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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?

Thanks for your help!

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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.

share|cite|improve this answer
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

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