Pullback of a strict normal crossings divisor by a smooth morphism is a strict normal crossings divisor.

Lemma 41.21.3. Let $X$ be a locally Noetherian scheme. Let $D \subset X$ be a strict normal crossings divisor. If $f : Y \to X$ is a smooth morphism of schemes, then the pullback $f^*D$ is a strict normal crossings divisor on $Y$.

Proof. As $f$ is flat the pullback is defined by Divisors, Lemma 31.13.13 hence the statement makes sense. Let $q \in f^*D$ map to $p \in D$. Choose a regular system of parameters $x_1, \ldots , x_ d \in \mathfrak m_ p$ and $1 \leq r \leq d$ as in Definition 41.21.1. Since $f$ is smooth the local ring homomorphism $\mathcal{O}_{X, p} \to \mathcal{O}_{Y, q}$ is flat and the fibre ring

$\mathcal{O}_{Y, q}/\mathfrak m_ p \mathcal{O}_{Y, q} = \mathcal{O}_{Y_ p, q}$

is a regular local ring (see for example Algebra, Lemma 10.140.3). Pick $y_1, \ldots , y_ n \in \mathfrak m_ q$ which map to a regular system of parameters in $\mathcal{O}_{Y_ p, q}$. Then $x_1, \ldots , x_ d, y_1, \ldots , y_ n$ generate the maximal ideal $\mathfrak m_ q$. Hence $\mathcal{O}_{Y, q}$ is a regular local ring of dimension $d + n$ by Algebra, Lemma 10.112.7 and $x_1, \ldots , x_ d, y_1, \ldots , y_ n$ is a regular system of parameters. Since $f^*D$ is cut out by $x_1 \ldots x_ r$ in $\mathcal{O}_{Y, q}$ we conclude that the lemma is true. $\square$

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