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this question might seem a bit special, but it came up at a crucial point of a proof I read and so I would be very obliged if someone could explain this to me:

given is a smooth projective variety $X$ with canonical sheaf $\omega_X$ and a closed point $x$ on $X$ with residue field $k(x)$. I want now denote with $k(x)$ also the skyskrapersheaf concentrated in $x$ with the field $k(x)$ as stalk.

Now in the proof occurs that one has an isomorphism

$k(x)\simeq k(x)\otimes \omega_X$ in the bounded derived category of coherent sheaves on X, i.e. in $D^{b}(X)$ in the usual notation.

I don't see where this Iso comes from.

Thank you very much!

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2 Answers

up vote 2 down vote accepted

I think there is a misunderstanding in Grigory's answer:

Fact: If $E$ is invertible and $F$ is a skycraper sheaf then there exists an isomorphism $E\otimes_{O_X} F \simeq F$. (In particular this applies to $E = \omega_X$, $F = k(x)$).

Proof: Just chose a neighborhood $U$ of $x$ and a trivialization $E|_U \simeq O_U$. Then $(E\otimes_{O_X} F)|_U = O_U\otimes_{O_U} F|_U = F|_U$ and extend this isomorphism by the $0$ map outside of $U$.

But this is false if $E$ is locally free of rank $\neq 1$. For example $E = O_X^2$, then $E\otimes F = F^2$ for any $F$ including a skycraper sheaf. Also there is no canonical map $F\to E\otimes F$ in general even if $E$ locally free. You have to chose a local section of $E$ to define such a map.

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Great, thank you! –  Descartes Aug 20 '11 at 13:37
    
Great, thank you! Just one more question: how do you technically do the extension of the morphism? To get a morphism of sheaves you have to give a morphism on every open set. So if your set doesn't contain x, then take the zero morphism. But if it contains x and is not trivializing, what then? –  Descartes Aug 20 '11 at 13:43
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Well, this has nothing to do with derived categories &c. Just note that if $F$ is any skyscraper sheaf and $E$ is any locally free sheaf of dim $n$, then $E\otimes F\cong F^{\oplus n}$ as sheaves. In particular, if $E$ is invertible $E\otimes F\cong F$.

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You mean: $F$ invertible? –  Descartes Aug 19 '11 at 11:07
    
@Descartes Oh, sorry. Fixed it. –  Grigory M Aug 20 '11 at 13:49
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