Given a smooth algebraic variety $X$ and a Q-Cartier Q-divisor $D$ which is semiample and big. Assume that $Y=Spec ( \oplus\_{m\geq 0}H^0 (X, mD))$ is Cohen-Macaulay. Do it follows that $H^i (X,mD)=0$ for $0 <i<\dim (X)$?


I know the answer is yes if $D$ is Q-ample and this characterize Cohen-Macaulayness for ring of sections of Q-ample divisors. 


If this is true, what is a reference in the literature?



Cohen-Macaulay ring of sections
This is not true in this form.


Think about it this way: For simplicity let us assume that $D$ is already basepoint-free (and to give a counter-example this is certainly enough), so there exists a proper surjective birational morphism $f:X\to Z$ and a very ample line bundle $\mathscr L$ on $Z$ such that $\mathscr N:=\mathscr O\_X(D)=f^\*\mathscr L$. We may assume that $Z$ is normal and $f\_\*\mathscr O\_X\simeq \mathscr O\_Z$. Then for any $m\in\mathbb Z$, 
$$
\tag{$\star$}
H^0(X, \mathscr N^{\otimes m})\simeq H^0(Z, \mathscr L^{\otimes m}),
$$
so in particular $Y\simeq \rm{Spec} (\oplus\_m H^0(Z, \mathscr L^{\otimes m}))$ and since $\mathscr L$ is very ample, this means that $Y$ is a cone over $Z$. 


Then you can connect the vanishing of the cohomology groups $H^i(Z, \mathscr L^{\otimes m}))$ to depth conditions on $Z$. In other words, it follows almost as you desire that
$$
H^i(Z, \mathscr L^{\otimes m})=0
$$
for all $m\in \mathbb Z$ and $0<i<\dim Z=\dim X$. However, this is exactly where the problem comes from.


While we do have $(\star)$ for $i=0$, in general we do not have that for arbitrary $i$'s. Instead what we have is a Leray spectral sequence, which essentially says that 


$$
%\tag{$\star$}
H^i(X, \mathscr N^{\otimes m})\simeq H^i(Z, \mathscr L^{\otimes m}\otimes Rf\_\*\mathscr O\_X).
$$


So, as long as $\mathscr O\_X$ have higher direct images, you're out of luck.


However, this tells you how to fix the statement. You need to assume that $Z$ has rational singularities. Based on the above you should be able to put together a proof of that.


Finally, here is an explicit simple example when your desired statement fails: Let $Z\subseteq \mathbb P^3$ be a normal projective surface with a non-rational singularity. For example a cone over an elliptic curve. Let $f:X\to Z$ be a resolution of singularities. Choose a very very ample line bundle $\mathscr L$ on Z such that it has no higher cohomology and let $D$ be the divisor corresponding to $f^\*\mathscr L$. Clearly this is semi-ample and big. Since $Z$ is a hypersurface, the cone over $Z$ (corresponding to the very ample line bundle $\mathscr L$) is Cohen-Macaulay, so your assumption on $Y$ is satisfied. Now let $\mathscr N:=\mathscr O\_X(D)=f^\*\mathscr L$ and consider the exact sequence provided by the obvious Leray spectral sequence:
$$
0 \to H^1(Z, \mathscr L) \to H^1(X, \mathscr N) \to H^0(Z, \mathscr L \otimes R^1f\_\*\mathscr O\_X) \to H^2(Z, \mathscr L) \to \dots 
$$
By assumption $H^1(Z, \mathscr L) =H^2(Z, \mathscr L) =0$, so 
$H^1(X, \mathscr N) \simeq H^0(Z, \mathscr L \otimes R^1f\_\*\mathscr O\_X)$, but the latter is strictly non-zero by the assumption that $Z$ has a non-rational singularity (which is necessarily isolated). 


