# blow-up and normal crossings

Hi friends,

Can anybody help me with the following question?

I start with a projective smooth variety $X$ over a field $k$ of a characteristic 0 and $D$ a normal crossing divisor on $X$, with irreducible components $D_i$, $i \in I$.

Now I take a smooth subvariety $Z$ of $X$ of codimension 2 meeting $$D_J:=\bigcap_{j \in J} D_j$$ transversally for all $J \subset I$.

Let $$\varphi: \tilde{X} \to X$$ be the blow-up of $X$ along $Z$. Then $\varphi^\ast D$ is a normal crossing divisor on $\tilde{X}$.

I am trying to relate $\mathcal{O}_{\tilde{X}}(\varphi^\ast D)$ on $\tilde{X}$ to the inverse images of $\mathcal{O}_X(D)$ and $\mathcal{O}_Z(D_Z)$, where $D_Z=D \cap Z$ (which is again a normal crossing divisor on $Z$).

I guess the start point is to look at the exact sequence $$0 \to \mathcal{I}_Z \to \mathcal{O}_X \to \mathcal{O}_Z \to 0$$ and tensor it with $\mathcal{O}_X(D)$ to get

$$0 \to \mathcal{I}_Z \otimes \mathcal{O}_X(D) \to \mathcal{O}_X(D) \to \mathcal{O}_Z \otimes \mathcal{O}_X(D) \to 0$$

The last term is $\mathcal{O}_X(D) _{| Z}=\mathcal{O}_Z(D_Z)$, isn't it?

How to continue?

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What is your definition of "meeting transversally"? Do you mean that the (scheme-theoretic) intersection of $Z$ and $D_J$ is empty or smooth of dimension $\text{dim}(Z) - \text{card}(J)$ –  Jason Starr Mar 25 '13 at 13:05
That's exactly my definition –  bupncd Mar 25 '13 at 13:09
No answer so far? Maybe the question is harder than I imagined... –  bupncd Mar 26 '13 at 11:20

Let $p$ be any point of $Z$. Denote by $J$ the collection of indices $j$ such that $p$ is contained in $D_j$, say $\{j_1,\dots,j_c\}$. Then there exists a Zariski open affine neighborhood $U$ of $p$, global sections $f_1,\dots,f_c \in \Gamma(U,\mathcal{O}_X)$, and global sections $g_1,\dots,g_d\in \Gamma(U,\mathcal{O}_X)$ such that $D_{j_k}$ is the zero scheme of $f_k$ for $k=1,\dots,c$, $Z$ is a complete intersection of the zero schemes of $g_1,\dots,g_d$ in $X$, and $(f_1,\dots,f_c,g_1,\dots,g_d)$ is a regular sequence.
Consider the morphism $(f_1,\dots,f_c,g_1,\dots,g_d):U \to \mathbb{A}^c \times \mathbb{A}^d$. Your transversality hypothesis implies that this morphism is smooth at $p$ (use the Jacobian criterion). Denote coordinates on $\mathbb{A}^c$, resp. $\mathbb{A}^d$, by $(x_1,\dots,x_c)$, resp. $(y_1,\dots,y_d)$. Then each $D_{j_k}$ is the inverse image of $Z(x_k)$, and $Z$ is the inverse image of $Z(y_1,\dots,y_d)$. Thus the blowing up of $Z$ in $U$ is smooth over the blowing up of $Z(y_1,\dots,y_c)$ in $\mathbb{A}^c\times \mathbb{A}^d$, and the inverse images of the $D_{j_k}$ in the blowing up are the inverse images of the $Z(x_k)$.
So now you are reduced to the "model" case where $X$ is $\mathbb{A}^c\times \mathbb{A}^d$, the divisors $D_{j_k}$ are the coordinate hyperplanes $Z(x_k)$, and $Z$ is the coordinate subspace $Z(y_1,\dots,y_d)$. In this model case, you can compute the blowing up explicitly to verify your claim.
Thank you so much Jason!! Could you just tell me which is the final result for the relation between $\mathcal{O}_X(\varphi^\ast D)$ and the inverse images of $\mathcal{O}_D$ and $\mathcal{O}_Z(D_Z)$ to be sure that I get the right answer when computing the model case? –  bupncd Mar 26 '13 at 15:47
Note also that $c=2$ in my original example. –  bupncd Mar 26 '13 at 15:51
By definition, $\mathcal{O}_{\widetilde{X}}(\phi^*D)$ equals $\phi^* \mathcal{O}_{X}(D)$. I thought you were asking whether or not $\phi^*D$ is a simple normal crossings divisor, which it is. –  Jason Starr Mar 26 '13 at 19:12