12. Let F be an extension of degree 4 over ℚ. Show that F can be generated over ℚ by an element with minimal polynomial of the form x4+ax2+b if and only if F contains a quadratic extension.
Proof: (11) Let F=ℚ(θ). We observe its minimal polynomial mθ(x) of degree 4. It is irreducible, and since simultaneously we have F⊆L and F is not Galois, by the procedure for quartics only the groups of order greater than 4 remain, which are those above.
(⇒) By the fundamental theorem, since every subgroup of order 2 in D8 is contained in a subgroup of order 4, we have F contains a subfield of index 4 in L, i.e. degree 2 over ℚ.
(⇐) Let ℚ(√D)⊆F be quadratic. Then we have F=ℚ(√D)(√a+b√D)=ℚ(√a+b√D) for any nonsquare a+b√D, which exists since quadratic extensions are generally extensions via square roots. Let α=√a+b√D. We observe x4−2ax2+a2−b2D is the minimal polynomial for α. The Galois closure is the splitting field for this polynomial. We observe the resolvent cubic for this quartic is x3+4ax2+4b2Dx which clearly has a factor of x. By the procedure the Galois group cannot be either S4 or A4, hence must be D8.
(12) (⇒) Let α have such a minimal polynomial. Factor x2+ax+b=(x−c)(x−d) so that x4+ax2+b=(x2−c)(x2−d). Necessarily c and d are not contained in ℚ as the minimal polynomial must be irreducible, so they must generate a quadratic extension. Hence α2 generates a quadratic extension in F.
(⇐) As we saw previously, without assuming F is non-Galois, we can arrive at a minimal polynomial x4−2ax2+a2−b2D for a generator of F. ◻
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