This is probably obvious, but I don't quite see it.
Archimedean completions of different number fields are always isomorphic to the same $\mathbb{R}$ or $\mathbb{C}$. Is the same true in the non-archimedean case?
More precisely, llet $K$, $L$ be non-isomorphic number fields, and $\mathfrak{p}$ a nonarchimedean prime of both $K$ and $L$.
When is it true that $K_\mathfrak{p}\cong L_\mathfrak{p}$?
As Qiaochu Yuan has said in his comments, maybe yes, maybe no. Just look at two quadratic number fields. Say $K=\Bbb Q(\sqrt{-3}\,)$ and $L=\Bbb Q(\sqrt5\,)$. Then their completions at $2$ are isomorphic, the unique quadratic unramified extension of $\Bbb Q_2$. At $3$, $K$ gives you a ramified extension of $\Bbb Q_3$, but $L$ gives you the unramified extension of $\Bbb Q_3$. It’s just the reverse at $5$: $K$ has unramified completion, $L$ has ramified completion. More interesting starting with $7$, for there, $-3\equiv4$, a square, so that the $7$-adic completion of $K$ is not even a field, it’s the product of two copies of $\Bbb Q_7$, while the $7$-adic completion of $L$ is again unramified.
In this description, I’ve kept secret the important role that Quadratic Reciprocity plays in the story, and that’s for you to find out on your own.