**Proof.**
Assume $D$ is a strict normal crossings divisor. Pick $p \in D$ and 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 $\mathcal{O}_{X, p}/(x_ i)$ is a regular local ring (and in particular a domain) we see that the irreducible components $D_1, \ldots , D_ r$ of $D$ passing through $p$ correspond $1$-to-$1$ to the height one primes $(x_1), \ldots , (x_ r)$ of $\mathcal{O}_{X, p}$. By Algebra, Lemma 10.106.3 we find that the intersections $D_{i_1} \cap \ldots \cap D_{i_ s}$ have codimension $s$ in an open neighbourhood of $p$ and that this intersection has a regular local ring at $p$. Since this holds for all $p \in D$ we conclude that (2) holds.

Assume (2). Let $p \in D$. Since $\mathcal{O}_{X, p}$ is finite dimensional we see that $p$ can be contained in at most $\dim (\mathcal{O}_{X, p})$ of the components $D_ i$. Say $p \in D_1, \ldots , D_ r$ for some $r \geq 1$. Let $x_1, \ldots , x_ r \in \mathfrak m_ p$ be local equations for $D_1, \ldots , D_ r$. Then $x_1$ is a nonzerodivisor in $\mathcal{O}_{X, p}$ and $\mathcal{O}_{X, p}/(x_1) = \mathcal{O}_{D_1, p}$ is regular. Hence $\mathcal{O}_{X, p}$ is regular, see Algebra, Lemma 10.106.7. Since $D_1 \cap \ldots \cap D_ r$ is a regular (hence normal) scheme it is a disjoint union of its irreducible components (Properties, Lemma 28.7.6). Let $Z \subset D_1 \cap \ldots \cap D_ r$ be the irreducible component containing $p$. Then $\mathcal{O}_{Z, p} = \mathcal{O}_{X, p}/(x_1, \ldots , x_ r)$ is regular of codimension $r$ (note that since we already know that $\mathcal{O}_{X, p}$ is regular and hence Cohen-Macaulay, there is no ambiguity about codimension as the ring is catenary, see Algebra, Lemmas 10.106.3 and 10.104.4). Hence $\dim (\mathcal{O}_{Z, p}) = \dim (\mathcal{O}_{X, p}) - r$. Choose additional $x_{r + 1}, \ldots , x_ n \in \mathfrak m_ p$ which map to a minimal system of generators of $\mathfrak m_{Z, p}$. Then $\mathfrak m_ p = (x_1, \ldots , x_ n)$ by Nakayama's lemma and we see that $D$ is a normal crossings divisor.
$\square$

## Comments (2)

Comment #2649 by David Hansen on

Comment #2668 by Johan on