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For a ringed space $(X,\mathcal{O}_X)$, one can define a sheaf of ideals $\mathcal{J}$ of $\mathcal{O}_X$. Then how can we see the $\mathcal{J}$ satisfies the conditions of sheaf? Especially, I cannot show the gluing property. For an open set $U$ and given open covering $\{U_i\}$, $s_i\in \mathcal{J}(U_i)$ with $s_i|_{U_i\cap U_j}=s_j|_{U_i\cap U_j}$ for all $i,j$, we can find $s\in \mathcal{O}_X(U)$ which satisfies $s|_{U_i}=s_i$. But how can I guarantee $s\in \mathcal{J}(U)$?

Can it be shown with just definition of sheaf of ideals?

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    $\begingroup$ I don't understand the question... a sheaf of ideals is a sheaf by definition. $\endgroup$
    – user314
    Commented Mar 14, 2013 at 10:32
  • $\begingroup$ I believe I misunderstood the definition of sheaf of ideals. In the definition which I saw, a sheaf of ideals $\mathcal{J}$ of $\mathcal{O}_X$ is defined by $\mathcal{J}(U)$ is just an ideal of $\mathcal{O}_X$. Is it right? $\endgroup$
    – User0829
    Commented Mar 14, 2013 at 10:37
  • $\begingroup$ Sorry to an errata, $\mathcal{J}(U)$ is an ideal of $\mathcal{O}_X(U)$. $\endgroup$
    – User0829
    Commented Mar 14, 2013 at 10:45
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    $\begingroup$ A sheaf of ideals of $\mathscr{O}_X$ is a sub-$\mathscr{O}_X$-module of $\mathscr{O}_X$ (EGA, 0, 4.1.3) (hence already a sheaf by assumption). If you just assign for each open $U \subset X$ an ideal $\Gamma(U, \mathscr{I}) \subset \Gamma(U, \mathscr{O}_X)$, you won't necessarily get a sheaf of ideals (or even a presheaf). $\endgroup$
    – user314
    Commented Mar 14, 2013 at 10:45
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    $\begingroup$ @Adeel: you can write it as an answer. $\endgroup$
    – Berci
    Commented Mar 14, 2013 at 13:09

2 Answers 2

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Just a remark: For ideal sheaves, and more generally subsheaves of a sheaf, the sheaf condition simplifies a little bit. If $F$ is a sheaf (say abelian groups), then a subsheaf $G$ of $F$ is given by subgroups $G(U) \subseteq F(U)$ for all opens $U$ with the following properties:

  • If $s \in G(U)$ and $V \subseteq U$, then $s|_V \in F(V)$ already lies in $G(V)$.
  • If $U = \cup_i U_i$ and $s \in F(U)$ has the property that $s|_{U_i} \in G(U_i)$ for all $i$, then it already follows that $s \in G(U)$.

If $F$ is the structure sheaf of a ringed space, this gives a description of ideal sheaves which is really useful in practice.

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A sheaf of ideals of $\mathscr{O}_X$ is a sub-$\mathscr{O}_X$-module of $\mathscr{O}_X$ (EGA, 0, 4.1.3), hence already a sheaf by definition. If you just assign to each open $U \subset X$ an ideal $\Gamma(U,\mathscr{I}) \subset \Gamma(U,\mathscr{O}_X)$, you won't necessarily get a sheaf of ideals (or even a presheaf).

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