This is Lemma 1.4 on Green and Lazarsfeld's Some results on the syzygies of finite sets and algebraic curves. Let $X$ be a closed subscheme of $\mathbb{P}^r$. Suppose the ideal sheaf $\mathcal{I}$ of $X$ is 3-regular, meaning that $H^i(\mathbb{P}^r,\mathcal{I}(3-i))=0$ for $i>0$. Then for $p\leq codim(X,\mathbb{P}^r)$, $(N_p)$ holds for $X\subset\mathbb{P}^r$ iff $Tor^S_p(R,k)_{p+2}=0$, where $S=k[x_0,\cdots,x_r]$, $R=\bigoplus\limits_{l\geq 0}H^0(\mathbb{P}^r,\mathcal{O}/\mathcal{I}(l))$. The proof is short. First observe that $m$-regularity implies that $\Gamma_*(\mathcal{I})$ is generated by elements of degree $\leq m$. Apply this to the sheafification $0\to\mathcal{E}_{r+1}\to\cdots\to\mathcal{E}_1\to\mathcal{I}\to 0$ of a minimal resolution of $I=\Gamma_*(\mathcal{I})$. One finds that $\mathcal{E}_i$ is $(i+2)$-regular, which means that $Tor_i^S(R,k)_j=0$ for $j\geq i+3$ and therefore $(N_p)$ only requires $Tor^S_p(R,k)_{p+2}=0$. I don't know how to deduce the $(i+2)$-regularity of $\mathcal{E}_i$. For example, let $r=2$. Consider the homogeneous ideal $I=(xy^2,z^3)$. Let $E_1=S(-3)\oplus S(-3)$, with the map $E_1\to I$ given by $(xy^2,z^3)$. Then the kernel is $S(-6)$ with the inclusion to $S(-3)\oplus S(-3)$ given by $(-z^3,xy^2)^T$. This is a minimal resolution but $\mathcal{O}(-6)$ is not 4-regular. Can anyone tell me what's wrong with my observation?

  • 1
    $\begingroup$ This just mean that your ideal $\mathscr{I}$ is not 3-regular -- in fact $H^1(\mathscr{I}(2))\cong H^2(\mathscr{O}(-4))\neq 0$. $\endgroup$
    – abx
    Apr 4 '20 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.