Let \(K\) be a Galois extension of a field \(F\) with \([K: F] = 2015\). Prove that \(K\) is an extension by radicals of the field \(F\).
If \(N \normal G\) is a normal subgroup and \(H\leq G\) is any subgroup containing \(N\), then \(N\) is normal in \(H\) since \(hN\inverseof{h} \subseteq gN\inverseof{g} = N\).
In characteristic zero, a polynomial is solvable by radicals iff its Galois group is a solvable group.
Solution
Let \(G\da \Gal(K/F)\), then it suffices to show that \(G\) is always a solvable group, i.e. any group of order \(n=2015\) is solvable. Factor \(2015 = 5\cdot 13\cdot 31\) – this is a \(pqr\) factorization, and in fact any group with exactly 3 prime factors (so \(n\) is squarefree in particular) will be solvable. Let \(p=5, q=13, r=31\) so that \(p<q<r\). We aim to construct a composition series whose successive quotients are simple groups. Applying Sylow 3 yields
\(n_p \divides qr, n_p \equiv 1 \mod p \implies n_5 \divides 13\cdot 31 = 403\) and \(n_5\equiv 1 \mod 5\).
- So \(n_5 \in \ts{1, 13, 31}\) by divisibility and imposing the congruence forces \(n_5\in \ts{1, 31}\) since \(13\not\equiv 1 \mod 5\).
\(n_q \divides pr, n_q \equiv 1 \mod q \implies n_{13} \divides 5\cdot 31 = 155\) and \(n_{13}\equiv 1 \mod 13\).
- So \(n_{13} \in \ts{1,5,31}\) by divisibility and the congruence imposes \(n_{13}\in \ts{1}\) since \(5,31 \not\equiv 1\mod 13\). In particular, there is one Sylow 13-subgroup which is normal.
\(n_r \divides pq, n_r \equiv 1 \mod r \implies n_{31} \divides 5 \cdot 13 = 65\) and \(n_{31}\equiv 1 \mod 31\).
- So \(n_{31}\in \ts{1,5,13}\) by divisibility and the congruence imposes \(n_{31}\in \ts{1}\) since \(5,13 \not\equiv 1\mod 31\). In particular, the one Sylow 31-subgroup is normal.
Let \(H_{13}, H_{31}\) be the two normal subgroups of \(G\). Taking the quotient \(\tilde G \da G/H_{31}\) yields a group of order \(5\cdot 13\), and a similar argument as above using the Sylow theorems shows that \(\tilde G\) has a normal subgroup of order 13. By the subgroup correspondence theorem, this yields a normal subgroup \(N_1\normal G\) containing \(H_{31}\) which has order \(13\cdot 31\). So define \(N_2 \da H_{31}\) to obtain \[ G \trianglerighteq N_1 \trianglerighteq N_2 \da H_{31} \trianglerighteq 0 .\] Since the quotients \(N_i/N_{i+1}\) have prime power order, they are cyclic and thus simple. We know \(N_1\) is normal in \(G\) since it came from extending a normal group in a quotient, and we know \(N_2\) is normal in \(N_1\) since it was normal in all of \(G\). So \(G\) is solvable.