Lemma 62.5.8. Let $f : X \to S$ be a morphism of schemes which is locally of finite type. Let $r \geq 0$ be an integer. Let $g : S' \to S$ be a surjective morphism of schemes. Set $S'' = S' \times _ S S'$ and let $f' : X' \to S'$ and $f'' : X'' \to S''$ be the base changes of $f$. Let $x \in X$ with $\text{trdeg}_{\kappa (f(x))}(\kappa (x)) = r$.

1. There exists an $x' \in X'$ mapping to $x$ with $\text{trdeg}_{\kappa (f'(x'))}(\kappa (x')) = r$.

2. If $x'_1, x'_2 \in X'$ are both as in (1), then there exists an $x'' \in X''$ with $\text{trdeg}_{\kappa (f''(x''))}(\kappa (x'')) = r$ and $\text{pr}_ i(x'') = x'_ i$.

Proof. Part (1) is Morphisms, Lemma 29.28.3. Let $x'_1, x'_2$ be as in (2). Then since $X'' = X' \times _ X X'$ we see that there exists a $x'' \in X''$ mapping to both $x'_1$ and $x'_2$ (see for example Descent, Lemma 35.13.1). Denote $s'' \in S''$, $s'_ i \in S'$, and $s \in S$ the images of $x''$, $x'_ i$, and $x$. Denote $k = \kappa (s)$ and let $Z \subset X_ k$ be the integral closed subscheme whose generic point is $x$. Then $x'_ i$ is a generic point of an irreducible component of $Z_{\kappa (s'_ i)}$. Let $Z'' \subset Z_{\kappa (s'')}$ be an irreducible component containing $x''$. Denote $\xi '' \in Z''$ the generic point. Since $\xi '' \leadsto x''$ we see that $\xi ''$ must also map to $x'_ i$ under the two projections. On the other hand, we see that $\text{trdeg}_{\kappa (s'')}(\kappa (\xi '')) = r$ because it is a generic point of an irreducible component of the base change of $Z$. $\square$

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