Order and simultaneous normal crossings are preserved by smooth morphisms

Work at an arbitrary $x^{\mathrm{\prime }}$ and its image $x$, with $A,B$ as in [F1], maximal ideals $\mathfrak{m},\mathfrak{n}$, and residue fields $\kappa =A\mathrm{/}\mathfrak{m}$, $\lambda =B\mathrm{/}\mathfrak{n}$. The map $A\to B$ is faithfully flat: for any proper ideal $J\subseteq A$, locality gives $JB\subseteq \mathfrak{n}$, so $B\mathrm{/}JB\mathrm{\ne }0$; any nonzero $A$-module contains a nonzero cyclic submodule $A\mathrm{/}J$, whose injection remains injective after flat tensoring, so its tensor with $B$ is nonzero. The closed-fibre local ring $B\mathrm{/}\mathfrak{m}B$ is regular by smoothness. This argument applies to nonclosed and nonrational points as well.

Choose a regular system of parameters $u_1,\dots ,u_d$ of $A$. Its image in $B$ is a regular sequence by [F3, F4]. Put $B_i=B\mathrm{/}(u_1,\dots ,u_i)B$; the terminal ring $B_d=B\mathrm{/}\mathfrak{m}B$ is regular. Backward induction using [F5] shows that every $B_i$ is regular and that the class of $u_{i+1}$ is outside the square of its maximal ideal. Thus $u_{i+1}\mathrm{\notin }(u_1,\dots ,u_i)B+{\mathfrak{n}}^2$, and the images of all $u_i$ are linearly independent in $\mathfrak{n}\mathrm{/}{\mathfrak{n}}^2$. Extend them to a cotangent basis; its lifts generate $\mathfrak{n}$ by the finite-generator Nakayama argument (if the quotient module equals its maximal-ideal multiple, a matrix $I-M$ with unit determinant annihilates its generators). They therefore form a regular system of parameters of $B$. When $d=0$ the tuple is empty and the same conclusion is immediate.

By [F6], the parameter systems in step 2.1 identify ${\mathrm{gr}}_{\mathfrak{m}}A$ with $\kappa [U_1,\dots ,U_d]$ and ${\mathrm{gr}}_{\mathfrak{n}}B$ with $\lambda [U_1,\dots ,U_d,V_1,\dots ,V_e]$. The induced graded map sends each $U_i$ to the corresponding parameter class and extends the residue-field embedding $\kappa \to \lambda$; it is therefore injective. In particular an element of ${\mathfrak{m}}^j\setminus {\mathfrak{m}}^{j+1}$ remains outside ${\mathfrak{n}}^{j+1}$.

At a point $x$ on an SNC union, choose its distinct component equations $u_1,\dots ,u_c$ as part of a regular system of parameters of $A$, as in [F8]. Step 2.1, applied with that system, shows that their images are part of a regular system of parameters of $B$. Hence the inverse image union is locally the product of those same distinct parameter equations, so it has SNC at $x^{\mathrm{\prime }}$. This applies to every subfamily and every point over it, independently of its residue field or its position in the fibre, proving simultaneous SNC. Where no component occurs, the equation is a unit and its inverse image is empty.

For an ideal $I\subseteq A$ and $N\ge 0$, the inclusion $I\subseteq {\mathfrak{m}}^N$ implies $IB\subseteq {\mathfrak{n}}^N$ because the map is local. Conversely, if $IB\subseteq {\mathfrak{n}}^N$ and $a\in I\setminus {\mathfrak{m}}^N$, choose the largest $j<N$ for which $a\in {\mathfrak{m}}^j$; step 3.1 gives $a\mathrm{\notin }{\mathfrak{n}}^{j+1}$, contradicting $a\in IB\subseteq {\mathfrak{n}}^N$. Thus $I\subseteq {\mathfrak{m}}^N⟺IB\subseteq {\mathfrak{n}}^N$ for every $N$, and taking suprema proves (1), including unit ideals of order zero and zero ideals of infinite order. Flatness identifies the ideal pullback with its extended ideal.

Each pulled-back member is a product of distinct parameters in the regular local domain $B$, hence a nonzerodivisor. Each individual parameter generates a prime ideal, since its quotient is regular by [F5] and a domain by [F3]. Their distinct principal prime ideals have intersection equal to their product: divisibility by one prime parameter and the fact that it divides none of the others prove this successively. The product ideal is consequently radical. Thus each inverse image is a reduced effective Cartier divisor, completing (2). AC is used only through the parameter and associated-graded suppliers [A1]; no completion isomorphism or equality of residue fields is used.