I was doing Professor Vakil's FOAG in exercise 9.1 F needs to prove:
that quasicoherent sheaf of ideals on $Y$ produce an closed subscheme on $Y$.
Here is my attempt, to construct it locally is not hard, for any $\text{Spec} B \subset Y$ be an open subset, then the section of the ideal sheaf denote as $I(B) = \mathcal{I}(\text{Spec}\ B)$ which is ideal of $B$ therefore we can construct the closed subscheme locally as $\text{Spec } B/ I(B)$.
The problem lies on that they glue together to a global closed subscheme. that is the inclusion $$\text{Spec } B/I(B) \to \text{Spec }B \subset Y\\\text{Spec } A/I(A) \to \text{Spec }A \subset Y$$
has same image on their overlap.
I guess this step needs to use the quasicoherent of the ideal sheaf, so that for $\text{Spec} A_f$, the section of the ideal sheaf is $$I(A_f) = (I(A))_f$$ due to quasicoherent of the sheaf. However I have no idea how to preceed then . How to glue them together.
Best Answer
Let $Y$ be covered by $\text{Spec } A_i$ such that $\text{Spec }A_i \cap \text{Spec }A_j$ is covered by some simutanuously distinguished open set, such that $$D_{A_i}(g) = D_{A_j}(f)$$ by the equivalence of the category of affine schemes and category of commutative ring, we have $$(A_i)_g \cong (A_j)_f$$ Also we have $I((A_i)_g) =(I(A_i))_g$ by the quasicoherent of the ideal sheaf. Therefore by the isomorphism of$$I(A_i)_g \cong I(A_j)_f$$ we have $$\text{Spec } (A_i/I(A_i))_g \cong \text{Spec } (A_j/I(A_j))_f$$
since schemes $\text{Spec } (A_i/I(A_i))$ has an isomorphic open subset say $\text{Spec } (A_i/I(A_i))\cap D_{A_i}(g) \cong \text{Spec } (A_j/I(A_j))\cap D_{A_j}(f)$, by the gluing lemma we can glue them together to a unique scheme, which can be checked to be a closed subscheme by definition.