Timeline for Which of these 4 definitions of Galois coverings of integral schemes are equivalent?
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7 events
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| Jun 7, 2022 at 5:41 | comment | added | Li Yutong | I had rethought (4) and felt that the lift may not always be possible: let $p \in X$ be a closed point, then $X-\{p\} \to Y$ also satisfies (4) (for most cases), but a lift is not always possible (because some point $q \in f^{-1}(f(p))$ may have to be sent to $p$). Certainly, if $Y$ is defined to be the normalization of $X$ in $K(Y)$, there is no such problem and your comments work. | |
| Jun 6, 2022 at 13:12 | comment | added | R. van Dobben de Bruyn | @LiYutong Oh yeah, it doesn't need to be as complicated as what I wrote. The point is that any $\eta_Y$-automorphism $f \colon \{\eta_X\} \to \{\eta_X\}$ comes from a (unique) $Y$-automorphism $f \colon X \to X$, because $X$ is the integral closure of $Y$ in $\{\eta_X\}$ and integral closure is functorial for isomorphisms. (Also I'm just realising that any fully faithful functor reflects isomorphisms, so I didn't need to mention that separately.) | |
| Jun 6, 2022 at 5:09 | comment | added | Li Yutong | The quoted result "Tag 0BN6" is a purely categorical result. I would like to see the reason why an isomorphism over the generic point can be lifted to the whole isomorphism. This is still unclear for me... | |
| Jun 19, 2016 at 22:58 | vote | accept | Quinlan Aktaş | ||
| Jun 18, 2016 at 19:13 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
The statement doesn't make sense without irreducibility.
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| Jun 18, 2016 at 7:38 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Added an example.
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| Jun 18, 2016 at 6:01 | history | answered | R. van Dobben de Bruyn | CC BY-SA 3.0 |