Does $\mathbb{Q}_p \cap \overline{\mathbb{Q}}$ depend on the embedding of $\overline{\mathbb{Q}}$ into $\overline{\mathbb{Q}_p}$? I think the title pretty much says it all. I'm getting confused in the subtle parts of a proof, and would appreciate some help.
 A: Once you fix an algebraic closure $\overline{\mathbb{Q}_p}$, the subfield $\overline{\mathbb{Q}}$ is uniquely determined as the smallest algebraically closed subfield containing the prime field, so all embeddings have the same image.
A: There's a subtlety here. 
$\overline{\mathbb{Q}}\cap \mathbb{Q}_p$ is well defined as a subfield of $\mathbb{Q}_p$ since it is simply the set of elements which are algebraic over $\mathbb{Q}$.
However, as a subfield of $\overline{\mathbb{Q}}$, this is not canonical. The choice of embedding $\phi: \overline{\mathbb{Q}}\rightarrow \overline{\mathbb{Q}_p}$ amounts to choosing a place of $\overline{\mathbb{Q}}$ lying over $p$. The induced subfield $\phi^{-1}(\mathbb{Q}_p)\subset \overline{\mathbb{Q}}$ will be the maximal subfield in which this place splits. This depends very much on the place we've chosen! This field is not Galois over $\mathbb{Q}$, and its Galois conjugates reflect the other choices of embeddings.
Here's another way of saying it. The embedding $\phi$ gives a choice of decomposition group
$D_p\subset$ Gal$(\overline{\mathbb{Q}}/\mathbb{Q})$. Then
$\phi^{-1}(\mathbb{Q}_p)=\mathbb{\overline{Q}}^{D_p}$.
