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Let $f\colon X \to Y$ be a morphism of schemes and $J \subseteq \mathcal{O}_Y$ a quasi-coherent ideal. Let $I$ denote the image of $f^* J \to f^* \mathcal{O}_Y = \mathcal{O}_X$. Then $I \subseteq \mathcal{O}_X$ is a quasi-coherent ideal and we have the equality of sets $f^{-1}(V(J)) = V(I)$.

This is easy and should be well-known. I would like to have a reference in the literature, preferred EGA. In EGAI I could only find the affine case, this is Proposition 1.2.2.

PS: Of course this can be also stated as follows: The scheme-theoretic pullback of a closed subscheme has as underlying set precisely the preimage of the underlying closed subset.

EDIT: Here is a deduction from EGA-results: Since $f^* J \to f^* \mathcal{O}_Y \to f^* (\mathcal{O}_Y/J) \to 0$ is exact, we have $\mathcal{O}_X/I = (f^* \mathcal{O}_Y) / I = f^* (\mathcal{O}_Y/J)$. Now apply EGA I, Chapitre 0, 5.2.4.1: $f^{-1}(V(J))=f^{-1}(\mathrm{supp}~ \mathcal{O}_Y/J ) \stackrel{!}{=} \mathrm{supp}~ f^*(\mathcal{O}_Y/J) = \mathrm{supp}~ \mathcal{O}_X/I = V(I)$

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    Yes, me too. There is a point in the book where they mention this fact as a consequence of the proposition, but they don't state it as an official result. I didn't even know this was true (the fact about the underlying topological space of the pullback being the set-theoretic pullback) until I read this book. You would think it would be a standard thing to state. I don't even think it is stated explicitly in Stacks.2012-12-16

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