The Stacks project

Proof. Let $x \in X$ with image $s \in S$. Since $f$ is smooth, there is a unique irreducible component $Z(x)$ of $X_ s$ which contains $x$. Then $\dim (Z(x)) = r$. Let $n_ x$ be the coefficient of $Z(x)$ in the cycle $\alpha _ s$. We will show the function $x \mapsto n_ x$ is locally constant on $X$.

Let $g : S' \to S$ be a morphism of locally Noetherian schemes. Let $X'$ be the base change of $X$ and let $\alpha ' = g^*\alpha $ be the base change of $\alpha $. Let $x' \in X'$ map to $s' \in S'$, $x \in X$, and $s \in S$. We claim $n_{x'} = n_ x$. Namely, since $Z(x)$ is smooth over $\kappa (s)$ we see that $Z(x) \times _{\mathop{\mathrm{Spec}}(\kappa (s))} \mathop{\mathrm{Spec}}(\kappa (s'))$ is reduced. Since $Z(x')$ is an irreducible component of this scheme, we see that the coefficient $n_{x'}$ of $Z(x')$ in $\alpha '_{s'}$ is the same as the coefficient $n_ x$ of $Z(x)$ in $\alpha _ s$ by the definition of base change in Section 62.3 thereby proving the claim.

Since $X$ is locally Noetherian, to show that $x \mapsto n_ x$ is locally constant, it suffices to show: if $x' \leadsto x$ is a specialization in $X$, then $n_{x'} = n_ x$. Choose a morphism $S' \to X$ where $S'$ is the spectrum of a discrete valuation ring mapping the generic point $\eta $ to $x'$ and the closed point $0$ to $x$. See Properties, Lemma 28.5.10. Then the base change $X' \to S'$ of $f$ by $S' \to S$ has a section $\sigma : S' \to X'$ such that $\sigma (\eta ) \leadsto \sigma (0)$ is a specialization of points of $X'$ mapping to $x' \leadsto x$ in $X$. Thus we reduce to the claim in the next paragraph.

Let $S$ be the spectrum of a discrete valuation ring with generic point $\eta $ and closed point $0$ and we have a section $\sigma : S \to X$. Claim: $n_{\sigma (\eta )} = n_{\sigma (0)}$. By the discussion in More on Morphisms, Section 37.29 and especially More on Morphisms, Lemma 37.29.6 after replacing $X$ by an open subscheme, we may assume the fibres of $X \to S$ are connected. Since these fibres are smooth, they are irreducible. Then we see that $\alpha _\eta = n[X_\eta ]$ with $n = n_{\sigma (\eta )}$ and the relation $sp_{X/S}(\alpha _\eta ) = \alpha _0$ implies $\alpha _0 = n[X_0]$, i.e., $n_{\sigma (0)} = n$ as desired. $\square$