Let \(F\) be a field and let \(f(x)\in F[x]\).
- Define a splitting field of \(f(x)\) over \(F\).
- Let \(F\) be a finite field with \(q\) elements and let \(E/F\) be a finite extension of degree \(n>0\). Exhibit an explicit polynomial \(g(x)\in F[x]\) whose splitting field over \(F\) is \(E\).
- Show that \(E/F\) is Galois.
Solution
1. A splitting field of \(f\) over \(F\) is a field extension \(L/F\) such that \(f\) splits into linear factors over \(L\) and \(L\) is generated over \(F\) by the roots of \(f\).
2. Since \(|F|=q\) and \([E:F]=n\), the field \(E\) has \(q^n\) elements. Take\[ g(x)=x^{q^n}-x\in F[x]. \]Then \(E\) is the splitting field of \(g\) over \(F\).
Every \(a\in E\) satisfies \[ a^{q^n}=a, \] so every element of \(E\) is a root of \(g\). The derivative is \[ g'(x)=q^n x^{q^n-1}-1=-1, \] so \(g\) has no repeated roots. Since its degree is \(q^n\) and \(E\) already supplies \(q^n\) distinct roots, its roots are exactly the elements of \(E\). Therefore \(g\) splits over \(E\) and its roots generate \(E\) over \(F\).
3. The extension \(E/F\) is Galois.
By 2, \(E\) is the splitting field over \(F\) of the separable polynomial \(x^{q^n}-x\). A splitting field of a separable polynomial is normal and separable, hence Galois.
Equivalently, \[ \Gal(E/F)=\langle x\mapsto x^q\rangle\cong C_n. \]