The Stacks project

Proof. Let $(S'', s'') \to (S', s')$ be any elementary étale neighbourhood with $S''$ affine. Let $Y'' \subset Y' \times _{S'} S''$ be any affine open neighbourhood containing the point $y'' = (y', s'')$. Then we obtain an affine $(Z'', z'')$ by (3). Moreover $Z_{S''} \to X_{S''}$ is a closed immersion and $Z'' \to Z_{S''}$ is an étale morphism. Hence Lemma 38.2.1 applies and we can find an étale morphism $X'' \to X_{S'}$ of affines such that $Z'' \cong X'' \times _{X_{S'}} Z_{S'}$. Denote $i'' : Z'' \to X''$ the corresponding closed immersion. Setting $x'' = i''(z'')$ we obtain a commutative diagram as in the lemma. Properties (1), (2), (3), and (4) hold by construction. Thus it suffices to show that (5) holds for a suitable choice of $(S'', s'') \to (S', s')$ and $Y''$.

We first list those properties which hold for any choice of $(S'', s'') \to (S', s')$ and $Y''$ as in the first paragraph. As we have $Z'' = X'' \times _ X Z$ by construction we see that $i''_*\mathcal{G}'' = \mathcal{F}''$ (with notation as in the statement of the lemma), see Cohomology of Schemes, Lemma 30.5.1. Set $n = \dim (\text{Supp}(\mathcal{F}_ s)) = \dim _ x(\text{Supp}(\mathcal{F}_ s))$. The morphism $Y'' \to S''$ is smooth of relative dimension $n$ (because $Y' \to S'$ is smooth of relative dimension $n$ as the composition $Y' \to Y_{S'} \to S'$ of an étale and smooth morphism of relative dimension $n$ and because base change preserves smooth morphisms of relative dimension $n$). We have $\kappa (y'') = \kappa (y)$ and $\kappa (s) = \kappa (s'')$ hence $\kappa (y'')$ is a purely transcendental extension of $\kappa (s'')$. The morphism of fibres $X''_{s''} \to X_ s$ is an étale morphism of affine schemes over $\kappa (s) = \kappa (s'')$ mapping the point $x''$ to the point $x$ and pulling back $\mathcal{F}_ s$ to $\mathcal{F}''_{s''}$. Hence

because dimension is invariant under étale localization, see Descent, Lemma 35.21.2. As $\pi '' : Z'' \to Y''$ is the base change of $\pi $ we see that $\pi ''$ is finite and as $\kappa (y) = \kappa (y'')$ we see that $\pi ^{-1}(\{ y''\} ) = \{ z''\} $.

At this point we have verified all the conditions of Definition 38.4.1 except we have not verified that $Y'' \to S''$ has geometrically irreducible fibres. Of course in general this is not going to be true, and it is at this point that we will use that $\kappa (s) \subset \kappa (y)$ is purely transcendental. Namely, let $T \subset Y'_{s'}$ be the irreducible component of $Y'_{s'}$ containing $y' = (y, s')$. Note that $T$ is an open subscheme of $Y'_{s'}$ as this is a smooth scheme over $\kappa (s')$. By Varieties, Lemma 33.7.14 we see that $T$ is geometrically connected because $\kappa (s') = \kappa (s)$ is algebraically closed in $\kappa (y') = \kappa (y)$. As $T$ is smooth we see that $T$ is geometrically irreducible. Hence More on Morphisms, Lemma 37.46.4 applies and we can find an elementary étale morphism $(S'', s'') \to (S', s')$ and an affine open $Y'' \subset Y'_{S''}$ such that all fibres of $Y'' \to S''$ are geometrically irreducible and such that $T = Y''_{s''}$. After shrinking (first $Y''$ and then $S''$) we may assume that both $Y''$ and $S''$ are affine. This finishes the proof of the lemma. $\square$