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Consider the closed subscheme $\text{Spec}(\mathbb{C})\sqcup\text{Spec}(\mathbb{C})$ of $\mathbb{P}_{\mathbb{C}}^{1}$. Let $i: \text{Spec}(\mathbb{C})\sqcup\text{Spec}(\mathbb{C}) \rightarrow \mathbb{P}_{\mathbb{C}}^{1}$ be the corresponding closed immersion. For simplicity we write $X=\mathbb{P}^{1}_{\mathbb{C}}$. I want to show that $i_{*}\mathcal{O}_{\text{Spec}(\mathbb{C})\sqcup\text{Spec}(\mathbb{C})}$ is isomorphic to $\widetilde{M}$, where $M = S/I$ with $S = \mathbb{C}[X_{0},X_{1}]$ and $I = (X_{0}X_{1})$. And that $\widetilde{I}$ is the ideal sheaf of $\text{Spec}(\mathbb{C})\sqcup\text{Spec}(\mathbb{C})$

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A hint: using the fact that $I$ is the homogeneous ideal defining your closed subscheme, try to prove directly that $\tilde I$ is the corresponding ideal sheaf. You can do this by looking at distinguished open affines and comparing the definition of the ideal sheaf to the definition of the quasicoherent sheaf $\tilde I$ - the sections and the restriction maps can be explicitly described nicely in this example.

(I don't have enough reputation to post this as a comment unfortunately)

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  • $\begingroup$ In general it is not enough to show that two sheaves are locally the same right? I mean if you have an isomorphism on the stalks of two sheaves, than it doesn't imply that you also have an isomorphism between the sheaves. $\endgroup$ – Bobby Apr 26 at 10:19
  • $\begingroup$ That's right (although if you have a morphism of sheaves which induces an isomorphism on all stalks, then it must be an isomorphism). What I mean is, you can write down a "morphism on a base" using the distinguished open affine base and show that this is an "isomorphism on a base", which gives an isomorphism of the sheaves. $\endgroup$ – Alex K Apr 26 at 13:16
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    $\begingroup$ When you restrict to $U_0\cap U_1 = {\rm Spec} (\mathbb C[X_0,X_1]_{X_0 X_1})_0$, you should end up with $\mathcal I(U_0\cap U_1) = (X_{10})_{X_{10}}$ and $\tilde I(U_0\cap U_1) = (I_{X_0 X_1})_0$. Now use the isomorphism $(\mathbb C[X_0,X_1]_{X_0 X_1})_0 \cong \mathbb C[X_{10}]_{X_{10}}$. $\endgroup$ – Alex K Apr 27 at 13:18

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