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FedeB
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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?

MyEDIT: 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 question is basically coming from the following situation: Let $C$ be an integral curve over a field $k$ 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 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."

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FedeB
  • 165
  • 8

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 $k$ 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.