**Proof.**
Set $S' = R' \otimes _ R S$ so that we have a commutative diagram of continuous maps of spectra of rings

\[ \xymatrix{ \mathop{\mathrm{Spec}}(S') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(S) \ar[d] \\ \mathop{\mathrm{Spec}}(R') \ar[r] & \mathop{\mathrm{Spec}}(R) } \]

Let $\mathfrak p' \subset R'$ be a prime ideal lying over $\mathfrak p \subset R$. If there is no prime ideal of $S$ lying over $\mathfrak p$, then there is no prime ideal of $S'$ lying over $\mathfrak p'$. Otherwise, by Remark 10.18.5 there is a unique prime ideal $\mathfrak r$ of $F = S \otimes _ R \kappa (\mathfrak p)$ whose residue field is purely inseparable over $\kappa (\mathfrak p)$. Consider the ring maps

\[ \kappa (\mathfrak p) \to F \to \kappa (\mathfrak r) \]

By Lemma 10.25.1 the ideal $\mathfrak r \subset F$ is locally nilpotent, hence we may apply Lemma 10.46.1 to the ring map $F \to \kappa (\mathfrak r)$. We may apply Lemma 10.46.7 to the ring map $\kappa (\mathfrak p) \to \kappa (\mathfrak r)$. Hence the composition and the second arrow in the maps

\[ \kappa (\mathfrak p') \to \kappa (\mathfrak p') \otimes _{\kappa (\mathfrak p)} F \to \kappa (\mathfrak p') \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak r) \]

induces bijections on spectra and purely inseparable residue field extensions. This implies the same thing for the first map. Since

\[ \kappa (\mathfrak p') \otimes _{\kappa (\mathfrak p)} F = \kappa (\mathfrak p') \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p) \otimes _ R S = \kappa (\mathfrak p') \otimes _ R S = \kappa (\mathfrak p') \otimes _{R'} R' \otimes _ R S \]

we conclude by the discussion in Remark 10.18.5.
$\square$

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