Let $H$ be a hypeplane in $\mathbb{P}^3$ containing a point $p$ and $I_p$ be the ideal sheaf corresponding to $p$. Consider the natural exact sequence :

$0 \to \mathcal{O} \to \mathcal{O}(H) \to \mathcal{O}(H) \mid_H \to 0$.

Is it true that the tensoring the exact sequence by $I_p$ remains exact ? I guess not, because if it is exact then we get the following exact sequence:

$0 \to I_p \to I_p(1) \to I_p(1) \mid_H \to 0$.

Note that $h^0(I_p) = h^1(I_p) = 0$. Thus considering the long exact sequence of cohomology of the above sequence, gives that $H^0(I_p(1)) \cong H^0(I_p(1)\mid_H)$, which is a contradiction as their dimensions are $3$ and $2$ respectively. Please correct me if i am wrong.


Essentially, you are asking whether the map $I_p \to I_p(1)$ induced by the multiplication with the equation of a hyperplane is injective. Since $I_p$ is a torsion-free sheaf, it is enough for this to check the map at the generic point of $\mathbb{P}^3$. But there $I_p$ agrees with $\mathcal{O}$, hence injectivity follows from injectivity of your first sequence, so the answer to your question is yes.

As for the contradiction you mention, it arises from your wrong interpretation of $I_p(1)\vert_H = I_p(1) \otimes \mathcal{O}_H$. One has, in fact, an exact sequence $$ 0 \to \mathcal{O}_p \to I_p(1) \otimes \mathcal{O}_H \to I_{p,H}(1) \to 0, $$ which shows that $\dim H^0(I_p(1) \otimes \mathcal{O}_H) = 3$.

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  • $\begingroup$ thank you very much for the answer $\endgroup$ – user130022 Mar 19 '19 at 8:03
  • $\begingroup$ In stead of taking one point, if we take a finite set of points $Z$ on $H$, then is the kernel of $I_Z(1) \otimes \mathcal{O}_H \to I_{Z, H} $ isomorphic to $\mathcal{O}_Z$ ? $\endgroup$ – user130022 Mar 19 '19 at 9:59
  • $\begingroup$ @user130022: Sure, the kernel is isomorphic to $Tor_1(\mathcal{O}_Z,\mathcal{O}_H) \cong \oplus Tor_1(\mathcal{O}_{p_i},\mathcal{O}_H) \cong \oplus \mathcal{O}_{p_i}$, where $Z$ is the union of points $p_i$. $\endgroup$ – Sasha Mar 19 '19 at 14:21

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