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?
 A: Note that in set theory, you cannot really take an intersection $X \cap Y$ of two sets $X$ and $Y$ unless you already assume that $X$ and $Y$ are subsets of a larger set $Z$.
The question "What is $\overline{\mathbb Q_p} \cap \mathbb C$?" does not really make sense as the sets $\overline{\mathbb Q_p}$ and $\mathbb C$ are not canonically contained in a larger set $Z$.
There are several, inequivalent definitions of both $\mathbb C$ and $\overline{\mathbb Q_p}$ and depending on which definition you use, your question will have different answers.  In the various ways you could define these sets, you could correctly say that the answer to your question is $\overline{\mathbb Q_p}$, or $\mathbb Q$, or the empty set!
So your question isn't really meaningful.  Note, however, that there are (infinitely many) field isomorphisms $\mathbb C \rightarrow \overline{\mathbb Q_p}$.  This is due to a standard result about transcendence bases that can be found in any graduate algebra textbook.
A: 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 $\overline{\Bbb{Q}_p},\Bbb{C}\to K$.

*

*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?
