Lemma 53.5.2 (Riemann-Roch). Let $X$ be a proper scheme over a field $k$ which is Gorenstein and equidimensional of dimension $1$. Let $\omega _ X$ be as in Lemma 53.4.1. Then

1. $\omega _ X$ is an invertible $\mathcal{O}_ X$-module,

2. $\deg (\omega _ X) = -2\chi (X, \mathcal{O}_ X)$,

3. for a locally free $\mathcal{O}_ X$-module $\mathcal{E}$ of constant rank we have

$\chi (X, \mathcal{E}) = \deg (\mathcal{E}) - \textstyle {\frac{1}{2}} \text{rank}(\mathcal{E}) \deg (\omega _ X)$

and $\dim _ k(H^ i(X, \mathcal{E})) = \dim _ k(H^{1 - i}(X, \mathcal{E}^\vee \otimes _{\mathcal{O}_ X} \omega _ X))$ for all $i \in \mathbf{Z}$.

Proof. Recall that Gorenstein schemes are Cohen-Macaulay (Duality for Schemes, Lemma 48.24.2) and hence $\omega _ X$ is a dualizing module on $X$, see Lemma 53.4.2. It follows more or less from the definition of the Gorenstein property that the dualizing sheaf is invertible, see Duality for Schemes, Section 48.24. By (53.5.0.3) applied to $\omega _ X$ we have

$\chi (X, \omega _ X) = \deg (c_1(\omega _ X) \cap [X]_1) + \chi (X, \mathcal{O}_ X)$

Combined with Lemma 53.5.1 this gives

$2\chi (X, \mathcal{O}_ X) = - \deg (c_1(\omega _ X) \cap [X]_1) = - \deg (\omega _ X)$

the second equality by (53.5.0.2). Putting this back into (53.5.0.3) for $\mathcal{E}$ gives the displayed formula of the lemma. The symmetry in dimensions is a consequence of duality for $X$, see Remark 53.4.3. $\square$

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