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I found the following definition.

A Weil divisor $D = \sum_{i}D_i \subset X$ on a smooth variety $X$ is simple normal crossing if for every point $p \in X$ a local equation of $D$ is $x_1\cdot...\cdot x_r$ for independent local parameters $x_i$ in $O_{p,X}$. A log resolution of the pair $(X,D)$ is a birational morphism $f:Y\rightarrow X$ such that $Y$ is smooth and $f^{-1}D \cup Exc(f)$ is simple normal crossing.

From this definition it seems to me that if $D = L\cup R\cup C\subset\mathbb{P}^2$ is the union of three lines passing through the same point, then $D$ is simple normal crossing.

On the other hand I found an example showing that if $C\subset\mathbb{P}^2$ is a cusp, in order to get a log resolution we have to blow-up three times. However after two blow-ups we end up with the strict transform $\tilde{C}$ of $C$ and two exceptional divisors $E_1, E_2$ intersecting transversally in a point $p = \tilde{C}\cap E_1\cap E_2$. Why is not the divisor $\tilde{C}\cup E_1\cup E_2$ simple normal crossing ?

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  • $\begingroup$ That does not look like the standard definition of simple normal crossing: the components of $D$ should also be smooth. What you have is just normal crossing. $\endgroup$ – user5117 May 23 '14 at 13:33
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You gave the definition of normal crossing divisor. The definition of simple normal crossing is the following.

A Weil divisor $D = \sum_{i}D_i \subset X$ on a smooth variety $X$ of dimension $n$ is simple normal crossing if any component $D_i$ is smooth and for every point $p \in X$ a local equation of $D$ is $x_1\cdot...\cdot x_r$ for independent local parameters $x_i$ in $O_{p,X}$ with $r\leq n$.

What goes wrong in the case of three line passing through the same point in $\mathbb{P}^2$ is that you need three local parameters in a neighborhood of $p$ and $3 > 2$. The same for the cusp.

More generally, let $D = \sum H_i\subset\mathbb{P}^n$ be a reduced divisor, where the $H_i$'s are hyperplanes. Consider the points $h_i\in\mathbb{P}^{n*}$ dual the the $H_i$'s. Then $D$ is simple normal crossing if and only if the $h_i$'s are in linear general position in $\mathbb{P}^{n*}$. You see that three lines through the same point in $\mathbb{P}^2$ correspond to three points on the same line in $\mathbb{P}^{2*}$. Therefore this case does not work. More generally for a curve $C$ on a surface simple normal crossing means any irreducible component is smooth and $C$ as at most nodes as singularities.

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