degree extension over filed of $p$-adic numbers Let $K = \mathbb{Q}(\theta)$ be a numberfield and $[K:\mathbb{Q}]=n$. When $\mathbb{Q}_p$ is the field of  $p$-adic numbers and $K_p=\mathbb{Q}_p(\theta)$, what about $[K_p : \mathbb{Q}_p]$?

I’ll give the results, with no hint of a proof:
The general situation is that if $G(X)=\text{Irr}(\theta,\Bbb Q[X])$ splits as a product of $\Bbb Q_p$ irreducibles $G=g_ig_2\cdots g_m$, then there are essentially $m$ ways of embedding $\Bbb Q(\theta)$ into some finite extension of $\Bbb Q_p$. Each of these embeddings has, associated to it, numbers $f_i$ and $e_i$, the residue field extension degree and the ramification index, respectively, and $\sum_ie_if_i=[\Bbb Q(\theta):\Bbb Q]=\deg(G)$.
You may look at this from the standpoint of the factorization of ideals in the integer-ring $I$ of $\Bbb Q(\theta)$ this way: $(p)=pI=\prod_0^m\mathfrak p_i^{e_i}$, where $\mathfrak p_1,\cdots,\mathfrak p_m$ are the distinct prime ideals of $I$ containing $p$. They’re all maximal ideals of $I$, and each $I/\mathfrak p_i$ is a finite extension $\kappa_i$ of $\Bbb F_p=\Bbb Z/p\Bbb Z$, with $[\kappa_i:\Bbb F_p]=f_i$, the number mentioned above.
For an example, let $\theta$ be a root of the $\Bbb Q$-irreducible polynomial $G=X^4 +3X^3+3X^2+6$. Then Hensel tells us that $G\equiv (X^2+9X-1)(X^2-6X-6)\pmod{16}$. The first factor is congruent to $X^2+X+1$ modulo $2$, so generates $\Bbb F_4$, giving $f=2$, while the second factor is Eisenstein for $2$, giving a ramification degree of $2$. Both quadratic factors are $\Bbb Q_2$-irreducible. We get the result that $f_1=2,e_1=1$ while $f_2=1,e_2=2$, so that $\Bbb Q(\theta)$ has two very different embeddings into $\Bbb Q_2$.
