Computation of $\mathrm{Tor}^1$ for two structure sheaves of subvarieties

Question

Let $X$ be a hyperplane defined by $(x_3=0)$ and let $p$ be a point $[1:0:0:0]$ in the projective 3-space $\mathbb{P}^3_{\mathbb{C}}$. Then I want to compute $\mathrm{Tor}^1_{\cal{O}_{\mathbb{P}^3}}({\cal{O}}_X,{\cal{O}}_p)$. First my attempt is : Start with an exact sequence $$ 0 \to {\cal{O}}_{\mathbb{P}^3}(-1) \to {\cal{O}}_{\mathbb{P}^3} \to {\cal{O}}_X \to 0 $$ taking a right exact functor $-\otimes{\cal{O}}_p$, we have $$ 0\to\mathrm{Tor}^1_{\cal{O}_{\mathbb{P}^3}}({\cal{O}}_X,{\cal{O}}_p)\to{\cal{O}}_p \overset{0}\to{\cal{O}}_p\overset{\cong}\rightarrow{\cal{O}}_p\to 0 $$ so that $$ \mathrm{Tor}^1_{\cal{O}_{\mathbb{P}^3}}({\cal{O}}_X,{\cal{O}}_p)\cong {\cal{O}}_p. $$ Is it correct? or how to compute the $\mathrm{Tor}$ in this case? On the otherhand, if we start with ${\cal{O}}_p$, I belive that the sheaf $\mathrm{Tor}^1_{\cal{O}_{\mathbb{P}^3}}({\cal{O}}_p,{\cal{O}}_X)$ may yield the same consequence of above computation. But if we start with a resolution $$ \mathrm{P}_* : 0\to{\cal{O}}(-3) \to {\cal{O}}(-2)^3 \to {\cal{O}}(-1)^3 \to {\cal{O}} \to {\cal{O}}_p \to 0, $$ from the definition of derived functor $\mathrm{Tor}^i({\cal{O}}_p,{\cal{O}}_X)\cong\mathrm{H}^i(\mathrm{P}_{*}\otimes{\cal{O}}_X)$ we may have $$ \mathrm{Tor}^i({\cal{O}}_p,{\cal{O}}_X)=0 $$ for $i\geq 1$. Where is my miss? Any comment will so help to me.

Answer 1

I think the problem is solved, thanks to the comment from Sasha. The mistaken of mine was simple, that "the given locally free resoultion after tensor product with $\mathcal{O}_X$ is still a locally free resolution, especially an (right) exact sequnce". Yes, it was a dumb misunderstood. There is no guarantee for exactness unless the resolution is given by a short exact sequence. Actually, in my case recall that a locally free resolution of a point $p=[1:0:0:0]$ in $\mathbb{P}^3$ : $$ \mathrm{P}_* : 0 \to \mathcal{O}_{\mathbb{P}^3}(-3) \overset{d_3}{\to} \mathcal{O}_{\mathbb{P}^3}(-2)^{\oplus{3}} \overset{d_2}{\to} \mathcal{O}_{\mathbb{P}^3}(-1)^{\oplus{3}} \overset{d_1}\to\mathcal{O}_{\mathbb{P}^3}\to\mathcal{O}_p\to 0, $$
But after tensor with a $\mathcal{O}_{\mathbb{P}^3}$-module $\mathcal{O}_X$, the sequence $$ (\mathrm{P}_*\otimes\mathcal{O}_X) : 0 \to \mathcal{O}_{X}(-3) \overset{d_3\otimes id_{X}}{\to} \mathcal{O}_{X}(-2)^{\oplus{3}} \overset{d_2\otimes id_{X}}{\to} \mathcal{O}_{X}(-1)^{\oplus{3}} \overset{d_1\otimes id_{X}}\to\mathcal{O}_{X}\to\mathcal{O}_p\to 0 $$ is no longer exact, even though $(d_{i+1}\otimes id_{X})\cdot(d_{i}\otimes id_{X})=0$ still holds. In fact, $$ \mathrm{Tor}^1_{\mathcal{O}_{\mathbb{P}^3}}(\mathcal{O}_p, \mathcal{O}_X) \cong \frac{\mathrm{Ker}(d_{1}\otimes id_{X})}{\mathrm{Im}(d_{2}\otimes id_{X})}\cong \frac{\mathcal{O}_{X}(-1)}{\mathcal{I}_{p,X}(-1)}\cong\mathcal{O}_p. $$ Thank you again for resolving my long-standing concerns, @Sasha.