I have a question ;

Let $ U \overset{i} \hookrightarrow Y \overset{j} \hookrightarrow Z $ , where i is closed open inclusion ( ; i.e, closed open subscheme), and j is closed inclusion, and $ k:= j \circ i : U \hookrightarrow Z $ is closed immersion. Then now define $ \mathcal{J} := ker(\mathcal{O}_{Z} \overset {j^{b}} \rightarrow j_{*} \mathcal{O}_{Y} ) $ and $ \mathcal{I} := ker(\mathcal{O}_{Z} \overset {k^{b}} \rightarrow k_{*} \mathcal{O}_{Y}|_{U} ) $ ; i.e. the sheaf of ideals defining j,k respectively.

($j^{b}, k^{b} $ are surjective since j, k are closed immersion)

Q. Then $\mathcal{J} = \mathcal{I}$ ?

This question originates from next proof

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I can't understand the above underlined statement. If my question is true, then it seems to be possible to deduce the underlined statement, since then $(f \times f)^{*} \ \mathcal{I}$ becomes also sheaf of ideal defining $\Delta _{X/S} = q\Delta_{X/Y}$ .

Is it really true?


If $U$ is the spectrum of $\mathbb{C}=\mathbb{C}[x]/(x)$, $Y$ that of $\mathbb{C}[x]/(x(x-1))$, $Z$ that of $\mathbb{C}[x]$, then (unless I’m mistaken) $\mathcal{I}$ (resp. $\mathcal{J}$) is the quasi-coherent sheaf of ideals on $Z$ whose global sections are the ideal $(x)$ (resp. $(x(x-1))$) – so they’re not equal.

If you watch the proof closely, what they’re using is that (in your setting) $k^*\mathcal{I}=k^*\mathcal{J}$.

This one is true, because obviously $\mathcal{J}$ is a subsheaf of $\mathcal{I}$ and if $x \in U$, $\mathcal{I}_x=\mathcal{J}_x$, morally because $U$ is an open subset of $Y$, and (more rigorously) because it holds when $Z$ is affine.

  • $\begingroup$ Confusion about notation. You meaning $\mathcal{I}_{k(x)} = \mathcal{J}_{k(x)}$ for all $x \in U $? I tried to check for the case that $U, Y, Z$ are affines, but I'm still stucked. Can you explain more detail? How can we use that $U$ is open in $Y$ ? $\endgroup$
    – Plantation
    Sep 11 at 13:46
  • $\begingroup$ Yes, I meant $\mathcal{I}_{k(x)}$. That $U$ is open in $Y$ means (when $Y$ is affine) that $\mathcal{O}(Y)=\mathcal{O}(U) \times \mathcal{O}(Y \backslash U)$. $\endgroup$
    – Mindlack
    Sep 11 at 15:43

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