When do we get free modules from Noether normalization

Let $X \subseteq \mathbb{P}_{\mathbb{C}}^n$ be an irreducible, projective, Cohen-Macaulay variety of dimension $k$. Let $L \subseteq \mathbb{P}_{\mathbb{C}}^n$ be a linear space of dimension $n-k-1$ that does not intersect $X$. Then the linear projection $\pi: X \to \mathbb{P}_{\mathbb{C}}^k$ from center $L$ is a finite morphism. Let $R=\mathbb{C}[x_0,\ldots,x_k]$ and let $S$ be the homogeneous coordinate ring of $X$. Then $S$ is a finitely generated graded $R$-module via $\pi$.

My question is: When is $S$ in fact a free $R$-module? In particular, I am interested in sufficient criteria and examples of when $S$ is not free.

• For a more general context and a purely algebraic approach see Bruns and Herzog, Cohen-Macaulay Rings, Proposition 2.2.11. – user26857 Jan 11 '15 at 10:09

$S$ is free if and only if it is a Cohen-Macaulay ring -- see e.g. Bourbaki, Algèbre commutative X, 4, no. 3, Corollaire (I am afraid this is not yet translated in english). In geometric terms, this means that $X$ is projectively (aka arithmetically) Cohen-Macaulay -- that is, $H^i(X,\mathcal{O}_X(n))=0$ for all $n$ and $0<i<\dim(X)$.
• thank you. is it true that since $X$ is Cohen-Macaulay, I just have to check whether $S$ localized at the irrelevant ideal is Cohen-Macaulay? And is it correct that $X$ is arithmetically Cohen-Macaulay if and only if the affine cone $\hat{X}$ is Cohen-Macaulay? – Hans Jan 8 '15 at 13:23
• Yes to both questions. Note that $\hat{X}= \mathrm{Spec}(S)$, so $\hat{X}$ Cohen-Macaulay $\Leftrightarrow \ S$ Cohen-Macaulay. – abx Jan 8 '15 at 14:50