The Stacks project

Proof. Let $\xi \in X$ with $s' = f(\xi )$ and $s' \leadsto s$ a specialization of $S$. If $\xi $ is an associated point of $X_{s'}$, then $\xi $ is the unique generic point because $X_{s'}$ is an integral scheme. Let $\xi _0$ be the unique generic point of $X_ s$. As $X \to S$ is flat we can lift $s' \leadsto s$ to a specialization $\xi ' \leadsto \xi _0$ in $X$, see Morphisms, Lemma 29.25.9. The $\xi \leadsto \xi '$ because $\xi $ is the generic point of $X_{s'}$ hence $\xi \leadsto \xi _0$. This means that $(\text{id}_ S, s' \to s, \xi )$ is not an impurity of $\mathcal{O}_ X$ above $s$. Since the assumption that $f$ is finite type, flat with geometrically integral fibres is preserved under base change, we see that there doesn't exist an impurity after any base change. In this way we see that $X$ is universally $S$-pure. $\square$