Counter-examples to $x \not \in K \implies \Bbb Q(x) \cap K = \Bbb Q$

Question

Let $K/\Bbb Q$ be a finite extension and let $x$ be an algebraic number such that $x \not \in K$.

1. Is it possible that $\Bbb Q(x) \cap K \neq \Bbb Q$, if the degree of $x$ over $\Bbb Q$ equals $[K:\Bbb Q]$ ?
2. Is it possible that $ \Bbb Q(x) \cap K \neq \Bbb Q$, if $K=\Bbb Q(a_1,\dots,a_n)$ and $x,a_1,\dots,a_n$ are conjugated over $\Bbb Q$ (i.e. have the same minimal polynomial)?

I know that if the degrees $\text{deg}_{\Bbb Q}(x)$ and $[K:\Bbb Q]$ are coprime, then $x \not \in K \implies \Bbb Q(x) \cap K = \Bbb Q$ holds. A simple counter-example to this implication is $K=\Bbb Q(\sqrt 2),x=\sqrt[4]{2}$ since the intersection is $K$. But if the degrees are the same, I did not find any counter-example.

As for $2$., we have in particular that $\text{deg}_{\Bbb Q}(x)$ divides $[K:\Bbb Q]$. However, I wasn't sure how to find a possible counter-example.

Thank you very much!

Answer 1

Let $x=\sqrt[4]{2}$ and $K=\mathbb Q(a_1)$ where $a_1=i\sqrt[4]{2}$.