[$R^1\theta_*(B_X^2(A_X))\otimes \mathcal O_C(A_C)$ is generated by global sections. ]{}

Then, it is sufficient to show $$H^1(X, B_X^2(A_X+\theta^*A_C))=0.$$ Since $X$ is a toric variety, this follows from Proposition \[F-split-B-vanish\].

\[step-H1-R1\] In this step, we prove that the following assertions are equivalent.

1.  [$R^1\theta_*(B_X^2(A_X))=0$.]{}

2.  [$H^1(X_c, B_X^2(A_X)|_{X_c})=0$ for every $c\in C$ where $X_c$ is the fiber.]{}

By [@Hartshorne Theorem 12.11], there exists an isomorphism $$R^1\theta_*(B_X^2(A_X))\otimes k(c)\simeq H^1(X_c, B_X^2(A_X)|_{X_c}).$$ By Nakayama’s lemma, if $$R^1\theta_*(B_X^2(A_X))\otimes k(c)=0,$$ then $R^1\theta_*(B_X^2(A_X))|_{U}=0$ for some open neighborhood of $b\in U\subset C$.

We can replace $C$ by a neighborhood of the point corresponding to the singular fiber. Then, we can find the following commutative diagram $$\begin{CD}
X'@<\gamma_X<< X \\
@VVf'V @VVf V\\
Y'@<\gamma_Y<< Y\\
@VV\pi'V @VV\pi V\\
C'@<\gamma_C<< C
\end{CD}$$ where $C'\simeq \mathbb A^1$, each square is a fiber product, $\gamma_C, \gamma_Y$ and $\gamma_X$ are finite étale morphisms. Let $\theta':=\pi'\circ f'.$ Note that $\gamma_X^*B_{X'}^2\simeq B_X^2$ by Lemma \[cartier-etale\].

Let $\tilde c\in C$ be the point corresponding to the singular fiber of $\theta$. Let $\tilde c':=\gamma_C(\tilde c).$ Then, $$\gamma:=\gamma_X|_{X_c}:X_c\to (X')_{c'}$$ is an isomorphism. We can find a $\theta'$-ample [$\mathbb{Z}$]{}-divisor $A_{X'}$ on $X'$ such that $\gamma^*(\mathcal O_{X'}(A_{X'})|_{X_{c'}})\simeq\mathcal O_X(A_X)|_{X_c}$. Then, we obtain $$\begin{aligned}
H^1(X_c, B_X^2(A_X))&=& H^1(X_c, \gamma^*B_{X'}^2(\gamma^*A_{X'}))\\
&=& H^1((X')_{c'}, B_{X'}^2(A_{X'}))\\
&=&0. \end{aligned}$$ The 