Timeline for A property of canonical singularity
Current License: CC BY-SA 4.0
17 events
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| Apr 24 at 8:06 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
two more typos of the same kind
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| Apr 24 at 7:22 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Apr 24 at 7:18 | answer | added | Francesco Polizzi | timeline score: 2 | |
| Apr 24 at 1:02 | comment | added | user267839 | essentially what I'm going to use is that by nature of canonicality if $g: X \to Y$ is an iso of regular schemes, then $f_*$ maps canonical sheaf to canonical (... of course modulo isom class, as canonicality is defined up to isom) | |
| Apr 24 at 0:56 | comment | added | user267839 | About where canonicality of the singularity goes in I have not idea. But concerning your last remark I would argue easier (in fact without using reflexitivity; but maybe I'm missing something): For big enough $m$ is $mK_X$ invertible. As $f$ induces isom $Y-f^{-1}(x) \cong X-x$ of regular schemes (...as you wrote that it is singular only in x) the pushforward functor $f_*$ maps outside $f^{-1}(x)$ as induced by an iso canonical sheaf to canonical, right? Then we have two isomorphic locally free sheaves over $X-x$. Doesn't locally freeness + normality suffice to apply Hartogs? | |
| Apr 24 at 0:29 | comment | added | George | In order to apply Hartog's argument, $f_*O_\mathcal{O}(mK_Y)$ and $\mathcal{O}(mK_X)$ need to be reflexive, right? Can we check? | |
| Apr 24 at 0:21 | comment | added | George | Amazing. Thanks for a lot of new technique for me. But I have some question. Where do we need the fact that $X$ has at worst cannonical singulairty? | |
| Apr 23 at 23:58 | comment | added | user267839 | so far I understand you correctly, you only want that projs are isomorphic,right? It is well known that if you start with a graded ring $A=\oplus_{n=0}A_n$ then for arbitrary two positive integers $ r,s>0$ the proj of another graded ring $R=\oplus_{n=0}R_n$ declared by (1) $R_0:=A_0$ ,(2) $R_d:=0$ for $d<r$ and (3) $R_k:=A_{ks}$ is isomorphic to it (think about Veronesse embedding). So maybe it suffice to twist with big enough $m$ (and it's multiples, such that all $msK$ become invertible, and then take proj of the new graded ring obtained via this Veronese trick. Maybe this helps? | |
| Apr 23 at 22:56 | history | edited | George | CC BY-SA 4.0 |
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| Apr 23 at 22:55 | comment | added | George | @user267839 Hartog's argument might be used for my porpose too. Thanks. | |
| Apr 23 at 22:53 | comment | added | George | @user267839 I mean for all non negative m. Could you see my motivation added in the question. | |
| Apr 23 at 22:52 | comment | added | George | @user267839 Thanks. You are right. | |
| Apr 23 at 22:51 | history | edited | George | CC BY-SA 4.0 |
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| Apr 23 at 22:29 | history | edited | George | CC BY-SA 4.0 |
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| Apr 23 at 22:11 | comment | added | user267839 | Do you really mean for all $m $ or big enough? Since $K_X$ is $\mathbb{Q}$-Cartier it's multiple for big enough $m$ becomes Cartier, so invertible. But as $X$ is normal, you can use Hartog's argument on the base as $x$ has codim $\ge 2$ (if your $X$ is not a curve) | |
| Apr 23 at 22:00 | comment | added | user267839 | Probably, you intended to write $K_Y=f^*K_X+\sum_{i=1}^r m_iE_i$ and the $a_i$ are $m_i$? | |
| Apr 23 at 21:49 | history | asked | George | CC BY-SA 4.0 |