For a prime $\mathfrak q$ of a finite separable extension $L/K$ above $\mathfrak p$, completion preserves the <ramification index> and <residue-field extension>:
$$
e(L_{\mathfrak q}/K_{\mathfrak p})=e_{\mathfrak q/\mathfrak p},
\qquad
k_{L_{\mathfrak q}}=k_{\mathfrak q}.
$$

Completion preserves ramification and residue data

The <completion of a number field at a prime ideal> does not change either invariant. More precisely, $\widehat B_q$ is the <valuation ring> of $L_q$, its maximal ideal is $q\widehat B_q$, and
$$
\boxed{e(L_q/K_p)=e_{q/p},\qquad
\widehat B_q/q\widehat B_q\cong B/q.}
$$
Thus the completed extension has the same <ramification index> and the same <residue-field extension>, hence the same <residue-field degree> $f_{q/p}$. This is <completion preserves ramification and residue data>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 123 / 1 / ii / Solution

A local homomorphism of local rings $(R,\mathfrak m)\to(S,\mathfrak n)$ induces the residue-field extension $R/\mathfrak m\to S/\mathfrak n$. For an extension of non-Archimedean local fields, its degree is the <residue-field degree>.

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 Incoming links: Residue-field extension