Restricting sheaves in projective space

Question

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.

Answer 1

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 $\dim H^0(I_p(1) \otimes \mathcal{O}_H) = 3$.