The Stacks project

Proof. For $X \to \mathop{\mathrm{Spec}}(A)$ as in the statement we have to bound $H^1(X, \mathcal{O}_ X)$. Let $\omega _ X$ be the dualizing module of $X$ as in the statement of Grauert-Riemenschneider (Proposition 54.7.8). The trace map is the map $Rf_*\omega _ X \to \omega _ A$ described in Duality for Schemes, Section 48.7. By Grauert-Riemenschneider we have $Rf_*\omega _ X = f_*\omega _ X$ thus the trace map indeed produces a map $\Gamma (X, \omega _ X) \to \omega _ A$. By duality we have $Rf_*\omega _ X = R\mathop{\mathrm{Hom}}\nolimits _ A(Rf_*\mathcal{O}_ X, \omega _ A)$ (this uses that $\omega _ X[2]$ is the dualizing complex on $X$ normalized relative to $\omega _ A[2]$, see Duality for Schemes, Lemma 48.20.9 or more directly Section 48.19 or even more directly Example 48.3.9). The distinguished triangle

is transformed by $R\mathop{\mathrm{Hom}}\nolimits _ A(-, \omega _ A)$ into the short exact sequence

(and $\mathop{\mathrm{Ext}}\nolimits _ A^ i(R^1f_*\mathcal{O}_ X, \omega _ A) = 0$ for $i \not= 2$; this will follow from the discussion below as well). Since $R^1f_*\mathcal{O}_ X$ is supported in $\{ \mathfrak m\} $, the local duality theorem tells us that

is the Matlis dual of $R^1f_*\mathcal{O}_ X$ (and the other ext groups are zero), see Dualizing Complexes, Lemma 47.18.4. By the equivalence of categories inherent in Matlis duality (Dualizing Complexes, Proposition 47.7.8), if $R^1f_*\mathcal{O}_ X$ is not annihilated by $d$, then neither is the $\mathop{\mathrm{Ext}}\nolimits ^2$ above. Hence we see that $H^1(X, \mathcal{O}_ X)$ is annihilated by $d$. Thus the required boundedness follows from Lemma 54.8.2 and (54.8.2.1). $\square$