Etale pullback commutes with derivative ideals

Affine-local reduction. The question is local on $X^{\mathrm{\prime }}$ and on $X$: it suffices to prove ${\phi }^{\ast }\mathcal{D}(\mathcal{I})=\mathcal{D}({\phi }^{\ast }\mathcal{I})$ on affine charts $\mathrm{Spec}B\to \mathrm{Spec}A$ of an étale ring map, and then to iterate for higher derivatives. So fix an étale map $A\to B$ and an ideal $I\subseteq A$; write $\phi$ also for the ring map.

The differential comparison. By [F4] one has ${\mathrm{\Omega }}_{B\mathrm{/}A}=0$, so the transitivity sequence of [F3] gives a surjection $B{\otimes }_A{\mathrm{\Omega }}_{A\mathrm{/}K}\twoheadrightarrow {\mathrm{\Omega }}_{B\mathrm{/}K}$. Both sides are finite locally free by [F5]. Their ranks agree at corresponding points because the relative dimensions of $X^{\mathrm{\prime }}\mathrm{/}K$ and $X\mathrm{/}K$ agree by [F5]; no identification with the local-ring dimension is needed. By [F6] the comparison map $B{\otimes }_A{\mathrm{\Omega }}_{A\mathrm{/}K}\to {\mathrm{\Omega }}_{B\mathrm{/}K}$ is an isomorphism.

Derivations under the étale map. Dualising the isomorphism of step 1.2 and using [F2, F7], ${\mathrm{Der}}_K(B)\cong {\mathrm{Hom}}_B({\mathrm{\Omega }}_{B\mathrm{/}K},B)\cong {\mathrm{Hom}}_B(B{\otimes }_A{\mathrm{\Omega }}_{A\mathrm{/}K},B)\cong {\mathrm{Hom}}_A({\mathrm{\Omega }}_{A\mathrm{/}K},B)\cong B{\otimes }_A{\mathrm{Der}}_K(A)$. Concretely, every $K$-derivation $D^{\mathrm{\prime }}$ of $B$ is a finite sum ${\sum }_i{b}_i{D}_i^{\mathrm{\prime }}$ where for each $i$ there is a $K$-derivation $D_i$ of $A$ with $D_i^{\mathrm{\prime }}(\phi (a))=\phi (D_i(a))$ for all $a$, and conversely each $D_i$ gives such a $D_i^{\mathrm{\prime }}$.

The ideals agree. The ideal $\mathcal{D}({\phi }^{\ast }I)$ is generated by ${\phi }^{\ast }I$ together with all $D^{\mathrm{\prime }}(g)$ for $g\in {\phi }^{\ast }I$ and $K$-derivations $D^{\mathrm{\prime }}$ of $B$ ([F1]). Write a generator $g$ of ${\phi }^{\ast }I=IB$ as ${\sum }_j{b}_j{\phi }^{\ast }{f}_j$ with $f_j\in I$. For $D^{\mathrm{\prime }}={\sum }_i{b}_i{D}_i^{\mathrm{\prime }}$ as in step 2.1, the Leibniz rule gives $D^{\mathrm{\prime }}(g)={\sum }_{i,j}{b}_i{D}_i^{\mathrm{\prime }}(b_j){\phi }^{\ast }{f}_j+{\sum }_{i,j}{b}_i{b}_j{\phi }^{\ast }(D_i{f}_j)$, which lies in ${\phi }^{\ast }\mathcal{D}(I)$. Conversely ${\phi }^{\ast }\mathcal{D}(I)$ is generated by the ${\phi }^{\ast }f$ and the ${\phi }^{\ast }(D_{i}f)=D_i^{\mathrm{\prime }}({\phi }^{\ast }f)$, which lie in $\mathcal{D}({\phi }^{\ast }I)$. Hence ${\phi }^{\ast }\mathcal{D}(\mathcal{I})=\mathcal{D}({\phi }^{\ast }\mathcal{I})$, and iteration gives ${\phi }^{\ast }{\mathcal{D}}^i(\mathcal{I})={\mathcal{D}}^i({\phi }^{\ast }\mathcal{I})$ for every $i\ge 0$. AC [A1] is used only through [F4] and [F5]; the derivation transport and ideal-generation computations are choice-free.