Let $y,z$ be two algebraic numbers such that $\Bbb Q(y)/\Bbb Q$ and $\Bbb Q(z)/\Bbb Q$ are Galois extensions (such numbers could be called "Galois numbers" in this post). Is it true that $yz$ and $\text{Re}(z)$ are also Galois numbers?
1) I think I solved the question for $y+z$: the answer is no. "In general", $\Bbb Q(y+z) = \Bbb Q(y,z)$ holds and hence is Galois over $\Bbb Q$ (because compositum of Galois extensions is Galois): a proof of the primitive element theorem shows that for infinitely many rational numbers $r$, the element $y+rz$ generates $\Bbb Q(y,z)$.
So I tried an example where $r=1$ doesn't work, as $y=\sqrt[3]{2}+e^{2\pi i / 3}$ and $z=-e^{2\pi i / 3}$. The only non-trivial fact is to show that $y$ is a Galois number; it is sufficient to show that it generates $K=\Bbb Q(\sqrt[3]{2},e^{2\pi i / 3})$. We have $[K:\Bbb Q]=[K:\Bbb Q(e^{2\pi i / 3})] [\Bbb Q(e^{2\pi i / 3}) : \Bbb Q] = 2 \cdot 3 = 6$, and $y$ has degree 6 over $\Bbb Q$ (actually I'm not how to do this by hand). Therefore the inclusion $\Bbb Q(y) \subset K$ is actually an equality.
2) For the product, this is probably very easy, but I have no good idea…
3) I know that $\Bbb Q(\text{Re}(z))$ is the fixed field of $\Bbb Q(z)$ under the complex conjugation. But in general if $L/K$ is Galois $L^H/K$ doesn't have to be Galois. To find a counter-example, I should at least avoid abelian extensions. The example $z = \sqrt[3]{2}+e^{2\pi i / 3}$ doesn't work.
Thank you for your help!
A counterexample for the product: Let $\omega=e^{2\pi i/3}$, $z=\root3\of2+\omega$, $y=\root3\of2+\omega^2=\overline{z}$. Then $$ yz=\root3\of4+(\omega+\omega^2)\root3\of2+\omega^3=\root3\of4-\root3\of2+1. $$ So $yz$ is an irrational element of the field $K=\Bbb{Q}(\root3\of2)$. Therefore $K=\Bbb{Q}(yz)$. But $K/\Bbb{Q}$ is well known not to be a Galois extension.