The question is basically coming from the following situation:
Let $C$ be an integral curve over a field $k$ (EDIT and assume that $k$ is not algebraically closed) and let $\phi\colon C^N\to C$ be the normalization morphism. Being finite (I like so much excellent schemes!) we can define a push-forward map on $0$-cycles
$$z\_0(C^N) \xrightarrow{\phi\_\*}z\_0(C).$$
Let $[P]\in z\_0(C^N)$ be a generator. How does $\phi\_\*([P])$ look like?
Generically this will be just $[P]$ itself, i.e. when $P$ is lying above a non-singular point of the original curve $C$. (I'm actually rephrasing in a stupid way the fact that the normalization induces an isomorphism between the local rings at regular points).
Suppose now that $P$ is actually in the fiber of a non regular point $Q$ of $C$. Then $$\phi\_\*([P]) = [k(P):k(Q)][Q].$$
Question: when is this degree extension $[k(P):k(Q)]$ equal to 1? 


This eventually brought me to the following slightly more general question, without geometrical background. 


Let $A$ be a local Noetherian domain, with maximal ideal $\mathfrak{p}$. Let $B$ be its normalization in $K=Frac(A)$ and suppose that $A\to B$ is finite. Then $B$ will be a semilocal ring. Let $\mathfrak{m}$ be a maximal ideal of $B$.


Question: under which conditions the residue field extension $A/\mathfrak{p}\to B/\mathfrak{m}$ is trivial? 


EDIT: I remove the following comment, due answer already given. 


"My - probably wrong feeling - is that this is pretty much true "in char 0". But maybe something funny happens when $p$ is in the story."



Does the normalization morphism induce isomorphism on residue fields?
The residue field extension is not always trivial in characteristic $0$. For instance, in $\mathbb{A}^2\_{\mathbb{R}}$, consider the plane curve $C$ of points $(x,y)$ satisfying the equation $x^2+y^2 = y^3$. The closed point $(0,0)$ of $C$ has residue field $\mathbb{R}$, yet the inverse image in the normalization is a single closed point with residue field $\mathbb{C}$.


