# Restricting sheaves in projective space

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 $$\dim H^0(I_p(1) \otimes \mathcal{O}_H) = 3$$.
• 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 $\mathcal{O}_Z$ ? – user130022 Mar 19 '19 at 9:59
• @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 $p_i$. – Sasha Mar 19 '19 at 14:21