# what are the singularities of a normal crossings divisor?

This is probably a very stupid question. I'm sorry.

Let $D$ be a simple normal crossings divisor on some smooth projective variety $D$. By this I mean that the irreducible components $D_i$ are smooth and all possible intersections $\bigcap D_{i_1} \cap \cdots \cap D_{i_k}$ are transversal.

I don't understand what people mean by the singular locus of $D$. In view of the definition, I would say $D$ is smooth. Could anybody help me clarify this point?

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The points of intersection of distinct reducible components are not smooth. Consider the equation $xy = 0$ in $A^2$, for example. The intersection of the coordinate axes is certainly transversal, but the variety is not smooth at the origin.
Ok, so what happens is that the intersections are smooth when considered as subvarieties (in your example the point is smooth) but $D$ is not smooth itself. That's it? – div90 Jun 20 '13 at 14:32
The points on $D$ are smooth as points on the ambient variety; in this example the ambient variety is some affine space, and the divisor $D$ is the union of the x-axis and y-axis. The origin is the only singular point on the divisor $D$ in this example. – Ariyan Javanpeykar Jun 20 '13 at 15:31
For a divisor with irreducible components having multiplicity one, we can identify it with a reducible subvariety. As a point on this subvariety, a point of intersection is singular, since the dimension of the cotangent space $\frac{m_x}{m_x^2}$ is not equal to the dimension of the subvariety. This is the same reason that an irreducible nodal curve is singular. – Jesse Silliman Jun 21 '13 at 3:18