Prove that the Galois Group Is $D_{10}$ I've been working on this problem for a couple days and I haven't heard back from any of my support team (TAs/Prof) at Uni. Not looking for an answer, but help and hints would be greatly appreciated. Thank you!
Problem: 
Let $f \in \mathbb{Q}[x]$ be an irreducible quintic. Let K be it's splitting field. $f$ has Galois group $D_{10}$ (Dihedral of order 10). Let $\mathbb{Q}(\sqrt{d})$ be the unique subfield of K of degree 2 over $\mathbb{Q}$. 
Let $\zeta = e^\frac{2\pi}{5}$. Show that the Galois group of the extension $K(\zeta):\mathbb{Q}(\zeta)$ is also Dihedral of order 10 unless $d = 5$. 
My thoughts so far:
So, I know that $\mathbb{Q}(\zeta)$ is a degree four extension over $\mathbb{Q}$. I also know that since $f$ has Galois group $D_{10}$, K must have degree 10 over $\mathbb{Q}$. 
I know that the subfield structure for K must mirror the subfield structure for $D_{10}$. So, first, $\mathbb{Q}(\sqrt{d})$ corresponds to $C_5$. Also, given this information about $d$, we know that $f$ has at least one real root. 
Basically, to solve the problem, we only need to show that $|K(\zeta):\mathbb{Q}(\zeta)| = 10$, since there is only one option for Galois groups of order 10. 
Now, our professor told us to look at the subfields of $K(\zeta)$, K and $\mathbb{Q}(\zeta)$ and to look at $K \cap \mathbb{Q}(\zeta)$. This is where I am stuck. I don't know how to think about this intersection because I have no clue how to find what else is in K given the information we have. 
 A: Let us fix notations: $K/\mathbf Q$ is Galois with group $\cong D_{10}$ and $\mathbf Q(\sqrt d)$ is its unique quadratic subfield. Denoting by $\zeta$ a primitive $5$-th root of unity, you want to show that $K(\zeta)/\mathbf Q(\zeta)$ is Galois with group $\cong D_{10}$ unless $d=5$ (more precisely, unless $5d^{-1}$ is a square in $\mathbf Q$).
It is well known that the cyclotomic extension $\mathbf Q(\zeta)/\mathbf Q$ is cyclic of degree $4$, hence admits a unique quadratic subfield, which is here its maximal real subfield $\mathbf Q(\zeta)^+=\mathbf Q(\zeta+{\zeta}^{-1})=\mathbf Q(\sqrt 5)$. Because $gcd (4,10)=2$, it follows that, unless $d=5$, the extensions $\mathbf Q(\zeta)$ and $K$ are linearly disjoint over $\mathbf Q$, hence $Gal(K/\mathbf Q) \cong Gal(K(\zeta)/\mathbf Q(\zeta))$, and we are done.
NB: For any odd prime $p$, it is known from the computation of the discriminant of $\mathbf Q(\zeta_p)/\mathbf Q$ that the unique quadratic subfield of $\mathbf Q(\zeta_p)$ is $\mathbf Q(\sqrt {p^*})$, where $p^*={(-1)}^{\frac {p-1}{2}}$, so your problem (and answer) can be generalized when replacing $5$ by $p$, and the condition $d\neq 5$ by $d\neq p^*$.
