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 $x^4+ax^2+b$ if and only if $F$ contains a quadratic extension.
Proof: (11) Let $F=ℚ(θ)$. We observe its minimal polynomial $m_θ(x)$ 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 $D_8$ 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 $ℚ(\sqrt{D})⊆F$ be quadratic. Then we have $F=ℚ(\sqrt{D})(\sqrt{a+b\sqrt{D}})=ℚ(\sqrt{a+b\sqrt{D}})$ for any nonsquare $a+b\sqrt{D}$, which exists since quadratic extensions are generally extensions via square roots. Let $α=\sqrt{a+b\sqrt{D}}$. We observe $x^4-2ax^2+a^2-b^2D$ 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 $x^3+4ax^2+4b^2Dx$ which clearly has a factor of $x$. By the procedure the Galois group cannot be either $S_4$ or $A_4$, hence must be $D_8$.
(12) ($⇒$) Let $α$ have such a minimal polynomial. Factor $x^2+ax+b=(x-c)(x-d)$ so that $x^4+ax^2+b=(x^2-c)(x^2-d)$. Necessarily $c$ and $d$ are not contained in $ℚ$ as the minimal polynomial must be irreducible, so they must generate a quadratic extension. Hence $α^2$ generates a quadratic extension in $F$.
($⇐$) As we saw previously, without assuming $F$ is non-Galois, we can arrive at a minimal polynomial $x^4-2ax^2+a^2-b^2D$ for a generator of $F$.$~\square$
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