Let \(f(x)\) be an irreducible polynomial of degree 5 over the field \(\mathbb Q\) of rational numbers with exactly 3 real roots.
Show that \(f(x)\) is not solvable by radicals.
Let \(E\) be the splitting field of \(f\) over \(\mathbb Q\). Construct a Galois extension \(K\) of degree 2 over \(\mathbb Q\) lying in \(E\) such that no field \(F\) strictly between \(K\) and \(E\) is Galois over \(\mathbb Q\).
Solution
1. Let\[ G=\operatorname{Gal}(E/\mathbb Q). \]Then \(G\) acts transitively on the five roots of \(f\).
The polynomial \(f\) is irreducible over \(\mathbb Q\) and, since the characteristic is zero, it is separable. In the splitting field of an irreducible separable polynomial, the Galois group acts transitively on the roots: any \(\mathbb Q\)-embedding sending one root to another extends to an automorphism of the normal splitting field \(E\).
2. The group \(G\) contains a \(5\)-cycle.
By 1, the action of \(G\) on five roots is transitive. Orbit-stabilizer therefore implies that \(5\) divides \(|G|\). By Cauchy’s theorem, \(G\) contains an element of order \(5\). Viewed inside \(S_5\), any element of order \(5\) is a \(5\)-cycle.
3. The group \(G\) contains a transposition.
Complex conjugation preserves \(E\) because \(f\) has rational coefficients. By hypothesis, three roots of \(f\) are real, so complex conjugation fixes those three roots. The remaining two roots are nonreal and, because the coefficients are real, form a conjugate pair; complex conjugation swaps them. Hence its permutation of the five roots is a transposition.
4. A \(5\)-cycle together with any transposition generates \(S_5\).
Let \[ \sigma=(0\,1\,2\,3\,4) \] be a \(5\)-cycle after relabeling the roots, and let \[ \tau=(a\,b) \] be any transposition. Conjugating by powers of \(\sigma\) gives the transpositions \[ \sigma^k\tau\sigma^{-k}=(a+k\,b+k) \qquad (k\in\mathbb Z/5\mathbb Z). \] Let \(d=b-a\not\equiv0\pmod5\). These are the edges \[ (k\,\,k+d) \] of a graph on \(\mathbb Z/5\mathbb Z\). Since \(5\) is prime and \(d\ne0\), repeated addition of \(d\) reaches every residue class, so this graph is connected. Transpositions along the edges of a connected graph generate the full symmetric group on its vertices. Hence \[ \langle\sigma,\tau\rangle=S_5. \]
5. Therefore\[ G=S_5. \]
By 2 and 3, \(G\) contains a \(5\)-cycle and a transposition. By 4, those two elements generate \(S_5\). Since \(G\le S_5\), it follows that \(G=S_5\).
6. The polynomial \(f\) is not solvable by radicals.
A polynomial over a field of characteristic zero is solvable by radicals only if the Galois group of its splitting field is a solvable group. By 5 this Galois group is \(S_5\). The subgroup \(A_5\trianglelefteq S_5\) is nonabelian simple, hence not solvable; therefore \(S_5\) is not solvable. Thus \(f\) is not solvable by radicals.
7. Let \(r_1,\dots,r_5\) be the roots of \(f\) and define\[ \delta=\prod_{1\le i<j\le5}(r_i-r_j). \]Then\[ \delta^2=\Delta(f)\in\mathbb Q, \]and\[ K=\mathbb Q(\delta)=\mathbb Q(\sqrt{\Delta(f)}) \]is the fixed field \(E^{A_5}\).
For \(\sigma\in S_5=G\), permuting the roots changes the Vandermonde product by the sign of the permutation: \[ \sigma(\delta)=\operatorname{sgn}(\sigma)\delta. \] Hence every element of \(A_5\) fixes \(\delta\), while every odd permutation sends \(\delta\) to \(-\delta\). Since \(G\) contains odd permutations, \(\delta\notin\mathbb Q\). Therefore the stabilizer of \(\delta\) in \(G\) is exactly \(A_5\), so by the Galois correspondence \[ \mathbb Q(\delta)=E^{A_5}. \] Also \(\delta^2\) is the discriminant \(\Delta(f)\), which lies in \(\mathbb Q\).
8. The extension \(K/\mathbb Q\) is Galois of degree \(2\).
By 7, \[ K=E^{A_5}. \] Since \(A_5\trianglelefteq S_5\) and \[ [S_5:A_5]=2, \] the Galois correspondence gives \[ [K:\mathbb Q]=2 \] and \[ \operatorname{Gal}(K/\mathbb Q)\cong S_5/A_5\cong C_2. \] Thus \(K/\mathbb Q\) is Galois.
9. No field \(F\) satisfying\[ K\subsetneq F\subsetneq E \]is Galois over \(\mathbb Q\).
Let \[ H=\operatorname{Gal}(E/F). \] Because \(K=E^{A_5}\), the strict containments \(K\subsetneq F\subsetneq E\) correspond to \[ 1<H<A_5. \] If \(F/\mathbb Q\) were Galois, then by the Galois correspondence \(H\) would be normal in \[ G=S_5. \] But a normal subgroup \(H\trianglelefteq S_5\) contained in \(A_5\) is also normal in \(A_5\). Since \(A_5\) is simple, such an \(H\) is either \(1\) or \(A_5\), contradicting \[ 1<H<A_5. \] Therefore no strict intermediate field between \(K\) and \(E\) is Galois over \(\mathbb Q\).