Proof.

As in Theorem B.1 the desired result is equivalent to $H^{q}(\mathcal{X},\mathcal{L}^{-1})=0$ for $q<n$ by relative duality. By flat base change we may assume that $k$ is algebraically closed.

Step 1. Assume first that $\mathcal{L}-D$ is ample. Let $E_{1},\dots,E_{N}$ be the components of $\mathcal{X}_{0}$ and set $a_{i}:=\ord_{E_{i}}D$ . Choose $m\in\mathbf{N}^{*}$ such that $b:=m\,a_{1}\in\mathbf{N}$ . By Lemma B.2 there exists an SNC $S$ -variety $\mathcal{X}^{\prime}$ with a finite surjective morphism $\rho:\mathcal{X}^{\prime}\to\mathcal{X}$ such that $\rho^{*}E_{i}$ is smooth (possibly disconnected) for each $i$ , $\sum_{i}\rho^{*}E_{i}$ is SNC and $\rho^{*}E_{1}$ is given as the zero divisor of a section $s\in H^{0}(\mathcal{X}^{\prime},\mathcal{M}^{m})$ for some $\mathcal{M}\in\Pic(\mathcal{X}^{\prime})$ . Note that $H^{q}(\mathcal{X},\mathcal{L}^{-1})$ is a direct summand of $H^{q}(\mathcal{X}^{\prime},\rho^{*}\mathcal{L}^{-1})$ thanks to the trace map. Now let

\mathcal{X}_{1}:=\spec_{\mathcal{X}^{\prime}}\left(\bigoplus_{0\leq j<m}\mathcal{M}^{-j}\right)

be the cyclic cover associated with $s\in H^{0}(\mathcal{X}^{\prime},\mathcal{M}^{m})$ , where $\bigoplus_{0\leq j<m}\mathcal{M}^{-j}$ is endowed with the $\mathcal{O}_{\mathcal{X}^{\prime}}$ -algebra structure induced by $s$ . By definition there is a finite surjective morphism $\tau:\mathcal{X}_{1}\to\mathcal{X}^{\prime}$ which satisfies

\tau_{*}\mathcal{O}_{\mathcal{X}_{1}}=\bigoplus_{0\leq j<m}\mathcal{M}^{-j}.

If we set

\mathcal{L}_{1}:=\tau^{*}\left(\rho^{*}\mathcal{L}\otimes\mathcal{M}^{-b}\right)

we thus have

H^{q}\left(\mathcal{X}_{1},\mathcal{L}_{1}^{-1}\right)\simeq\bigoplus_{0\leq j<m}H^{q}\left(\mathcal{X}^{\prime},\rho^{*}\mathcal{L}^{-1}\otimes\mathcal{M}^{b-j}\right).

But $b/m=a_{1}$ is less than $1$ by assumption, and we thus see that $H^{q}(\mathcal{X}_{1},\mathcal{L}_{1}^{-1})$ contains $H^{q}(\mathcal{X}^{\prime},\rho^{*}\mathcal{L}^{-1})$ , hence also $H^{q}(\mathcal{X},\mathcal{L}^{-1})$ , as a direct summand.

Since $\rho^{*}{E_{i}}$ is smooth for each $i$ and $\sum_{i=2}^{N}\rho^{*}E_{i}$ has normal crossings with $\mathrm{div}(s)=\rho^{*}E_{1}$ , one sees as in [KM98, Claim 2.65] that $E_{i}^{(1)}:=\tau^{*}\rho^{*}E_{i}$ is smooth for each $i$ and $\mathcal{X}_{1,0}$ has SNC support, so that $\mathcal{X}_{1}$ is an SNC $S$ -scheme. Finally $\mathcal{L}_{1}-\sum_{i=2}^{N}a_{i}E_{i}^{(1)}$ is $\mathbf{Q}$ -linearly equivalent to $\tau^{*}\rho^{*}(\mathcal{L}-D)$ , hence is ample.

We now use Lemma B.2 to find $c_{1}:\mathcal{X}_{1}^{\prime}\to\mathcal{X}_{1}$ such that $c_{1}^{*}E_{i}^{(1)}$ is smooth for all $i$ , $\sum_{i}c_{1}^{*}E_{i}^{(1)}$ is SNC and $c_{1}^{*}E_{2}^{(1)}$ is divisible in $\Pic(\mathcal{X}_{1}^{\prime})$ by the denominator of $a_{2}$ . We then perform the same cyclic cover construction as above. Iterating the whole process finally yields an SNC $S$ -variety $\mathcal{X}_{N}$ with an ample line bundle $\mathcal{L}_{N}$ such that $H^{q}(\mathcal{X},\mathcal{L}^{-1})$ is a direct summand of $H^{q}(\mathcal{X}_{N},\mathcal{L}_{N}^{-1})$ , and we conclude by Theorem B.1.

Step 2. We now consider the general case where $\mathcal{L}-D$ is merely nef. Since $\mathcal{L}|_{X}$ is ample by assumption there exists a vertical blow-up $\pi:\mathcal{Y}\to\mathcal{X}$ with $\mathcal{Y}$ SNC and a vertical $\pi$ -exceptional effective $\mathbf{Q}$ -divisor $E\in\Div_{0}(\mathcal{Y})_{\mathbf{Q}}$ such that $\pi^{*}\mathcal{L}-E$ is ample. This condition implies in particular that $-E$ is $\pi$ -ample. If we fix $0<\varepsilon\ll 1$ rational so that $\varepsilon E$ has coefficients $<1$ then $\pi^{*}\mathcal{L}-\varepsilon E=(1-\varepsilon)\pi^{*}\mathcal{L}+\varepsilon(\pi^{*}\mathcal{L}-E)$ is also ample since $\pi^{*}\mathcal{L}$ is nef, and we get

H^{q}(\mathcal{Y},\omega_{\mathcal{Y}}\otimes\pi^{*}\mathcal{L})=0\,\,\text{ for all }q\geq 1

by Step 1.

We are next going to show that $R^{q}\pi_{*}(\omega_{\mathcal{Y}}\otimes\pi^{*}\mathcal{L})=0$ for each $q\geq 1$ . Since we have $\pi_{*}\omega_{\mathcal{Y}}=\omega_{\mathcal{X}}$ (the relative canonical bundle $K_{\mathcal{Y}/\mathcal{X}}$ is $\pi$ -exceptional and effective since $\mathcal{X}$ is regular), the degeneration of the Leray spectral sequence of $\pi$ will then yield as desired

H^{q}(\mathcal{X},\omega_{\mathcal{X}}\otimes\mathcal{L})\simeq H^{q}(\mathcal{Y},\omega_{\mathcal{Y}}\otimes\pi^{*}\mathcal{L})=0

for $q\geq 1$ . Let us now prove the claim. Given $q\geq 1$ choose $\mathcal{A}\in\Pic(\mathcal{X})$ sufficiently ample to guarantee that $\mathcal{A}\otimes R^{q}\pi_{*}(\omega_{\mathcal{Y}}\otimes\pi^{*}\mathcal{L})$ is globally generated on $\mathcal{X}$ and

H^{p}\left(\mathcal{X},\mathcal{A}\otimes R^{m}\pi_{*}(\omega_{\mathcal{Y}}\otimes\pi^{*}\mathcal{L})\right)=0\,\,\text{ for all }p\geq 1\text{ and }m\geq 0

(note that we are only imposing finitely many non-trivial conditions). The degeneration of the Leray spectral sequence yields

H^{0}\left(\mathcal{X},\mathcal{A}\otimes R^{q}\pi_{*}(\omega_{\mathcal{Y}}\otimes\pi^{*}\mathcal{L})\right)\simeq H^{q}\left(\mathcal{Y},\omega_{\mathcal{Y}}\otimes\pi^{*}(\mathcal{L}\otimes\mathcal{A})\right)=0

for $q\geq 1$ by Step 1 again, since $\pi^{*}(\mathcal{L}\otimes\mathcal{A})-\varepsilon E$ is also ample. It follows that $\mathcal{A}\otimes R^{q}\pi_{*}(\omega_{\mathcal{Y}}\otimes\pi^{*}\mathcal{L})=0$ by global generation, which proves the claim since $\mathcal{A}$ is invertible. ∎

Proof. [01I0]