Let X be an intersection of 2 surfaces of degree $d_1,d_2$ in $\mathbb P^3$. Is it true that there is a short exact sequence $$ 0\to\mathcal{O}_{\mathbb{P}^3}(-d_1-d_2)\to\mathcal{O}_{\mathbb{P}^3}(-d_1)\oplus\mathcal{O}_{\mathbb{P}^3}(-d_2)\to\mathcal{I}_X\to0, $$ where $\mathcal{I}_X$ is ideal sheaf of $X$. And if it is true, how it could be proven?

  • $\begingroup$ I assume $X$ is a complete intersection. Say the surfaces are zero sets of sections $f_1, f_2$. Consider the map $O(-d_1) \oplus O(-d_2) \rightarrow O_{\mathbf P^3}$ given by $(a,b) \mapsto af_1+bf_2$. What is the image and kernel? $\endgroup$
    – user64687
    May 8, 2015 at 0:51
  • $\begingroup$ This is true. Let $R=k[x_0,x_1,x_2,x_3]$. Then $\mathbb{P}^3=\text{Proj}\,R$. Let the surfaces are given by equations $f=0$ and $g=0$. Then the ideal of the intersection is given by $I=(f,g)$. You have to prove the exactness of the sequence $0\to R(-d_1-d_2)\to R(-d_1)\oplus R(-d_2)\to I\to0$, where the first map is given by $(-g\,f)$ and the second by $\binom{f}{g}$. $\endgroup$
    – vitaliy
    May 8, 2015 at 1:06

1 Answer 1


If $X$ is a so-called complete intersection, meaning that the ideal sheaf of $X$ is generated by the generators, say $f$ and $g$, of the ideals of the two surfaces, and the two surfaces have no components in common, then yes, it pops out very naturally: Anything in the ideal of $X$ is locally a linear combination of the two generators, hence the surjection on the right, given by $(a,b)\mapsto af+gb$. The kernel is as advertised, because since the surfaces have no common components, $f$ and $g$ are relatively prime, so the relations they satisfy are generated by the obvious one $gf-fg$ and the map on the left is accordingly given by $c \mapsto (cg,-cf)$.


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