Noncyclicity of Negative-Discriminant Quartic Fields over Q (14.6.19)

Dummit and Foote Abstract Algebra, section 14.6, exercise 19:

Let $f(x)$ be an irreducible polynomial of degree $4$ in $ℚ[x]$ with discriminant $D$. Let $K$ denote the splitting field of $f(x)$, viewed as a subfield of $ℂ$.
(a) Prove $ℚ(\sqrt{D})⊂K$.
(b) Let $τ$ denote complex conjugation and let $τ_K$ denote complex conjugation restricted to $K$. Prove $τ_K$ is an element of $\text{Gal}(K/ℚ)$ of order $1$ or $2$ depending on whether $K⊆ℝ$.
(c) Prove that if $D < 0$ then $K$ cannot be cyclic of degree $4$ over $ℚ$.
(d) Prove generally that $ℚ(\sqrt{D})⊈K$ for any $D < 0$ when $K$ is a cyclic quartic field.

Proof: (a) $\sqrt{D}$ is an expression in the roots of $f(x)$ and hence is clearly within $K$. Since $f(x)$ is irreducible we have $[K~:~ℚ]≥4$ and so the proper containment is evident.

(b) When $K⊆ℝ$, $τ_K$ is merely the identity. So assume $K⊈ℝ$: Then $[K~:~K∩ℝ]=2$ as $≥$ clearly holds, and $≤$ must hold as $K$ is Galois over $K∩ℝ$ and the only nonidentity automorphism must be $φ(a+bi)=a+bφ(i)=a-bi$, i.e. $τ_K$.

(c,d) Assume $D < 0$ so that $\sqrt{D}∉ℝ$. Then $τ_K≠1$ and it must be the sole automorphism of degree 2 generated by either of the two other nonidentity automorphisms. But now $ℚ(\sqrt{D})$ is not fixed by anything but the identity, a contradiction as $ℚ(\sqrt{D})≠K$. Nothing more was needed than the fact that $D < 0$.$~\square$