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