Lemma 38.4.4. Let $S$, $X$, $\mathcal{F}$, $s$ be as in Definition 38.4.1. Let $(Z, Y, i, \pi , \mathcal{G})$ be a one step dévissage of $\mathcal{F}/X/S$ over $s$. Let $(S', s') \to (S, s)$ be any morphism of pointed schemes. Given this data let $X', Z', Y', i', \pi '$ be the base changes of $X, Z, Y, i, \pi $ via $S' \to S$. Let $\mathcal{F}'$ be the pullback of $\mathcal{F}$ to $X'$ and let $\mathcal{G}'$ be the pullback of $\mathcal{G}$ to $Z'$. If $S'$ is affine, then $(Z', Y', i', \pi ', \mathcal{G}')$ is a one step dévissage of $\mathcal{F}'/X'/S'$ over $s'$.

**Proof.**
Fibre products of affines are affine, see Schemes, Lemma 26.17.2. Base change preserves closed immersions, morphisms of finite presentation, finite morphisms, smooth morphisms, morphisms with geometrically irreducible fibres, and morphisms of relative dimension $n$, see Morphisms, Lemmas 29.2.4, 29.21.4, 29.44.6, 29.34.5, 29.29.2, and More on Morphisms, Lemma 37.27.2. We have $i'_*\mathcal{G}' \cong \mathcal{F}'$ because pushforward along the finite morphism $i$ commutes with base change, see Cohomology of Schemes, Lemma 30.5.1. We have $\dim (\text{Supp}(\mathcal{F}_ s)) = \dim (\text{Supp}(\mathcal{F}'_{s'}))$ by Morphisms, Lemma 29.28.3 because

This proves the lemma. $\square$

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