Timeline for When is the morphism to a variety from its normalization a homeomorphism?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| May 10, 2017 at 10:01 | comment | added | R. van Dobben de Bruyn | @FriedrichKnop: ah, I somehow did not make that connection. It's never explicitly stated that unibranch boils down to this, but it seems to follow immediately from the definition (plus the fact that localisation commutes with normalisation). | |
| May 10, 2017 at 7:47 | comment | added | Friedrich Knop | In EGA varieties whose normalization morphism is bijective are called (geometrically) unibranch. There is also a wikipedia entry on that with references. | |
| May 10, 2017 at 4:28 | comment | added | R. van Dobben de Bruyn | @AviSteiner: no, it's somehow opposite to that. Seminormal means if $a \in \operatorname{Frac}(A)$ satisfies $a^2, a^3 \in A$, then $a \in A$. In our case, we get $b_1^2, b_1^3 \in A$, but $b_1 \not\in A$. I think the condition is probably that the seminormalisation of $A$ is normal, but I don't know the terminology well enough to say this with full conviction. | |
| May 10, 2017 at 4:04 | comment | added | Avi Steiner | If I understand correctly, your proposition implies that if B is the integral closure of an affine $\Bbb C$-algebra A, then $Spec B \to Spec A$ is a universal homeomorphism iff A is seminormal. Is this right? | |
| May 10, 2017 at 3:55 | vote | accept | Avi Steiner | ||
| May 10, 2017 at 3:55 | history | answered | R. van Dobben de Bruyn | CC BY-SA 3.0 |