Timeline for Does the normalization morphism induce isomorphism on residue fields?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 4, 2014 at 18:56 | comment | added | Karl Schwede | @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 16:21 | comment | added | Jason Starr | @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 16:16 | comment | added | Karl Schwede | For instance, if you do what ACL suggested on $\mathbb{A}_{\mathbb{R}}^1$ you get $\text{Spec} \mathbb{R}[x, ix]$. | |
| Feb 4, 2014 at 15:20 | vote | accept | FedeB | ||
| Feb 4, 2014 at 15:14 | comment | added | ACL | 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 $\mathrm{Spec}(L)$. | |
| Feb 4, 2014 at 15:08 | comment | added | FedeB | Ok, this gives pretty much the answer I was looking. Would you elaborate on this? | |
| Feb 4, 2014 at 15:04 | comment | added | Jason Starr | @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 14:50 | comment | added | FedeB | Yes, but this has nothing to do with the normalization. It's just because you're taking closed points in a scheme of finite type over an algebraically closed field. The closed points of $C$ do have already $k$ as residue field. | |
| Feb 4, 2014 at 14:46 | comment | added | Jason Starr | 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 14:44 | comment | added | FedeB | Ok, this show that my feeling was wrong as expected. Still: is there some condition under which the residue field extension is trivial? | |
| S Feb 4, 2014 at 14:34 | history | answered | Jason Starr | CC BY-SA 3.0 | |
| S Feb 4, 2014 at 14:34 | history | made wiki | Post Made Community Wiki by Jason Starr |