Let \(F\) be a field and let \(f(x) \in F[x]\).
State the definition of a splitting field of \(f(x)\) over \(F\).
Let \(F\) be a finite field with \(q\) elements. Let \(E/F\) be a finite extension of degree \(n>0\). Exhibit an explicit polynomial \(g(x) \in F[x]\) such that \(E/F\) is a splitting field of \(g\) over \(F\). Fully justify your answer.
Show that the extension in \((b)\) is a Galois extension.
Solution
1. A splitting field of \(f\in F[x]\) is an extension \(L/F\) such that \(f\) splits into linear factors over \(L\) and \(L\) is generated over \(F\) by the roots of \(f\).
Equivalently, if \(\alpha_1,\ldots,\alpha_r\) are all roots of \(f\) in an algebraic closure, then a splitting field is \[ F(\alpha_1,\ldots,\alpha_r). \] The minimality condition rules out adjoining unrelated elements.
2. If \(|F|=q\) and \([E:F]=n\), then \(|E|=q^n\) and every element of \(E\) is a root of\[ g(x)=x^{q^n}-x\in F[x]. \]
The field \(E\) is an \(n\)-dimensional vector space over the \(q\)-element field \(F\), so \(|E|=q^n\). For \(a\in E^\times\), Lagrange’s theorem in the finite group \(E^\times\) gives \[ a^{q^n-1}=1, \] so \(a^{q^n}=a\). The same equation is trivial for \(a=0\).
3. The polynomial \(g(x)=x^{q^n}-x\) has exactly the elements of \(E\) as its roots, so \(E\) is its splitting field over \(F\).
Its derivative is \[ g'(x)=q^n x^{q^n-1}-1=-1 \] in characteristic \(p\), where \(q\) is a power of \(p\). Thus \(g\) has no repeated roots. It has degree \(q^n\) and, by 2, already has \(q^n\) distinct roots in \(E\). Hence it splits completely over \(E\) and has no other roots. Since the set of roots is all of \(E\), the field generated over \(F\) by the roots is exactly \(E\).
4. The extension \(E/F\) is Galois.
By 3, \(E\) is the splitting field of \(g\). The polynomial \(g\) is separable because \(g'=-1\). Therefore its splitting field over \(F\) is normal and separable, hence Galois.