Lemma 42.67.2. In Situation 42.67.1 let $X \to S$ be locally of finite type. Denote $X' \to S'$ the base change by $S' \to S$. If $X$ is integral with $\dim _\delta (X) = k$, then every irreducible component $Z'$ of $X'$ has $\dim _{\delta '}(Z') = k + c$,

**Proof.**
The projection $X' \to X$ is flat as a base change of the flat morphism $S' \to S$ (Morphisms, Lemma 29.25.8). Hence every generic point $x'$ of an irreducible component of $X'$ maps to the generic point $x \in X$ (because generalizations lift along $X' \to X$ by Morphisms, Lemma 29.25.9). Let $s \in S$ be the image of $x$. Recall that the scheme $S'_ s = S' \times _ S s$ has the same underlying topological space as $g^{-1}(\{ s\} )$ (Schemes, Lemma 26.18.5). We may view $x'$ as a point of the scheme $S'_ s \times _ s x$ which comes equipped with a monomorphism $S'_ s \times _ s x \to S' \times _ S X$. Of course, $x'$ is a generic point of an irreducible component of $S'_ s \times _ s x$ as well. Using the flatness of $\mathop{\mathrm{Spec}}(\kappa (x)) \to \mathop{\mathrm{Spec}}(\kappa (s)) = s$ and arguing as above, we see that $x'$ maps to a generic point $s'$ of an irreducible component of $g^{-1}(\{ s\} )$. Hence $\delta '(s') = \delta (s) + c$ by assumption. We have $\dim _ x(X_ s) = \dim _{x'}(X_{s'})$ by Morphisms, Lemma 29.28.3. Since $x$ is a generic point of an irreducible component $X_ s$ (this is an irreducible scheme but we don't need this) and $x'$ is a generic point of an irreducible component of $X'_{s'}$ we conclude that $\text{trdeg}_{\kappa (s)}(\kappa (x)) = \text{trdeg}_{\kappa (s')}(\kappa (x'))$ by Morphisms, Lemma 29.28.1. Then

This proves what we want by Definition 42.7.6. $\square$

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