# Does a dualizing sheaf $\omega_X$ give rise to a dualizing module?

Let $X = \text{Proj } R$ be a projective equidimensional Cohen-Macaulay scheme, where $R$ is a finitely generated graded Cohen-Macaulay $\mathbb{C}$-algebra and $\mathcal{O}_X(1)$ is ample. Suppose that the induced homomorphism $R \to H^0(X,\Gamma_*(\mathcal{O}_X))$ is an isomorphism, where $\Gamma_*(\mathcal{O}_X) = \bigoplus_{d \in \mathbb{Z}} \mathcal{O}_X(d)$.

Let $\omega_X$ be a dualizing sheaf for $X$. Is it true that $H^0(X,\Gamma_*(\omega_X))$ is a dualizing module for $R$?

Work out the case where $R$ is a polynomial ring over $\mathbb{C}$ (if you haven't already). Then use that an arbitrary $R$—Cohen-Macaulay or not—is finite over a polynomial ring, and that if $f: X \to Y$ is a finite map of schemes and $\omega_Y$ is dualizing for $Y$ then the sheaf$$f^! \omega_Y := \text{Hom}_Y(f^* \mathcal{O}_X, \omega_Y)$$is dualizing for $X$, as is the case for the corresponding statement about graded $\mathbb{C}$-algebras.
First of all, I don't assume the ring to be graded by elements of $R_1$. So the finite morphism $X \to Y$ would be into a weighted projective space $Y$.
• First of all, I don't assume the ring to be graded by elements of $R_1$. So the finite morphism $X \to Y$ would be into a weighted projective space $Y$. Second, (assuming it works for weighted projective space) in this case we would have $H^0(X,\Gamma_*(\omega_X)) = H^0(Y,\bigoplus_{d \in \mathbb{Z}} \mathcal{H}om_Y(f_*\mathcal{O}_X,\omega_Y)(d))$. How do you compare this with $Hom_P(R,H^0(Y,\Gamma_*(\omega_Y))$, where $P = H^0(Y,\Gamma_*(\mathcal{O}_Y))$? Commented Nov 23, 2016 at 23:51
• I think I see the second point actually, using that $f_*\mathcal{O}_X$ is free as an $\mathcal{O}_Y$-module (by CM of R) anyway. Commented Nov 24, 2016 at 0:04