Timeline for Complete intersection argument
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2021 at 18:06 | vote | accept | Spencer Leslie | ||
| Oct 5, 2018 at 8:59 | answer | added | abx | timeline score: 1 | |
| Oct 4, 2018 at 21:53 | comment | added | inkspot | If $Y=Z$ then $T\times_ZY=T$ so you need $T$ to be a CI in T\times_kY$. That is, you need a point to be CI in $Y$, so it is necessary that $Y$ be smooth. | |
| Oct 4, 2018 at 21:33 | comment | added | Spencer Leslie | Yes of course, thanks for the simple counterexample. I have edited the question to include a flatness assumption. | |
| Oct 4, 2018 at 21:33 | history | edited | Spencer Leslie | CC BY-SA 4.0 |
edited to include a flatness assumption on one of the morphisms
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| Oct 4, 2018 at 21:23 | comment | added | Alexander Braverman | The answer is obviously "no" in general. For example, consider the case when $Y$ is one point in $Z$. Then $T\times _Z Y$ is the preimage of this point in $T$ which is not necessarily a complete intersection in $T$ in general. However, I think that the statemet is true if one of the maps is flat (and the map ${\mathfrak t}\to {\mathfrak t}/W$ is flat. | |
| Oct 4, 2018 at 20:50 | history | asked | Spencer Leslie | CC BY-SA 4.0 |