# Does the normalization morphism induce isomorphism on residue fields?

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?

"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."

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 $\mathbb{C}$.
• If $C$ is a (locally) finite type scheme over an algebraically closed field $k$, then for every closed point of $C^N$, the extension of residue fields is closed. Feb 4, 2014 at 14:46
• @FedeB: What you say is true, yet, nonetheless, I expect that is the only "general" answer to your question. For a field $K$ and for every finite field extension $L/K$, you can realize this extension as the residue field extension for the normalization of a $K$-point on a $K$-variety $C$. If $L/K$ is separable, then you may assume that $C$ is a curve. Feb 4, 2014 at 15:04
• Take the affine line $A^1=\mathrm{Spec} (K[T])$, take a closed point $a$ with residue field $L=K[T]/(P)$, and pinch it to a rational point. (Glue $A^1$ and $\mathrm{Spec}(K)$ along $a$.) Over the algebraic closure of $K$, this amounts to identifying all conjugates of the points $a$ to a single point. On the new curve, the point becomes $K$-rational, and its fiber in the normalization is just $\mathrm{Spec}(L)$.
• @KarlSchwede: "... you get $\text{Spec} \mathbb{R}[x,ix]$." I think not. If $P(T)$ is $T^2+1$, then you get $\text{Spec} \mathbb{R}[T(T^2+1),T^2+1]$, which equals $\text{Spec} \mathbb{R}[x,y]/\langle x^2+y^2-y^3 \rangle$ for $x=T(T^2+1)$ and $y=T^2+1$. Of course that is the equation I first wrote down. Feb 4, 2014 at 16:21
• @Jason You are right of course. I was taking the pullback of the diagram of rings $$\Big(\mathbb{C}[x] \to \mathbb{C} \leftarrow \mathbb{R} \Big).$$ Which gives a counter-example in a different way... (this pinches a point of $\mathbb{A}^1_{\mathbb{C}}$ to an $\mathbb{R}$-point). Feb 4, 2014 at 18:56