Simple extensions in splitting fields with Galois group \(S_4\) (D&F 14.2.11)

Suppose \(f(x) \in \mathbb{Z}[x]\) is an irreducible quartic whose splitting field has Galois group \(S_4\) over \(\mathbb{Q}\). Let \(\theta\) be a root of \(f(x)\) and set \(K = Q(\theta)\). Prove that \(K\) is an extension of \(\mathbb{Q}\) of degree 4 which has no proper subfields.

Let the splitting field be \(L\). We have \(\text{Gal}(L/K) \cong S_3\) as one root is fixed, but the other 3 roots can be freely permuted. If \(M \subset K\) is a proper subfield, it must have degree 2 and hence \(\text{Gal}(L/M) \cong A_4\), since \(A_4\) is the unique index 2 subgroup in \(S_4\). However, we show that this leads to a contradiction as \(A_4\) does not contain an isomorphic copy. Simply, note that there are 3 elements of order 2 in both \(S_3\) and \(A_4\), hence any embedding of \(S_3\) and \(A_4\) gives a one-to-one correspondence between these. But only the order 2 elements of \(A_4\), together with the identity, form a subgroup (namely \(Z_4\)), meaning there is no such embedding.