Equivalent definitions of arithmetically Cohen-Macaulay varieties - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:48:34Z http://mathoverflow.net/feeds/question/95385 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95385/equivalent-definitions-of-arithmetically-cohen-macaulay-varieties Equivalent definitions of arithmetically Cohen-Macaulay varieties Fei YE 2012-04-27T18:35:46Z 2012-04-27T23:08:14Z <p>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?</p> http://mathoverflow.net/questions/95385/equivalent-definitions-of-arithmetically-cohen-macaulay-varieties/95386#95386 Answer by Piotr Achinger for Equivalent definitions of arithmetically Cohen-Macaulay varieties Piotr Achinger 2012-04-27T18:48:58Z 2012-04-27T21:31:25Z <p>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: </p> <p>$H^i(X, \mathcal{O}_X(-q)) = 0$ for $i &lt; n$ and $q\gg 0$,</p> <p>$H^i(P, \mathcal{O}_X(-q)) = 0$ for $i &lt; n$ and $q\gg 0$,</p> <p>$Ext_P^j(\mathcal{O}_X(-q), \omega_P) = 0$ for $j>r$ and $q\gg 0$,</p> <p>$H^0(P, \mathcal{E}xt_P^j(\mathcal{O}_X(-q), \omega_P)) = 0$ for $j > r$ and $q\gg 0$ ,</p> <p>$\mathcal{E}xt_P^j(O_X, \omega_P) = 0$ for $j > r$,</p> <p>$\mathcal{E}xt^j_{O_{P, x}} (O_{X, x}, \omega_{P, x}) = 0$ for $j > r$ and $x\in X$,</p> <p>$H^i_{\mathfrak{n}} (O{X, x}) = 0$ for $i &lt; n$ and $x\in X$, where $\mathfrak{n}$ is the maximal ideal of $O_{P, x}$,</p> <p>$H^i_{\mathfrak{m}} (O_{X, x}) = 0$ for $i &lt; n$ and $x\in X$, where $\mathfrak{m}$ is the maximal ideal of $O{X, x}$.</p> <p>$X$ is Cohen-Macaulay.</p> http://mathoverflow.net/questions/95385/equivalent-definitions-of-arithmetically-cohen-macaulay-varieties/95402#95402 Answer by Karl Schwede for Equivalent definitions of arithmetically Cohen-Macaulay varieties Karl Schwede 2012-04-27T22:37:32Z 2012-04-27T23:08:14Z <p>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.</p> <p>Now, either condition implies that $X$ itself is Cohen-Macaulay, let me explain this.</p> <p>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. </p> <ol> <li><p>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 $\text{Spec} S$. </p></li> <li><p>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 $i &lt; \dim X$ 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 $0 &lt; i &lt; \dim X$ (here $\omega_X^{\bullet}$ is the dualizing complex).</p></li> </ol> <p>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).</p> <p>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.</p> <p><strong>It sounds like you already know what follows:</strong></p> <p>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. </p> <p>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.</p> <p>Now, for the next part:</p> <p>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.</p> <p>This is described in somewhat more detail in <a href="http://www.math.lsa.umich.edu/~kesmith/santa.ps" rel="nofollow">THIS</a> paper by Karen Smith.</p>