Completion of an extension $K|\mathbb{Q}$ w.r.t. a non archimedean abs. value, isomorphic to $K\cdot \mathbb{Q}_p$ As the title suggests I want to prove the following result:

given $\mathfrak{p}|(p)$, prime ideal of $\mathcal{O}_K$ over $p$, we have the canonical absolute value induced by it on $K$, and we can complete $K$ and obtain $K_{\mathfrak{p}}| \mathbb{Q}_p$ a new field extension. I want to prove that $K_{\mathfrak{p}}\cong \mathbb{Q}_p\cdot K$.

the inclusion $\mathbb{Q}_p\cdot K \subseteq K_{\mathfrak{p}}$ is clear. But I don't know how to show the other. I know by Krasner's lemma that I can assume $K_{\mathfrak{p}}=\mathbb{Q}_p(\alpha)$, for $\alpha \in \mathbb{Q}^{alg}$ (algebraic closure), but I don't know how to prove that $\alpha \in K$.
 A: To form a compositum $K\Bbb Q_p$, the two fields must be injected in some way into a larger field. You want something like an algebraic closure $\Omega$ of $\Bbb Q_p$, and then try to map $K$ into it. But there will be $[K:\Bbb Q]$ such embeddings into $\Omega$. Which one?
Let’s take a specific example, $K=\Bbb Q[a]/(a^3-10)$, what you get by adjoining a cube root of $10$ to $\Bbb Q$. If your $p$ is $3$, then $K$ has three embeddings into an algebraic closure $\Omega$ of $\Bbb Q_3$. As it happens, $\Bbb Q_3$ already has a cube root of $10$, call it $b$, and $a\mapsto b$ gives one embedding of $K$ into $\Bbb Q_3$. But in $\Bbb Q_3(\omega)$ ($\omega$ a primitive cube root of unity), there are two more cube roots of $10$, so there are two embeddings of $K$ into that extension of $\Bbb Q_3$. One way to look at this is to take the minimal $\Bbb Q$-polynomial for $a$, namely $X^3-10$, and look it as a $\Bbb Q_3$-polynomial, in which case it factors as $(X-b)(X^2+bX+b^2)$, linear times irreducible quadratic. Each factor gives you an equivalence class of embeddings of $K$ into $\Omega$.
But which embedding is the right one? You’ve framed the question in terms of a prime over $p$; in this case, $(3)=\mathfrak p_1\mathfrak p_2^2$, both primes being of degree one. If you complete with respect to $\mathfrak p_1$, the completion is (isomorphic to) $\Bbb Q_3$; if $\mathfrak p_2$, the completion is $\Bbb Q_3(\omega)$.
With all this, I think you can see your way to your desired conclusion.
A: Try to show directly that $\mathbb Q_p \cdot K$ is complete.
