Let X be a deminormal variety (over Char 0). Then is it true that the dualizing sheaf is divisorial?
Please provide a reference..
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It is. In any characteristic. It is torsion-free and $S_2$ on any excellent scheme. On a demi-normal scheme it is free in codimension $1$, so it is reflexive by [Stacks Project, Tag 0AVB] and hence it is a rank $1$ reflexive sheaf.
answered Sep 20, 2017 at 16:55
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$\begingroup$ How do we find this Weil divisor D(say), whose ideal sheaf is the dualizing sheaf of the deminormal variety X? I guess we have a Weil divisor D' on the smooth locus of X which gives the dualizing sheaf. Then we just take the closure of D'. Is it D? It is clear that if this is correct then D cannot intersect the codimension 1 nodal locus of the deminormal variety X..............the reference that you have shared is very helpful...is there some article which says how to find this Weil divisor? $\endgroup$– KumarCommented Sep 20, 2017 at 17:25
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$\begingroup$ On non-normal schemes,you cannot consider divisors as cycles.If it is $S_2$, then the singular set itself is codimension $1$ and that messes up dealing with divisors the "usual" way.You should just stick to dealing with refelxive sheaves of rank $1$. If you really want "divisors" as opposed to a "linear system", then take sections, but the zero set of those sections will not behave exactly the same way.Alternatively, you can consider the associated divisors on the Gorenstein locus and insist that they never contain an irreducible component of the singular set.That might give you what you want. $\endgroup$ Commented Sep 20, 2017 at 19:09
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$\begingroup$ : Please see the picture i have shared in the edited question ...this is from en.wikipedia.org/wiki/Canonical_bundle .......this says that there exist a Weil divisor class for dualizing sheaf if the variety is Gorenstein in codimension 1 and $S_2$. Also can you please elaborate "Alternatively, you can consider the associated divisors on the Gorenstein locus and insist that they never contain an irreducible component of the singular set.That might give you what you want."? $\endgroup$– KumarCommented Sep 20, 2017 at 19:25
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$\begingroup$ also it is clearly written here (see the notation and conventions) books.google.co.in/… $\endgroup$– KumarCommented Sep 20, 2017 at 19:37
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1$\begingroup$ I couldn't see the link in your last comment (google is random about this), but otherwise, the point is that the best you can do is what I wrote after "alternatively". In other words, Consider divisors that do not contain a codim $1$ comp of the sing locus. The Gor in codim $1$ assumption tells you that the canonical divisorial sheaf can be represented by such a divisor. However, this will not be true for a general divisorial sheaf. $\endgroup$ Commented Sep 20, 2017 at 20:40