Equivalent definitions of arithmetically Cohen-Macaulay varieties Let $X\subset \mathbb{P}^n$ be a projective algebraic variety with coordinate ring $R$. 
$X$ is said to be arithmetically Cohen-Macaulay if $R$ is a Cohen-Macauly ring. A equivalent definition is that the natural morphism 
$$R\to \oplus_{k\in\mathbb{Z}}\mathcal{O}_X(k)$$ is bijictive and $H^i(X,\mathcal{O}_X(k))=0$ for all $k\in\mathbb{Z}$ and $1\leq i\leq \text{dim} X-1$. Does anyone know a proof for the equivalence? Without assuming that $X$ is Cohen-Macaulay, I can only prove that $R$ is Cohen-Macauly at $0$, the vertex of the affine cone. How to prove $R$ or the section ring is Cohen-Macauly at other points of the affine cone? Did I miss something from the idea? Or is there an complete alternative proof?
 A: Here's a proof.  Let me assume that $X$ is not zero dimensional, but instead is equidimensional.  I'll actually prove the whole thing (more than you want), but I'll prove the Cohen-Macaulay thing first.
Now, either condition implies that $X$ itself is Cohen-Macaulay, let me explain this.
Set $S$ to denote the section ring 
$$
\bigoplus_{k \in \mathbb{Z}} O_X(k)
$$
and use $m = S_+$ to denote the irrelevant ideal at the origin. 


*

*In the first condition, we notice that the punctured spectrum of $\text{Spec} S$ is an $\mathbb{A}^1$-bundle over $X$.  Thus if $X$ is not Cohen-Macaulay, neither is $\text{Spec} S$. 

*In the second condition, choose $k \gg 0$.  Then $0 = H^i(X, O_X(-k))$ is Grothendieck-dual to $\mathbb{H}^{-i}(X, \omega_X^{\bullet}(k)) \cong H^0(X, {\bf h}^{-i} \omega_X^{\bullet}(k))$ which is non-zero for some $i < \dim X$ if $X$ is not Cohen-Macaulay (for the isomorphism, we used a spectral sequence). Indeed an equidimensional variety $X$ is Cohen-Macaulay if and only if ${\bf h}^{-i} \omega_X^{\bullet} = 0$ for $0 < i < \dim X$ (here $\omega_X^{\bullet}$ is the dualizing complex).
On the other hand the fact that $X$ is Cohen-Macaulay implies that $S$ is also Cohen-Macaulay except possibly at the origin (again use the $\mathbb{A}^1$-bundle description).
Certainly the map $R \to S$ is bijective in high degree and so birational.   Then $\text{Spec} S \to \text{Spec} R$ is an isomorphism everywhere except possibly at the origin.  Thus $R$ is also Cohen-Macaulay except possibly at the origin.
It sounds like you already know what follows:
Suppose now that $R \subsetneq S$.  But then if $U \subseteq \text{Spec}R$ is the punctured spectrum, we have that $H^0(U, O_{Spec R}) = H^0(U, O_{\text{Spec} S}) \supseteq S$.  In particular, $H^0(U, O_{\text{Spec} R}) \neq R$ and so $R$ cannot be S2 because $H^1_m(S) \neq 0$.  This follows from the exact sequence:
$$0 \to R \to H^0(U, O_{\text{Spec} R}) \to H^1_m(S) \to 0.$$   In particular, this shows that the equality $R = S$ is necessary.  
Conversely suppose that $R = S$.  Additionally since $X$ is Cohen-Macaulay, it is S2, then the section ring $S$ is also S2 away from the origin by the $\mathbb{A}^1$-bundle thing again.  Furthermore, its trivial to see that $S = H^0(U, O_{\text{Spec} S})$ and so the depth of $S$ is at least 2 at the origin.  This implies that $R = S$ is S2.  Thus the isomorphism theorem also gives us some vanishing of local cohomology.  In particular, at least the $H^1_m(S) = 0$ by the above exact sequence.
Now, for the next part:
It follows from straightforward computations with Cech cohomology that 
$$
H^i(X, O_X(k)) = [H^{i+1}_{m}(S)]_k
$$
for $i > 0$ (here $[\bullet]_k$ means the $k$th graded piece of the module).  Thus the vanishing over the other $H^i(X, O_X(k))$ implies the vanishing in all remaining degrees of $H^{i+1}_m(S)$.  This completes the proof.
This is described in somewhat more detail in THIS paper by Karen Smith.
A: Maybe this is helpful. Let $X$ be an $n$-dimensional equidimensional closed subscheme of $P = \mathbb{P}^N$. Let $r = N - n$ be the codimension. The following are easily seen to be equivalent:
$H^i(X, \mathcal{O}_X(-q)) = 0$ for $i < n$ and $q\gg 0$,
$H^i(P, \mathcal{O}_X(-q)) = 0$ for $i < n$ and $q\gg 0$,
$Ext_P^j(\mathcal{O}_X(-q), \omega_P) = 0$ for $j>r$ and $q\gg 0$,
$H^0(P, \mathcal{E}xt_P^j(\mathcal{O}_X(-q), \omega_P)) = 0$ for $j > r$ and $q\gg 0$ ,
$\mathcal{E}xt_P^j(O_X, \omega_P) = 0$ for $j > r$,
$\mathcal{E}xt^j_{O_{P, x}} (O_{X, x}, \omega_{P, x}) = 0 $ for $j > r$ and $x\in X$,
$H^i_{\mathfrak{n}} (O_{X, x}) = 0$ for $i < n$ and $x\in X$, where $\mathfrak{n}$ is the maximal ideal of $O_{P, x}$,
$H^i_{\mathfrak{m}} (O_{X, x}) = 0$ for $i < n$ and $x\in X$, where $\mathfrak{m}$ is the maximal ideal of $O_{X, x}$.
$X$ is Cohen-Macaulay.
