What is the intersection of algebraic closure of $\mathbb{Q}_p$ (that is, $\overline{\mathbb{Q}_p}$) and $\mathbb{C}$? What is the intersection of algebraic closure of $\mathbb{Q}_p$ (that is, $\overline{\mathbb{Q}_p}$) and $\mathbb{C}$?
I guess that is just $\mathbb{Q}$, but why?

For any field $K$ containing both $E\cong \overline{\Bbb{Q}_p}$ and $F\cong \Bbb{C}$ you'll have that $E\cap F$ (which makes sense, in contrary to your question) contains $\overline{\Bbb{Q}}$.
For the remaining part it depends on $K$ and the embeddings $\overline{\Bbb{Q}_p},\Bbb{C}\to K$.

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*Taking an isomorphism (given by the axiom of choice) $\overline{\Bbb{Q}_p}\to\Bbb{C}$ you'll have $E=F=K$,


*taking $K=Frac(\overline{\Bbb{Q}_p}\otimes_{\overline{\Bbb{Q}}} \Bbb{C})$ you'll have $E\cap F=\overline{\Bbb{Q}}$.
(the tensor product is a bit sloppy, we need to fix an embedding $\overline{\Bbb{Q}}\to \overline{\Bbb{Q}_p},\overline{\Bbb{Q}}\to \Bbb{C}$ first, but the resulting field and $E\cap F$ doesn't depend on it)


*Can you construct the intermediate cases?
