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.

  • $\begingroup$ 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/… $\endgroup$ – Francesco Polizzi Dec 22 '11 at 10:47
  • $\begingroup$ 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? $\endgroup$ – tarosano Dec 23 '11 at 11:25

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.