Prove that if \(F\) is a field and \(g(x) \in F[x]\) has a root \(\alpha \in F\), then its Galois group is the same as the Galois group of \(g(x)/(x - \alpha)\).
Solution
The Galois group of a nonzero polynomial means the group of \(F\)-automorphisms of its splitting field. As usual this notation presupposes \(g\ne0\). Fix an algebraic closure \(\overline F\).
1. The quotient \(h(x)=g(x)/(x-\alpha)\) belongs to \(F[x]\).
Division by the monic linear polynomial \(x-\alpha\) gives \(g=(x-\alpha)h+c\), with \(h\in F[x]\) and \(c\in F\). Evaluation at \(\alpha\) gives \(c=g(\alpha)=0\). Since \(g\ne0\), the quotient \(h\) is nonzero and \(\deg h=\deg g-1\).
2. The two splitting fields inside \(\overline F\) are equal.
Let \(R_g\) and \(R_h\) be the sets of roots in \(\overline F\), without counting multiplicities. The identity \(g=(x-\alpha)h\) implies \[ R_g=R_h\cup\{\alpha\}. \] Indeed, for any \(b\in\overline F\), the product \((b-\alpha)h(b)\) vanishes exactly when one of its factors does. Since \(\alpha\in F\), adjoining it does not enlarge any extension of \(F\). The generated fields therefore satisfy \[ F(R_g)=F(R_h,\alpha)=F(R_h). \] These are exactly the two splitting fields. If \(\deg g=1\), then \(h\) is a nonzero constant, \(R_h\) is empty, and both fields are \(F\); the same equality covers that case.
3. The Galois groups coincide.
Writing the common splitting field as \(L\), both groups equal \(\operatorname{Aut}_F(L)\), with the same composition law.