Tell me a condition on the Galois group which is implied by irreducibility of the polynomial. What happens when the polynomial has a root in the base field?
Solution
Theorem.
Let \(f\in F[x]\) be irreducible of
degree \(n\ge1\) and let \(L\) be its splitting field. Then \(G=\mathrm{Gal}(L/F)\) acts transitively on
the \(n\) roots of \(f\). If \(f\) has a root in \(F\), then \(n=1\), so \(L=F\) and \(G\) is trivial.
Let \(\alpha,\beta\) be roots of \(f\) in \(L\). Each generates an \(F\)-embedding \(F(\alpha)\hookrightarrow L\) with \(\alpha\mapsto\beta\).
Because \(L/F\) is normal, any such embedding extends to an \(F\)-automorphism \(\sigma\in G\).
Then \(\sigma(\alpha)=\beta\), so every root is in the \(G\)-orbit of every other root. Hence the action is transitive.
If \(f\) has a root \(\alpha\in F\), then \(x-\alpha\) divides \(f\) in \(F[x]\). Irreducibility forces \(f(x)=x-\alpha\) (degree \(1\)).
For degree \(1\) there is no nontrivial splitting field extension, so \(L=F\) and \(G=\{1\}\).