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Let $X$ be a smooth projective variety with two different reduced subschemes $Z,Y\subset X$ which induced from ideal sheaves $\mathscr{I}_Z,\mathscr{I}_Y$. If codimensions of them are both larger than $2$, why $\mathrm{Hom}(\mathscr{I}_Z,\mathscr{I}_Y)=0$?

I think this may be a very stupid question but I can't understand here...

My situation, we consider:

Fix a complex vector space $V$ of dimension $6$. Pick a general linear subspace $L\subset \mathbb{P}\left(\bigwedge^2V\right)$ of dimension $8$. Then we consider $$S:=\mathrm{Pf}(2,V)\cap L,\quad W:=\mathrm{Pf}(4,V^{\vee})\cap L^{\bot}.$$ It's easy to see that $S$ is a smooth K3 surface of degree $14$ and $W$ is a smooth cubic fourfold.

Consider $\Gamma:=\{(s,w)\in S\times W:s\cap\ker w\neq0\}$ with projections $p_W:\Gamma\to W$ and $p_S:\Gamma\to S$. Let $\Gamma_p=p_S^{-1}(p)$. For $p\neq q$ we can show that $\Gamma_p\neq\Gamma_q$ are two distinct surface which have codimension $2$. Why $\mathrm{Hom}(\mathscr{I}_{\Gamma_p},\mathscr{I}_{\Gamma_q})=0$?

Actually I don't know what is the real meaning of the morphisms of ideals. Maybe we can associated them into the sequence $0\to\mathscr{I}_{\Gamma_p}\to\mathscr{O}_W\to\mathscr{O}_{\Gamma_p}\to0$?

Thank you for your any help!

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    $\begingroup$ Hint: Show that $\operatorname{Hom} (\mathcal{I}_Y, \mathcal{O}_X)=k$, a one dimensional vector space. $\endgroup$
    – Mohan
    Commented Nov 12, 2023 at 16:38
  • $\begingroup$ @Mohan Thank you very much! $\endgroup$
    – DVL-WakeUp
    Commented Nov 14, 2023 at 10:35

1 Answer 1

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The main observation is that, if $Y \subset X$ and $\operatorname{codim}(Y) \ge 2$ then $$ (I_Y)^\vee \cong \mathcal{O}_X, \qquad\text{hence}\qquad (I_Y)^{\vee\vee} \cong \mathcal{O}_X. $$ Now assume $\phi \colon I_Z \to I_Y$ is a nontrivial morphism. The by functoriality of the double dual, it gives a morphism $$ \mathcal{O}_X \cong (I_Z)^{\vee\vee} \stackrel{\phi^{\vee\vee}}\longrightarrow (I_Y)^{\vee\vee} \cong \mathcal{O}_X, $$ which is compatible with the natural morphisms $$ I_Z \to (I_Z)^{\vee\vee} \cong \mathcal{O}_X \qquad\text{and}\qquad I_Y \to (I_Y)^{\vee\vee} \cong \mathcal{O}_X $$ (because this is a natural transformation). Since any endomorphism of $\mathcal{O}_X$ is a multiplication by a constant, it follows that $\phi$ is an embedding of ideals $I_Z \subset I_Y$, hence $Y \subset Z$.

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    $\begingroup$ Is the assertion that $ (I_Y)^{\vee} = \mathcal{O}_X $ a consequence of the following steps: (1) $ (I_Y)^{\vee} $ is reflexive by a short lemma (2) $ (I_Y)^{\vee \vee} = \mathcal{O}_X $ because the double dual is a line bundle which is trivial outside a codimension $ 2 $ closed subset, hence is trivial by Hartog's lemma. I don't immediately see how you get the first dual trivial and then the second. $\endgroup$
    – Cranium Clamp
    Commented Nov 13, 2023 at 9:04
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    $\begingroup$ @CraniumClamp: Start with the standard exact sequence $$0 \to I_Y \to \mathcal{O}_X \to \mathcal{O}_Y \to 0$$, apply the dualization functor, and use the fact that $$\mathcal{H}om(\mathcal{O}_Y, \mathcal{O}_X) = \mathcal{E}xt^1(\mathcal{O}_Y, \mathcal{O}_X) = 0$$ since the codimension of $Y$ is at least 2. $\endgroup$
    – Sasha
    Commented Nov 13, 2023 at 14:04
  • $\begingroup$ @Sasha Thank you for your nice solution! $\endgroup$
    – DVL-WakeUp
    Commented Nov 14, 2023 at 3:20
  • $\begingroup$ @Sasha Sorry for my stupid questions, why $\mathcal{E}xt^1(\mathscr{O}_Y,\mathscr{O}_X)=0$? Is this just a direct calculation? $\endgroup$
    – DVL-WakeUp
    Commented Nov 14, 2023 at 3:46
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    $\begingroup$ One way to see that $\mathcal{E}xt^1(\mathcal{O}_Y, \mathcal{O}_X) = 0$ is by using the local-to-global spectral sequence and Serre duality on $X$. $\endgroup$
    – Sasha
    Commented Nov 14, 2023 at 22:01

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