Timeline for Relate Codimensions of Integral Schemes and their Generic Fibers
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2019 at 22:54 | comment | added | user267839 | ...I see it now. Dense sets of irreducible sets are irreducible (with induced subspace topology) so restricting the sequence to generic fibers preserves dimension. Thank you | |
| Sep 17, 2019 at 12:40 | comment | added | user267839 | Or do I make the life here too awkward for me and the fact that $(X_i)_{\zeta}$ are dense in $(X_i)$ provide immediately the desired result (maybe by a pure topological lemma which I don't know)? | |
| Sep 17, 2019 at 12:35 | comment | added | user267839 | members; the $(X_i)_{\zeta}$ retain only the property beeing closed in $Z_{\zeta}$ (since $X_i$ closed in $Z$ and using base change argument which preserves closed immesions) and dense in $X_i$ (since it contains the generic). So the chain need to be modified,right? I think that at this point it requires more carefull analysis of it? Like construct a chain of $K_i$s where $K_i$ is defined as closure of $\zeta_i$ in $(X_i)_{\zeta}$ then check that $K_i$ is closed contained in $K_{i+1}$ and blabla... | |
| Sep 17, 2019 at 12:34 | comment | added | user267839 | The remaining point that stays unclear is that following your argument that "...there is a chain of closed irreducibles of length 2 in between X and Z but then this when restricted to the generic fiber gives contradiction (everything is closed irreducible so generic fiber can recover the whole thing by taking closure)..." when you restict the chain $X:= X_0 \subset X_1 \subset ... \subset X_n =:Z$ to generic fibers then the resulting chain $X_{\zeta} \subset (X_1)_{\zeta} \subset ... \subset (X_n)_{\zeta} =:Z_{\zeta}$ not more consists of irreducible | |
| Sep 17, 2019 at 12:26 | comment | added | user267839 | @GTA: ah yes I think I see (partially) the point. So the essential observation is that if we have such chain $X:= X_0 \subset X_1 \subset ... \subset X_n =:Z$ of irreducibles with unique generic points $\zeta_i \in X_i$ for each $i$ then the fact that $X \to Y$ is dominant implies that all induced maps $\phi_i:X_i \to Y$ are dominant and therefore again for every $i$ the generic fiber $(X_i)_{\zeta}$ always contain the generic point $\zeta_i$ of $X_i$ but not $\zeta_{i+1}$. | |
| Sep 17, 2019 at 7:02 | comment | added | GTA | Morphism being dominant in this case means there is a point in X sent to the generic point of Y so one could pick an irreducible component of X containing such point and show that the codimension of the irreducible component is <=1. So we are reduced to the case when X is irreducible. If codim of X is bigger than one then there is a chain of closed irreducibles of length 2 in between X and Z but then this when restricted to the generic fiber gives contradiction (everything is closed irreducible so generic fiber can recover the whole thing by taking closure). | |
| Sep 16, 2019 at 22:39 | history | edited | user267839 | CC BY-SA 4.0 |
added 194 characters in body
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| Sep 16, 2019 at 21:10 | review | Close votes | |||
| Sep 22, 2019 at 20:51 | |||||
| Sep 16, 2019 at 15:44 | history | asked | user267839 | CC BY-SA 4.0 |