Let $k$ be a field. Let $X$ be a smooth projective equidimensional variety over $k$. How to show the following proposition?

For any coherent sheaf $F_1$, $F_2$ on $X$, $\operatorname{Ext}^i(F_1,F_2)=0$ if $i\notin [0,n]$.

This question is from Lemma 5.6 in this paper

Lemma 5.6. Suppose that $X$ is a smooth projective equidimensional variety over $k$. Then $$ \overline{\operatorname{Sdim}} D^{b}(\operatorname{coh} X)=\underline{\operatorname{Sdim}} D^{b}(\operatorname{coh} X)=\operatorname{dim} X. $$

Proof. Let $n=\operatorname{dim} X$. Recall that the Serre functor on $D^{b}(\operatorname{coh} X)$ is isomorphic to $(-) \otimes \omega_{X}[n] .$ Recall also that for any coherent sheaves $F_{1}, F_{2}$ on $X$ we have $\operatorname{Ext}^{i}\left(F_{1}, F_{2}\right)=0$ if $i \notin[0, n]$. ...

Thank you very much!


1 Answer 1


First, $\mathrm{Ext}^i(F_1,F_2)$ for $i < 0$ vanish by definition. Second, for $i > n$ the vanishing follows from Serre duality $$ \mathrm{Ext}^i(F_1,F_2) \cong \mathrm{Ext}^{n-i}(F_2,F_1 \otimes \omega_X)^\vee $$ and the above vanishing.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.