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."


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 $\mathbb{C}$.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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. $\endgroup$ – Jason Starr Feb 4 '14 at 14:46
  • 1
    $\begingroup$ @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. $\endgroup$ – Jason Starr Feb 4 '14 at 15:04
  • 3
    $\begingroup$ 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 $\mathrm{Spec}(L)$. $\endgroup$ – ACL Feb 4 '14 at 15:14
  • 1
    $\begingroup$ @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. $\endgroup$ – Jason Starr Feb 4 '14 at 16:21
  • 1
    $\begingroup$ @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). $\endgroup$ – Karl Schwede Feb 4 '14 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.