Lemma 108.4.6. Let B be an algebraic space. Let \pi : X \to Y be a closed immersion of algebraic spaces which are separated and of finite presentation over B. Then the morphism \mathcal{C}\! \mathit{oh}_{X/B} \to \mathcal{C}\! \mathit{oh}_{Y/B} of Lemma 108.4.4 is a closed immersion.
Proof. Let \mathcal{I} \subset \mathcal{O}_ Y be the sheaf of ideals cutting out X as a closed subspace of Y. Recall that \pi _* induces an equivalence between the category of quasi-coherent \mathcal{O}_ X-modules and the category of quasi-coherent \mathcal{O}_ Y-modules annihilated by \mathcal{I}, see Morphisms of Spaces, Lemma 67.14.1. The same, mutatis mutandis, is true after base by T \to B with \mathcal{I} replaced by the ideal sheaf \mathcal{I}_ T = \mathop{\mathrm{Im}}((Y_ T \to Y)^*\mathcal{I} \to \mathcal{O}_{Y_ T}). Analyzing the proof of Lemma 108.4.4 we find that the essential image of \mathcal{C}\! \mathit{oh}_{X/B} \to \mathcal{C}\! \mathit{oh}_{Y/B} is exactly the objects \xi = (T \to B, \mathcal{F}) where \mathcal{F} is annihilated by \mathcal{I}_ T. In other words, \xi is in the essential image if and only if the multiplication map
is zero and similarly after any further base change T' \to T. Note that
Hence the vanishing of the multiplication map on T' is representable by a closed subspace of T by Flatness on Spaces, Lemma 77.8.6. \square
Comments (0)