Lemma 71.4.8. With notation and assumptions as in Lemma 71.4.7. Assume g is locally quasi-finite, or more generally that for every y' \in |Y'| the transcendence degree of y'/g(y') is 0. Then \text{Ass}_{X'/Y'}(\mathcal{F}') is the inverse image of \text{Ass}_{X/Y}(\mathcal{F}).
Proof. The transcendence degree of a point over its image is defined in Morphisms of Spaces, Definition 67.33.1. Let x' \in |X'| with image x \in |X|. Choose a scheme V and a surjective étale morphism V \to Y. Choose a scheme U and a surjective étale morphism U \to V \times _ Y X. Choose a scheme V' and a surjective étale morphism V' \to V \times _ Y Y'. Then U' = V' \times _ V U is a scheme and the morphism U' \to X' is surjective and étale. Choose u \in U mapping to x. Denote v \in V the image of u. Then x \in \text{Ass}_{X/Y}(\mathcal{F}) is equivalent to u \in \text{Ass}(\mathcal{F}|_{U_ v}) by definition. Choose a point u' \in U' mapping to x' \in |X'| and to u \in U (possible by Properties of Spaces, Lemma 66.4.3). Let v' \in V' be the image of u'. Then x' \in \text{Ass}_{X'/Y'}(\mathcal{F}') is equivalent to u' \in \text{Ass}(\mathcal{F}'|_{U'_{v'}}) by definition. Now the lemma follows from the discussion in Divisors, Remark 31.7.4 applied to u' \in \mathop{\mathrm{Spec}}(\kappa (v') \otimes _{\kappa (v)} \kappa (u)). \square
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