Smooth base change of multiple test blow-ups

The base case and the invariants of the induction. The pure dimension of $X^{\mathrm{\prime }}$ is preserved by blowing up regular centers: standard charts over positive-codimension centers retain the ambient dimension, whereas whole-component centers simply delete those components. For $i=0$ we have $X_0^{\mathrm{\prime }}=X^{\mathrm{\prime }}{\times }_{X}X=X^{\mathrm{\prime }}$ and $({\mathcal{I}}_0^{\mathrm{\prime }},E_0^{\mathrm{\prime }},\mu )={\phi }^{\ast }(\mathcal{I},E,\mu )=({\phi }^{\ast }\mathcal{I},{\phi }^{-1}E,\mu )$ by definition of the pullback of a marked ideal; assertion (1) for $i=0$ is the smoothness of $\phi$ itself. We prove by induction on $i$ that ${\phi }_i\text{\hspace{0.17em}⁣}:X_i^{\mathrm{\prime }}\to X_i$ is smooth, that $(X_j^{\mathrm{\prime }}{)}_{j\le i}$ is a multiple test blow-up of $({\mathcal{I}}^{\mathrm{\prime }},E^{\mathrm{\prime }},\mu )$, and that ${\mathcal{I}}_j^{\mathrm{\prime }}={\phi }_j^{\ast }{\mathcal{I}}_j$, $E_j^{\mathrm{\prime }}={\phi }_j^{-1}{E}_j$ with the induced order, omitting empty members.

The inductive step: center and blowup. Assume the induction hypothesis for $i$. If ${\sigma }_{i+1}$ is an inserted isomorphism step, identify $X_{i+1}^{\mathrm{\prime }}$ with $X_i^{\mathrm{\prime }}$ along that step and ${\phi }_{i+1}={\phi }_i$; all assertions follow by transport. Otherwise, even if the blowup morphism is an isomorphism, let $C_i\subseteq \mathrm{supp}({\mathcal{I}}_i,E_i,\mu )$ be the center and put $C_i^{\mathrm{\prime }}:={\phi }_i^{-1}(C_i)$. By [F2] the base change $X_{i+1}^{\mathrm{\prime }}:=X_i^{\mathrm{\prime }}{\times }_{X_i}{X}_{i+1}\to X_i^{\mathrm{\prime }}$ of the blowup is the blowup of $C_i^{\mathrm{\prime }}$ when $C_i^{\mathrm{\prime }}\mathrm{\ne }\mathrm{\varnothing }$ and an isomorphism otherwise; it is smooth over $X_{i+1}$ because ${\phi }_i$ is smooth and smoothness is stable under base change [F4]. The center $C_i^{\mathrm{\prime }}$ is regular (the fibre product of the smooth morphism ${\phi }_i$ with the regular closed subscheme $C_i$ is smooth over $K$ by [F4], hence regular over the characteristic-zero field), and it has SNC with $E_i^{\mathrm{\prime }}={\phi }_i^{-1}{E}_i$ by [F3].

The inductive step: supports. For $x^{\mathrm{\prime }}\in C_i^{\mathrm{\prime }}$ with $x={\phi }_i(x^{\mathrm{\prime }})$ one has $x\in C_i\subseteq \mathrm{supp}({\mathcal{I}}_i,\mu )$ and hence, by [F3], ${\mathrm{ord}}_{x^{\mathrm{\prime }}}({\phi }_i^{\ast }{\mathcal{I}}_i)={\mathrm{ord}}_x({\mathcal{I}}_i)\ge \mu$; since ${\mathcal{I}}_i^{\mathrm{\prime }}={\phi }_i^{\ast }{\mathcal{I}}_i$ this says $C_i^{\mathrm{\prime }}\subseteq \mathrm{supp}({\mathcal{I}}_i^{\mathrm{\prime }},\mu )$, so the pulled-back sequence is admissible at step $i+1$.

The inductive step: transform identities. The exceptional divisor $D_{i+1}^{\mathrm{\prime }}$ of the pulled-back blowup is ${\phi }_{i+1}^{-1}(D_{i+1})$ and satisfies $\mathcal{I}(D_{i+1}^{\mathrm{\prime }})={\phi }_{i+1}^{\ast }\mathcal{I}(D_{i+1})$ by [F2]. Therefore ${\mathcal{I}}_{i+1}^{\mathrm{\prime }}=\mathcal{I}(D_{i+1}^{\mathrm{\prime }}{)}^{-\mu }({\sigma }_{i+1}^{\mathrm{\prime }}{)}^{\ast }{\mathcal{I}}_i^{\mathrm{\prime }}={\phi }_{i+1}^{\ast }(\mathcal{I}(D_{i+1}{)}^{-\mu }{\sigma }_{i+1}^{\ast }{\mathcal{I}}_i)={\phi }_{i+1}^{\ast }{\mathcal{I}}_{i+1}$, using the commutativity of pullback with products and with the controlled transform of ideals; and $E_{i+1}^{\mathrm{\prime }}=({\sigma }_{i+1}^{\mathrm{\prime }}{)}^{\mathrm{c}}(E_i^{\mathrm{\prime }})\cup \{D_{i+1}^{\mathrm{\prime }}\}={\phi }_{i+1}^{-1}(({\sigma }_{i+1}{)}^{\mathrm{c}}(E_i)\cup \{D_{i+1}\})={\phi }_{i+1}^{-1}{E}_{i+1}$ by [F5]; the order is the induced one after omitting empty members. If the pulled-back center is empty, its exceptional inverse image is empty and the step transports the existing boundary; a nonempty Cartier-center blowup retains its nonempty exceptional divisor despite having an isomorphic underlying morphism. This closes the induction and proves assertions (1) and (2).

Resolution case. Suppose $(X_i)$ is a resolution of $(\mathcal{I},E,\mu )$, so $\mathrm{supp}({\mathcal{I}}_r,\mu )=\mathrm{\varnothing }$. By the induction identity ${\mathcal{I}}_r^{\mathrm{\prime }}={\phi }_r^{\ast }{\mathcal{I}}_r$, and by [F3] every point $x^{\mathrm{\prime }}$ of $X_r^{\mathrm{\prime }}$ satisfies ${\mathrm{ord}}_{x^{\mathrm{\prime }}}({\mathcal{I}}_r^{\mathrm{\prime }})={\mathrm{ord}}_{{\phi }_r(x^{\mathrm{\prime }})}({\mathcal{I}}_r)$ if ${\phi }_r(x^{\mathrm{\prime }})$ lies in the locus where ${\mathcal{I}}_r$ is defined; since $\mathrm{supp}({\mathcal{I}}_r,\mu )=\mathrm{\varnothing }$, no point satisfies $\mathrm{ord}\ge \mu$, so $\mathrm{supp}({\mathcal{I}}_r^{\mathrm{\prime }},\mu )=\mathrm{\varnothing }$ as well. Hence the steps of $(X_i^{\mathrm{\prime }})$ with nonempty centers form a resolution of $({\mathcal{I}}^{\mathrm{\prime }},E^{\mathrm{\prime }},\mu )$, and the full sequence $(X_i^{\mathrm{\prime }})$ is an extension of it in the sense of [F1], which is assertion (3).