The Stacks project

Proof. It is clear that we may replace $S$ by an open neighbourhood of $\eta$ and $X$ by the restriction to this open. Thus we may assume $S = \mathop{\mathrm{Spec}}(A)$ where $A$ is a reduced ring and $\eta$ corresponds to a minimal prime ideal $\mathfrak p$. Recall that the local ring $\mathcal{O}_{S, \eta } = A_\mathfrak p$ is equal to $\kappa (\mathfrak p)$ in this case, see Algebra, Lemma 10.25.1.

Apply Varieties, Lemma 33.25.11 to the scheme $X_\eta$ over $k = \kappa (\eta )$. Denote $k'/k$ the purely inseparable field extension this produces. In the next paragraph we reduce to the case $k' = k$. (This step corresponds to finding the morphism $V \to U$ in the statement of the lemma; in particular we can take $V = U$ if the characteristic of $\kappa (\mathfrak p)$ is zero.)

If the characteristic of $k = \kappa (\mathfrak p)$ is zero, then $k' = k$. If the characteristic of $k = \kappa (\mathfrak p)$ is $p > 0$, then $p$ maps to zero in $A_\mathfrak p = \kappa (\mathfrak p)$. Hence after replacing $A$ by a principal localization (i.e., shrinking $S$) we may assume $p = 0$ in $A$. If $k' \not= k$, then there exists an $\beta \in k'$, $\beta \not\in k$ such that $\beta ^ p \in k$. After replacing $A$ by a principal localization we may assume there exists an $a \in A$ such that $\beta ^ p = a$. Set $A' = A[x]/(x^ p - a)$. Then $S' = \mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A) = S$ is finite locally free, surjective, and universally injective. Moreover, if $\mathfrak p' \subset A'$ denotes the unique prime ideal lying over $\mathfrak p$, then $A'_{\mathfrak p'} = k(\beta )$ and $k'/k(\beta )$ has smaller degree. Thus after replacing $S$ by $S'$ and $\eta$ by the point $\eta '$ corresponding to $\mathfrak p'$ we see that the degree of $k'$ over the residue field of $\eta$ has decreased. Continuing like this, by induction we reduce to the case $k' = \kappa (\mathfrak p) = \kappa (\eta )$.

Thus we may assume $S$ is affine, reduced, and that we have a $t \geq 0$ and closed subschemes

such that $Z_{\eta , 0} = (X_\eta )_{red}$ and $Z_{\eta , i} \setminus Z_{\eta , i + 1}$ is smooth over $\eta$ for all $i$. Recall that $\kappa (\eta ) = \kappa (\mathfrak p) = A_\mathfrak p$ is the filtered colimit of $A_ a$ for $a \in A$, $a \not\in \mathfrak p$. See Algebra, Lemma 10.9.9. Thus we can descend the diagram above to a corresponding diagram over $\mathop{\mathrm{Spec}}(A_ a)$ for some $a \in A$, $a \not\in \mathfrak p$. More precisely, after replacing $S$ by $\mathop{\mathrm{Spec}}(A_ a)$ we may assume we have a $t \geq 0$ and closed subschemes

such that $Z_ i \to X$ is a closed immersion of finite presentation, such that $Z_0 \to X$ is a thickening, and such that $Z_ i \setminus Z_{i + 1}$ is smooth over $S$. In other words, the lemma holds. More precisely, we first use Limits, Lemma 32.10.1 to obtain morphisms

over $S$, each of finite presentation, and whose base change to $\eta$ produces the inclusions between the given closed subschemes above. After shrinking $S$ further we may assume each of the morphisms is a closed immersion, see Limits, Lemma 32.8.5. After shrinking $S$ we may assume $Z_0 \to X$ is surjective and hence a thickening, see Limits, Lemma 32.8.15. After shrinking $S$ once more we may assume $Z_ i \setminus Z_{i + 1} \to S$ is smooth, see Limits, Lemma 32.8.9. This finishes the proof. $\square$