Find the Galois groups, over \(\mathbb{Q}\), of the following polynomials.
\(x^4 - 25\).
\(x^3 - x + 1\).
Solution
1. The Galois group of \(x^4-25\) over \(\mathbf Q\) is \(C_2\times C_2\).
The factorization \((x^2-5)(x^2+5)\) gives the four roots \(\sqrt5,-\sqrt5,i\sqrt5,-i\sqrt5\). Hence the splitting field is \[ L=\mathbf Q(\sqrt5,i): \] it contains all the roots, and the field generated by the roots contains \(i=(i\sqrt5)/\sqrt5\). The polynomial \(x^2-5\) is irreducible over \(\mathbf Q\) by Eisenstein at \(5\) [@DF04]. Since \(\mathbf Q(\sqrt5)\subset\mathbf R\) does not contain \(i\), the tower law gives \([L:\mathbf Q]=2\cdot2=4\).
This is a splitting field in characteristic zero, so it is Galois and has four automorphisms. Every automorphism sends \(\sqrt5\) to one of \(\pm\sqrt5\) and \(i\) to one of \(\pm i\), and these images determine it. The resulting injective homomorphism \[ \operatorname{Gal}(L/\mathbf Q)\longrightarrow \{\pm1\}\times\{\pm1\} \] is a bijection because both groups have order \(4\). Thus independent sign changes of \(\sqrt5\) and \(i\) give the claimed Klein four group.
2. The Galois group of \(h(x)=x^3-x+1\) over \(\mathbf Q\) is \(S_3\).
2.1. A rational root of the monic polynomial \(h\) must be \(1\) or \(-1\),but \(h(1)=h(-1)=1\). Thus \(h\) is irreducible, since a reducible cubicover a field has a linear factor. In characteristic zero its rootsare distinct. The Galois group of its splitting field acts faithfullyon those three roots, and irreducibility makes that action transitive[@DF04]. Consequently it is a subgroup of \(S_3\) whose order isdivisible by \(3\), by orbit-stabilizer.
2.2. There is exactly one real root. Indeed,\(h'(x)=3x^2-1\) has critical points \(\pm1/\sqrt3\), and\[ h(-1/\sqrt3)=1+\frac{2}{3\sqrt3}>0,\qquad h(1/\sqrt3)=1-\frac{2}{3\sqrt3}>0. \]The function is strictly increasing from \(-\infty\) to a positive valueon \((-\infty,-1/\sqrt3]\), so has precisely one root there. On theremaining two monotonicity intervals its minimum is the positivenumber \(h(1/\sqrt3)\), so it has no further real root.
2.3. The other two roots form a nonreal conjugate pair. Complexconjugation preserves the splitting field, fixes the real root, andinterchanges the other two roots. It is therefore a transposition inthe Galois group, so that group’s order is divisible by \(2\) as wellas \(3\). Being a subgroup of \(S_3\), it has order at most \(6\). Its orderis thus \(6\), and it is all of \(S_3\).