Let $\alpha\in\Bbb C$ be algebraic over $\Bbb Q$. Consider $K=\Bbb Q(\alpha)$. Let $p$ be a prime number of $\Bbb Z$ and $P$ a prime of $K$ lying above $p$.
Q: does it make sense to say $K_P=\Bbb Q_p(\alpha)$, i.e. does 'adjoining elements' commute with completion?
I think the answer to this is no, because there might be several primes in $K$ lying over $p$ with different inertia/ramification degrees. Thus the degrees $[K_P:\Bbb Q]$ could be different. However the degree $[\Bbb Q_p(\alpha):\Bbb Q]$ does not depend on the prime $P$ chosen.
On the other hand we have $\Bbb Q\subset \Bbb Q_p(\alpha)$ and thus $K\subset \Bbb Q_p(\alpha)$. As $\Bbb Q_p(\alpha)$ is complete this should imply $K_P\subseteq \Bbb Q_p(\alpha)$ in a somewhat canonical way, right? (I think here lies the/a problem, $\Bbb Q_p(\alpha)$ is complete with the unique extension of the absolute value of $\Bbb Q_p$, but there can be different extensions of the $p$-adic absolute value on $\Bbb Q$ to $K$.)
What would be a way to solve this, e.g. if we have only one prime in $K$ lying over $p$, would we then get the equality $K_P=\Bbb Q_p(\alpha)$?
Best Answer
I think the root of all your problems comes from the fact that $\mathbb Q_p(\alpha)$ doesn't actually make any sense. In general if $L/K$ is an extension of fields (e.g. if $L = \overline K$), then if $\alpha\in L$, we define $K(\alpha)$ to be the subfield of $L$ generated by $K$ and $\alpha$. In your case, by definition, $\alpha$ is an element of $\mathbb C$ (or really $\overline{\mathbb Q}$). But there is no canonical embedding of $\overline{\mathbb Q}$ into $\overline{\mathbb Q}_p$. In other words, $\alpha$ is not automatically an element of any extension of $\mathbb Q_p$, so the notation $\mathbb Q_p(\alpha)$ doesn't make sense.
In order to make sense of $\mathbb Q_p(\alpha)$, you are forced to implicitly fix an embedding $\overline{\mathbb Q}\to\overline{\mathbb Q}_p$ (or at least $K\to \overline{\mathbb Q}_p$). The set of such embeddings is naturally isomorphic to the set of primes $\mathfrak p$ over $\overline{\mathbb Z}$ above $p$ (resp. the set of primes of $K$ above $p$).
Hence, to define $\mathbb Q_p(\alpha)$, you have implicitly chosen a prime of $K$ above $p$ already. Thus, when you then choose another prime $P$ of $K$ above $p$, unless this prime was the prime you already implicitly chose (for example, if there was only one possible prime to choose), there's no reason that $K_P$ should be isomorphic to $\mathbb Q_p(\alpha)$. Indeed, it's not even true that $[K_P:\mathbb Q_p] = [\mathbb Q_p(\alpha) : \mathbb Q_p]$: just consider the case that $p$ splits as a product of a residue degree $1$ prime and a residue degree $2$ prime (e.g. $5$ in $\mathbb Q(\sqrt[3]2)$).
Here's what does make sense: $\alpha$ is the root of a polynomial $f\in\mathbb Q[X]$, and there is a unique embedding $\mathbb Q\hookrightarrow\mathbb Q_p$, so we can view $f$ as being in $\mathbb Q_p[X]$. You can then consider the ring $\mathbb Q_p[X]/(f)$. In general, $f$ is not irreducible over $\mathbb Q_p$, and there is a natural isomorphism $$K\otimes_\mathbb Q\mathbb Q_p\cong\mathbb Q_p[X]/(f)\cong \prod_{\mathfrak p\mid p}K_\mathfrak p.$$ Fixing an embedding $\colon K\hookrightarrow\overline{\mathbb Q}_p$ is the same as fixing a prime $\mathfrak p$ above $p$. If $K= \mathbb Q(\alpha)$, then what's true is that $K_\mathfrak p = \mathbb Q_p(\alpha)$ for the prime $\mathfrak p$ defined by that specific choice of embedding.