Is ideal sheaf left exact

Question

Let $P$ be a zero dimensional scheme on a projective surface $X$. One can consider its ideal sheaf $I_P$ in $X$ to be the kernel of the morphism between $\mathcal O_X$ and pushforward of $\mathcal O_P$.

Is the following true : the injection $\mathcal O_X \to \mathcal O_X(1)$ also stays an injection when we tensor it with $I_P$, i.e. is $I_P \to I_P(1)$ also injective?

Any argument or counterexample is welcome

The title of your question looks very strange. – Sasha Sep 15 '22 at 05:42 — Sasha, Sep 15 '22 at 05:42

score 2 · Accepted Answer · answered Sep 15 '22 at 05:41

2

The composition $I_P \to \mathcal{O}_X \to \mathcal{O}_X(1)$ is injective (because both arrows are), hence the composition $I_P \to I_P(1) \to \mathcal{O}_X(1)$ is injective, and hence the map $I_P \to I_P(1)$ is.

answered Sep 15 '22 at 05:41

Sasha

17,011

that would also then induce $I_P(m) \subset I_P(n)$ for $m \leq n$ right? This then also means if no section of $\mathcal O_S(n)$ passes through $P$, then same is true for sections of $\mathcal O_S(m)$ right? – New Sep 15 '22 at 06:41

Is ideal sheaf left exact

1 Answers1