# $\Bbb Q(\sqrt{D_f})$ is an intermediate field with degree of extension $2$

Let $$K |\Bbb Q$$ be a cyclic extension of even degree.Let $$f \in \Bbb Q[X]$$ be an irreducible polynomial of degree $$[K:\Bbb Q]$$ having a root in $$K$$ .Then show that the unique field $$F$$ such that $$\Bbb Q \subset F \subset K$$ with $$[F:\Bbb Q]=2$$ is given by $$F=\Bbb Q(\sqrt{D_f})$$ where $$D_f ( =\prod_{i and $$\alpha_i$$ are all the roots of $$f$$ ) is the discriminant.

Since, $$\rm{Gal(K|\Bbb Q)}$$ is cyclic of even order there exists a unique subgroup of order 2, so that takes care of uniqueness. Since, $$K|\Bbb Q$$ is Galois, $$\alpha_i \in K, \forall i$$ .

So I need to show that

(i) $$\sqrt{D_f}\in K$$ and (ii) $$[\Bbb Q(\sqrt{D_f}):\Bbb Q]=2$$

First of all, there exists a unique intermediate field $$F$$ such that $$[F : \mathbb Q] = 2$$ because $${\rm Gal}(K : \mathbb Q)$$ has a unique subgroup of index $$2$$.
(i) $$\sqrt{D_f} = \prod_{i < j}(\alpha_i - \alpha_j)$$ is clearly generated by the $$\alpha_i$$'s over $$\mathbb Q$$, and each $$\alpha_i$$ is in $$K$$, so $$\sqrt{D_f}$$ is in $$K$$.
(ii) First of all, the discriminant $$D_f$$ is in $$\mathbb Q$$. So $$[\mathbb Q(\sqrt{D_f}) : \mathbb Q]$$ is at most $$2$$, being a root of the quadratic polynomial $$X^2 - D_f \in \mathbb Q[X]$$.
So we just need to rule out the possibility that $$[\mathbb Q(\sqrt{D_f} ) : \mathbb Q] = 1$$. By the Galois correspondence, this means ruling out the possibility that $$\sqrt{D_f} = \prod_{i < j}(\alpha_i - \alpha_j)$$ is fixed by all automorphisms in $${\rm Gal}(K : \mathbb Q)$$. Let $$n := {\rm deg}(f) = [K : \mathbb Q]$$. Then $${\rm Gal}(K : \mathbb Q)$$ is a cyclic subgroup of order $$n$$ within $$S_n$$, the group of permutations on the roots $$\{ \alpha_1, \dots, \alpha_n \}$$. Moreover, it is a transitive subgroup of $$S_n$$ (since $$K$$ is a normal extension over $$\mathbb Q$$, and the $$\alpha_i$$'s are roots of a single irreducible polynomial $$f \in \mathbb Q[X]$$). An automorphism fixes $$\sqrt{D_f}$$ if and only if it is an even permutation of the roots. So you need to prove that the alternating group $$A_n$$ does not contain a transitive cyclic subgroup of order $$n$$. Indeed a transitive cyclic subgroup of order $$n$$ can only be generated by an element with cycle structure $$(123\dots n)$$. Since $$n$$ is even, an element with this cycle structure must be odd, which is a contradiction.