# Sheaf of ideals of composition of closed immersion with clopen immersion

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

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.
• 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 $Y$ ? Sep 11 at 13:46
• Yes, I meant $\mathcal{I}_{k(x)}$. That $U$ is open in $Y$ means (when $Y$ is affine) that $\mathcal{O}(Y)=\mathcal{O}(U) \times \mathcal{O}(Y \backslash U)$. Sep 11 at 15:43