Let $X\subset \mathbb{P}^n$ be a projective algebraic variety with coordinate ring $R$. $X$ is said to be arithmetically CohenMacaulay if $R$ is a CohenMacauly 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 $1\leq i\leq \text{dim} X1$. Does anyone know a proof for the equivalence? Without assuming that $X$ is CohenMacaulay, I can only prove that $R$ is CohenMacauly at $0$, the vertex of the affine cone. How to prove $R$ or the section ring is CohenMacauly at other points of the affine cone? Did I miss something from the idea? Or is there an complete alternative proof?
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 CohenMacaulay thing first. Now, either condition implies that $X$ itself is CohenMacaulay, 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.
On the other hand the fact that $X$ is CohenMacaulay implies that $S$ is also CohenMacaulay 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 CohenMacaulay 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 CohenMacaulay, 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. 


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 CohenMacaulay. 

