots,v_k\) of \(g\). It also fixes \(K\), so it fixes \[ E=K(v_0,\ldots,v_k) \] pointwise. Therefore \[ \operatorname{Aut}_K(F)\subseteq\operatorname{Aut}_E(F). \]
Combining steps 3 and 4, \[ \boxed{\operatorname{Aut}_E(F)=\operatorname{Aut}_K(F).} \]