Link between $\mathbb{Q}_p(\alpha)/\mathbb{Q}_p$ and $\mathbb{Q}(\alpha)/\mathbb{Q}$ for $\alpha \in \overline{\mathbb{Q}}$ Let $\mathbb{Q}(\alpha)$ be a finite extension of $\mathbb{Q}$. We can consider $\mathbb{Q}_p$ the completion of $\mathbb{Q}$ for $p \in \mathbb{P}$. Then we also have $\mathbb{Q}_p(\alpha) / \mathbb{Q}_p$ a finite extension. Now, if we are considering the norm $|N_{\mathbb{Q}(\alpha)/\mathbb{Q}}(.)|_{p}^{1/[\mathbb{Q}(\alpha):\mathbb{Q}]}$ over $\mathbb{Q}(\alpha)$, it's a norm which divide $p$. Then, I want to say that if we take the completion, we have $(\mathbb{Q}(\alpha))_p/\mathbb{Q}_p$ - where $(\mathbb{Q}(\alpha))_p$ is the completion for the norm defined above- (cause $\mathbb{Q}(\alpha)_p$ contains $\mathbb{Q}_p$).
Do we have : $(\mathbb{Q}(\alpha))_p = \mathbb{Q}_p(\alpha)$ ? 
As $(\mathbb{Q}(\alpha))_p$ is a finite extension of $\mathbb{Q}_p$, the norm over $(\mathbb{Q}(\alpha))_p$ is uniquely determined, ans is : $|N_{\mathbb{Q}(\alpha)_p/\mathbb{Q}_p}(.)|_{p}^{1/[\mathbb{Q}(\alpha)_p:\mathbb{Q}_p]}$. 
Now, I think we have : $\mathbb{Q}_p(\alpha) \subset \mathbb{Q}(\alpha)_p$ as $\mathbb{Q}(\alpha)_p$ contains $\mathbb{Q}_p$ and $\alpha$. On the other side, $\mathbb{Q}_p(\alpha)$ contains $\mathbb{Q}(\alpha)$. Now, I would like to say that as this last fields is complete, we should have the other inclusion. But my problem is that the norm we are considering on this last field is : $|N_{\mathbb{Q}_p(\alpha) / \mathbb{Q}_p}(.)|_{p}^{1/[\mathbb{Q}_p(\alpha) : \mathbb{Q}_p]}$, so as the norm seems different, we can't use argument dealing with completion.
So, the statement that $(\mathbb{Q}(\alpha))_p = \mathbb{Q}_p(\alpha)$ is it true ?
Thank you !
 A: I'd dare say that you ought to be a little more careful.
One important thing to understand is that the p-adic absolute valute of an algebraic number is not well defined in general, since the extension of the p-adic absolute value to an algebraic extension of $\mathbb{Q}$ is not unique because you must choose a prime ideal in $\mathcal{O}_{\mathbb{Q}[\alpha]}$ lying over p. See this https://math.stackexchange.com/a/1131025/917010 and comments for an example of how they may differ.
So $(\mathbb{Q}(\alpha))_{p}$ is not well defined.
Then also the main equality is not true, because it suffices to take an extension of the p-adic valuation and a root alpha in $\overline{\mathbb{Q}_{p}}$ whose valuation do not coincide. Again, the example given in the comments of the above post by KCD should easily give you this.
To conclude, I think that the main point here is that the algebraic numbers over $\mathbb{Q}$ and the elements of $\overline{\mathbb{Q}_{p}}$ which are algebraic over the subfield $\mathbb{Q}$ of $\mathbb{Q}_{p}$ are not the same. You always have to choose embeddings, and with these, extension of the valuation.
