退出

Existence of a dualizing sheaf for projective schemes

代数几何 Math StackExchange 3 票 0 回答 99 浏览 提问者: Fung San Gaan 2026-08-17 04:13
geometry algebraic-geometry proof-explanation sheaf-theory sheaf-cohomology

问题内容

I'm studying theorem III.7.5 from Hartshorne's book. There are a few things I don't understand in this proof.

Theorem. Let $X$ be a projective scheme over a field $k$. Then $X$ has a dualizing sheaf $\omega_X^\circ$.

Proof. Let $\dim(X) = n$. Embed $X$ as a closed subscheme of $P:=\mathbb{P}^N$ for some $N$. Let $r: = N-n = \text{codim}_P(X)$. We propose $\omega_X^\circ:= \mathcal{E}xt^r_P ({\cal O}_X, \omega_P)$. We want to show this $\omega_X^\circ$ is indeed the dualizing sheaf we are looking for. For any ${\cal O}_X$-module ${\cal F }$, we have an isomorphism $\hom_X({\cal F }, \omega_X^\circ) \simeq \text{Ext}_P^r ({\cal F }, \omega_P)$ from a previous lemma. By the duality theorem for $P$, (when ${\cal F}$ is coherent) we have $\text{Ext}_P^r({\cal F}, \omega_P) \simeq H^{N-r} (P, {\cal F})^*$. Since $N-r = n = \dim (X)$ and ${\cal F}$ is a sheaf on $X$, so we get a functorial isomorphism for coherent ${\cal F}$: \begin{align*} \hom_X({\cal F}, \omega_X^\circ) \simeq H^n(X, {\cal F})^* \tag{$\alpha$} \end{align*} Take ${\cal F} = \omega_X^\circ$. The element $1\in \hom (\omega_X^\circ, \omega_X^\circ)$ gives us the trace morphism $t: H^n(X, \omega_X^\circ) \to k$. Then it is clear by functoriality that $(\omega_X^\circ,t)$ is a dualizing sheaf.


Question 1. Where did ($\alpha$) come from? It looks like the functorial isomorphism $\text{Ext}^i({\cal F}, \omega_P) \simeq H^{n-1}(P, {\cal F})^*$ from the duality for $P$ (theorem III.7.1 of Hartshorne). I tried to follow through the proof of this with $\omega_P$ swapped out for $\omega_X^\circ$ and $P$ swapped out for $X$. The only difference is this time I had to use Serre vanishing to conclude the coeffaceability of the LHS. Is this right?

Question 2. How exactly does $1\in \hom (\omega_X^\circ, \omega_X^\circ)$ give us the trace morphism? Initially I thought this came from the pairing in the duality for $P$. Now that I think about it, it's probably not because of this.

Question 3. Why does functoriality prove $(\omega_X^\circ, t)$ is a dualizing sheaf?

Any help is appreciated!

回答 (0)

暂无回答记录。