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?



Equivalent definitions of arithmetically Cohen-Macaulay varieties
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. 


1. 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 $\text{Spec} S$.
2. 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](http://www.math.lsa.umich.edu/~kesmith/santa.ps) paper by Karen Smith.


