Let \(L\) be a Galois extension of a field \(K\) of degree 4. What is the minimum number of subfields there could be strictly between \(K\) and \(L\)? What is the maximum number of such subfields? Give examples where these bounds are attained.
How do these numbers change if we assume only that \(L\) is separable (but not necessarily Galois) over \(K\)?
<1>2. In the cyclic case \(C_4\), there is exactly one field strictly between \(K\) and \(L\).
<1>3. In the Klein-four case \(C_2\times C_2\), there are exactly three fields strictly between \(K\) and \(L\).
<1>4. Thus, among Galois quartic extensions, the minimum number of proper intermediate fields is \(1\) and the maximum is \(3\).
<1>5. For any separable extension \(L/K\) of degree \(4\), every proper intermediate field has degree \(2\) over \(K\).
<1>6. A separable quartic extension has at most three proper intermediate fields.
<1>7. The minimum under the assumption “separable, not necessarily Galois” is \(0\).
<1>8. Therefore, among all separable quartic extensions, the minimum number of proper intermediate fields is \(0\) and the maximum is \(3\).
Solution
1. If \(L/K\) is Galois of degree \(4\), then\[ \operatorname{Gal}(L/K)\cong C_4 \quad\text{or}\quad C_2\times C_2. \]
By the Galois correspondence, proper intermediate fields correspond to proper nontrivial subgroups of \(C_4\). A cyclic group of order \(4\) has exactly one such subgroup, namely its unique subgroup of order \(2\).
The Klein four group has exactly three proper nontrivial subgroups, all of order \(2\). Galois correspondence therefore gives exactly three intermediate quadratic fields.
3. They are attained, for example, by\[ \mathbb Q(\zeta_5)/\mathbb Q, \]whose Galois group is \((\mathbb Z/5\mathbb Z)^\times\cong C_4\), and by\[ \mathbb Q(\sqrt2,\sqrt3)/\mathbb Q, \]whose Galois group is \(C_2\times C_2\).
If \(K\subsetneq F\subsetneq L\), then by the tower law \[ 4=[L:K]=[L:F][F:K]. \] Both factors are integers greater than \(1\), so both equal \(2\).
Suppose \(F_1 e F_2\) are two intermediate fields. By 5 each is quadratic over \(K\), hence separable and therefore Galois over \(K\). Their compositum lies in \(L\), and since \(F_1\ne F_2\), \[ [F_1F_2:K]=4. \] Thus \(F_1F_2=L\). A compositum of Galois extensions is Galois, so \(L/K\) is then Galois. By 3, a Galois quartic with more than one intermediate field has exactly three. Hence no separable quartic can have more than three.
Let \(\alpha\) be a root of \[ f(x)=x^4-x-1\in\mathbb Q[x], \] and put \(L=\mathbb Q(\alpha)\). The reduction modulo \(2\) is \(x^4+x+1\), which is irreducible over \(\mathbb F_2\); hence \(f\) is irreducible over \(\mathbb Q\), so \([L:\mathbb Q]=4\).
Let \(N\) be the splitting field of \(f\). The factorization modulo \(2\) is irreducible, so the Galois group of \(N/\mathbb Q\) contains a \(4\)-cycle. Modulo \(7\), \[ f(x)\equiv (x-3)(x^3+3x^2+2x-2), \] and the cubic factor has no root in \(\mathbb F_7\), so the group contains a \(3\)-cycle. Modulo \(17\), \[ f(x)\equiv (x+2)(x+5)(x^2-7x+5), \] and the quadratic factor is irreducible because its discriminant \(29\equiv12\pmod{17}\) is not a square, so the group contains a transposition. A transitive subgroup of \(S_4\) containing a \(4\)-cycle, a \(3\)-cycle, and a transposition is \(S_4\). Hence \[ \operatorname{Gal}(N/\mathbb Q)\cong S_4. \] The subgroup fixing \(\alpha\) is a point stabilizer \(S_3\), which is maximal in \(S_4\). By the Galois correspondence for \(N/\mathbb Q\), intermediate fields between \(\mathbb Q\) and \(L=N^{S_3}\) correspond to subgroups between \(S_3\) and \(S_4\). Since there are none, \(L/\mathbb Q\) has no proper intermediate fields.
The upper bound is 6, the example in 7 attains \(0\), and the biquadratic example in 4 attains \(3\).
If one restricts specifically to separable quartic extensions that are not Galois, then 6 shows there can be at most one intermediate field. Both possibilities occur: 7 gives none, while \(\mathbb Q(\sqrt[4]{2})/\mathbb Q\) has the intermediate field \(\mathbb Q(\sqrt2)\), and cannot have a second one because two distinct quadratic intermediate fields would make the quartic extension Galois.