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Let $X \rightarrow Y$ be a birational projective morphism between smooth varieties over $\mathbb{C}$. I think that the exceptional locus $E \subset X$ of $f$ is codimension $1$. Assume that $\dim X = \dim Y =3$.

Question Is $E$ normal crossing?

If you know counterexample, please let me know.

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There is the case of the birational contraction of a quadric cone in X, that is not a normal crossing divisor (If I remember correctly, in the definition of normal crossing divisor all components must be smooth). See mathoverflow.net/questions/31426/… –  Francesco Polizzi Dec 22 '11 at 10:47
    
Thank you for the helpful comment. I assumed that both $X,Y$ are smooth. The target space of the contraction have an ordinary double point. Is there a further contraction to a smooth point? –  tarosano Dec 23 '11 at 11:25

1 Answer 1

up vote 4 down vote accepted

The answer is no. Here's an example.

Fix $Y$ and a closed point $p \in Y$. Blow up $p$. This is clearly smooth. Call the resulting scheme $X'$ and let $F$ denote the exceptional divisor of $f' : X' \to Y$. I am actually going to assume that $Y = \mathbb{A^3}$ in my mind.

Here's the idea then, within $X'$, choose a smooth curve $C$ which is tangent to $F$ at some point $p'$ in some "sufficiently funny way" and blow up the entire curve $C$.

Explicitly, if $X'$ has a chart $\text{Spec} k[x,y,z]$ and $F$ is given by the equation $z = 0$, then you can set $C = V(y, x^2 + z)$. This gives us $g : X \to X'$. Certainly $X$ is smooth and $f : X \xrightarrow{g} X' \xrightarrow{f'} Y$ is birational.

Now, I claim that the exceptional divisor of $f$ is not simple normal crossing. It is enough to show that the strict transform of $F$ in $X$ is not smooth. In our particular example, the strict transform of $F$ corresponds to the blowup of $(y,x^2)$ within $k[x,y]$. It is therefore easy to see that the strict transform of $F$ has a quadric cone singularity.

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Thank you for the example. –  tarosano Dec 27 '11 at 5:09

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